47429190 unit 5 ans excellent loads of questions physics edexcel

1. D[1]

2. C[1]

3. D[1]

4. C[1]

5. D[1]

6. B[1]

7. A[1]

8. B[1]

9. B[1]

10. B[1]

11.

density, mass (1)

tough, energy (1)

compressive, tensile (1)

[3]

12. (a) (i) Use of Ek = 2

1mv2

Correct answer [0.44] (1)

Example of calculation:

44.0)30(

)20(2

2

30

20 ==E

E1

(ii) Collision energy is more than halved (1), so claim is justified (1) 2

(b) Calculation of collision energy [60 kJ] (1)Use of W = Fx (1)Correct answer [500 kN] (1)


Ek = 2

1mv2 = 0.5 × 1200 × (10)2 = 60,000 J

W = Fx so kNmx

WF 500

12.0

000,60 === 3

(c) Crumple zone increases displacement of car during crash so collision forceis reduced or crumple zone increases collision time and so decreases theacceleration (and force) (1) 1

[7]

13. (a) As skydiver speeds up, air resistance will increase (1)Net force on skydiver will decrease, reducing acceleration (1) 2

(b) Parachute greatly increases the size of the air resistance (1)When air resistance = weight of skydiver, skydiver is in equilibrium (1) 2

(c) Use of as v2 = u2 + 2as or 2

1mv2 = mg∆ h (1)

Correct answer [7.7 m s–1] (1)


v = 381.92 ×× = 7.7 m s–1 2[6]

14. Spelling of technical terms must be correct and the answer must beorganised in a logical sequence (QWC)Mixing of layers leading to eddies/whorls (1)Air circulates around at edge of platform (1)Passenger may be pushed over due to eddies/whorls (1)

[3]

15. (a) Proportional / Hooke’s law limit (1) 1

(b) Stiffness of sample (1) 1

(c) Work done / strain energy (1)To stretch (OR strain) wire to fracture (1) 2

[4]

16. (a) Use of ρπ 3

3

4r (1)

Correct answer [1.44 × 10–6 kg] (1)


kg 1044.11000)107.0(3

4

3

4 6333 −− ×=××== πρπrm 2

(b) Use of mg = 6π η rv (1)



134

6

s m 2.1107.01090.86

81.91044.1

6−

−−

−

=××××

××==ππ ηr

mgv (2) 2

[4]

17. (a) Reference to free fall whilst bungee is slackIdea of KE increasing as GPE is transformedIdea of work being done against frictional forcesGPE converted into EPE (and KE) once bungee stretchesKE (and GPE) converted into EPE beyond equilibrium pointAt lowest point all of the KE has been converted into EPE Max 4

(b) Use of F = m a (1)



2s m 25.680

285785 −=−==m

Fa 2

[6]

18. (a) Use of s = ut + ½ at2 (1)Correct answer [1.1 s] (1)


s 1.16.1

122 =×==a

st 2

(b) Use of v = u + at (1)



v = u + at = 1.6 × 1.1 = 1.8 m s–1 2[4]

19. (a) Length and breadth of the thin cross section to calculate area (1)Digital callipers / micrometer (1) 2

(b) • Small extensions can be measured accurately (1)• Large data set / easy processing of data (1) 2

[4]

20. (a) Use of v sin θ (1)Correct answer [4.2 ms-1] (1)


v sin θ = 10 sin 25 = 4.2 m s–1 2

(b) Use of at v = u + at (1)Correct answer [0.43 s] (1)

Example of calculation:v = u + at 0 = 4.2 – 9.8 × t

81.9

2.4=t = 0.43 s 2

(c) Use of 2

2

1atuts += or t

vus .

2

)( += (1)

Correct answer [0.90 m] (1)


m 90.091.081.1)43.0(81.95.043.02.42

1 22 =−=××−×=+= atuts

or m 90.043.02

02.4.

2

)(=×

+=

+= t

vus 2

[6]

21. (a) (i) Assumption: spring obeys Hooke’s Law (1)Use of F = kx (1)Correct answer [60N] (1)


1

2

1

2

x

x

F

F=

F2 = 5

50 × 6 = 60 N 3

(ii) Use of W = Fav.x (1)Correct answer [75J] (1)

Example of calculation:W = Fav.x

W = 5 × 2

60 × 0.5 = 75 J 2

(b) (i) Attempt at estimation of area under graph / average force (1)0.5 m extension used (1)Correct answer [53 → 57J] (1)

Example of calculation:Energy represented by 1 square =10 × 0.02 = 0.2 J280 squares × 0.2 J = 56 J

Treating the area as a large triangleW = 0.5 × 205 × 0.5 = 51 J 3

(ii) Spelling of technical terms must be correct and the answer must be

organised in a logical sequence (QWC)Energy returned is less than the work done in stretching the cords (1)Energy must be conserved, so internal energy of cords must increase (1)Rubber cords will get warmer (1) 3

[11]

22. (a) Idea that no resultant force acts (e.g. forces are balanced / cancel) (1) 1

(b) Use of w = mg (1)Correct answer [640 N] (1)

Example of calculation:w = mg = 65 × 9.81 = 638 N 2

(c) (i) Tension in rope marked (1)Push from rock face marked (1)Weight marked (1) 3

(ii) Use of T = w.sin40 (1)Correct answer [410 N] (1)

Example of calculation:T = w.sin40 = 640 × sin40 = 410 N 2

(d) Spelling of technical terms must be correct and the answer must beorganised in a logical sequence (QWC)Rigid/stiff exterior to resist deformation under small forces (1)Must undergo plastic deformation under large forces (1)so that collision energy can be absorbed (1)Low density so that helmet is not uncomfortably heavy (1) 4

[12]

23. D[1]

24. C[1]

25. B[1]

26. A[1]

27. B[1]

28. C[1]

29. B[1]

30. C[1]

31. D[1]

32. (a) D 1

(b) Wavelength

Use of v = f λ (1)Use of f = 1/T (1)Answer T = [0.002 s] (1)[give full credit for candidates who do this in 1 stage T = /v ]

Example of answerv = fλf = 330 / 0.66T = 1/f = 0.66 / 330T = 0.002 s 3

[4]

33. Direction of travel of light is water → air (1)Angle of incidence is greater than the critical angle (1) 2

[2]

34. (a) Transverse waves oscillate in any direction perpendicular to wave direction (1)Longitudinal waves oscillate in one direction only OR parallel to wavedirection. (1)Polarisation reduces wave intensity by limiting oscillations and wavedirection to only one plane OR limiting oscillations to one direction only. (1)(accept vibrations and answers in terms of an example such as a ropepassing through slits) 3

(b) Light source, 2 pieces of polaroid and detector e.g. eye, screen, LED ORlaser, 1 polaroid and detector (1)Rotate one polaroid (1)Intensity of light varies (1) 3

[6]

35. Frequency unaltered (1)Wavelength decreases (1)Speed decreases (1) 3

[3]

36. (a) Use of sensorEvent happens very quickly OR cannot take readings fast enough (1)Sampling rate: 50+ samples per second (1) 2

(b) Initially the temperature is low so current is highResistance of filament increases as temperature increasesCurrent falls to steady value when temperature is constantMaximum heating is when lamp is switched on / when current is highestFilament breaks due to melting caused by temperature rise Max 4

[6]

37. The answer must be clear and the answer must be organised in a logicalsequence (QWC

• It was known that X penetrated (1)

• It was not known that X rays were harmful (1)

• Doctors died because of too much exposure (1)

• Lack of shielding (1)

• New treatments may have unknown side effects (1)

• Treatments need to be tested / time allowed for side effects to appear (1) Max 4[4]

38. (a) [1.0 m] (1) 1

(b) Ratio of (5 or 6 / 3 ) × 60 (1)Answer [f = 100 Hz] (1) 2

[3]

39. Use of sin i / sin r = µ (1)Use of either 80° or 1.33 (1)[r = 48°] (1) 3

Example of answersin 80 / sin r = 1.33[r = 48°]

Both rays refracted towards the normalViolet refracted more than red 2

[5]

40. (a) Calculation of energy required by atom (1)Answer [1.8 (eV)] (1)

Example of answer:Energy gained by atom = 13.6 eV – 3.4 eV = 10.2 eVKE of electron after collision = 12 eV – 10.2 eV = 1.8 eV 2

(b) Use of E = hf and c = fλ (1)Conversion of eV to Joules (1)

Answer = [1.22 × 10–7 m] (1)

Example of answerE = hf and c = fλ E = hc/λλ = (6.63 × 10–34 J s × 3 × 108 m s–2) ÷ (10.2 eV × 1.6× 10–19 C)

λ = 1.21 × 10–7 m 3[5]

41. The answer must be clear, use an appropriate style and be organised in alogical sequence. (QWC)Reference to I = nqvA (1)

For the lampIncreased atomic vibrations reduce the movement of electrons (1)Resistance of lamp increases with temperature (1)

For the thermistorIncreased atomic vibrations again reduce movement of electrons (1)But increase in temperature leads to a large increase in n (1)Overall the resistance of the thermistor decreased with increase intemperature. (1) Max 5

[5]

42. (a) Diffraction is the change in direction of wave or shape orwavefront (1)when the wave passes an obstacle or gap (1) 2

(b) The energy of the wave is concentrated into a photon (1)One photon gives all its energy to one electron (1) 2

(c) Energy of photon increases as frequency increases OR reference toE = hf (1)Electrons require a certain amount of energy to break free and thiscorresponds to a minimum frequency (1) 2

[6]

43. (a) (i) Use of speed = distance over time (1)Distance = 4 cm (1)

Answer = [2.7 × 10–5 s] (1)

Example of answer

t = 4 cm ÷ 1500 m s–1

t = 2.7 × 10–5 s 3

(ii) Use of f = 1/T (1)Answer = [5000 Hz] (1) 2

(iii) Time for pulse to return greater than pulse interval (1)All reflections need to reach transducer before next pulse sent. (1)Will result in an inaccurate image. (1) (Max 2)Need to decrease the frequency of the ultrasound. (1) (Max 3) Max 3

(iv) X-rays damage cells/tissue/foetus/baby but ultrasound doesnot (need reference to both X-rays and ultrasound) (1) 1

(b) The answer must be clear, use an appropriate style and be organisedin a logical sequence (QWC)

Doppler shift is the change in frequency of a wave when the source or thereceiver is moving (1)Requirement for a continuous set of waves (1)Two transducers required (one to transmit and one to receive) (1)Change in frequency is directly related to the speed of the blood (1) 4

[13]

44. (a) Voltmeter is across resistor should be across cell (1) 1

(b) (i) Plot of graphCheck any three points (award mark if these are correct) (3)Line of best fit 3

(ii) e.m.f. = [1.36 – 1.44 V] (1) 1

(iii) Attempt to find gradient (1)Answer [0.38 – 0.42 Ω ] (1) 2

(c) Intercept would twice value above (1) (accept numerical value 2× value(b)(ii))Gradient would be twice value above (1) (accept numerical value 2× value(b)(iii)) 2

[9]

45. C[1]

46. D[1]

47. C[1]

48. B[1]

49. D[1]

50. A[1]

51. D[1]

52. C[1]

53. A[1]

54. B[1]

55. (a) They act on the same body or do not act on different bodies (1)They are different types of, or they are not the same type of, force(1) 2

(b) As the passenger or capsule or wheel has constant speed (1)there is No resultant tangential force (acting on the passenger) (1) 2

(c) Friction between seat & person or push of capsule wall on person (1) 1[5]

56. (a) A baryon is a (sub-atomic) particle made up of 3 quarks(1) 1

(b) n (ddu) → (1)p (duu) (1) 2

[3]

57. (a) High frequency or high voltage(1)Alternating or square wave voltage(1) 2

(b) No electric field inside cylinders (due to shielding) (1)so no force (on electrons) (1) 2

(c) As speed increases (along the accelerator), (1)cylinders are made longer so that time in each stays the same(1) 2

[6]

58. The answer must be clear, use an appropriate style and be organised in alogical sequence (QWC)α -particles fired at (named) metal (film) (1)in a vacuum (1)Most went straight through or suffered small deflections. (1)A few were reflected through large angles or some were reflected alongtheir original path (1)suggesting the mass or charge of the atom was concentrated in a very smallvolume (1) 5

[5]

59. (a) (i) Use of E = ½ CV2 (1)Answer [0.158 J] (1)

E = ½ CV2 = 0.5 × 2200 × 10–6 F × (12 V)2

E = 0.158 J 2

(ii) Correct substitution into ∆ Ep = ∆ mgh (1)Answer 0.75 [75%] (1)

∆ Ep = 0.05 kg × 9.8 N kg–1 × 0.24 m [= 0.12 J]Efficiency = 0.12 ÷ 0.16 J = 0.75 [75%] 2

(b) (i) (t = CR) = 2200 × 10–6 (F) × 16 (Ω ) = 35.2 (ms) (1) 1

(ii) Curve starting on I axis but not reaching t axis (1)

I0 = 1.6 V / 16Ω = 100 mA shown on axis (1)

Curve passing through about 37 mA at t = 35 ms (1) 3

(c) (i) The vibrations of the air particles (1)are parallel to the direction of travel of the wave (energy) (1) 2

(ii) T = 1/f = 50 ms (1)

Sensible comment related to time constant of 35 ms (1) 2[12]

60. (a) 3Li7 + 1p1 = 2He4 + 2He4

completing LHS (1)completing RHS(1) 2

(b) (i) Charge (1)(mass/) energy (1) 2

(ii) Mass of Li + p = 7.0143 u + 1.0073 u = 8.0216 u (1)Mass of 2 α -particles = 2 × 4.0015 u = 8.0030 u (1)∆ m = 8.0216 u – 8.0030 u = 0.0186 u

= 0.0186 × 1.66 × 10–27 kg = 3.09 × 10–29 kg (1)

∆ E= c2∆ m = (3.00 × 108 m s–1)2 × 3.09 × 10–29 kg = 2.78 × 10–12 J (1)

[Allow ecf from equation] 4

(iii)119

12

eV J 1060.1

J 1078.2−−

−

××= = 1.74 × 107 eV = 17.4 MeV (1)

The incoming proton has an energy of 300 keV = 0.30 MeV (1)So total energy = 17.4 MeV + 0.3 MeV = 17.7 MeV (1)The calculated energy differs by

%3%100MeV)2.177.17(

MeV 17.2-Mev 7.17

21

≈×+ (1)

The experiment therefore provides strong evidence for Einstein’sprediction (1) 5

[13]

61. (a) Total (linear) momentum of a system is constant, (1)

provided no (resultant) external force acts on the system (1) 2

(b) The answer must be clear, use an appropriate style and be organised in alogical sequence

Use of a light gate (1)Use of second light gate (1)

Connected to timer or interface + computer (accept ‘log-it’) (1)

Cards on gliders (1)

Measure length of cards (1)Velocity = length ÷ time (1) 6

(c) Multiplies mass × velocity to find at least one momentum (1)

1560 g cm s–1 (0.0156 kg m s–1) before and after (1) 2[10]

62. (a) (i) Use of A = π r2 leading to 0.87 (m) (1) 1

(ii) Correct use of ω = 2π /t leading to 62.8 (rad s–1) (1) 1

(iii) Correct use of v = rω = 55 m s–1 [allow use of show that value] (1) 1

(b) (i) Substitution into p = ½ ρ Av3 (1)

3047 (W) (1) 2

(ii) Air is hitting at an angle/all air not stopped by blades (1)Energy changes to heat and sound (1) 2

(c) (i) Attempts to find volume per second (A × v) (1)

44 kg s–1 (1) 2

(ii) Use of F = ∆ mv/∆ t (1)

F = 610 N (1) 2

(d) Recognises that 100 W is produced over 24 hours (1)Estimates if this would fulfil lighting needs for a day(1)Estimates energy used by low energy bulbs in day(1)Conclusion (2)The answer must be clear and be organised in a logical sequence

Example:

The 100 W is an average over the whole day. Most households would uselight bulbs for 6 hours a day in no more than 4 rooms, so this would meanno other energy was needed for lighting.4 low energy bulbs would be 44 W for 6 each hours so would require energyfrom the National grid.

[Accept an argument based on more light bulbs/longer hours that leads tothe opposite conclusion] 5

[16]

63. C[1]

64. B[1]

65. B[1]

66. C[1]

67. C[1]

68. D[1]

69. A[1]

70. B[1]

71. D[1]

72. C[1]

73. (a) as ideal gases do not have forces between molecules so no potential energy (2) 2

(b) use of pv = NkT

conversion of T to kelvin and answer = 5.8 × 1022 molecules 2[4]

74. (a) use of counter (+GM tube)determine background count in absence of sourceplace source close to detector and:place sheet of paper between source and counter (or increase distancefrom source 3-7 cm of air) reduces count to background 4

(b) alpha radiation only has range of 5 cm in air / wouldn’t get through casing (1) 1[5]

75. reason (1)and consequence(1)reason (1)and consequence(1)

2 from:Could be different sizes; So larger one could be further awayCould be different temperatures; So hotter one could be further awayDifferent luminosities; So more luminous one could be further away 4

[4]

76. (a) attempt to find gradient eg evidence of triangle on graph or values ofdy/dx (1)value of n (1) 2

(b) needs T4 (1)(electrical) power P related to luminosity (1) 2

[4]

77. (a) difficult to measure distances to “far” objects accurately / difficult tomeasure speeds of far objects accurately (1) 1

(b) appreciate 1/H is age of universe (1) 1

(c) fate of universe depends on the density of the universe (1)link between gravity and density (1)Hubble “constant” is changing due to gravitational forces (1) 3

[5]

78. (a) 19 protons identified (1)calculation of mass defect (1)Conversion to kg(1)use of E = mc2 (1)divide by 40(1)

= 1.37 × 10–12 J(1)

[eg 19 × 1.007276 = 19.138244 + 21 × 1.008665 = 40.320209 – 39.953548 =0.36666

× 1.66 × 10–27 = 6.087 × 10–28

× c2 = 5.5 × 10–11

/40 = 1.37 × 10–12 J] 6

(b) cannot identify which atom/nucleus will be the next to decaycan estimate the fraction /probability that will decay in a given time /cannot state exactly how many atoms will decay in a set time 1

(c) (i) conversion of half life to decay constant

[eg λ = ln 2 / 1.3 × 109 = 5.3 × 10–10 y–1] 1

(ii) add both masses to find initial mass (1)

use of N = Noe–λ t (1)rearrange to make t subject (1)

Answer = 4.2 × 109 years (1)(if 0.84 used instead of 0.94 3 max)[eg total initial mass 0.94

t = ln 0.1 / 0.94 / 5.3 × 10–10

= 4.2 × 109] 4[12]

79. (a) (i) Use of F = –kx (1)F = ma (1) 2

(ii) cf with a = –ω 2x ie ω 2 = k/m (1)use of T = 2π /ω to result (1) 2

(b) (i) resonance (1) 1

(ii) natural freq = forcing frequency (1) 1

(iii) use of c = fλ (1)

answer 9.1 × 1013 Hz (1) 2

(iv) use of T = 1/f eg T = 1.1 × 10–14 (1)rearrange formula (1)to give 550 N/m(1) 3

[11]

80. (a) equates F = GMm/ r2 and mv2 /r (1)Use of v = 2π r/T (1)

Cancel m’s To give GMT2 = 4π 2 r3 in any form (1) 3

(b) (i) remove constants or cancel G 4π 2 (1)

use of idea MT2 / r3 = Constant (1)substitution 27 /36 = Mstar / Msun (1) 3

(ii) both will complete orbit in same time period (1)star covers small distance / orbit radius smaller compared to planet (1) 2

(c) The answer must be clear, use an appropriate style and be organised in alogical sequence (QWC)Larger planets will move centre of mass towards planet / away from starcentre (1)Star moves faster (1)Doppler shift greater for larger speeds (1) 3

[11]

81. (a) top row : 17 1 (14) 4 (1)bottom row: 8 1 7 2 (1)other product – helium (1) 3

(b) The answer must be clear, use an appropriate style and be organised in alogical sequence (QWC)dead star / no longer any fusion (1)small dense hot / still emitting radiation/light (1)consisting of products of fusion such as carbon / oxygen / nitrogen (1) 3

(c) (i) use of 3/2 kT (1)conversion to eV (1)answer [1.3 (keV)] (1) 3

(ii) gravitational force does work on hydrogen (1)increases internal energy of gas (1) 2

(d) The answer must be clear, use an appropriate style and be organised in alogical sequence (QWC)A standard candle (in astronomical terms) produces a fixed amount of light/luminosity (1)Quantity of hydrogen (1)and fusion temperature (1) must be similar for various novae. 3

[14]

82. Add missing information

For three correct responses in the ‘vector or scalar’ column (1)For the ‘base unit’ column:3 correct responses (2)2 correct responses (1) 3

Quantity Base unit Vector or scalar

m vector

kg m2 s–2 scalar

kg m2 s–3 scalar

kg m s–1 vector

[3]

83. (a) (i) Describe motion

Constant / uniform acceleration or (acceleration of) 15 m s–2 (1)

(Followed by) constant / uniform speed / velocity (of

90 m s–1) (1) 2

(ii) Show that distance is approximately 800 mAny attempt to measure area under graph or select appropriateequations of motion required to determine total distance (1)

Correct expression or value for the area under the graph betweeneither 0-4 s [240 m] or 4-10 s [540 m] (1)

Answer : 780 (m) (1)

Eg distance = 60 m s–1 × 4 s + 90 m s–1 × 6 s= 240 m + 540 m= 780 (m)

Eg distance in first 4 s

s = 2

s m 30s m 90

2

uv 11 −− +=+t 4 s = 240 m

Distance in final 6 s

s = ut = 90 m s–1 × 6 s = 540 mTotal distance = 240 m + 540 m = 780 (m) 3

(b) Sketch graphGraph starts at 760 m – 800 m/their value and initially showsdistance from finishing line decreasing with time (1)The next two marks are consequent on this first mark being awarded

Curve with increasing negative gradient followed by straight line (1)

Graph shows a straight line beginning at coordinate (4 s, 540 m)and finishes at coordinate (10 s, 0 m) (1) 3

[8]

84. (a) (i) Give expressionW = R + F (1) 1

(ii) Complete statements……. surface / ground (1)

……. Earth (–s mass) [Only accept this answer] (1)

……. gardener(–s hands) / hand(s) (1) 3

(b) (i) Add to diagramLine inclined to the vertical pointing to the left and upwards (1) 1

(ii) Explain change in direction and magnitudeThe force (at X) will have a magnitude greater than For the force (at X) must increase. (1)

This is because the wheelbarrow / it has to be lifted /tilted/ supported/ held up (by the vertical component) (1)

And also because the wheelbarrow / it has to be moved(forward by the horizontal component) (1) 3

[8]

85. (a) Show that rate of decay of radium is about 7 × 10 13 Bq Power divided by alpha particle energy (1)

Answer [(7.1 – 7.2) × 1013 (Bq)] (1)

[Give 2 marks for reverse argument ie7 × 1013 Bq × 7.65 × 1013 J (1)(53.5 – 53.6) (W) (1)]

Eg Rate of decay = J 107.65

W5513−×

= 7.19 × 1013 (Bq) 2

(b) Show that decay constant is about 1.4 × 10–11 s–1

Use of λ = 1/2T

69.0 (1)

Answer [(1.35 – 1.36) × 10–11 (s–1)] (1)

Eg λ = s 103.15 years 1620

69.07××

= 1.35 × 10–11 (s–1) 2

(c) The number of radium 226 nucleiUse of A = λN (1)

Answer [(5.0 – 5.4) × 1024] (1)

Eg 7.19 × 1013 Bq = 1.35 × 10–11 s–1 × N

N = 5.33 × 1024 2

(d) The mass of radium

Divides number of radium 226 nuclei by 6.02 × 1023 andmultiplies by 226 (1)

Answer [1870 – 2040 g]

Eg Mass of radium = 226 g × 23

24

106

1033.5

××

= 2008 g 2

(e) Why mass would produce more than 50 WThe (daughter) nuclei (radon) formed as a result of the decay ofradium are themselves a source of (alpha)radiation / energy (1)Also accept(having emitted alpha) the nucleus[allowsample/radium/atom] (maybe left excited andtherefore also) emits gammaAlso accept(daughter) nucle(us)(i) recoil releasing (thermal) energyDo not acceptNucleus may emit more than one alpha particleNucleus may also emit beta particle 1

[9]

86. (a) Paths of alpha particlesPath A drawn less deflected than B (1)

Path A drawn as a straight line (1) 2

(b) (i) Why alpha source inside containerAlpha would be absorbed by [accept would not get through]container (material) (1) 1

(ii) Why the same kinetic energy?EitherTo restrict observation to two variables / closeness of approachand deflectionor so that speed / velocity / (kinetic) energy does not havean effect (on the observation / deflection /results /contact time) 1

(iii) Why an evacuated container?Eitherso that alphas do not get absorbed by / collide with / getdeflected by / stopped by / scattered by / get in theway of / ionise / lose energy to atoms / molecules (ofair) [Do not accept ‘particles’ of the air]

or so that all alphas reach the foil with the same (kinetic) energy 1[5]

87. (a) Diode or LED (1) 1

(b) (i) Use of R = V / I current between 75 and 90 ignoring powers of 10 (1)answer 6.7 – 8.0 Ω (1)

Example of answer

R = 0.60 V ÷ (85 × 10–3) AR = 7.06 Ω 2

(ii) Infinite OR very high OR ∞ 1

(c) ANY ONERectification / AC to DC / DC supply [not DC appliances]Preventing earth leakageStabilising power outputTo protect componentsA named use of LED if linked to LED as component in (a)(egcalculator display / torch)A voltage controlled switch(Allow current in only one direction) 1

[5]

88. (a) Resistivity definition

Resistivity = resistance × (1)× cross sectional area / length (1)

ρ= RA/l with symbols defined scores 2/2equation as above without symbols defined scores ½equation given as R = ρl/A with symbols defined scores 1/2

(1st mark is for linking resistivity to resistance with some other terms) 2

(b) (i) Resistance calculationConverts kW to W (1)Use of P = V2/R OR P = VI and V = IR (1)Resistance = 53 Ω (1)

Example of answer

R = (230 V)2 ÷ 1000 WR = 53 Ω 3

(ii) Length calculationRecall R = ρl/A (1)Correct substitution of values (1)Length = 6.3 m (accept 6.2 m) (1)ecf value of R

Example of answer

l = (52.9 Ω × 1.3 × 10–7 m2) ÷ (1.1× 10–6 Ω m)l = 6.3 m 3

(iii) Proportion methodIdentifies a smaller diameter is needed (1)Diameter = 0.29 mm (1)ORCalculation methodUse of formula with l = half their value in (b)(ii) (1)Diameter = 0.29 mm (1)(Ecf a wrong formula from part ii for full credit)

Example of answerdnew = 0.41 mm ÷ √2dnew = 0.29 mm 2

[10]

89. (a) Definition of E.M.F.Energy (conversion) or work done (1)Per unit charge (1)[work done/coulomb 1/2, energy given to a charge 1/2, energygiven to a charge of a coulomb 2/2]OR ORE = W/Q (1) E = P/ISymbols defined (1) Symbolsdefined(E = 1 J/C scores 1) (E = 1 W/A scores 1)

((Terminal) potential difference when no current is drawn 1/2) 2

(b) (i) Internal resistance calculationAttempt to find current (1)Pd across r = 0.2 V (1)r = 0.36 (Ω) (1)[You must follow through the working, I have seen incorrectmethods getting 0.36 Ω]

Example of answerI = 2.8 V ÷ 5.0 Ωr = (3.0 – 2.8) V ÷ 0.56 A = 0.36 Ω 3

(ii) Combined resistanceUse of parallel resistor formula (1)Resistance = 3.3 Ω [accept 3 1/3 but not 10/3] (1) 2

(iii) Voltmeter reading(ecf bii)Current calculation using 3 V with either 3.3 Ω or 3.7 Ω (1)Total resistance = 3.7 Ω [accept 3.66 to 3.73 Ω]OR use of V = E – Ir (1)Voltmeter reading = 2.7 V (1)

ORPotential divider method, ratio of resistors with 3.7 Ω on bottom (1)Multiplied by 3.0 V (1)2.7 V (1)

Example of answerRtotal = 3.7 ΩI = 3 V ÷ 3.7 Ω = 0.81 AVvoltmeter = 3.3 Ω × 0.81 A = 2.7 V 3

(c) Ideal voltmeterIdeal voltmeter has infinite resistance OR extremely high resistanceOR highest possible R OR much larger resistance than that ofcomponent it is connected across OR quotes value > 1 M Ω (1)

Current through voltmeter is zero (negligible) OR doesn’t reduce theresistance of the circuit OR doesn’t reduce the p.d. it is meant tobe measuring. (1) 2

[12]

90. (a) Circuit diagramPotentiometer correctly connected i.e potential divider circuit (1)Ammeter in series and voltmeter in parallel with bulb (1)

(light bulb in series with resistance can score second mark only) 2

(b) (i) Graph+I, +V quadrant; curve through origin with decreasing gradient (1)

[do not give this mark if curve becomes flat and then starts goingdown i.e. it has a hook]

–I, –V quadrant reasonably accurate rotation of +I,+V quadrant (1) 2

(ii) Shape of graphAs current/voltage increases, temperature of the lamp increases /lamp heats up (1)Leading to increase in resistance of lamp (1)Rate of increase in current decreases OR equal increases in Vlead to smaller increases in I (1)Qowc (1)

Ecf if a straight line graph is drawn max 3R constant (1)V α I (1)Qowc (1) 4

[8]

91. (a) Absolute zero of temperature(Temperature at which) pressure / volume (of a gas) is zero. (1)OR(Temperature at which) the kinetic energy of the molecules is zero) 1

(b) (i) Number of moles show that calculation

Recall pV = nRT (1)Addition of air pressure (1)Conversion to kelvin (1)Number of moles = 0.52 (mol) (1)

Reverse calculations using n = 0.5 to arrive at one of the othervalues can score maximum 3

Example of answer

K)10273(mol K J 31.8

)m 105.8 Pa 10)1.10.1(11

335

+××××+

=−−

−

n

n = 0.52 mol 4

(ii) Mass of air

Mass = 1.5 × 10–2 kg (1)

Example of answer

mass = 0.52 mol × 0.029 kg mol–1 = 0.015 kg 1

(iii) Temperature calculation

Use of P1/T1 = P2/T2 (1)

Correct P2 1.6 × 105 Pa (1)Lowest temperature = 216 K (–57 °C) (1)

ORUse of pV = nRT (must see correct value of R) (1)

Correct P2 1.6 × 105 Pa (1)Lowest temp 215K – 223K (–58 to –50 °C) (1)

Example of answer

Pa 101.2

K) 283Pa 10)6.00.1((5

5

2 ×××+

=T

T2 = 216 K 3[9]

92. Core remnant stars

All core remnants ticked AND no main sequence (1)

< 1.4 M column: White dwarf only (1)

> 2.5 M column: Black hole only (1) 3[3]

93. (i) Hydrogen burning

Quality of written communication (1)

Nuclear fusion reaction [accept nuclei, nucleus, fusing] (1)

Hydrogen / deuterium /protons turn into He [penalise contradictions,e.g. molecules atoms; accept symbols ] (1)

Release of energy (1) 4

(ii) Sun as red giant calculation

Attempted use of L = σT4A (accept r substituted as A) (1)

A = 4 π r2 [or A α r2 if ratios calculated directly] (1)

3.85 × 1026 (W) or 1.13 × 1030 (W) [or substitution as ratio] (1)

2930 [accept 2900 – 2940] (1)

L = σT4A = 4 π σT4r2

Lbefore = 4 π × 5.67 × 10–8 W m–2 K4 × (5780 K)4 × (6.96 × 108 m)2

= 3.85 × 1026 W

Lafter = 4 π × 5.67 × 10–8 W m–2 K4 × (3160 K)4 × (1.26 × 1011 m)2

= 1.13 × 1030 W

Hence ratio = 1.13 × 1030 W ÷ 3.85 × 1026 W = 2930 4

(iii) H-R diagram plots

X at 100 on main sequence [± 1 mm by eye] AND between 5000 K andcentre of 5000 – 10 000 K box (1)

Y above and to right of actual X (1)

Attempt to plot Y at 3160 K [between 5000 K and 2500 K] (1)

Attempt to plot Y between 103 L and 104 L [ecf] (1) 4[12]

94. (i) Sun as white dwarf

Any 2 [comparative statements] of

Higher temperature / hotterLower luminosity [accept Power, not E or I]No fusion in core [or equivalent; not just “not on main sequence”]More dense (1) + (1) 2

(ii) Future of white dwarf

Cools / T decreases (1)

Dims / fades / correct colour change [not brown dwarf] / Luminositydecreases [accept intensity here] (1) 2

[4]

95. (i) Distance to Sirius

Substitution in v × t /s [ignore 8.6, accept 365 or 365¼ days] (1)

8.1 × 1016 (m) [8.13, 8.14] (1)

d = v t

= 8.6 × 3.00 × 108 m s–1 × (60 × 60 × 24 × 365¼) s

= 8.1 × 1016 m 2

(ii) Sirius A intensity calculation

Use of I = L / 4 π D2 (1)

Correct substitution (1)

1.2 × 10–7 W m–2 [1.20 – 1.24] (1)

I = L / 4 π D2

= 1.0 × 1028 W / 4 π (8.1 × 1016 m)2

= 1.2 × 10–7 W m–2 3

(iii) Mass rate conversion

E = m c2 seen [or implied] (1)


1.1 × 1011 kg (s–1) (1)

1.0 × 1028 W = 1.0 × 1028 J s–1

∆m = ∆E / c2

= 1.0 × 1028 J / (3.00 × 108 m s–1)2

= 1.1 × 1011 kg 3

(iv) Peak wavelength calculation

Use of Wien’s law (1)

2.93 × 10–7 m (1)

λmax = 2.90 × 10–3 m K / 9900 K

= 2.93 × 10–7 m 2[10]

96. (i) Fence wire cross-section

Use of π r2 and 10–3 m (1)

4.9 × 10–6 (m2) [do not accept m] (1)

A = π r2

= π × (0.5 × 2.50 × 10–3)2 2

(ii) Stress calculation

Substitution: 1500 N / 4.9 [or 5] × 10–6 m2 (1)

310 MPa [accept 300, ecf] (1)

σ = F / A

= 1500 N / 4.9 × 10–6 m2

= 3.1 × 108 Pa 2

(iii) Extension calculation

E = σ / ε and ε = ∆l / l (or E = F l / A ∆l) (1)

Substitution in E = σ / ε and ε = ∆l / l [or in E = F l / A ∆l, ecf,ignore 10n] (1)

0.048 (m) [ecf] (1)

48 mm [accept 47 – 49 mm, bald answer scores 4/4] (1)

E = F l / A ∆l

∆l = (1500 N × 33 m) / (210 × 109 Pa × 4.9 × 10–6 m2)= 0.048 m = 48 mm 4

[8]

97. (i) Young modulus experiment

(G–) clamp [vice], wire, pulley, mass / weight / load

three correct (1)

all four correct (1) 2

(ii) Labelling of l

Accurate indication of l [to 1 mm] (1)

Length 2 m to 6 m (1) 2

(iii) Additional apparatus

Micrometer (screw gauge) / (digital) callipers (1)

Ruler or similar [e.g. tape measure, metre stick] (1) 2[6]

98. Particle classification

Neutron: baryon and hadron (1)

Neutrino: lepton (1)

Muon: lepton (1) 3[3]

99. (i) Binding energy

Energy required to separate a nucleus into nucleons (1) 1

(ii) 8n + 6p (1)

Substitution / m = 0.1098 u (1)

Multiply by 930 [only, or E = m c2 route] (1)

102 MeV [or 103 MeV] (1)

∆m = (6 × 1.007 28 u) + (8 × 1.008 67 u) – 14.003 24 u = 0.1098 u∆E = 0.1098 u × 930 MeV/u = 102 MeV 4

(iii) More stable isotope

Binding energy per nucleon attempted (1)

7.4 (MeV) and 7.3 (MeV) [accept 7.1, ecf] (1)

Hence carbon–12 [based on two values, ecf] (1)

BE / A (14C) = 102 MeV / 14 = 7.3 MeV

BE / A (12C) = 89 MeV / 12 = 7.4 MeV 3[8]

100. (i) Conservation laws

First reaction, Q: 0 + 0 ≠ 1 + 1 (1)

Second reaction B: 1 = 1 + 0 AND Q: –1 = –1 + 0 (1)

Hence only Ω– decay possible [based on B and Q conservation for thisdecay, accept simple ticks and crosses] (1) 3

(ii) Quark charges

Use of sss = –1 to show s = –⅓ (1)

Hence correct working (from baryons) to show u = ⅔ and d = –⅓ (1) 2[5]

101. (a) (i) Why speed is unchanged

Force/Weight [not acceleration] is perpendicular tovelocity/motion/direction of travel/instantaneous displacement[not speed]OR no component of force/weight in direction of velocity etc (1)

No work is doneOR No acceleration in the direction of motion (1) 2

(ii) Why it accelerates

Direction (of motion) is changing (1)

Acceleration linked to a change in velocity (1) 2

(b) Speed of satellite

Use of a = v2/r (1)

Correct answer [3.8 to 4.0 × 103 m s–1] (1)

Example calculation:

v = √(2.7 × 107 m × 0.56 m s–2)

[Allow 1 mark for ω = 1.4 × 10–4 rad s–1] 2[6]

102. (a) (i) Demonstrating the stationary wave

Move microphone between speaker and wall OR perpendicular to wallOR left to right OR towards the wall [could be shown by labelledarrow added to diagram] (1)

Oscilloscope/trace shows sequence of maxima and minima (1) 2

(ii) How nodes and antinodes are produced

Superposition/combination/interference/overlapping/crossingof emitted/incident/initial and reflected waves (1)

Antinodes: waves (always) in phase OR reference to coincidenceof two compressions/rarefactions/peaks/troughs /maxima/minima,hence constructive interference/reinforcement (1)

Nodes: waves (always) in antiphase/exactly out of phase ORcompressions coincide with rarefactions etc, hence destructiveinterference / cancellation (1) 3

(iii) Measuring the speed of sound

Measure separation between (adjacent) nodes / antinodes anddouble to get λ/this is ½λ [not between peaks and troughs] (1)

Frequency known from/produced by signal generator ORmeasured on CRO / by digital frequency meter (1)

Detail on measurement of wavelength OR frequencye.g. measure several [if a number is specified then ≥3] nodespacings and divide by the number [not one several times]OR measure several (≥3) periods on CRO and divide by the numberOR adjust cro so only one full wave on screen (1)

Use v (allow c) = fλ 4

(b) (i) Application to concert hall

Little or no sound /amplitudeOR you may be sat at a node (1)

(ii) Sensible reason

Examples:Reflected wave not as strong as incident waveOR walls are covered to reduce reflectionsOR waves arrive from elsewhere [reflections/different speakers]OR such positions depend on wavelength / frequency (1) 2

[11]

103. (a) (i) Amplitude and frequency

0.17 m (1)

0.8(3) Hz or s–1 (1) 2

(ii) Maximum velocity

Use of vmax = 2πfx0 (1)

Correct answer (1)

Example calculation:vmax = 2π × 0.83 Hz × 0.17 m

ORUse of maximum gradient of h versus t graph

Answer to 2 sig fig minimum 2

(iii) Velocity-time graph

Wave from origin, period 1.2 s (1)

Inverted sine wave with scale on velocity axis &initial peak value

0.9 m s–1 (1) 2

(b) (i) Definition of SHM

Acceleration / resultant force proportional to displacementOR Acceleration / resultant force proportional to distance from a

fixed point [not just distance from equilibrium but –distance from

equilibrium position– is acceptable]OR a = (–) constant × x [with a and x defined]OR F = (–) constant × x [with F and x defined] (1)

Acceleration /resultant force directed towards the fixed point / inopposite direction (to displacement)OR negative sign in equation explained [e.g a and x in oppositedirections] (1) 2

(ii) Verifying SHM

Read off h value and use it to get displacement (1)[only penalise the first mark if h used for displacement throughout]Plot acceleration-displacement graph OR calculate ratios eg a÷x (1)Straight line through the origin OR check ratios to see if

constant (1)Negative gradient / observe acceleration OR constant is negative (1)and displacement have opposite signs

OR

Use x = xocos(2πft) for a range of t OR Read off h and get x (1)

Use values of xo and f from part (a) OR Use a = –(2πf)2x for range of x (1)Add equilibrium value to x to get h OR Use value of f from part (a) (1)If results agree with values of h (or a) from graph it is SHM (1) 4

[12]

104. (a) (i) Line B

Knot T at 2.4 m [±½ small square, no label needed] (1)

(ii) Knots Q, R, S at 0.6, 1.2, 1.8 m [±½ small square, no labels needed][ecf from wrong position of knot T i.e. Q at ¼T, R at ½T & R at ¾T](1) 2

(b) How model represents the Universe and its behaviour

Knots/letters/points represent galaxies (1)

Reference to expansion of Universe / galaxies moving apart[NOT galaxies move away and stay same distance apart] (1) 2

(c) How model illustrates Hubble–s law

Stating or showing velocities are different for 2 of the knots (1)[Shown by either calculating speeds or comparing distances movedbetween diagrams A and B]

Calculation of velocity for at least 2 of the knots [other than T] (1)

Use of their data to show speed (of knot) ∝ distance (from P) (1)Examples:determine values of v÷d [allow v÷∆d]sketch graph of v against d [allow v against ∆d] 3

(d) Defects of the model

Any 2 sensible points (1)(1)

Examples:Galaxies are not evenly spacedInitial spacing of knots is not zeroNo force pulling galaxies/Universe apartRate of expansion of Universe OR speed of galaxies increasing/not constant [not speed decreasing]Relative sizes of knot and spacing are unrealisticUniverse is 3 dimensional/galaxies are not in a straight line 2

[9]

105. (a) Meaning of statement

(5.89 × 10–19 J / work function) is the energy needed to remove anelectron [allow electrons] from the (magnesium) surface/plate

Consequent markMinimum energy stated or indicated in some way [e.g. at least /or more] (1) 2

(b) (i) Calculation of time

Use of P = IA (1)

Use of E = Pt (1)

[use of E = IAt scores both marks]

Correct answer [210 (s), 2 sig fig minimum, no u.e.] (1)[Reverse argument for calculation leading to either intensity,energy or area gets maximum 2 marks]

Example calculation:

t = (5.89 × 10–19 J)/(0.035 W m–2 × 8 × 10–20 m2) 3

(ii) How wave-particle duality explains immediate photoemission

QOWC (1)

Photon energy is hf / depends on frequency / depends on wavelength (1)

One/Each photon ejects one/an electron (1)

The (photo)electron is ejected at once/immediately (1)[not just ‘photoemission is immediate’] 4

[9]

106. (a) (i) GMS/R2 (*)

(ii) GME/r2 (*) (1) 1

(*) (symbols must be as given in the Q, though allow lower case m)

(b) (i) Evidence of equating of GMS/R2 and GME/r2 (ecf from part a) (1)

Correct answer 570 – 580 (1)

Example of answer:

E

SESES

M

M

r

R

r

M

R

M

r

GM

R

GM=→=→=

2

2

2222

5771033.3kg 100.6

kg 100.2 524

23

=×=××

==∴E

S

M

M

r

R2

(ii) 1.5 × 108 km × 1/601 [ignore powers of 10 in distance value] (1)

Correct answer 2.5 – 2.6 × 105 km (or 2.5 – 2.6 × 108 m) (1) 2

(c) Letter L on or against line to left of point P(coming within one Earth radius of dotted line) (1)

Reason*:[*Consequent marks; allow only if L position correct or not shown]

Reference to centripetal force/centripetal acceleration/(net) force towards Sun (1)

Force due to Sun must be > force due to Earth (1) 3[8]

107. (a) (i) W = QV (1)

Correct answer 3.2 nJ [3.2 × 10–9 J, etc.] (1)

Example of answer:

W = QV = 0.8 × 10–9 C × 4.0 V = 3.2 × 10–9 J 2

(ii) +0.8 (nC) on top plate and –0.8 (nC) on bottom plate (1)(both needed) 1

(b) Statement (E =) ‘Area’ or (E =) ½ QV (1)

See calculation ½ × 4.0 × 0.8 or ½ × base × height (1)

ORC found from graph (1)

Use of W = ½ CV2 (1)

Example of answer:

F 100.2V 0.4

C 100.8 109

−−

×=×+=V

QC

∴ W = ½CV2 = 2

V) (4.0 F 100.2 210 ×× −

= 1.6 × 10–9 J 2

(c) (i) Correct answer 0.2 nC (1) 1

(ii) Graph is straight and through origin (1)

ends at 3.0V and their Q (1) 2

(iii) Attempt to use C = Q/V or C = ∆Q/∆V (1)

Correct answer 0.067 nF / 67 pF (1)

Example of answer:

V 0.3

C 102.0 9−×==V

QC = 6.7 × 10–11 F 2

[10]

108. (a) (i) 1.2 keV = 1.2 × 103 × 1.6 × 10–19 JOR

Use of e∆V with e as 1.6 × 10–19 C and V as 1200 V (1)

Use of ∆(½mev2) with me as 9.1(1) × 10–31 kg. (1)

Correct answer 2.0 – 2.1 × 107 m s–1 (1) 3

(ii) 1200 × 8/100 = 96 (eV delivered per electron) (1)96/2.4 = 40 (1)

Or2.4 × 100/8 = 30 (incident eV needed per photon) (1)1200/30 = 40 (1)

Or1200 / 2.4 = 500 (photons per electron, ideally) (1)500 × (8/100) = 40 (1) 2

(b) Electrons on screen repel electrons in beam / force opposeselectron motion/decelerating force (1)

Electrons (in beam) decelerated /slowed /velocity reduced/ work done by electrons (against force) (1)

Electron (kinetic) energy reduced (not –shared–) (1)

Fewer photons (per electron, stated or implied) (1)

Trace less bright (1)

QoWC (1) Max 4[9]

109. (a) Scale interval is 0.1 (V) (1) 1

(b) (i) Use of ε = (–)N∆φ/∆t (1)

Correct answer 9.6 × 10–7 (Wb) / 0.96 (µWb) [ignore +/–] (1)

Example of answer:

∆ φ = ε × N

t∆ = 0.12 V ×

5000

s 1040 3−× = 9.6 × 10–7 Wb 2

(ii) Use of ‘φ’ or ‘∆φ’ or ‘flux’ = BA, or B = ε ∆t/NA (1)

Correct answer 0.012 T / 0.013 T (1)

Example of answer:

ϕ = BA

∴ B = 22

7

2

m 100.1

Wb106.9

××

×=−

−

π

ϕA = 0.012 T

[N.B. φ = 0.96 µWb → 0.012T, φ = 1µWb → 0.013T] 2[5]

e t110. (a) pair of values of k.e. and υ2 read from graph / gradient (1)

υ2 > 5 × 1016 m s–2 (1)

⇒ mp = 1.62 – 1.69 × 10–27 (kg) to 3 s.f. (1) 3

(b) (i) (values 1.3 – 1.7, 3.1 – 3.5, 6.0 – 6.5) any two correct (1)(1)

(ii) ∆E = c2∆m / E = mc2 (1)

⇒ one value for ∆m (× 10–28 kg) (1)

use of mp from (i) [no mark]

⇒ one value of ∆m/mp : about 10%, 20%, 40% (1) 5[8]

111. (i) appreciation that area of (first) rectangle / at gives speed υ (1)

∆υaccel = (3 m s–2)(8 s) / 30 small squares each worth 0.8 m s–1 (1)

⇒ 24 m s–1 (1)

(ii) appreciation that area of second is of same area as first /

∆υdecel = (4 m s–2)(6 s) [negative idea not needed] (1) 4

(iii) use of P = IV / E = IVt (1)

use of P = Fυ / E = Fυt (1)

(3000 N)υ = (96 A)(750 V) / equating the Ps or Es (1)

⇒ υ = 24 m s–1 3[7]

112. (a) Comment on use of weighing

Clear statement correctly identifying weight or mass (or theirunits)e.g. kg a unit of mass, not weight (1) 1

(b) Calculation to check statement

Use of equation of motion to show time or distance (1)Answer to 2 sig figs [120 m or 4.5 s] [no ue] (1)


s = ut + ½ at2

s = 0 + ½ × 9.81 m s–2 × (5s)2 OR 100 = 0 + ½ × 9.81 m s–2 × t2

s = 123 m OR t = 4.5 s 2

(c) Calculation of kinetic energyEitherUse of equation(s) of motion which allow(s) v2 or v to be found (1)

Recall of ke = ½ mv2 (1)Answer [69 000 J] (1)

ORRecall of Ep = mgh (1)Substitution (1)Answer [69 000 J] (1)


υ2 = u2 + 2as

υ2 = 0 + 2 × 9.81 m s–2 × 100 m

υ2 = 1962 m2 s–2

ke = 1/2 mv2

= 69 000 J (68 670 J)

ORgpe = mgh

gpe lost = 70 kg × 9.81 N kg–1 × 100 mgpe lost = 69 000 J (68 670 J)[so ke = 69 000 J because ke gained = gpe lost] 3

[6]

113. (a) (i) Condition for reflection

Angle of incidence greater than critical angle [accept i > c] (1) 1

(ii) Description of path of ray

Any two from:• Ray refracted at A and C• Description of direction changes at A and C• Total internal reflection at B (1)(1) 2

(b) (i) Things wrong with the diagram

Angle of refraction can–t be 0 / refracted too much (1)

No refraction on emergence from prism (1)[Allow 1 mark for correct reference to partial reflection] 2

(ii) Corrected diagram• emergent ray roughly parallel to the rest of the emergent rays (1)• direction of refraction first surface correct (1)• direction of refraction second surface correct (1) 3

[8]

114. (a) Formula for C6

v = u + at OR v = 10.7 – (9.81 × 0.2) [units need not be given]OR C6 = C5 – 9.81*A6 (1) 1

(b) Explain B5 to B16 constantg affects vertical motion only / no horizontal force (1) 1

(c) Significance of negative valuesThe ball moving downwards (1) 1

(d) (i) Completion of diagramVertical arrow has 6.8 added, horizontal arrow has 10.7 added (1) 1

(ii) Calculation of velocity at time t = 0.4 s

Use of Pythagoras

Answer for magnitude of v [12.7 m s–1] [ecf from diagram]

Use of trigonometrical function [ecf from magnitude]

Answer for direction [32.4°] [ecf from diagram]

Example of answer:

v2 = (6.8 m s–1)2 + (10.7 m s–1)2

v = 12.7 m s–1

tan θ = 6.8 m s–1 ÷ 10.7 m s–1

θ = 32.4°

[For scale drawing– components drawn correctly to scale(1),

resultant shown correctly (1), answer for v ± 0.5 m s–1 (1), angle to± 2°(1)] 4

(e) (i) Calculation of components for new angle

Answer for vertical component [8.7 m s–1] (1)

Answer for horizontal component [12.5 m s–1] (1)

[1 mark only if answers reversed]

Example of answer:

vertical component = v sinθ = 15.2 m s–1 × sin 35° = 8.7 m s–1

horizontal component = v cosθ = 15.2 m s–1 × cos35° = 12.5 m s–1 2

(ii) Suggest reason for greater distance

Examples – greater horizontal component of velocity; easier tothrow at higher speed closer to the horizontal; launching fromabove ground level affects the range; force applied for longer;more force can be applied (1) 1

[11]

115. (a) Show that heat energy supplied at about 500 W

Recall of power = energy/time (1)

Answer to 2 sig figs [470 [W]] [no ue] (1)


power = energy/time= 1.63 × 105 J/ 347 s= 470 W 2

(b) (i) Show that heat energy gained is about 1 × 10 5 J

Use of ∆Q = mc∆θ (1)

Correct answer [1.4 × 105 [J]] [no ue] (1)


∆Q = mc∆θ

= 0.44 kg × 3800 J kg–1 °C–1 × (96 °C – 12 °C)

= 1.4 × 105 J 2

(ii) Calculate the time taken to reach 96 °C

Use of time = energy/power (1)

Correct answer [300 s] (1)


time = energy/power

= 1.4 × 105 J/ 470 W= 299 s 2

(c) (i) Explain why it might take longer

Heat supplied to milk at a lower rate / expansion onmechanism of heat loss /destination of heat lost (1) 1

(ii) Suggest why time the same

Power calculation includes a heat loss factor / rate of heat gain thesame as for water / appropriate mechanism to reduce heat loss(Allow 1 for heat losses already taken into account whenwarming the water) (1) 1

[8]

116. (a) Calculation of adaptor–s input

Recall of: power = IV (1)

Correct answer [0.01 A] (1)


power = IVI = P/V = 25 W / 230 V= 0.01 A 2

(b) (i) Explain why VA is a unit of power

Power = voltage × current so unit = volt × amp (1) 1

(ii) Calculation of efficiency of adaptor

Use of efficiency equation (1)

Correct answer [24%] (1)


efficiency = (0.6 VA / 2.5 W) × 100%= 24 % [0.24] 2

(iii) Reason for efficiency less than 100%

Resistance (accept explanations beyond spec, e.g. eddy currents) (1)

hence heat loss to surroundings (1) 2

(c) (i) Calculation of charge

Recall of: Q = It (1)

Correct answer [4000 C] (1)


Q = It= 0.2 A × 6 h= 0.2 A × (6 × 60 × 60) s= 4000 C (4320 C) 2

(ii) Calculation of work done

Recall of: W = QV OR Recall of W = Pt (1)


Correct answer [13 000 J] (1)


W = QVW = 4320 C × 3 V [ecf]= 13 000 J (12 960 J)ORW = PtW = 0.6 W × 6 hW = 0.6 W × (6 × 60 × 60) s= 13 000 J 3

[12]

117. (a) (i) Add standing waves to diagrams

Mark for each correct diagram (1)(1) 2

(ii) Mark place with largest amplitude of oscillation

antinode marked [allow clear indication near centre of wave otherthan an X, allow correct antinode shown on diagrams B or C] (1) 1

(iii) Name of place marked

(Displacement) Antinode [allow ecf from (a) (ii)] (1) 1

(b) (i) Calculation of wavelength

Correct answer [5.6 m]

Example of calculation:= 2 × 2.8 m= 5.6 m (1) 1

(ii) Calculation of frequency

Recall of v = f λ (1)

Correct answer [59 Hz] [ecf] (1)

Example of calculation:v = f λ

f = 330 m s–1 / 5.6 m= 58.9 Hz 2

(c) (i) Explanation of difference in sound

as the room has a standing wave for this frequency / wavelength /it is the fundamental frequency(allow relevant references to resonance) (1) 1

(ii) Suggest another frequency with explanation

Appropriate frequency [a multiple of 59 Hz] [ecf] (1)

Wavelength 1/2, 1/3 etc (stated or used) (1) 2

(d) Explain change in frequencies

wavelengths (of standing waves) bigger / f = v/2l (1)

hence frequencies smaller/lower (1) 2[12]

118. (a) Blue light:Wavelength / frequency / (photon) energy 1

(b) (i) Frequency:Conversion of either value of eV to JoulesUse of f = E / h

Correct frequency range [4.8 × 1014 – 8.2 × 1014 Hz or range =

3.4 × 1014 Hz][no penalty for rounding errors]

eg.2 eV = 2 × 1.6x 10–19 = 3.2 × 10–19 J

= 6.63 × 10–34 × f

f = 4.8 × 1014 Hz

3.4 eV = 3.4 × 1.6 × 10–19 = 5.4 × 10–19 J

f = 8.2 × 1014 Hz 3

(ii) Diagrams:Downward arrow from top to bottom levelOn larger energy gap diagram 2

(c) (i) Resistivity drop:Less heating / less energy lost / greater efficiency / lowervoltage needed / less power lost 1

(ii) Resistance:Recall of R = ρL/AUse of R = ρL/ACorrect answer [80(Ω)] [allow 80–84 (Ω) for rounding errors]

Eg.

R = (2 × 10–2 × 5.0 × 10–3) / (3.0 × 10–3 × 4.0 × 10–4)= 83 Ω 3

[10]

119. (a) (i) Type of behaviour:PlasticCorrect definition of circled word:Ductile: can be pulled into a long thin shapeElastic: returns to original shape/size (once force removed)Plastic: does not return to original shape/size (once force removed)

Tough: can withstand dynamic loads / shocks / impacts / absorbsa lot of energy before breaking 2

(ii) Brittle:Snaps / cracks / shatters / breaks without (plastic) deformation(when subjected to a force) 1

(iii) Strong:Large force / stress required to break it 1

(b) (i) Breaking stress:Use of σ = ε × E

Correct answer [2 × 108 Pa]

Eg.

σ = 2 × 1011 × 0.001

= 2 × 108 Pa 2

(ii) Force to break wire:

Use of A = πr2

Use of F = σ × ACorrect answer [157 (N)][allow 156 – 157 (N) for rounding errors – no u.e]

Eg.

A = π × (1 × 10–3/2)2 = 7.9 × 10–7 m2

F = 2 × 108 × 7.9 × 10–7 m2

Weight (= F) = 157 N 3

(iii) Force to break Biosteel fibre:

3.1 × 103 N [allow 3.1 × 103 N – 3.2 × 103 N]

eg.20 × 157 = 3140 N (3200 N if 160 N used) 1

(iv) Assumption:Elastic limit (of both materials) not reached / elastic behaviour /Hooke’s law obeyed / Young modulus still holds at breaking point/ Area remains constant / best Biosteel scenario / 20 × stronger 1

[11]

120. (a) Angles:Normal correctly added to raindrop (by eye)

An angle of incidence correctly labelled between normal andincident ray and an angle of refraction correctly labelledbetween normal and refracted ray 2

(b) Angle of refraction:Use of µ = sin i / sin rCorrect answer [20°][allow 20°–21° to allow for rounding errors]

eg.sin r = sin 27°/ 1.3r = 20° 2

(c) (i) Critical angle:The angle beyond which total internal reflection (of the light)occurs [allow T.I.R] / r = 90° 1

(ii) Critical angle calculation:

Use of µ = 1 / sin CCorrect answer [50.3°] [allow 50° – 51°]

Eg.Sin C = 1/1.3C = 50.3° 2

(d) Diagram:i = 35° [allow 33° –37°]

Ray of light shown refracting away from normal on leavingraindropSome internal reflection of ray also shown with i = r [by eye]

Reflected ray shown refracting away from the normal as it leavesthe front of the raindrop / angle of refraction correctlycalculated at back surface 4

(e) Refractive index:(Red light has) lower refractive index (than violet light) 1

[12]

121. (a) Graph:Line of best fit drawn as straight line through origin

Stiffness:Use of k = F/x

k = 22 N m–1 (allow 21 –23 N m–1)[allow ecf from graph]

eg.

k = 110/5.0 = 22 N m–1 3

(b) Energy stored in rope:Use of E = area under graph = ½ Fx

[also allow E = ½ kx2 / E = F2/2k]Correct answer [514 J (511J if k = 22 Nm–1 is used][allow 505 – 522 J]

E = ½ × 150 × 6.85= 514 J 2

(c) Less extension with child:Any one of:• Rope connected to sheet as (many) parallel sections

(making it stiffer)• Each section of rope supporting (much) less than full weight• Work done / energy lost to friction with trampoline fram Max 1

(d) New rope dimensions:Rope needs to be thicker / shorterAs stiffness would have to be increased / reference toE = Fl/Ax / as it would have to withstand a greater stress /otherwise extension would be (much) greater 2

[8]

122. (a) Show sum of quark charges in proton = +1+2/3 +2/3 –1/3 = (+) 1 (1)

Show sum of quark charges in neutron = 0+2/3 –1/3 –1/3 = 0 (1)

[ignore references to e] 2

(b) (i) • baryon (1)• meson (1) 2

(ii) baryon: 3 quarks (1)meson: quark/antiquark (1)[1 for answers reversed or baryon/meson not specified] 2

(c) any 4 marks from the following examples:high speed means high energy/momentum (1)may need to overcome (electrostatic) repulsion (1)more energy available for creating particles (1)higher energy/momentum/speed means shorter wavelength (1)reference to λ = h/mv or λ = h/p (1)for diffraction/scattering (1)need λ approx equal to particle spacing/internal structure (1) max 4

(d) Speeds near the speed of light (1)[11]

123. Recall speed = s/t (1)Use of s = π D (1)Answer for speed (1)Conclusion (1)

ORUse of v = rωUse of ω = 2π × 20 000Answer for speedConclusion

v = s/ts = π × 8000 (m)v = π × 8000 × 20 000 (m/s)

v = 5 × 108 m/s

inaccurate/not possible since speed > c[4]

124. (a) Lines (1)[not crossing; minimum 2 lines starting from S pole of magnet]Correct arrow(s) (1)[minimum 1 arrow pointing towards S pole, any incorrect arrow scores 0] 2

(b) (i) Use of F = BIl rearranged to B = F/Il OR with two correct subs (1)Leading to correct answer (1)

B = F/Il = 0.008/(5.8 × 0.012) (T)B = 0.11 (T) 2

(ii) Assumption:parallel field/uniform field/constant field for 12 mm then falls to zero /assume wire perpendicular to field (at all points)/θ = 90° (where F=Bilsinθ given earlier)/force same at all points on the wire (1) 1

(c) Experimental value less because field divergesOR field strength decreases with distanceOR field could be 0.3 T at magnet surface and only 0.1 T at wire (1) 1

(d) Wire would levitate (again) (1)Two reversals cancel/applying FLHR (1)[wire moves downwards due to current OR field reversed scores 1] 2

[8]

125. (a) capacitors need d.c. (1)OR Mains is a.c. / mains current changes direction constantly

Charge given in one half of cycle is removed the next half (1)OR C charged then discharges 2

(b) (i) Voltage value for initial voltage × 1/e [or use of 37%] (1)

OR use 2 values where 2nd is 1/eth of firstOR draw tangent at time = 0 s

OR V = Vo e–t/RC with correct substitution of (t,V) from graph

T = 0.07 s [allow 0.065 s – 0.075 s] (1) 2

(ii) Recall time constant = CR (1)Answer for R [allow ecf for T] (1)

R = T/C = 0.07 s/(100 × 10–9 F)

R = 7 × 105 Ω 2

(c) (i) Recall Q = CV [equation or substitution] (1)Answer for Q (1)

Q = CV = 100 × 10–6 × 300Q = 0.03 C 2

(ii) Recall W = 1/2 CV2 OR W = 1/2 Q2/C (1)OR 2 correct subs into W = 1/2 QV [allow ecf]Answer (1)

Eg: W = 1/2 QV = 0.5 × 0.03 × 300 (J) = 4.5 J

OR W = 1/2 CV2 = 0.5 × 100 × 10–9 × 300 × 300 (J) = 4.5 J

OR W = 1/2 Q2/C = 0.5 × 0.03 × 0.03/(100 × 10–9) (J) = 4.5 J 2[10]

126. Any 8 marks from:Recall of p = mv (1)Use of momentum before collision = momentum after collision (1)Correct value for speed (1)

Example:1250 × 28.0 + 3500 x25.5 = (1250 + 3500) v

v = 26.16 m s–1

Recall of ke = 1/2 mv2 or ke = p2/2m (1)Total ke before (1)Total ke after (1)Loss in ke (1)Recall of work = force × distance (1)Correct answer for force to 2 SF (1)

Example:Total ke before = 1 138 000 J + 490 000 J = 1 627 938 JTotal ke after = 1 625 059 JLoss of KE = 2879 JBraking force = 2879/5 = 576 N Max 8

[8]

127. (a) Name of effect• Resonance• Idea that step frequency = natural frequency of bridge 2

(b) (i) Why oscillations are forced• E.g. amplitude is increasing OR oscillations are driven by the

wind 1

(ii) Calculation of maximum acceleration• Use of ω = 2 π / T to obtain value for ω

• Correct answer for acceleration [14 m s–2]


ω = 2 π / T = 2 π /(60 s / 38) = 4.0 s–1

amax = ω2 A = (4.0 s–1)2 × 0.90 m = 14 m s–2 2

(iii) Show that car loses contact• Required amax is greater than g• So, at a position (of 0.61 m) above the equilibrium

position, vehicle loses contact with the road


x = g/ ω2 = 9.81 m s–2/(4 s–1)2 = 0.61 m 2[7]

128. (a) Name of reaction

• (Nuclear) fusion 1

(b) (i) How to determine distance of star

• Explanation using F = L / 4 π d2 (Use known luminosity withmeasured flux at Earth to determine d) 1

(ii) How to determine velocity of star• Mention of Doppler Effect OR red shift• Identify (pattern of) lines and compare with lab

frequency• ∆f ∝ (relative) v OR the greater the velocity of the star

(relative to Earth), the greater the change infrequency/wavelength observed Max 2

(c) (i) Line of best fit• Insertion of line of best fit, through origin ± 1 square, with

approx. the same number of points each side of line• Idea that the greater the distance to the galaxy, the

greater its velocity (relative to Earth) 2

(ii) Use of gradient to calculate age of universe• Use of v = Ho d to argue that Ho is the gradient of the graph• Correct answer for age of universe

[4.6 × 1017 s, accept 4.0 × 1017 s → 5.2 × 1017 s]


1/gradient = 120 × 106 pc × 3.09 × 1016 m pc–1/ 8,000 × 103 m s–1 =

4.6 × 1017 s 2[8]

129. (a) (i) Meaning of symbols• m = mass of a gas molecule

• <c2> = mean square speed of gas molecule• T = absolute temperature [accept kelvin temperature] 3

(ii) Physical quantity represented• (mean) kinetic energy (of a gas molecule) 1

(iii) Calculation of velocity

• Use of ½ m< c2 > = 3/2 k T with T = 223 K

• Correct answer for velocity [410 m s–1]


c = √(3 × 1.38 × 10–23 J K–1 × 223 K / 5.4 × 10–26 kg) = 413 m s–1 2

(b) (i) Obtain expression for escape velocity• Idea that total energy must be zero for molecule just to

escape

• So, ½ m vesc2 – GMm/r = 0, leading to required equation 2

(ii) Show that escape velocity > 10 km s–1

• Use of vesc = √(2GM/r) with r = (6.37 + 0.01) × 106 m

• Correct answer for escape velocity [11.1 km s–1, at least 2sig. figs. required]

Example of calculation:vesc = √(2GM/r)

vesc = √(2 × 6.67 × 10–11 N m2 kg–2 × 5.98 × 1024 kg /(6.37 + 0.10) × 106 m)

= 1.11 × 104 (m s–1)

= 11.1 (km s–1) 2

(iii) Use of graph to explain whether molecules are likely to escape• Idea that only a tiny fraction of molecules have a very

high velocity• Any quantitative attempt to compare the r.m.s. velocity

with the escape velocity leading to the conclusion thatmolecules are not likely to escape. e.g. 410 is much lessthan 11,000 2

[12]

130. (a) Change in nuclear composition• Nucleus has one less neutron OR nucleus has one more proton) 1

(b) (i) Calculation of age of skull• Use of λ = ln2/t½ to obtain value for λ

• Use of N = Noe–λt

• Correct answer for age of skull [1.2 × 104 y; 3.83 × 1011 s]


λ = ln 2/t½ = ln 2/5730 y = 1.2 × 10–4 y–1 [3.84 × 10–12 s–1]ln(N/No) = –λt

ln(2.3 × 10–11/1.0 × 10–10) = –(1.2 × 10–4 y–1)t

t = 1.2 × 104 y

Alternative mark scheme• Use of half life rule• Correct answer for number of half lives [2.12]

• Correct answer for age of skull [1.2 × 104 y]

Example of calculation:N/ No = (0.5)n

(2.3 × 10–11)/(1 × 10–10) = (0.5)nlog(0.23) = n log(0.5)n = log(0.23)/log(0.5) = 2.12

t = 2.12 × 5730 = 1.2 × 104 y 3

(ii) Reason for inaccuracy• Idea that it is impossible to know the exact proportion of

14C in the atmosphere when the bones were formed ORreference to the difficulty of measuring such small

percentages of 14C. 1

(iii) Why 210Pb is more suitable:• Idea that the half life of 210Pb is closer to the age of

recent bones [e.g. a greater proportion of 210Pb will havedecayed as the time elapsed is one or more half lives] 1

[6]

131. (a) 18 1 18 1 (1) O + p/H equals F + n (1)8 1 9 0 (1)[omitting the n with everything else correct = 1] 3

(b) Accelerated through 19 × 106 V / MVUsing linear accelerator / cyclotron / particle accelerator / (1)recognisable description (1) 2

(c) Time taken for half the original quantity/ nuclei /activity to decay (1)

Long enough for (cancer/tumour/body to absorb) and still beactive/detected (1)

Will not be in body for too long (1) 3

(d) Use of E = mc2 (1)Use of E = hf (1)Use of v = f λ (1)

λ = 2.4 × 10–12 m (1)

eg 9.11 × 10–31 × 9 × 1016 (×2)

f = 8.2 × 10–14 / 6.6 × 10–34 ecf

λ = 3 × 108 / 1.2 × 1020 ecf 4

(e) Conservation of momentum (1)

Before momentum = 0 (1)

so + for one photon and – for other (1) 2 max[14]

132. (a) Show that the speed is approximately 30 m s –1 Sets EK = mg∆ h (1)

Substitution into formulae of 9.8(1) m s–2 or 10 m s–2 and 50 m. (1)[Also allow substitution of 60 m for this mark]

Answer [31 m s–1. 2 sig fig required. No ue.] (1)

Eg2

1mv2 = mg∆ h

v2 = 2gh = 2 × 9.81 m s–2 × 50 m

v = 31.3 (m s–1) Answer is 31.6 m s–1 if 10 m s–2 is used

Also allow the following solution although this is notuniformly accelerated motion.

v2 = u2 + 2as

v2 = 0 + 2 × 9.81 m s–2 × 50m

v = 31.3 (m s–1) 3

(b) (i) Average braking force [ecf value of vl

For the equation 2

1mv2 (1)

[give this mark if this is shown in symbols, words or values]Attempts to obtain the difference between two energyvalues that relate to with and without the braking system orfor setting an energy value equal toForce × 80 m (1)

Answer [800 N if 30 m s–1 used; If 31.3 m s–1 or 31.6 m s–1 are

used accept answers in the range 1100 N – 1300 N; If 34 m s–1

is used answer is 2000 N] (1)

Eg (367875 J)ke after free fall – (273375 J)ke at 27m/s = 94500 JF × 80 m = 94500 JF = 1180 N

Also allow the following solution.

Selects v2 = u2 + 2as and F = ma (1)Attempts to obtain the difference between two forces /accelerations that relate to with and without the brakingsystem. (1)

Answer [800 N if 30 m s–1 used; If 31.3 m s–1 or 31.6 m s–1 are

used accept answers in the range 1100 N – 1300 N; If 34 m s–1

is used answer is 2000 N] (1)

Eg Braking force = (–) 750(m 802

)0()s m 27(

m 802

)0()s m 3.31( 221221

×−

−×

− −−

)

= (–) 1175 N 3

(ii) Why braking force of this magnitude not requiredAir resistance (would also act to reduce speed) (1)Or Number and/or mass of passengers will varyOr Friction [ignore references to where forces act for thismark. A bald answer ie ‘friction’ is acceptable]Or Accept some (kinetic) energy is transferred [not ‘lost’] tothermal energy [accept heat] (and sound)Or Work is done against friction 1

(iii) Explain whether braking force would changeQWOC: (1)EitherThe kinetic energy will be greater (because the mass of thepassengers has increased) (1)(hence) more work would have to be done(by the brakingsystem) (1)(The distance travelled, P to Q, is the same therefore)greater (braking) force is required (1)

OrMomentum (of the truck) will be greater (because the massof the passengers has increased) (1)Rate of change of momentum will be greater or [allow] thetime taken to travel (80 m) will be the same [if thecandidate writes ‘constant’ allow this if you feel they mean‘same’] (1)(Therefore) greater (braking) force is required (1)

Or(Allow) Change in velocity and the time taken (for the truckto travel 80 m) will be the same or (Average) deceleration /acceleration will be the same [accept ‘constant’ if theymean ‘same’. Also accept any fixed value for acceleration

eg 9.8 m s–2] (for greater mass of passengers) (1)(since) F = ma and mass has increased (1)A greater (braking) force is required (1) 4

[11]

133. (a) (i) Additional heightAnswer [ 5 (m)] (1)

Eg distance = area of small triangle = 0.5 × 1 s × 10 m s–1 = 5 m 1

(ii) Total distance travelled [Allow ecf of their value]Distance travelled between 1 s and 4s [45 m] (1)Answer [ 50 m] (1)

Eg distance fallen = area of large triangle

= 0.5 × 3 s × 30 m s–1

= 45 m total distance =45m + 5m = 50m 2

(b) Objects displacement40 m (1)Below (point of release) or minus sign (1)[Ecf candidates answers for additional height and distance ieuse their distance – 2 × their additional height] 2

(c) Acceleration time graphLine drawn parallel to time axis extending from t = 0 (1)[Above or below the time axis]The line drawn parallel to the time axis extends from 0 s to 4 s (1)[If line continues beyond or stops short of 4 s do not give this mark]

Acceleration shown as minus 10 m s–2 (1)[This mark is consequent on the second mark being obtained] 3

[8]

134. (a) Account for the forceWhen the flea pushes (down) on the surface the surface[accept ground, not earth] pushes back / upwards (1)with an equal (magnitude of) force (1)

[A statement of Newton’s 3rd law gets no marks – it must be applied] 2

(b) (i) Show acceleration is about 1000 m s –2

Either Selects v2 = u2 + 2as Or two appropriate equations ofmotion (1)Correct substitution into the equation (1)[Do not penalise power of ten error. Allow 0.4 mm and 0.9 m

s–1 substitutions for this mark.]

Answer [in range (1025 – 1060) m s–2, must be given to atleast 3 sig fig. No ue] (1)

Eg (0.95 m s–1)2 = 2 × a × 0.44 (× 10–3) m

a = 1026(ms–2)

OrSets changing Ke = work done (as legs expand) (1)Correct substitution into the equation (1)

Answer [1030 m s–2, must be given to at least 3 sig fig. No ue] (1)

Eg ∆ Ke = average F × height

½ m (0.95 m s–1)2 = m × a × 0.44 × (10–3) m

a = 1026 (m s–2) 3

(ii) Resultant force(Allow ecf)

Answer [4.1 × 10–4 N. 4(.0 ) × 10–4 N if 1000 m s–2 used. Ue.] (1)

Eg Force = 4 × 10–7 kg × 1030 m s–2 = 4.12 × 10–4 N 1

(c) (i) What constant force opposes upward motionThe weight of / gravitational attraction / gravitational force /gravitational pull / force of gravity / accept pull of earth(on flea) (1)[Not just ‘gravity’. Accept bald answers ie ‘weight’] 1

(ii) Change in height

Selects s = (2

uv +)t or uses v = u + at (to find a) then either

v2 = u2 + 2as or s = ut + 2

1at2 (1)

Correct substitution (1)[If two equations are used ‘a’ is negative]Answer [4.4(2) cm. Do not accept 4.5 cm] (1)[Nb the correct answer can be obtained from omitting ut andusing +a – this would get 1 /3]

[Use of s = ut + 2

1at2 or v2 = u2 + 2as with lal = g and u = 0.95

m s–1 will get 1/3 if no attempt is made to find ‘a’. For

candidates who use a = 1000 m s–2 from (b)(i) give no marks]

Eg s = )2

s m 95.00(

1−+9.3 × 10–2 s

= 0.0442 m

Or

a = 2

2

1

s m 2.10m 1093.0

s m 95.0 −−

−

−=×

−

s = 0.95 m s–1 × 9.3 × 10–2 s + 2

1 – 10.2 m s–2 (9.3 × 10–2 s)2

= 0.0442 m

or 0 = (0.95 m s–1)2 + 2 – 10.2 m s–2 s hence s = 0.0442m 3[10]

135. (a) Identify particle

Alpha (particle) / Helium nucleus/ He42 / 4

2He / α42 /

42

42

42 aalpha/alph/α /α 1

(b) Momentum of particleMomentum equation [In symbols or with numbers] (1)

Either

Correct substitution into 2

1mv2 = energy (1)

Use the relationship to determine the mass [6.6 × 10–27 kg] (1)

Answer [9.3 × 10–20 (kg m s–1) Must be given to 2 sig fig. Nounit error] (1)

Or

Rearrangement of Ek = 2

1mv2 to give momentum ie

v

E2 K (1)


Answer [9.3 × 10–20 kg m s–1. Must be given to 2 sig fig. Nounit error] (1)

Eg 2

1m(1.41 × 107 m s–1)2 = 6.58 × 10–13 J

m = 217

13

)s m 1041.1(

J 1058.62−

−

×××

= 6.6 × 10–27 kg

momentum = 6.6 × 10–27 kg × 1.41 × 107 m s–1

= 9.3 × 10–20 (kg m s–1)

Or

Momentum = 17

13

s m 1041.1

J 1058.62−

−

×××

= 9.3 × 10 (kg m s–1) 4

(c) Consistent with the principle of conservation of momentum(Since total) momentum before and after (decay) = 0 (1)State or show momentum / velocity are in opposite directions (1)[Values of momentum or velocity shown with opposite signswould get this mark]

Calculation ie 3.89 × 10–25 kg × 2.4 × 105 m s–1 = 9(.3) × 10–20 (kg m s–1) (1)

Eg 3.89 × 10–25 kg × 2.4 × 105 m s–1 = 9(.3) × 10–20 kg m s–1 3[8]

136. (a) Calculate the ratio the densities of the atom and the nucleusDensity equation [In symbols or numbers] (1)Show the relationship between density and radius. (1)[Candidates who start by stating that density is inverselyproportional to the radius cubed would get both these marks.Candidates who show an expression where the mass is

divided by 3

3

4rπ would set both these marks. Candidates who

write Ratio = (1/105)3 would get both of these marks.]

Factor 10–15 established. [Some working must be shown forthis mark] (1)

Eg (Density)atom = 3atom3

4r

m

π or Density

3

1

rα

(Density)nucleus = 3nucleus3

4r

m

π

3

atom

nucleus

nucleus

atom )(Density)(

(Density)

r

r=

= (10–5)3

Assumption – (entire) mass of the atom is concentrated in thenucleus[there must be a reference to the nucleus] (1)[eg mass of the atom =/approx – mass of the nucleus; most /majority of the atom’s mass is in the nucleus. The followingwould not be awarded marks; The atom is mostly emptyspace; mass of the electrons is negligible; the nucleus is avery dense.] 4

(b) ObservationA very small percentage of particles [accept ‘very few’ notjust ‘a few’. Do not accept ‘some’] are deflected throughangles greater than 90° / are back-scattered / deflected back. (1)[Allow; nearly all / most (alpha) particles pass through(the atom) without being deflected (showing the atom isvirtually empty space).][Accept ‘nearly all’, not ‘many’ for the word ‘most’.] 1

[5]

137. (a) How a beta-minus particle ionisesWhen a beta particle removes [accept repel] an electronfrom an atom / molecule (1) 1

(b) How ionisation determines rangeState that each ionisation requires energy (1)The energy (to ionise) is obtained from the (transfer of)(kinetic) energy of the beta particle (which is therefore reduced) (1)Along its path it produces many ionisations until all its(kinetic) energy is used up (1)The more ionising a particle the shorter its range or the lessionising the greater the range (1)[Candidates may give the wrong reason for ionisation or evencompare alpha and beta but still award this mark.]Max 3 marks from 4

[Note that the word –kinetic– is not essential for marks 2 and 3] 3

(c) Why more ionisation is produced towards the end of its range(Towards the end of its range) the beta particle is travellingslower or has less kinetic energy (than at the beginning of its range) (1)(as a result it takes longer travelling a given length) andtherefore has more (close) encounters with atoms / moleculesor more opportunities to ionise (atoms / molecules)or will remain in contact (with atoms / molecules) longeror will collide with more (atoms / molecules per unit length)or ionisation (of atoms/molecules) is more frequent (towards end of range) (1) 2

[6]

138. (a) n is (number of) charge carriers per unit volume or

number density or (number of) charge carriers m–3 orcharge carrier density(1)

[allow electrons]

v is drift speed or average velocity or drift velocity(of the charge carriers) (1)

[just speed or velocity scores zero] 2

(b) / A and Q A s or / Cs–1 and Q C (1)

n m–3 (1)

A m2 and v m s–1 (1)[If no equation written assume order is that of equation] 3

(c) (i) n 1 and Q Need all three 1

(i) Ratio vA/ vB less than 1 following sensible calculation (1)Ratio = ¼ // 0.25 // 1:4 (1)(ratio 4:1 scores 1)[4vA:1vB scores 1] 2

[8]

139. (a) Use of P = IV (1)Current in lamp A – 2 A (1)

[0.5 A scores zero unless 24 = I × 12 seen for 1st mark] 2

Example of answerI = P ÷ V = 24 W ÷ 12VI = 2A

(b) (i) Voltmeter reading = 12 V (1) 1

(ii) p.d. across R2 = 6 V or their (b)(i) minus 6V (1)Use of R = V/I (1) conditional on first markR2

Answer to this part must be consistent withvoltmeter reading and if voltmeter reading is wrongthis part has a max 2. If (b)(i) = 15 V then need to see

If (b)(i) = 6V or less they are going to score zero for this section. 3

(iii) current through R1 = 5 A (1) ecf answers from (a) 1

Example of answerCurrent through R1 = 2 A + 3 A = 5 A

(iv) p.d. across R1 = 3 V (1) ecf (15V minus their (b)(i)) 1

Example of answerp.d. across R1 = 15 V – 12 V = 3 V

(v) R1 1

Example of answerR1 = 3 V ÷ 5A = 0.6[accept fraction 3/5]

[9]

140. (a) (i) EI (1) 1

(ii) I2R (1) 1

(iii) I2r (1) 1

(b) EI = I2R + I2r or E = IR + Irecf Must use values (a)(i)-(iii) 1

(c) I for circuit given by Imax = E / r or substitution of5000V into the equation (1)(for safety) need I to be as small as possible (1) 3

[7]

141. (a) (i) Reference to a temperature related gas law (1)[V/T= constant or p/T= constant or pV/T = constantor pV = nRT; just symbols acceptable or word equivalentbut not Pressure law or Charles’ law]At absolute zero, V = zero or p = zero or pV = zero (1) 2

(ii) temperature or at absolute zero the molecules haveno kinetic energy (1)[do not accept depends on, is related to etc]

(implies) at absolute zero molecules stationary ornot moving or still or speed / rms of molecules is zero (1)[If particles/atoms used for both statements 1/2] 2

(b) Kelvin [absolute thermodynamic scale] 1

(c) (i) Use of pV = nRT (1)Use of 300 K (1)n = 4.0 moles (1)mass of air = 0.12 kg (1) ecf their n[If no temp conversion n = 45 moles, mass = 1.3 kg scores 2/4] 4

Example of answer

n = (1.0 × 105 Pa × 0.10 m3) ÷ (8.31 J K–1 mol–1 × 300 K)n = 4.0 moles

mass of air = 4.0 moles × 0.029 mole kg–1 = 0.12 kg

(ii) nT = constant or or calculation ofinitial density (1)correct use of above equations using Kelvintemperatures or calculation of final density (1)ratio = 3/5 or 0.6 (1) consequent on gaining method marks

[ratio of 1.7 or 5/3 could score 2][calculation of new mass in oven = 0.072 kg scores 1]

[If Kelvin not used in (c)(i) do not penalise here.Ratio is 0.12 – again consequent on method marks] 3

Example of answer227 °C = 500K 27 °C = 300 K 500 300 = 300 ÷ 500 = 0.6

[12]

142. (a) (i) Volume of gas (1)amount of gas or mass of gas or number of moles of gas (1) 2

(ii) Suitable diagram to include following labelled items(Trapped) mass of gas (1)method of indirectly heating gas (1)pressure gauge/reader/scale/mercury manometer (1)thermometer (1)

[wrong experiment e.g. Boyle’s Law 0/4] 4

(iii) precaution;Minimise amount of gas not in water bath, stirring,allowing time for gas to reachtemp, parallax errors,ANY ONE

[not insulating the beaker or the water bath][not repeat readings] 1

(b) Axes labelled with variables and units (1)Straight line graph with positive gradient (1)+ve intercept on pressure axis and meeting temp axisat –273 °C OR graph through origin if Kelvin scaleused and zero written where axes cross. (1)

[if variables other than p and T used 0/3] 3[10]

143. (i) Sun now & as red giant

Similarity: (nuclear) fusion / burning (in core) OR mass (1)

Difference: H vs. He fusion / r.g. lower (surface) T/ r.g. higher core T /r.g. lower (mean) density [assume r.g. referred to if not specified] (1)

(ii) White Dwarf star terms

Hot, low, (surface) area, off (the main sequence)

Any three correct (1)

All four correct (1) 4[4]

144. (i) Large mass star fusion rate


(30 M star) fuses at a greater rate AND spends less time on m.s. (1)[accept power, luminosity]

(30 M star) has greater temperature / (gravitational) forces /pressure (1)

leaves main sequence after hydrogen (and/or He) burning ceases /H fuel depleted (in core) [accept H “used up”] (1) 4

(ii) 2.2 M core remnant

Neutron star (1) 1

(iii) Sun evolutionary phases

Red giant (1)

White dwarf (1)

Black dwarf (1)

[–1 mark per error only if more than three phases circled] 3[8]

145. (i) Stress-strain graph regions

Neckings C or D (1)

Elastic deformation = A or B (1)

Plastic flow = E [ignore extra C or D] (1) 3

(ii) Young modulus calculation

Attempt at gradient / stress ÷ strain (1)

Sensible pair of values [from linear region, ignore × 10n] (1)

1.35 [allow 1.30– 1.40] (1)

× 1011 Pa [or N m2] (1) 4

(iii) Second material

Straight line [allow slight curvature at end] (1)

Less steep than original line (1)

Stops at ε = 2.6 × 10–3 (1) 3[10]

146. (a) Calculation of angular speed

Use of ω = 2π /T (1)

7.27 × 10–5 [2 sig fig minimum] (1) 2

2π /(24h × 3600 s h–1) = 7.27 × 105 rad s–1

(b) (i) Calculation of acceleration

Use of a = rω 2 OR v = rω and a = v2/r (1)

0.034/031 m s–2 (1) 2

(6400 × 10J m)(7.27 × 10–5 rad s–1)2

= 0.034 m s–2

(ii) Direction of accelerationArrow to the left (1)[No label needed on arrow. If more than one arrow shown, nomark unless correct arrow labelled acceleration] 1

(iii) Free-body diagramArrow to left labelled Weight/W/mg/pull of Earth/gravitationalforce (1)Arrow to right labelled Normal reaction/N/R/push of Earth (ORground)/(normal)contact force (1)

[Don’t accept “gravity” as label][More than two forces max 1][Diagram correct except rotated gets 1 out of 2] 2

(iv) How the acceleration is producedN is less than W (1)Resultant (OR net OR unbalanced) force towards centre (1)

[Accept downward / centripetal for towards the centre, butnot as an alternative to “resultant”] 2

[9]

147. (a) Experiment

Scheme for timing methods:

QOWC (1)Use f = 1/T OR f = (number of cycles/time taken) (1)Apparatus (1)Principle of method (1)One precaution for accuracy (1) Max 4

Examples for last three marks:

Stopclock / stopwatch (1)Measure time taken for a number of cycles (1)

ensure vertical oscillations / not exceeding elastic limit (1)OrMotion (OR position) sensor and datalogger (OR computer) (1)Read time for one (or more) cycles from displacement-time graph (1)Read time for several cycles / ensure vertical oscillations (1)OrLight gate and datalogger (OR computer) (1)Computer measures time interval between beam interruptions (1)Use narrow light beam / position gate so beam cut at equilibriumposition / ensure vertical oscillations (1)OrVideo Camera (1)Read time for one (or more) cycles from video (1)Read time for several cycles / ensure vertical oscillations (1)

[Mark other reasonable techniques on the same principles]

Scheme for strobe method:

QOWC (1)Use stroboscope [Accept strobe] (1)Adjust frequency until mass appears at rest (1)Find highest frequency at which this happens (1)Repeat and average / ensure vertical oscillations (1) Max 4

Scheme for use of T m / k )/ T e / g QOWC (1)Calculate T from measured (OR known) m and k usingT m/k) (1) or calculate T from measured e known g usingT e/g (1)Use f = 1/T (1) Max 2[e is the extension of the spring produced by weight of mass m]

[Do not give any credit for experiments to measure the resonantfrequency of the system]

(b) (i) GraphAxis labels and single peak (1)Rounded top and concave sides (1)f0 marked on the frequency axis at, or just to right of, peak.[Amplitude at f0 should be at least 75% of maximum amplitude] (1)

[A sharp kink loses mark 2 only][Graphs with multiple peaks lose marks 1 and 2; f0markedcorrectly on lowest frequency peak for mark 3][Ignore whether or not curve goes to origin] 3

(ii) Name of phenomenonResonance (1)

[Mark this independent of whether graph is correctDo not accept “resonant frequency”] 1

(iii) Footbridge applicationPeople walking / wind / earthquake can cause vibration / act as adriver / apply regular impulses (1)If resonance occurs OR if frequency equals / is close to f0 we mayget large / dangerous / violent oscillations OR large energytransfer OR damage to bridge (1) 2

[10]

148. (a) (i) How we know the speed is constant

Crest spacing constant / circular crestsOr wavelength constant / equal wavelength (1)

[Accept wavefront for crests][Don’t accept wave] 1

(ii) Calculation of speedλ is 10 mm (1)[Allow 9 to 11]Use of v = fλ (1)

0.40 m s–1 (1)

[Allow 0.36 to 0.44Allow last two marks for correct calculation from wrong wavelength] 3

(40Hz)(10 × 10–3 m)

= 0.40 m s–1

(b) Line X

1st constructive interference line below PQ, labelled X (1)

[Accept straight lineIgnore other lines provided correct one is clearly labelled X] 1

(c) (i) Superposition along PQConstructive interference / reinforcement / waves of largeramplitude / larger crests and troughs (1)Crests from S1 and S2 coincide / waves are in phase / zero phasedifference / zero path difference (1)Amplitude is the sum of the individual amplitudes (OR twice theamplitude of the separate waves) (1) 3

(ii) TableA constructive (1)B destructive (1) 2

[10]

149. (a) Amplitude

(i) Amplitude remains constant (1) 1

(ii) Amplitude decreases then increases (1)Amplitude is zero at node (OR half way between X and Y) (1) 2

(b) Phase difference

(i) Phase difference increases / is proportional to distance XP (1) 1

(ii) Up to node phase difference is zero / in phase (1)Beyond the node phase difference is / 180° / half a cycle / inantiphase (1)

[Do not allow completely out of phase] 2[6]

150. (a) Part of spectrum

Light / Visible / red (1) 1

Calculation of work functionUse of ϕ = hc/λ (1)

3.06 × 10–19 [2 sig fig minimum] (1) 2

(6.63 × 10–34 J s)(3.00 × 108 m s–1)/(6.5 × 10–7 m)

= 3.06 × 10–19 J

(b) (i) Meaning of stopping potentialMinimum potential difference between C and A / across thephotocell (1)Which reduces current to zero OR stops electrons reaching A /crossing the gap / crossing photocell (1) 2

(ii) Why the graphs are parallelCorrect rearrangement giving Vs = hf/e – ϕ /e (1)Gradient is h/e which is constant / same for each metal (1)

[Second mark can be awarded without the first if norearrangement is given, or if rearranged formula is wrong butdoes represent a linear graph with gradient h/e] 2

[7]

| €151. (a) Calculation of recession speedλ = 684 – 656 (1)Use of v/c λ /λ (1)

1.28 × 107 m s–1 (1)

[Substituting 684 for λ , leading to 1.23 × 107, loses last two marks] 3

(3.00 × 108 m s–1)(28 × 10–9 m)/(656 × 10–9 m)

= 1.28 × 107 m s–1

(b) Calculation of wavelength received from YBy Hubble’s law / v = Hd / v proportional to d, as d is doubled,v is doubled (1)λ = 56 / is doubled (1)712 nm (1)

[Bald answer of 712 nm, with no working or explanation, gets 2marks only]

[If candidate gets part (a) wrong, accept EITHER 56, 712 for thelast two marks (if they have avoided reusing the formula or madethe same mistake again) OR ecf (if they have repeated thecalculation but avoided the original mistake)] 3

[6]

152. (a) (i) P.d. across capacitor

Use of VR = I × R (1)

[allow one error of 103 in individual substitutions; disallow ifVR value is 6V]

VC = 6.0 V – 4.0 V (= 2.0V) (1)[No ecf] 2

Example of answer:

VR = 20 × 10–6 A × 2.0 × 105 Ω = 4.0 V

Hence Vc = 6.0 V – 4.0 V = 2.0 V

(b) Calculation of charge

Use of Q = C × V with 560 µ F & 2.0 V (1)

[Check correct equation is being used; allow power of 10 error incapacitance value. If capacitance value mis-transcribed, allow thisfirst mark only]

Answer 1.1 (2)mC (1120µ C) [no ecf] (1) 2

(c) Calculation of energy stored

Use of W = ½CV2 with given values, or W = ½ V with their Q,to get 1.1(2) mJ (1120µ J) or their correct answer. (1)[same numerical value as in (b)] 1

(d) Calculation of energy transferred

Use of E = QV, with their Q and V = 6.0 V, to get 6.7(2) mJ (6720µ J)or their answer [6 × value at part c] correctly found. (1) 1

(e) Main reason for energy difference

Energy is transferred to thermal / heat energy in / work is doneagainst, the resistance of the resistor in the circuit [NOT just ‘the resistance of the wires’, nor the ‘components’] (1)

[Do not credit vague reference to energy dissipation, nor ‘energyis lost to the surroundings’] 1

[7]

153. (a) Calculation of potential difference

Use of ½mpv2 with v = 2.77 × 105 m s–1

and mp = 1.67 × 10–27 kg (1)

Use of eV with e = 1.60 × 10–19 C (1)[beware confusion of v and V]

Answer = 400(.4) / 401 V (1)[If data used to 2 sf, → 380V, 384V or 364V, allow 2/3] 3

Example of answer:

eV = ½ mpv2

V = C 106.12

)s m 1077.2( kg 1067.1

2 19

215272

−

−−

×××××

=e

vm p

= 400V[beware unit error of eV here]

(b) Add second path to diagram

Path at B stays equidistant from that at A [gauge by eye] (1) 1

(c) (i) Add α path to diagram

Added path at A [allow through letter A] also curves upwards (1)

But is less curved than the original, straight beyond plates andcontinues to diverge from it (1) 2

(ii) Explanation

Charge on a is double that on proton / α has 2 protons /force on a is double force on proton. (1)

Mass of a particle is (approx) 4 times / more than double that of theproton. (1)

[hence acceleration is approximately halved].

[Ignore reference to F = Bqv; do not credit reference

to He42 unless implication of numbers 4 and 2 is made clear]. 2

[8]

154. (a) Two deductions [not simplv word descriptions of features of the diagram]

The gravitational potential is increasing with height / when movingaway from the Earth / Work must be done to move away from the Earth (1)

[Ignore idea that V ∝ r

1; in words or symbols]

The field is non-uniform / radial / Field strength decreases withheight / when moving away from the Earth (1) 2

(b) Entry speed at Earth’s atmosphere

| –1MJkg–1 – (– 61MJkg–1) = 60MJkg–1 [accept ± 60MJkg–1] (1)

Loss of GPE/ Gain in KE of spacecraft = m (1)

Statement / use of ½ mv2 = m∆ V OR ½v2 = ∆ V/(= 60 MJ kg–1) (1)[either of these statements also earns the second mark, if not already awarded]

[See v2 = 1.2 × 108

Answer 1.095 × 103 m s–1 /10950 m s–1/11.0 km s–1 [more than 2 sf] (1) 4

(c) Showing relative distance

Use of Newton’s Law; FE = 2E

E

r

mGM or FM = 2

M

M

r

mGM (1)

M

E

M

E

M

E2

M

2E

M (or)

M

r

r

M

M

r

r== (1)

[or equivalent re-arrangement]

)1(.81kg104.7

kg100.622

24

2M

2E =

××

=r

r [or equivalent] (1)

[correct relationship, expressed in terms of numerical values][If inverted, then MM:ME = 0.0123]

So ]11.0[or 981E

M

M

E ===r

r

r

r (independent mark) (1)

[Stating 81 =100 at the third mark stage does not, alone, earn the 4th mark]

[Beware ambiguity or transposition of r values at steps 2 or 3] 4[10]

155. (a) (i) Direction of current

Position 1 = Q to P / anticlockwise / to the left (1)Position 3 = P to Q / clockwise / to the right [both needed; arrows added to diagram may givecurrent directions at 1 & 3]

Position 2 = no current (1) 2

(ii) Current calculation

Use of t

BA

t ∆∆

=∆∆ )(φ

, or ε = Blv, = 2 × 10–2 T × 0.12 m × 0.05 m s–1 (1)

[ignore power of 10 errors in dimension and velocity values]

(Emf =) 1.2 × 10–4 V (1)

I = R

V or I =

R

E seen or used (1)

Answer = 6.0 × 10–5 A or 60µ A [ecf their emf] (1) 4

(b) Uniform acceleration?

QoWC (1)

Magnitude of current would be increasing as frame movesthrough position 1 (or position 3) (1)

Magnitude of current would be greater for position 3 than 1[Beware comparison of position 3 with position 2 here] (1)

Reference to increased rate of flux cutting / increased rate of flux change / increased area swept out per second (1)(Beware suggestion that B or flux density is changing)

So induced emf is greater (1)

Current for position 2 is zero [Do not credit equal and opposite currents cancelling] Both needed (1)

Since flux linkage is constant / (net) rate of flux cutting is zero / Emfs in PS and QR are equal and opposite Max 4

[10]

156. (a) (i) Calculate maximum current

Recall of P = IV (1)


Example of calculation:P = IVI = 5.9 W / 12.0 V= 0.49 A 2

(ii) Show that resistance is about 24 Ω

Recall of V = IR (1)

Correct answer to 3 s.f. [24.5 Ω] [no u.e.] (1)

Example of calculation:R = 12 V / 0.49 A= 24.5 Ω 2

(b) (i) Calculate current

Use of correct circuit resistance (1)


Example of calculation:I =V / R= 12 V ÷ (24.5 Ω + 2 Ω)= 0.45 A 2

(ii) Calculate power

Recall of P = IV and V = IR (accept P = I2R) (1)

or P = R

V 2

Correct answer [5.0 W] (1)


P = I2R

= (0.45 A)2 × 24.5 Ω= 5.0 W 2

(c) Increase in power available to pump

e.g. lower resistance in wire thicker wire, panel nearer to motor (1)(accept relevant answers relating to panels, e.g. more panels) 1

[9]

157. (a) (i) Explain upward force is about 0.1 N

Correct answer for force to 2 s.f. [(–)0.092 N] [no ue] (1)

Explanation that negative means upwards (1)

Example of calculation:W = mg

= 0.0094 kg × 9.81 N kg–1

= –0.092 N 2

(ii) Label balloon diagram ands show that weight is about 0.07 N

Tension + arrow (1)Weight + arrow (1)Weight = 0.068 N (1)

(Do not accept ‘gravity’ for ‘weight’) 3

(b) (i) Label 2 nd balloon diagram

Weight (1)Air resistance (1) 2

(ii) Expression for vertical component

T cos 43° / upthrust – weight / 0.16 N – 0.068 N / (1)(accept T sin 47°) 1

(iii) Calculate tension in string

Correct expression showing vertical forces on balloon (1)

Correct answer (0.13 N) (1)

Example of calculation:T cos 43° = 0.16 N – 0.068 NT cos 43° = 0.092 NT = 0.13 N 2

(c) Explain change in angle

Air resistance increases (1)

Horizontal component of tension increases (while verticalcomponent stays the same) (1) 2

[12]

158. (a) (i) Name process

Refraction (1) 1

(ii) Explanation of refraction taking place

change in speed / density / wavelength (1) 1

(b) (i) Draw ray from butterfly to fish

refraction shown (1)

refraction correct (1) 2

(ii) Explain what is meant by critical angle

Identify the angle as that in the denser medium (1)

Indicate that this is max angle for refraction OR total internalreflection occurs beyond this (1)[angles may be described in terms of relevant media] 2

(iii) Explain two paths for rays from fish A to fish B

direct path because no change of medium/refractive index/density (1)

(total internal) reflection along other path /angle of incidence > critical angle (1)

direct ray correctly drawn with arrow (1)

total internal reflection path correctly drawn with arrow (1)

[lack of ruler not penalised directly] [arrow penalised once only] 4[10]

159. (a) Show that Ep lost is about 37 000 J

Recall of Ep = mgh (1)

Correct answer to 3 s.f. [37 300 J] [no ue] (1)

Example of calculation:Ep = mgh

Ep = 760 kg × 9.81 N kg–1 × 5 m= 37278 J 2

(b) (i) Show that Ek of projectile and counterweight is about 26 000 J

Correct calculation of Ep gained by projectile [10 800 J] [no ue] (1)

Correct calculation of Ek to 3 s.f. [26 200 J] [no ue] (1)


Ep gained by projectile = 55 kg × 9.81 N kg–1 × 20 m = 10 800 JEk = 37 000 J – 10 800 J= 26 200 J 2

(ii) State assumption

All lost gpe → ke of projectile and counterweightOR Mass of moving arms negligibleOR No loss of energy to /work done against friction/air resistance (1) 1

(iii) Explain term 1/2 × 760 kg × (v/4)2

2 points from:

Ek of counterweight

Ek = ½ mv2

Counterweight has speed v/4Due to lever arm ratio 1:4 (2) 2

(c) (i) Calculate time of flight

Use of s = ut + ½ at2 (1)

Correct answer [2.1 s]


for vertical motion, s = ut + ½ at2

21 m = 0 + ½ × 9.81 m s–2 × t2 (1)

t = √(21 m × 2 / 9.81 m s–2)t = 2.07 s (1) 2

(ii) Calculate distance travelled

Recall of s = vt (1)

Correct answer [46.6 m] (1)

Example of calculationhorizontal motion, s = vt

= 22.5 m s–1 × 2.07 s= 46.6 m (1) 2

[11]

160. (a) Ultrasound:High frequency sound / sound above human hearing range / soundabove 20 kHz / sound too high for humans to hear (1) 1

(b) (i) Pulses used:to prevent interference between transmitted and reflected signals /allow time for reflection before next pulse transmitted / to allow forwave to travel to be determined (1)

(ii) High pulse rate:

Greater accuracy in detection of prey–s motion / position / continuousmonitoring / more frequent monitoring (1) 2

(c) Size of object:Use of λ = v/f (1)Correct answer (0.0049 m or 4.9 mm) (1)[accept 0.0048 m or 0.005 m]

example:

λ = 340 m s–1/ 70000 Hz= 0.0049 m = 4.9 mm (accept 5 mm) 2

(d) Time interval:Use of time = distance / speed (1)

Correct answer (2.9 × 10–3 s) [allow 3 × 10–3 s][allow 1 mark if answer is half the correct value ie. Distance = 0.5mused] (1)

example:

time = 1 m / 340 m s–1

= 2.9 × 10–3 s 2

(e) Effect on frequency:Frequency decreases (1)Greater effect the faster the moth moves / the faster themoth moves the smaller the frequency (1) 2

[9]

161. (a) Diffraction diagram:Waves spread out when passing through a gap / past an obstacle(1)λ stays constant (1) 2

(b) Diagrams:Diagram showing 2 waves in phase (1)Adding to give larger amplitude (1) 2

(c) Information from diffraction pattern:Atomic spacing (similar to λ)Regular / ordered structureSymmetrical structureDNA is a double helix structure (2) Max 2

(d) Electron behaviour:(Behave) as waves (1) 1

[7]

162. (a) Force arrow diagram:Weight and upthrust correctly labelled (1)Tension in string shown downwards (1) 2

(b) Upthrust on balloon:Knowledge of: upthrust = weight of displaced air (1)Use of upthrust = ρgV (1)Correct answer (0.18 N) [allow 0.2 N] (1)

Example:

Upthrust =1.30 kg m–3 × 9.81 m s–2 × 4/3 π(0.15 m)3

= 0.18 N 3

(c) (i) Airflow diagram:Diagram showing at least three continuous lines around the balloon (1) 1

(ii) Type of airflow:Streamline / laminar (1) 1

(d) (i) Word equation:Weight + (viscous) drag = upthrust (1) 1

(ii) Terminal velocity:

gr ρπ 3

3

4 = upthrust = value obtained in (b) [or 0.2 N] (1)

correct substitution into mg + 6πr ηv = gr ρπ 3

3

4 (1)

Correct answer (202 m s–1) [196 – 202 m s–1 to allow for rounding

errors] [if 0.2N is used v = 590 m s–1] (1)

Example:

v = (0.18 – 0.17) / (6π × 1.8 × 10–5 × 0.15)

= 202 m s–1 3

(iii) Comment:Any one of:Air pressure also acts on balloon / becomes less with heightAir becomes less dense with heightUpthrust becomes less with heightRelationship only valid for small objects (1) Max 1

[12]

163. (a) (i) Diagram:i and r correctly labelled on diagram (1)i = 25 +/– 2° (1)r = 38 +/– 2° (1)[allow 1 mark if angles measured correctly from interfaceie. i = 65 +/– 2°, r = 52 +/– 2°] (1) 3

(ii) Refractive index:

Use of gµa = sin i / sin r [allow ecf] (1)Use of aµg = 1/gµa (1)

example:gµa = sin 25 / sin 38 = 0.686

aµg = 1/ gµa = 1.46 2

(b) Ray diagram:Ray added to diagram showing light reflecting at interface withangles equal (by eye) (1) 1

(c) Observation:Incident angle > critical angle (1)T.I.R occurs (1) 2

(d) largest angle:sin C = 1/1.46 (allow ecf) (1)

C = sin–1 (1/1.46) = 43° (1) 2[10]

164. (a) Deformation of spring:As spring must return to original length when (compressive) forceis removed (1)Elastic (conditional on 1st mark) (1) 2

(b) Graph:4 points plotted correctly (1)all points plotted correctly (to within +/– ½ square) (1)straight line of best fit through points and origin (1) 3

(c) Stiffness:Use of stiffness = F/x taking any pair of values from the table orgraph (1)

= 0.53 N mm–1 (530 Nm–1) [allow 0.52 – 0.54 N mm–1] (1) 2

(d) Force exerted:Correct reading from graph= 3.2 N [allow 3.1 – 3.3 N] (1)

ORF = kx = 0.53 × 6 = 3.2 N [allow ecf] (1) 1

(e) Elastic energy:Energy stored = area under graph OR Energy stored = ½ Fx OREnergy stored = ½ kx2 (1)Correct values substituted [ignore powers of 10] (1)

Correct answer (9.6 × 10–3 J) [allow 9.3 – 9.9 × 10–3 J] (1) 3

Example:

Energy stored = ½ (3.2 × 6 × 10–3) = 9.6 × 10–3 JOR

Energy stored = ½ × 530 × (6 × 10–3)2 = 9.6 × 10–3 J

(f) New force to compress:Half of the original force / 1.6 N [allow ecf] (1) 1

[12]

165. (a) recall of p = mv [eqn or sub] (1)answer (1)

p = mv

= 2 × 0.024 × 0.88 (N s) = 0.042(24) N s OR kg m s–1 2

(b) recall of KE = 1/2mv2 OR p2/2m [eqn or sub] (1)answer (1)

KE = 1/2mv2

= 0.5 × 2 × 0.024 × 0.88 × 0.88 (J) = 0.0185(856) (J) 2

(c) (i) provided no external force acts (1)OR balls do not interact with/transfer momentum to anything else 1

(ii) v = momentum/mass = 0.042(24)/0.072 = 0.5833 (0.5867) (m s–1) (1) 1

(iii) 0.5 × 0.096 × 0.442 OR 0.5 * B9 * C9 * C9 (1) 1

(iv) 3 points from:can’t be one ball as too much KE (1)collision pretty elastic/not much loss of energy (1)so won’t be 3 or 4 or 5 balls (1)2 balls gives same energy (1) Max 3

(d) 2 points fromkinetic energy is lost (as sound/through deformation/to heat) (1)OR collisions not perfectly elasticMomentum still conserved (1)as the total ke decreases (column D) more balls are in motion (1) Max 2

[12]

166. (a) use of Q = CV OR statement or use of W = CV2/2 OR Q2/2C (1)answer (1)

W = CV2/2= 0.5 × 2500 × 2 × 2 (J) = 5000 J 2

(b) 1 correct value (1)All correct values; 1.62, 1.39, 1.16 (1)(1 mark for one correct or inappropriate sig figs) 2

(c) graph of ln (y) v. time (x) (1)appropriate scales and both axes labelled fully (1)points plotted properly (+/– 1 mm) (1)best fit line drawn (1) 4

(d) recognise that gradient = (–)1/RC (1)evaluate gradient (1)conversion days to seconds (1)obtain appropriate value for R (1)

gradient = (–) 0.92/(40 (days))

R = 40 × 24 × 3600 (s) / 0.92 × 2500 (F)= 1500 Ω

(OR method using graph of V v. t)recognise that time to Vo/e = RC (1)this time estimated (42–45 days) (1)conversion days to seconds (1)obtain appropriate value for R (1) Max 4

[12]

167. (a) energy (of proton) converts to mass (1)7 TeV > 251 GeV, (so enough energy present to create Higgs particle) (1) 2

(b) (i) calculate rest–mass energy of proton in J (1)comparison with 7 TeV (1)

rest mass energy of proton – E = mc2 = 1.67 × 10–27 × c × c

= 1.5 × 10–10 J

= 1.5 × 10–10 / 1.6 × 10–19 (eV) = 9.4 × 108 (eV)much less than 7 TeV.

OR 7 TeV = 7 × 1012 × 1.6 × 10–19 (J)

= 1.12 × 10–6 (J)

>> 1.5 × 10–10 J 2

(ii) Appropriate use of 1.6 × 10–19 OR energy from above in J (1)Answer (1)

momentum = energy/c = 7 × 1012 × 1.6 × 10–19 (J)/(3 × 108 (m/s)) =

3.73 × 10–15 (kg m s–1) 2

(iii) Attempt to use r = p/Bq (1)two correct subs into formula OR rearrangement (1)circumference => radius (1)answer (1)

r = p/BqB = p/rQ

= 3.73 × 10–15 / [(27000/2_) × 1.6 × 10–19] (T)= 5.4 T 4

(iv) Yes (stated or clearly implied) (1)because motion and force both horizontal OR motion/force/B must allbe perpendicular (1) 2

[12]

168. (a) 1 equating PE and KE (1)2 recall of mv2/r (1)3 find centripetal force = 2mg (1)4 force on rider = centripetal force + weight OR force = 3mg (1)5 hence “g-force” = 3 (1) 5

(b) height not a factor, so B is correct (1)(some will reach this conclusion via much longer routes) 1

[6]

169. (a) Displacement and distance?

Displacement has direction distance doesn’t or displacement isa vector, distance is a scalar or an explanation in terms of an example. (1) 1

[Candidates who describe displacement as “measured from a point”but do not mention direction or equivalent do not get this mark]

(b) (i) Position of train relative to A

300 m (1)

West (of) or a description[Do not accept backwards, behind or negative displacement] (1) 2

(ii) Velocity against time graph

Constant velocity shown extending from t = 0, positive / negative (1)[Above mark awarded even if graph does not reach or stop att = 4 min]

Constant velocity shown beginning at t = 4 min and ending at t =8 min, negative/positive (respectively) (1)

Values 2.5 (m s–1) or 3.75 (m s–1) or 3.8 (m s–1) seen[either calculated or on graph] (1)

Both values [allow their values] correctly plotted using a scale (1)[Only give this fourth mark if marking points 1 and 2 are correct.Also a clear scale must be seen eg 1, 2, 3, –1, –2, –3.The plot must be accurate to about half a small square.] 4

[7]

170. (a) (i) Speed of spade at impact with soil

Selects correct equation ie v = u + at or 2 appropriate equations (1)

Correct substitution into equation (1)

[Accept a substitution of –9.81 m s–2, only if it fits their definedpositive convention]

Answer

[to at least 2 sig. fig., 2.8 m s–1, no unit error. (1)

Allow use of g = 10 m s–2 giving 2.9 m s–1][Check that all working is correct for marks 2 and 3]

Eg v = 9.81 m s–2 × 0.29 s

= 2.84 m s–1

[This would get 3 marks even though the equation is not stated]

[Allow 2/3 for reverse argument – gives t = 0.3(05) s with

9.81 m s–2 and 0.3 s with 10 m s–2] 3

(ii) Acceleration in soil [Apply ecf]

Use of equation v2 = u2 + 2as or use of two appropriate equations (1)

[ignore power of 10 error and allow this mark even if theysubstitute the velocity value as v and not u]

[If acceleration of freefall used for acceleration, award 0/3]

Magnitude of acceleration [78.4 (m s–2), 80.7 (m s–2) or 81(m s–2)

if 2.84 m s–1 is used; (1)

84.1 (m s–2) if 2.9 m s–1 is used; 90 (m s–2) if 3 m s–1 is used]

[Check that all working is correct for mark 2]

Correct sign [minus] and unit (1)

[Only award this mark if there has been correct substitution intoequation or equations]

Eg 0 = (2.8 m s–1)2 + 2a5 × 10–2 m

a = – 78.4 m s–2 3

(b) Change in impact speed and acceleration in soil

Speed – the same (1)

Acceleration – a lower (1) 2[8]

171. (a) How constant measurable force is applied

(i) Newtonmeter/forcemeter (pulled to constant reading) or elasticband (pulled to fixed extension).[Allow a mass on the end of a string as the force, even if they donot make it clear that the mass being accelerated includes this mass] (1) 1[Do not allow a ramp at a fixed angle]

(ii)

Ticker tape Lightgate/sensor

Motion sensor Video / strobe

Ticker timer timer /datalogger /PC

Datalogger/PC

Metre rule /markings onthe track

[A labelled diagram can get both these marks.] (1)(1)[Do not give first 2 marks for ruler and stopwatch]

Description of distance measured and corresponding time or

any mention of v = t

d (1) 3

[Give this mark even if they have not obtained the first two marks]

(b) Additional measurements required for acceleration

Another velocity [accept ‘final velocity’] measurement or (zero)velocity at start (1)[Accept mention of double interrupter for first mark]

Either distance between velocity measurements / distance tosingle velocity measurement [If zero velocity is given for firstmarking point] (1)

Or time between velocity measurements / time to single velocitymeasurement from start (1)

[It must be clear what distance or time they are using to award thismark] 2

(c) How relationship is shown

Divide onaccelerati

Force (Applied) for each pair of measurements or Plot

graph of (applied) force v acceleration (1)

Ratio should give same value or graph gives straight line throughorigin[Could obtain these marks from a sketch graph] (1)

[A statement “force is proportional to acceleration” would not getthese marks] 2

(d) Why effect of friction must be eliminated

(In Newton’s law) the force referred to is the resultant force /unbalanced force / accelerating force acting on an object / adescription of the resultant force (1)

(If friction is not compensated for) the (measured) force would begreater than/not equal to the resultant force (by an amount equalto that needed to overcome friction) or the (measured) force wouldalso have to overcome friction (1) 2

[Accept ‘friction will reduce the acceleration’ for this mark][10]

172. (a) Complete statements

(i) ……... tyre/ wheel ……………… road(surface) (1)

(ii) ……...road(surface) ……………… tyre/wheel (1) 2

(b) Power

(i) Use of power = Fv (1)

Answer [4000W]

Eg Power = 400 N × 10 m s–1 (1)

= 4000 W [or J s–1 or N m s–1] 2

(ii) Work done (ecf their value of power)

Answer [1.2 × 106 J] (1)

Eg Work done = 4000 W × 5 × 60 s) = 1.2 × 106 J [or N m] 1

(c) Why no gain in Ek

Either(All the)Work (done)/energy is being transferred [not lost or through]to thermal energy [accept heat] / internal energy (and sound) (1)Overcoming friction (within bearings, axle, gear box but not roadsurface and tyres) / air resistance / resistive force/ drag (1) 2

[The information in the brackets is, of course, not essential for themark. However, if a candidate refers to friction between the roadsurface and the tyre do not give this mark]

Or (allow the following)

Driving force is equal to resistive force / friction / air resistance /drag or unbalanced force is zero or forces in equilibrium (1)

(Therefore) acceleration is zero (hence no change in speedtherefore no change in ke) (1) 2

[7]

173. (a) MeaningsSpontaneous: Happens independently of/cannot be controlled by/unaffected by chemical conditions/physical conditions/temperature/pressure or without stimulation/without trigger. (1)

[Do not accept random/cannot be predicted]

Radiation: alpha, beta and gamma and positron[give the mark if they name one of these] (1)

Unstable: (Nuclei) [not atoms] are (liable) to break up / decay /disintegrate or nucleus has too much energy or too many nucleons[not particles]/may release radiation/[Accept] binding force isnot sufficient/[Accept] binding energy is not sufficient/[Accept] too many/too few protons/neutrons (1)

[For this mark do not accept ‘nucleus has high energy’ or ‘..hasmany particles’] 3

(b) (i) Half life

Evidence of an average calculated ie have used more than justone value (1)

[Make sure to look at graph, if 2 sets of lines are seen,award this mark, even if there is no evidence in written answer]

Answer [(5.6 – 6) hours (20160 s – 21600 s)] (1) 2

(ii) Decay constant

Answer [Accept answers in the range 3.1 – 3.5 × 10–5 s–1 /

0.11(5) – 0.12(3) h–1] (1) 1

[ecf their value of half life]

[Do not accept Bq for the unit]

Eg λ = h 6

69.0/

s 36006

69.0

× = 3.19 × 10–5 s–1 / 0.12 h–1

(iii) Number of atomsUse of A = λN

Answer [in range (1.50 – 1.65) × 1011]

Eg N = 15

7

s102.3

Bq 105.0−−×

×

= 1.56 × 101 2[8]

174. (i) J C –1 Potential difference (1)

(ii) Product of two quantitiesPotential difference (1)

(iii) Rate of changecurrent (1)

(iv) Base quantitycurrent (1)

(for any part if two answers are given score is zero) 4[4]

175. (a) (As temperature of thermistor increases) its resistancedecreases [Do not credit the converse] (1)any TWO(slight) decrease in v (symbol, velocity or drift velocity)Large increase in n increases [accept electrons/charge carriers for n]A, Q and (pd) remain constant (1)(1)[ignore any reference to v staying constant] 3(n constant, can’t score mark for 3,4)

(b) (i) ammeter reading decreases (1)voltmeter reading unaltered (1)

(ii) ammeter is used to indicate temperature (1)

(iii) Assumption: ammeter; ideal/ has zero/negligible resistance (1)(Reference to meters is zero mark) 4

[7]

176. (a) Tungsten filamentQowc (1)I is not (directly) proportional to VTemperature of filament increases/ filament heats up/gets hotter as current/pd increases[accept bulb or lamp but not wire]Links temperature increase to resistance increasestungsten filament does not obey Ohm’s law/not anOhmic conductor or resistor. (1)(1)(1)Any THREE 4

(b) (i) Reading current from graph 1.5 A (1)answer 5.3 Ω (1)(misread current → 0/2)

Example of answerV = IRR = 8.0 ÷ 1.5 = 5.3 Ω 2

(ii) Addition of two currents (1)OR use of R = V/I and resistors in parallel formula1.5 + 1.2 = 2.7 A (1)ecf candidates’ current from above[If you see 2.7 A give 2marks] 2

[8]

177. (a) (i) Use of P = V2 / R OR P = IV and V = IR (1)Total R = 4.5 Ω (1) 2

Example of answer

R = V2 ÷ P = 12 V × 12 V ÷ 32 WR = 4.5 Ω

(ii) Use of 1/R = 1/R1 + 1/ R2……………… OR ΣR = 1/5R (1)[OR find total current, divide that by 5 and use V = IR]Resistance of strip = 22.5 Ω (1)ecf candidates’ R. 2

[common error is to divide by 5 → 0.9 Ω scores 0/2 butecf to next part gives l = 0.033 m which will then score 3/3]

(b) R = ρl / A or correct rearrangement (1)Correct substitution (1)Length = 0.82 m (1)ecf candidates’ R 3

Example of answer

l = RA/ρ = (22.5 Ω × 4.0 × 10–8 m2) ÷ 1.1 × 10–6 Ω ml = 0.82 m

(c) See P = V2 / R OR P = IV leading to increase in currentor decrease in resistance (1)more strips in parallel / material of lower resistivity (1)[not greater conductivity] 2

[9]

178. (a) E.M.F. = work done / charge OR energy transferred / charge (1)

OR power / current

[There is only one mark here and this is consistent withspecification but it must not be Joules or coulombs] 1

(b) (i) Use of V = IR (1)I = 2.0 A (1) 2

Example of answerI = V / R = 8.0 V / 4.0 ΩI = 2.0 A

(ii) Uses p.d. = 4.0 V (1)r = 2.0 Ω ecf their I (1) 2

Example of answerr = V / I = 4.0 V / 2.0 Ar = 2.0 Ω

(iii) Use of P = VI // I2R // V2/R (1)P = 16 W ecf their I (1) 2

Example of answerP = VI = 8 V × 2 AP = 16 W

(iv) Uses 4V or 2A × 2Ω or their I × r (1)see 5 × 60 s in an energy equation (1)energy = 2400 J (1) 3

Example of answerE = VIt = 4 V × 2 A × 5 × 60 sE = 2400 J

[10]

179. (a) (i) Specific heat capacity is the energy required when a kg /unit mass (1)undergoes a temp change of 1°C / 1K (1) 2OR equation(1 mark) terms defined (1 mark)[there must be a reference to a temperature so “energy to

raise a kg by 1°C” does not get the 2nd mark.

This is a definition which should be learnt. Also 2nd marklost if 1 C or 1°K]

(ii) It is the sum/total of the molecular/atomic (1)Potential and kinetic energies (1)[not particles, or gravitational potential energy] 2[it is not enough to say it is the KE and PE……….]

(b) Use of ∆E = mc∆T there must be a temperature (1)difference for this mark∆T = 50 °C or K (1)

Energy = 4.4 × 107 J (1)[Ignore any negative signs] 3

Example of answer

∆E = 800 kg × 1.1 × 103 J kg–1 K–1 × 50 °C

∆E = 4.4 × 107 J[7]

180. (a) Intensity and Luminosity

Luminosity = power [or energy / time, accept “per second”] (1)

Intensity = power (or energy / time) [e.c.f. from first mark] perunit area [accept per square metre] (1)

Luminosity: measured at star OR Intensity: measured at Earth /

depends on distance (from star) / observed OR W with W m–2

OR I = L ÷ 4 π D2 (1) 3

(b) (i) Wavelength of Sun

Use of Wien’s law [accept any attempted use] (1)

5.0 × 10–7 m (1) 2

(ii) Surface area of Sun

Use of 4 π r2 (1)

6.1 × 1018 (m2) (1) 2

(iii) Luminosity of Sun

L = σ A T4 [or L = σ T4 4 π r2] (1)

Correct substitution [e.c.f.] (1)

3.9 × 1026 W [accept 3.8 or 3.84 × 1026 W from 6 × 1018 m2] (1) 3[10]

181. (a) Main sequence mass requirement


(Main sequence requires) hydrogen fusion / burning (1)

Mass linked to gravitational forces / field [/energy] (1)

High forces [or temperature, pressure] required for fusion / burning / m.s. (1) 4

(b) Hertzsprung-Russell diagram

(i) Axes change in (fixed) multiples [accept exponential changes] (1)

x-axis multiple: × ½ OR × 2 (1) 2

(ii) L on diagonal falling line in lower right quadrant (1)

W indicated mostly in lower left quadrant (1)

R indicated mostly in upper right quadrant [not on main sequence] (1)

S in line with 10° [± 2 mm to centre of S, to left of 5000 K. on m.s.] (1) 4[10]

182. Parallax analogy

(i) 5 tan 84° [beware 5 / cos 84° = 47.8 m] (1)

47.6 m (1) 2

(ii) 2 AU / Earth orbital radius × 2 / Earth orbital diameter /

distance between Earth at a six month interval / 3 × 1011 m (1) 1

(iii) Inaccurate readings / difficult to measure AND small angles /movement relative to background (stars) (1) 1

[4]

183. (a) Elastic and Plastic behaviour

Plastic = permanent AND elastic = reversible [may be impliedanywhere] (1)

Elastic: bonds stretch but not broken / atoms move apart butthen return (1)

Plastic: bonds broken (when stressed) / atoms do not return tooriginal position (after stress) (1) 3

(b) (i) Ultimate tensile strength

(3.6 – 3.7) × 108 N m–2 / Pa (1) 1

(ii) Young modulus calculation

Attempt at gradient / stress ÷ strain [ignore 10n] (1)

Valid pair of readings taken from graph [108 and 10–3 required] (1)

8.0 to 9.0 × 1011 N m–2 / Pa (1) 3

(iii) Tough or brittle explanation

Tough (1)

Any reference to plastic behaviour (1)

(Large area under) non-linear part of graph referred to (1) 3[10]

184. Definitions

(i) Stress = force ÷ area AND strain = extension ÷ original [initial] length (1) 1

(ii) E = stress ÷ strain [accept symbols here] (1)

E = ll

AF

/

/

∆ (1) 2

(iii) Radius “show that” calculation

Correct substitution in E = lA

Fl

∆ / A = 2.7 × 10–7 (m2) (1)

A = π r2 (1)

2.9 × 10–4 m / 0.29 (mm) (1) 3[6]

185. (a) Experiment

[Marks may be earned on diagram or in text]

Named light source plus polaroid (OR polariser ORpolarising filter) / Laser / Named light source andsuitable reflector (e.g. bench) (1)

2nd Polaroid plus means to detect the transmitted light (1)(i.e. eye OR screen OR LDR OR light detector ORinstruction to e.g. look through polaroids)Rotate one Polaroid [Only award if expt would work] (1)Detected intensity varies / No light when polaroids are at 90° (1)Maxima and minima 90° apart / changes from dark to light every 90° (1)[Use of microwaves, slits or “blockers”: 0/5Use of filters or diffraction gratings: lose first two marksUse of “sunglasses” to observe: lose mark 2] 5

(b) Why sound can’t be polarised

They are longitudinal / They are not transverse / Only transversewaves can be polarised / Longitudinal waves cannot be polarised /Because the (*) is parallel to the (**) (1)

(*) = vibration OR displacement OR oscillation OR motion of particles

(**) = direction of travel OR direction of propagation OR motion ofthe wave OR direction of energy transfer 1

[6]

186. (a) Table

[Ignore crosses. If more than one tick in a line, no mark.]Top line: To the left (1)Bottom line: Downwards (1) 2

(b) Calculation of rotation period

Use of T = 2πr/v or T = 2π/ω and ω = v/r (1)Correct answer [0.084 s] (1)

e.g.

2π (0.28 m)/(21 m s–1)= 0.084 s 2

(c) (i) How the angular speed is affected

ω is increased, plus correct supporting argument in formula or words (1)i.e. Since v = rω / T decreases / f increases / wheel must turn faster 1

(ii) Speedometer reading

Speedometer reading is too high because frequency (1)

(OR ω OR revs per second OR rate of rotation of wheel)is increased[Allow ecf from “ω decreased” in c (i)] 1

[6]

187. (a) (i) Table

λ f

2.4 (110)

1.2 220

0.8 330

All wavelengths correct (2)[One or two wavelengths correct gets 1]Both frequencies correct (1)[Accept extra zero following wavelength figure, e.g. 2.40.Accept units written into table, e.g. “2.4 m”, “220 Hz”] 3

(ii) Why nodes

String cannot move / no displacement / zero amplitude /no oscillation / phase change of π on reflection / two wavescancel out / two waves are exactly out of phase (1)(OR have phase difference of π OR half a cycle) /destructive interference 1

(b) Why waves with more nodes represent higher energies

More nodes means shorter wavelength (1)Momentum will be larger (1)[OR Allow 1 mark for “More nodes means higher frequency and E = hf”] 2

[6]

188. (a) Diagram

At least 3 crests drawn, with correct even spacing (1)[Judge by eye. Allow +/– 20 %. Check in centre of pattern.]Crests approximately straight opposite harbour entrance andcurved in the “shadow” region (1)Wavefronts get longer, but diffraction at the edges throughno more than 45° (1) 3

(b) (i) Values from graph

Period: 3.0 (1)[Accept 3]Maximum acceleration: 1.2[Accept 1.17, 1.18, 1.19, 1.20]

[Both values are needed for the mark. Ignore written

units, e.g. “3.0 s”, “1.2 m s–2”.] 1

(ii) Calculation of amplitude

Use of acceleration = (–)(2πf)2x (1)Use of f = 1/T (1)

Correct answer [0.27 m] (1) 3[Negative amplitude loses third mark]

e.g.

ω = 2π / (3.0 s)

= 2.09 rad s–1

A = (1.2 m s–2) / (2.09 rad s–1)2

= 0.27 m

(iii) Displacement graph

Cosine curve [Correct way up] (1)Axis labelled ( i.e. displacement / x / y / z) plus all (1)amplitudes consistent with previous answer ( i.e. within 1 square)plus unit 2 cycles shown with T = 3 s (1) 3[All zero crossings correct within 1 square]

[10]

189. (a) Why statement correct

Blue photon has more energy than red photon (1)

Why statement incorrect

Blue beam carries less energy per unit area per second / Blue beamcarries less energy per second / Blue beam carries less energy perunit area / Blue beam has lower intensity and intensity = energy per unitarea per second

Additional explanation

[Under “correct”] Blue has a higher frequency (OR shorter wavelength) /[Under “incorrect”] Blue beam has fewer photons (1)

[Allow reverse statements about Red throughout part a] 3

(b) (i) Meaning of work function

Energy to remove an electron from the surface (ORmetal OR substance) (1)[Don’t accept “from the atom”. Don’t accept “electrons”.]Minimum energy… / Least energy… / Energy to just…/ …without giving the electron any kinetic energy (1) 2

(ii) Calculation of threshold frequency

Use of φ = hf0 (1)

Correct answer [6.00 × 1014 Hz] (1)

e.g.

(3.98 × 10–19 J)/(6.63 × 10–34 J s) = 6.00 × 1014 Hz 2[7]

190. (a) Which transition

Use of (∆)E = hc/λ OR (∆)E = hf and f = c/λ (1)

Use of 1.6 × 10–19 (1)Correct answer [1.9 eV] (1)C to B / –1.5 to – 3.4 (1)[Accept reverse calculations to find wavelengths]

e.g.

(6.63 × 10–34 J s)(3.00 × 108 m s–1)/

(656 × 10–9 m)(1.6 × 10–19 J eV–1)= 1.9 eV 4

(b) Explanation of absorption line

QOWC (1)Light of this wavelength is absorbed by hydrogen (1)In the outer part of the Sun (OR Sun’s atmosphere) (1)

Absorbed radiation is reemitted in all directions (1)Transition from B to C (OR –3.4 to –1.5) (1) Max 4

(c) Why galaxy receding

Wavelength increased (OR stretched) / red shift /frequency decreased 1

[9]

191. (a) Show that

See ‘v = T

rπ2’ OR ‘ω =

T

rπ2’ (1)

Substitution of (60 × 60 × 24)s or 86400s for T (giving 7.27 × 10–5, no u.e.) (1)

Unit of ω

s–1/rad s–1 (1) 3

(b) Height above Earth’s surface

Statement / use of 2

2

2 OR

r

mGM

r

mv

r

mGM EE = = mrω2 (1)

[Equations may be given in terms of accelerations rather than forces][Third mark (from below) may also be awarded here if (rE + h) is used for r]

Correct value for r, i.e. 4.2(3) × 107 m (1)Use of h = their r – RE (1)

Correct answer = (3.58 – 3.60) ×107 m [no ecf] (1)

Example of answer:

r

mv

r

mGM E2

2=

rr

r

r

v

r

GM E 222

2

)( ωω===→

∴ GME = ω 2r3

3215

242211

32

)s 1027.7(

kg 1098.5kg m N 1067.6−−

−−

××××

==∴ω

EGMr

= 4.23 × 107 m

∴ h = 4.23 × 107 m – 6.38 × 106 m

= 3.59 × 107 m 4[7]

192. (i) Add to diagram.

Arrows at A and B, both pointing directly away from the nucleus. (1)[Arrow end (head or tail) need not touch A /B, but direction must be correct.Gauge by eye, accept dotted construction lines as indication of intent] 1

(ii) Calculation of force

Use of F = 20

21

4 r

QQ

π ε or F = 221

r

QkQ (1)

[ignore error/omission of ‘2’ and/or ‘79’ or ‘e’ or ‘1.6 × 10–19’ for this first mark,providing numerator clearly has a product of charges and denominator a distancevalue squared. Ignore power of 10 errors in values of Q or r]

2 × 1.6 × 10–19 C and 79 × 1.6 × 10–19 C seen (consequential mark, dependentupon correct use of equation previously) (1)

Correct answer = 1.6 – 1.7 N (1)

Example of answer:

213112

1919

20

21

)m 105.1(m F 1085.84

C) 106.12(C) 101679(

4 −−−

−−

×××××××××

==ππ εr

QQF

= 1.62 N 3

(iii) Effect on motion of α

Slows down [decelerates] and then speeds up again [accelerates].(both needed)[accept ‘slows down at A and speeds up at B] (1) 1

[5]

193. (a) (i) Direction of e.m.f.?Hub ‘–’ and Rim ‘+’.[Allow mark for either on its own, but not if contradicted.] (1) 1

(ii) Why a constant e.m.f.?

Reference to flux cutting / rate of change of flux / change offlux linkage due to spoke motion / spokes moving at rightangles to field / Reference to Faraday’s Law (1)

Constant rate of spin implies constant rate of flux cutting.[Link made clear] (1)[continuous process does not mean constant rate] 2

(iii) The time for one revolution

Use of ε = t

BA with ‘A’ recognisable as area of a circle (1)

[ignore power of 10 errors for e.m.f. and radius values, andinclusion of N = 24]

Correct substitution of all values [ but only N = 1 acceptable here] (1)

Correct answer 0.31 – 0.32 s (1)[t = 7.6s scores 1/3; t = 1.12s scores 0/3, t = 0.64s scores 1/3 here]

Example of answer:

εϕε BA

tt

BA

t=→==

s 317.0V 1025

m) 10(30 T 108.26

225

=×

××××=∴

−

−− πt 3

Alternative answer

Use of ε = Blv with v = (mean) velocity of spoke. ((1))

→ v = 2.98 m s–1 ((1))

Hence rim velocity = 2.98 × 2 = 5.96 m s–1.

1s m 96.5

m 3.022−

×==→ ππRIMv

rt = 0.316s. ((1))

[t = 0.63s scores 2/3 here]

(b) What effect?

(i) Reduced [accept ‘halved’] ANDRate of flux cutting is reduced / Fewer field lines are being cut /Component of Earth’s field perpendicular to the wheel is less /Flux through wheel is less / Area of wheel perpendicular tofield is less / Wheel is no longer perpendicular to the field (1)[do not credit answers suggesting changes in the field strength itself] 1

(ii) Increased / increasing ANDRate of flux cutting [etc.] would be increasing (1) 1

(iii) (Reduced to) zero [but not ‘very small’ / ‘negligible’, etc.] AND

No flux cut by spoke(s) / No component of the Earth’s fieldperpendicular to the wheel / No flux through wheel / Wheelis spinning parallel to the field / in plane of field (1)[but not just ‘∆Ф = 0’, nor ‘motion is not perpendicular to field’]

[Allow 1/3 for three correct statements of ‘ε’ outcome without anyexplanation, but only if score would otherwise be zero.]

[Disallow ‘breaking’ for ‘cutting’ on first occasion in entirequestion, but allow, ecf, thereafter] 3

[9]

194. (a) Describe propagation of longitudinal waves

Particles oscillate / compressions/rarefactions produced (1)

oscillation/vibration/displacement parallel to direction of propagation (1) 2

(b) Calculation of wave speed


Correct answer [7.2 km s–1] (1)


v = f λ

v = 9 Hz × 0.8 km

= 7.2 km s–1 [7200 m s–1] 2

(c) Determine if elephants can detect waves more quickly

Recall of v = s / t (1)

Correct answer for t in minutes or hours [about 6 minutes] orrelevant comment with 347 s or calculation of tidal wave speed

[0.35 km s–1] with comment [allow ecf] (1) 2


v = s / t

t = 2500 km ÷ 7.2 km s–1 OR v = 2500 km ÷ (2 × 60 × 60 s)

t = 347 s OR v = 0.35 km s–1

t = about 6 minutes (stated) / much less than hours / 2 h is 7200 s

OR 7.2 km s–1 >> 0.35 km s–1

[6]

195. (a) (i) Show that acceleration is about 1.7 m s –2

Use of appropriate equation(s) of motion (1)

Correct answer [a = 1.73 m s–2] [no ue] (1) 2


s = ½ at2

1.35 m = ½ × a × (1.25 s)2 OR a = 2 × 1.35 m / (1.25 s)2

a = 1.73 m s–2

(ii) Explain constant acceleration

No air resistance (1)

Accelerating force on each is constant / Resultant force remainsjust weight (1) 2

(b) Calculate weight

Recall of W = mg (1)

Correct answer [179 N] (1) 2


W = mg

= 105 kg × 1.7 N kg–1

= 179 N

(c) (i) Time of flight of ball

Recall of trigonometrical function (1)

Recall of v = u + at (1)

Correct answer [t = 18.1 s] (1) 3


vertical component of velocity = 45 m s–1 × sin 20°

= 15.4 m s–1

v = u + at

15.4 m s–1 = –15.4 m s–1 + 1.7 m s–2 × t

t = 30.8 m s–1 ÷ 1.7 m s–2

t = 18.1 s

(ii) Horizontal distance

Use of trigonometrical function (1)

Correct answer [766 m] [ecf] (1) 2


horizontal component of velocity = 45 m s–1 × cos 20°

= 42.3 m s–1

distance = 42.3 m s–1 × 18.1 s

= 766 m

(iii) Comment on this distance

[766 m ÷ 1600 m/mile = 0.48 mile] [ecf] – This is only about half amile (N.B. answer for (c)(ii) required to get this mark) (1) 1

[12]

196. (a) Meaning of superposition

When vibrations/disturbances/waves from 2 or more sources coincideat same position (1)

resultant displacement = sum of displacements due to individual waves (1) 2

(b) (i) Explanation of formation of standing wave

description of combination of incident and reflected waves/waves in opposite directions (1)

described as superposition or interference (1)

where in phase, constructive interference / antinodesOR where antiphase, destructive interference / nodesOR causes points of constructive and destructive interferenceOR causes nodes and antinodes (1) 3

(ii) Calculate wavelength

Identify 2 wavelengths (1)

Correct answer [2.1 × 10–9 m] (1) 2


(NANANANAN) X to Y is 2 × λ

λ = 4.2 × 10–9 m ÷ 2

= 2.1 × 10–9 m

(iii) Explain terms

amplitude – maximum displacement (from mean position)(can use diagram with labelled displacement axis) (1)

antinode – position of maximum amplitudeOR position where waves (always) in phase (1) 2

[9]

197. (a) (i) Calculate resistance

Recall of R = V/I (1)

Correct answer [8.65 Ω] (1) 2


R = V/I

R = 2.68 V ÷ 0.31 A

= 8.65 Ω

(ii) Show that internal resistance is about 0.4 Ω

Recall of relevant formula [V = ε – Ir OR lost volts = (ε – V) (1) ORε = I(R + r)] including emf

Correct answer [0.39 Ω] [no ue] [allow ecf if ε = I(R + r)] (1) 2


V = ε – Ir

r = (ε – V)/I

= (2.8 V – 2.68 V)/0.31 A

= 0.39 Ω

(iii) Comment on match to maximum power

Not matched [ecf for R in (a) (i) and r in (a)(ii)] (1)

Max power when internal resistance = load resistance (1) 2

(b) (i) Show that charge is about 14 000 C

Recall of Q = It (1)

Correct answer [14 400 C] [no ue] (1) 2


Q = It

= 2 × 2 A × 60 × 60 s

= 14 400 C

(ii) Calculate time for which battery maintains current

Use of Q = It OR use of W = Pt (1)

Correct answer [46 450 s or 12.9 h] (1) 2


t = Q/I

= 14 400 C / 0.31 A

= 46 450 s

(c) Explain effect on efficiency

Efficiency = I2R / I2(r + R) / Efficiency depends on R /(r + R) /more heat dissipated in cells / Efficiency is V/ and V decreases (1)

so efficiency is less (1)

[Must attempt explanation to get 2nd mark] 2[12]

198. (a) (i) Calculate ave speed from D8

Use of equations of motion to find correct answer

[15.2 m s–1][no ue] (1) 1


v = 7.6 m / 0.5 s

= 15.2 m s–1 [No ue]

(ii) Formula for E7

E6 + B7 OR 35.5 + 9.1 OR B4 + B5 + B6 + B7OR sum(B4:B7) OR 35.3 + 9.1 (1) 1

(iii) Use graph to find ave deceleration

line drawn – full width, 0 s to 2 s (1)

substitution of values in gradient formula (1)

correct answer [5.5 m s–2 (± 0.3 m s–2)] (1) 3


gradient = (28 m s–1 – 17 m s–1) / 2 s

= 5.5 m s–2 (± 0.3 m s–2) [ignore any negative sign]

(b) (i) Calculate average braking force

Recall of F = ma (1)

Correct answer [3300 N] [ecf] (1) 2


F = ma

= 600 kg × 5.5 m s–2

= 3300 N

(ii) State origin of force

friction between brake pad and disc (1)

[frictional force of road on tyres] 1

(c) (i) Calculation of kinetic energy from F6

Recall of Ek = ½ mv2 (1)

Correct answer [132 kJ] [no ue] (1) 2


Ek = ½ mv2

Ek = ½ × 600 kg × (21 m s–1)2

= 132 kJ

(ii) Explain gradient = braking force

Change in kinetic energy = work done by braking force (1)

work/distance = force (1)

OR

gradient = change in kinetic energy / distance (1)

= work done by braking force / distance = force (1)

(Showing units/dimensions of gradient consistent withforce gains 1 mark) 2

[12]

199. (a) Material properties:Strength – Force/ load/stress required to break / Strong– large force required to break (1)

Brittle – shatters/snaps/fractures / cracks (under force) /breaks with no/little plastic deformation/breaks with littlestrain [ignore reference to size of force needed eg.cracks easily – mark awarded for ‘cracks’] (1)

Plastic – does not return to original length (when load removed) /deformation is permanent (1) 3

(b) Maximum force:

Use of A = πr2 (1)Use of F = σ × A (1)Correct answer [193 (N)] (1) 3[accept 190 – 193 N to allow for rounding errors] [no u.e][2 out of 3 for correct reverse working out]

Example:F = σ × A

A = πr2 = (0.127 × 10–3)2 × π

= 5.07 × 10–8 m2

F = 3.8 × 109 Pa × 5.07 × 10–8 m2

= 193 N

(c) Extension calculation:

Use of ε = σ / E (1)Correct answer for ε [0.015] (1)Correct answer for extension [0.017 m] (1)[allow 0.016 – 0.017 m to allow for rounding errors]

[allow 1st 2 marks for correct substitution into E = Fl/xA] 3

Example:ε = σ / E

= 2.00 × 109 Pa / 1.31 × 1011 Pa= 0.015extension = ε × length = 0.015 × 1.1m = 0.017 m (1.7 cm)

(d) Polymer:Long chain (1)Of repeating units / of monomers / molecule /atoms (1)[1 mark only for long chain of molecules] 2

[11]

200. (a) Work function:Energy needed for an electron to escape the surface /to be released (from the metal) (1) 1

(b) How current produced:Any 3 from:Photon of light passes energy to an electronIf energy above the work function/frequency above threshold (1)(1)Electron released as a photoelectron / photoelectron released /surface electron released (1)Moving electrons produce a current 3

(c) (i) Intensity of light increased:More electrons released (1)

(ii) Frequency of light increased:Electrons gain more (kinetic) energy (1) 2

(d) Photon energy:Use of f = v/λ or E = hc/λ (1)

Correct answer for E (4.7 × 10–19 J or 2.96 eV) (1)[allow 3.0 eV] 2

Example:

f = v/λ = 3 × 108 / 4.2 × 10–7 = 7.1 × 1014 Hz

E = hf = 4.7 × 10–19 J or 2.96 eVOR

E = hc/λ = 3 × 108 × 6.63 × 10–34/ 4.2 × 10–7

= 4.7 × 10–19 J or 2.96 eV

(e) Max kinetic energy:Knowledge that kemax = energy calculated in (d) – φ (1)

Correct answer for kemax (0.26 eV or 4.2 × 10–20 J)

[allow 0.25–0.26 eV or 4.1 – 4.2 × 10–20 J and allow ecf from (d)] (1) 2

Example:kemax = 2.96 eV – 2.7 eV= 0.26 eV

(f) (i) Why current reduced:Many / some electrons will not have enough (kinetic) energyto reach the anode / only electrons with large (kinetic) energywill reach the anode (1) 1

(ii) Stopping potential:eV = (–) keV = ke / e = 0.26V (1) 1

[12]

201. (a) Plane polarised:Vibrations / oscillations (1)in one plane (1)ORdouble-headed arrow diagram (1)with vibrations / oscillations labelled (1) 2

(b) Polarising filter:• Intensity goes from maximum to minimum (1)• Twice per rotation / after 90° (1)• As filter only lets through vibrations in a particular plane (1) 3[marks may be gained from a clearly labelled diagram]

(c) Response of beetle:Changed direction by 90° / turned through a right-angle (1) 1

(d) No moon:Beetle moves in a random direction / in circles / appears disorientated (1) 1

[7]

202. (a) Circuit:Potential divider (1) 1

(b) Relay potential difference:4 V (1) 1

Example:5/15 × 12 = 4V

(c) (i) Resistance:Recall of R = ρL/A (1)Correct substitution of values into formula (1)Correct answer [98(Ω)] (1)[allow 97 – 98Ω to allow for rounding errors] [no u.e.] 3

Example:

R = (3.4 × 102 × 1.44) / (100 × 0.05)= 98 Ω

(ii) Combined resistance:Use of 1/RTot = 1/R1 + 1/R2 (1)Correct answer for R [4.8Ω] (1)[allow 4.7Ω – 4.8Ω to allow for rounding errors] 2

1/R = 1/98 + 1/5 (or = 1/100 + 1/5)R = 4.8 Ω

(iii) Relay voltage:P.d. across relay with ballast very similar to p.d acrossthe relay alone / p.d. = 3.9 V / p.d. lower (slightly) (1) 1

(iv) Train on track:Relay voltage becomes very small / zero (1) 1

(v) Wet ballast:

Any two–• Combined resistance now small / RT = 0.45 Ω• Relay voltage now small / V = 0.52 V• Relay voltage too small to trigger green light /

signal remains red (1)(1) 2[11]

203. (a) High viscosity flow:Slower (than low viscosity flow) / greater time taken toflow the same distance / flows less distance in the same time (1) 1

(b) Measurement of viscosity:Distance travelled by lava in a set time / time taken to travel aset distance / speed of lava flow (1) 1

(c) Effect of cooling:(viscosity) increases (1) 1

(d) Laminar/Turbulent flow:Any 3 points –• Laminar – smooth• Shown by at least 2 straight-ish lines• Turbulent – flow causes whirlpools /eddies (or

explanation involving energy)• Turbulent flow shown on diagram with at least 3

lines resulting in eddies (1)(1)(1) Max 3

(e) Viscosity graph:Use a log scale / powers of 10 scale (1) 1

[7]

204. (a) (i) Not matter/antimatter pair [stated or implied] (1)particle/antiparticle have same mass OR electron/proton not samemass OR other correct reason (eg electron is fundamental, protonis quarks) (1)antiparticle to proton is antiproton OR antiparticle to electron ispositron/antielectron (1) 3

(ii) Not matter/antimatter pair [stated or implied] (1)anti to up is anti-up OR anti to down is anti-down (1)up and down have different charge (1) any 5

(b) particles/antiparticles carry opposite charge (1)(component of) field perpendicular to travel (1)(magnetic/LH rule) forces act in opposite directions (1)some pairs uncharged so no separation/deflection (1)[not annihilation] any 2

(c) number = 5000 × 10–12 kg / 9.11 × 10–31 kg = 5.5 (5.488) × 1021 (1) 1

(d) (i) correct use of E = mc2 [subs] (1)correct use of E = hf and c=fλ [rearranged or subbed] (1)correct answer [ue] (1)

E = mc2 = 9.11 × 10–31 × (3 × 108)2 J (= 8.199 × 10–14 J) (1)E = hf = hc/λ ⇒ λ = hc/E (1)

= 6.63 × 10–34 × 3 × 108/8.199 × 10–14 m

= 2.4 (2.426 or 2.42 or 2.43) × 10–12 m [ignore omissionof both factors of 2] (1)[factor of 2 wrong is a.e. = –1][use of λ = h/p scores 0] 3

(ii) this wavelength is not visible lightOR this is x-ray or gamma or high energy photon so need shielding (1) 1

[12]

205. (a) (i) arrow towards centre of curvature (1) 1

(ii) Use of formula with correct q OR v subbed (1)correct answer (1)

F = Bqv

= 0.5 × 1.6 × 10–19 × 800 000 N (correct q or v) (1)

= 6.4 × 10–14 N (1) 2

(iii) Use of formula: EITHER correct m subbed OR d identified with r (1)correct answer

r = p/Bq = 1.67 × 10–27 × 800 000/0.5 × 1.6 × 10–19 (m) (1)= 0.017 m (1)[Penalise factor 1000 error once only in question] 2

(iv) derive formula for T (1)correct answer (1)

T = π r/v (OR T = 2πr/v for (1)x) (1)= π × 0.017/800 000 (s) (ecf)

= 6.6 (6.5 – 6.7) × 10–8 s (1) 2

(v) correct statement of force = change of momentum/time (1)correct use of factor 2 (1)correct answer (1)

F = change of momentum/time (1)

= 2 × 1.67 × 10–27 × 800 000/6.7 × 10–8 (N) (ecf) (1)

= 4.1 (4.0) × 10–14 N [errors in m are self-cancelling] (1) 3

(b) Recall of formula (1)correct answer (1)

F = k q1 q 2/r2 OR F = q1 q 2/4πε0r2 OR k = 1/4πε0 (1)

= 1.6 × 10–19 × 1.9 × 10–6/4 × π × 8.85 × 10–12 × 5 × 5 (N)

= 1.1 × 10–16 N (1) 2[12]

206. (i) magnetic field changing (1)field cuts across conductor/flux linkage changes (1)Faraday/V induced (1) (any 3)V causes I (1) 3

(ii) Direction of induced current has an effect tending to cancel itscause OR [reasonable attempt at putting Lenz into words –not just “Lenz”] (1) 1

[4]

207. (a) (i) recall of formula (1)correct answer (1)

C = Q/V (stated or implied) [this way round] (1)= (appropriate pair of values, eg 4 C/4.8 V)= 0.83 (0.82 – 0.84) F (1) 2

(ii) strip width ∆Q (1)1∆W = V. ∆ Q (1)2add strips => area under graph (1)3area = 1/2Q V (1)4energy stored = work done (1)5showing 1/2QV has unit J/joule (1)6 (any 3)[integration answer – max (1)(1)][answer in words – max (1)(1)] 3

(iii) derive or recall E = 1/2 C V2 OR use correct Q value from graph (1)OR line across graph at 4 Vcorrect answer (1)

E = 1/2 Q V= 1/2 × 3.3 (3.3–3.35) C × 4 V (1)= 6.6 (6.6–6.7) J (1)

OR E = 1/2 C V2 (1)

= 1/2 × 0.83 F × (4 V)2 = 6.6 J (1) 2

(b) (i) Q decreases ⇒ V decreases OR I decreases (1)mention of P = VI (1) 2

(ii) 125–145 s (1) 1[10]

208. B in accelerators:changes direction of motion of charged particles ORforce/B perpendicular to motion of charged particlesOR ref to LHR (1)1(moving) charged particles stored in circles/circularpath/spirals (1)2

Bqv = mv2/r (1)3cyclotron: T = 2 π m/Bq (1)4fixed frequency voltage for acceleration (1)5diag/construction detail [probably on diag] (1)6synchrotron: r fixed, B adjusted as needed (1)7 (up to 4)

B in detectors:charged particles ⇒ (detectable) curved paths (1)8find sign of charge from sense of curvature (1)9find momentum/speed/energy/mass from r (= p/Bq) (1)10

[5]

209. (i) two correct arrows [ignore labelling] (1) 1

(ii) Some use of mv2/r with v correctly subbed OR

mrω2 with v correctly used (1)[subbing may happen later in answer]

T cos θ = mg mg

OR T sin θ = mv2/r [either gains (1)] (1)

⇒ tan θ = v2/rg (1)

⇒ r = v2/g tan θ

= 30 × 30/9.81 × tan 20 m

= 252 m (1) 4[5]

210. (a) Meaning of resonance:Parts of building have about the same natural frequency asdriving vibrator (1)so vibrate with increased/large amplitude or maximum/largeenergy transfer occurs (1) 2

(b) (i) Why springs reduce vibration:Spring deforms (instead of building) (1)absorbing energy (1) 2

(ii) Less damage during earthquake:vibration of building is damped / amplitude of vibration isreduced (so damage to building would be less) (1) 1

[5]

211. (a) Expression for gravitational force:

F = GMm/r2 (1) 1

(b) Expression for gravitational field strength:

g = force on 1 kg, so g = GM/r2, or g = F/m so g = GM/r2 1

(c) Radius of geostationary orbit:Idea that a = g, and suitable expression for a quoted [can bein terms of forces] (1)substitution for velocity in terms of T (1)algebra to obtain required result (1) 3

Example of derivation:

g = v2/r or g = ω2rand v = 2πr/T or ω = 2π/T

so (2πr/T)2/r = GM/r2 or (2π/T)2r = GM/r2, leading to expression given

(d) Calculation of radius:Substitution into expression given (1)

Correct answer [4.2 × 107 m] (1) 2


r3 = 6.67 × 10–11 N m2 kg–2 × 6.0 × 1024 kg × (24 × 60 × 60 s)2 /4π2

= 7.6 × 1022 m3

So r = 4.2 × 107 m

(e) (i) Satellite with greater mass:Yes – because, in geostationary orbit, r constant soacceleration remains the same, regardless of mass (1)

(ii) Satellite with greater speed:No + suitable argument (1) 2[e.g. for geostationary orbit, T and r are fixed, so v cannotincrease (v = 2πr/T)]

(f) Why satellite must be over equator:Idea that centre of satellite’s orbit must be the centre of theEarth (can be shown on diagram) (1)there must be a common axis of rotation for the satellite andthe Earth / the satellite’s orbit must be at right angles to thespin axis of the Earth (1) 2

[11]

212. (a) Description of fusion:Two light nuclei combine to form a single (heavier) nucleus (1)Energy is released (1) 2

(b) Need high temperature and high density / pressure (1)

and one point from:

• to overcome electrostatic repulsion / for a large collision rate

• a reference to containment problems (1) 2

(c) When hydrogen nuclei fuse, there is a loss of mass

This is converted into energy, according to: ∆E = c2 ∆m 2[6]

213. (a) Estimate of volume of helium:Estimate of temperature at sea level [0 °C – 30 °C] (1)

Conversion of temperatures to kelvin (1)

Correct substitution for V in expression V = 4/3 π r3 (1)


V = 4/3 π (105 m)3

V = 4.8 × 106 m3

Recall of general gas equation (pV/T = constant) (1)Use of general gas equation (1)

Correct answer [6 × 103 m3 → 7 × 103 m3 ] to 2 s.f. [since estimate] (1) 6

Example of calculation:V1 = p2 / p1 × T1/T2 × V2

V1 = 1/1000 × 293/213 × 4.8 × 106 m3

V1 = 6.7 × 103 m3

(b) Labelling of graph:Graph with lower peak labelled “sea level” (1)At sea level, temperature is higher, so average K.E. ofmolecules is higher (1) 2

[8]

214. (a) Time 10 oscillations then divide by 10 / keep eye in thesame position each time (1)[do not accept light gates etc] 1

(b) Nearest 0.01 m / 1 cm (1)Either suitable because a 1 m length is sensibly measuredto nearest 1 cm (1)Or could measure to nearest mm with a metre rule (1) 2 max

(c) (i) Column headed T2 / 10.5 / loglog or lnln (1)

Units s2 / m0.5 / no units (1)

Correct values [T2 check last fig 5.02 / 10.5 0.60 row = 0.78 (1)Log 0.6 row : –0.22 0.210]Scales: points occupy more than half page (1)Points (1)Best fit straight line [not thro origin] (1)[T v l graph marks 4 and 5 only 2 max] 6

(ii) Line does not go through origin (1)

Therefore T2 not proportional to l / T not proportional to (1)

l1/2 (1)[LogLog : Need to find gradient (1)

Round off = 0.5] 2 max

(d) Line does not go thro origin / When T = 0 there appears to bea value of length (1)Intercept is about .. cm this shows an error in l (1)The actual length of pendulum is longer than measured (1)The intercept is long enough to be a possible (systematic) error (1) 3 max[No marks for log log graphs ]

(e) Gradient from large triangle (1)

= 3.9 – 4.1 for T2 / = 1.9 – 2.05 for root l (1)

Equate with either 4π2/g / 2π / root g (1)

value 9.6 – 10.5 m s–2 (1)[log log intercept (1)= log (2π / root g) (1)(1)

value 9.6 – 10.5 m s–2] (1) 4 max[18]

215. How A and B changeForce BFor ticking ‘no change’ in all 4 boxes (1)

Force A4 ticks right (1)3 ticks right (1)2 ticks right (1)

increases no change decreases

[4]

216. (a) Path of coinCurved line that must begin to ‘fall’ towards the ground immediately (1) 1

(b) (i) Show that..

Selects s = (ut +)2

1at2 or selects two relevant equations (1)

Substitution of physically correct values into equation or both (1)equations.Answer [0.37 s – 0.38 s] (1)

[Allow use of g = 10 m s–2. Must give answer to at least 2 sig. fig.,bald answer scores 0. No ue.] 3

eg 0.7 m = 2

1(9.81 m s–2)t2

(ii) Horizontal distance [ecf their value of t]

Use of v = t

d with correct value of time. [s = t

uv

2

+ is sometimes (1)

used. In this case v and u must be given as 1.5 m s–1 and t must

be correct. Also s = ut + 0.5at2 OK if ‘a’ is set = 0.]Answer [0.55 m – 0.60 m] (1)

eg d = 1.5 (m s–1) × 0.38 (s)= 0.57 m 2

(c) A coin of greater mass?QWOC (1)It will follow the same path [accept ‘similar path’,do not accept ‘same distance’] (1)All objects have the same acceleration of free fall / gravity oracceleration of free fall / gravity is independent of mass / it will takethe same time to fall (to the floor) (1)Horizontal motion / velocity is unaffected by any force or (gravitational)force (acting on coin) has no horizontal component or horizontalmotion/velocity is the same/constant. (1) 4

[10]

217. (a) Meaning of 0.8 sReaction time (of cyclist and car driver) (1) 1[Accept descriptions of reaction time eg ‘time it takes both to takein that the lights have changed to green’. Accept response time]

(b) (i) Same speed timeAnswer [6.8 s –6.9 s] [Accept any value in the range] (1) 1

(ii) How much further ahead?EitherFor measuring area under car graph at 6.8 s (1)

eg = 2

s m 9 s 6 1−× = 27 m [27.5 m if 6.9 s used]

For measuring area under cyclist graph at 6.8 s (1)

eg 2

s m 9 s 2 1−× + 4 s × 9 m s–1 = 45 m [45.9 m if 6.9 s used]

[For candidates who read the velocity 9 m s–1 as 8.5 m s–1 butotherwise do their calculation(s) correctly give 2/3][Allow one mark to candidates who attempt to measure anappropriate area]Answer [(45 m – 27 m =) 18 m] (1)

OrFor recognising the area enclosed by cyclist and car graphsas the difference in distance travelled (1)Using values from the graph to determine this area (1)Answer [(45 m – 27 m =) 18 m] (1)

eg distance = 2

1 × (6.8 – 2.8) s × 9 m s–1

= 18 m 3

(c) Relationship between average velocitiesThey are the same (1) 1

[6]

218. (a) Momentum at impactp = mv seen or used (1)

Answer [11 kg m s–1] (1) 2

eg momentum = 0.42 kg × 27 m s–1

= 11.34 kg m s–1

(b) Momentum at releaseMinus (1)

8.4 kg m s–1 (1) 2

(c) (i) Average force(ecf momenta values)

Use of F = t

p

∆∆

ie for using a momentum value divided by (1)

0.22Adding momentum values (1)

Answer [88.0 N – 89.8 N] (1)

F = s 22.0

s m kg 3.11s m kg 4.8 11 −− −−

F = (–) 89.5 N

Or

Use of F = ma (1)Adding velocities to calculate acceleration (1)Answer [88.0 N – 89.8 N] (1)

Eg acceleration = s 22.0

s m 27s m 20 11 −− −− (= –213.6 m s–2)

Force = 0.42 kg × –213.6 m s–2 = (–)89.7(2) N 3

(ii) Direction of force on diagramRight to left (1)[Accept arrow drawn anywhere on the diagram. Label not required] 1

(d) Difference and similarityDifference: opposite direction / acts on different object (1)

Similarity: same type of force / same size / acts along same line / (1)act for same time / same size impulse[‘Magnitude’ and ‘size’ on their own is sufficient. ‘They are equal’is OK. Accept; they are both contact forces; they are bothelectrostatic forces] 2

[10]

219. (a) EK of helium nucleus

Use of EK = 2

1mv2 (1)

Answer [ 3.1 × 10–15 J. No ue. Min 2 sig fig required] (1) 2

eg EK = 2

1 × 6.65 × 10–27 kg × (9.65 × 105 m s–1)2

= 3.096 × 10–15 J

(b) (i) Loss of EK of proton [ecf their value for EK of helium nucleus]

3 × 10–15 J or 3.1 × 10–15 J (1) 1

(ii) Speed of proton after collision[ecf their value for loss of EK of proton, but not if they havegiven it as zero]Calculation of initial EK of proton (1)

Subtraction of 3.1 × 10–15 J [= 1.7 × 10–15 J] (1)

Answer [(1.40 – 1.50) × 106 m s–1] (1)

eg EK = 2

11.67 × 10–27 kg × (2.4 × 106 m s–1)2 (= 4.8 × 10–15 J)

EK after collision = 4.8 × 10–15 J – 3.1 × 10–15 J (= 1.7 × 10–15 J)

v =

×××

−

−

kg 1067.15.0

J 107.127

15

= 1.43 × 106 m s–1

OrUse of the principle of conservation of momentum. (1)Correct expression for the total momentum after the collision (1)

Answer [ (1.40 – 1.50) × 106 m s–1] (1)

Eg 1.67 × 10–27 kg × 2.4 × 106 m s–1

= 6.65 × 10–27 kg × (9.65 × 105 m s–1) + 1.67 × 10–27 kg × V

V = –1.44 × 106 m s–1

[For both these solutions allow the second marking point to 3candidates who incorrectly write:

the mass of the proton as 1.6 × 10–27 kg or 1.7 × 10–27 kg,

or the mass of the helium as 6.6 × 10–27 kg or 6.7 × 10–27 kg

or the velocity as 9.6 × 105 m s–1 or 9.7 × 105 m s–1]

(c) Other factor conservedMomentum / mass / charge / total energy (1) 1

[7]

220. Deductions

(a) (i) The atom is mainly empty space (1)[The atom must be referred to. The words ‘empty’ and ‘space’ mustbe qualified eg ‘there is a large amount of space in the atom’ is notsufficient]

(ii) Within the atom there is an area / the nucleus which is positive/ charged or more massive than the alpha particle[If they choose to describe only the mass it must be a comparison ie‘the nucleus is (much) more massive than the alpha’. ‘The atom hasa dense centre,’ ‘the nucleus has a large mass’ are both insufficient.] (1) 2

(b) Explain(Deflection could have been) repulsion from positive nucleus (1)(Deflection could have been) attraction towards negative nucleus (1)[The words repulsion and attraction can be described eg ‘α deflectedaway from positive nucleus’, ‘α is deflected towards a negativenucleus’] [Diagrams showing the path of an alpha deflected by botha negatively and a positively charged nucleus would get both marks] 2

(c) Value of n (4 – 6) (1)[Allow minus values] 1

[5]

221. (a) Meaning of ‘random’Impossible to predict which atom/nucleus (in a given sample) willdecay (at any given moment)/ unable to predict when a given atomwill decay (1)[Mention of atom(s), or nucleus, or nuclei is essential because theword ‘random’ is to be described in context. Do not accept for atomor nucleus; substance; material; particle; molecule; sample.] 1

(b) Nuclear equation

He NpAm 42

23793

24195 +→

α42

42 or He (1)

Np / Am 23793

24195 / both proton numbers correct / both mass numbers

correct (1)Entirely correct equation (1) 3

(c) Absorbtion experimentDiagram [must include the source, detector and an indication ofwhere the absorber is placed (maybe written in their account ratherthan on the diagram) – none of these need to be labelled] (1)(Record) background count (1)

Source – detector distance must be close / less than or equal to 2 cm (1)

Insert eg paper between source and detector = no change in countrate or increase (by a small amount) the separation of the detectorand source = no change in count rate (therefore no α present) (1)

Insert aluminium/brass (a few mm thick) or lead (≈ 2mm thick) / concrete (1)

Count reduced to background (therefore no gamma present)[Do not give this mark if only paper or card or plastic is used asthe absorber. Accept ‘0’ in place of ‘reduced to background’ ifcandidate has deducted background from their measurements.] (1) Max 5

[9]

222. Tungsten filament bulb

(a) Resistance

Use of P = V2/R or P = VI with V = IR (1)answer 960 Ω (1) 2

Example of answerR = (240 V × 240 V) ÷ 60 WR = 960 Ω

(b) Drift speedrearrangement of I = nAvQ (1)

Use of Q = 1.6 × 10–19 (C) (1)

answer 0.15/0.148 m s–1 (1) 3

Example of answer

v = 0.25 A ÷ (3.4 × 1028 m–3 × 1.6 × 10– 19 C × 3.1 × 10–10 m2)

(c) ExplanationQowc (1)

Any THREE• Resistance due to collisions between electrons & ions/atoms/particles• (as T increases) ions/atoms/particles have more energy• (as T increases) ions/atoms/particles vibrate through larger

amplitude /vibrate faster OR amplitude if lattice vibrationincreases.

• more chance/increased frequency of collision/interactionOR impedes the flow of electrons (1)(1)(1) 4

[9]

223. Emf and Internal resistance

(a) DerivationE = I (R + r) OR E = IR + Ir (1) 1

(b) (i) Correct working (allow even if evidence of working backwards) (1)

Example of answerE/I = R + rRearranging R = E/I – r

(ii) EmfAttempt to use gradient (1)answer 1.5 V (bald answer 1.5 V scores 0/2) (1) 2

(iii) PowerFrom graph find value of 1/I when R = 5 Ω (1)

Use of P = I2R (1)answer 0.31 (W) (1) 3

Example of answer

1/I = 4 A–1 → I = 0.25 AP = 0.25 A × 0.25 A × 5 Ω = 0.3125 W

(c) GraphIntercept at –2 (ohms) (1)Graph steeper than original (1)Gradient is 3.0 V i.e. line passes through [10, 27-29] [no ecf] (1) 3

[10]

224. Potential divider

(a) First circuitMiddle terminal MOuter terminals L and K (any order) (1) 1

(b) (i) P.d across lamp.External resistance in circuit is 25 or (20+5) ohms (1)See ratio of resistances (denominator larger) × 6.0V (1)OR current = 6/25 Aanswer 4.8 V (1) 3

(ii) AssumptionThe resistance of the ammeter is zero/negligible. (1) 1

(c) Second circuitSee 2 resistors in parallel with supply (1)Supply across ends of variable resistor (10 Ω) (1)Fixed resistor across one end and slider (consequent mark) (1) 3

[8]

225. Gas laws

(a) Boyle’s lawUses pV = constant (1)

See V2 = 15 or (20 – 5) cm3 (1)

Pressure = 267/270 × 103 Pa /Nm–2 (1) 3

Example of answerp1V1 = p2V2

p2 = 200 × 103 Pa × 20 cm3 ÷ 15 cm3

p2 = 266667 Pa

(b) ForceUses F = pA (1)15.8/16.0 N (1) 2

Example of answer

F = 200 × 103 Pa × 7.9 × 10–5 m2

F = 15.8 N

(c) (i) Pressure lawUses p/T = constant Kelvin or Celsius. (1)At least one conversion to Kelvin (295K or 308K) (1)

209 × 103 Nm–2/Pa (1) 3

Example of answerp1/T1 = p2/T2

p2 = 200 ×103 Pa × 308 K ÷ 295 Kp2 = 208813 Pa

(ii) GraphCurve reasonably similar to one given (1)Curve above first. (no ecf for their p2 less than p1) (1) 2

[10]

226. (a) Wien’s law

(i) • Wavelength of peak [or maximum] intensity [allow brightness, butnot power / energy / L] (1)

• Absolute or Kelvin (surface) temperature (of star) (1) 2

Spectrum

(ii) Microwave / infra–red [accept i.r.] (1) 1

(iii) λmax = 1.05 (mm) (1)

Substitution in T = 2.90 × 10–3 m K ÷ their λmax with ×10–3 (1)

[2.90 × 10–3 m K ÷ 1.05 × 10–3 m, with mm → m conversionrequired]

T = 2.76 [accept range 2.6 – 2.9] K (1) 3

(b) (i) Supernova minimum mass

8 M [ accept 1.6 × 1031 kg] (1) 1

(ii) Energy from Sun

1 × 1010 × 365(¼) × 24 × 60 × 60 / 3(.15) × 1017 (1)

Use of E = P × t (1)

1(.2) × 1044 J [Beware E = t

P ⇒ 1.24 × 109 J] (1)

E = P × t

= 3.9 × 1026 W × 1 × 1010 × 365(¼) × 24 × 60 × 60 s= 1.2 × 1044 J 3

(iii) 1046 ÷ (1.2 ×) 1044 [ecf for any E] (1)

80 – 100 [ecf] [accept 83:1 or 1:0.012] (1)[inverted answer scores zero, unless values identified for 1/2] 2

(iv) Supernova future

neutron star / pulsar (1)

black hole (1)

n.s. if > 1.4 M OR b.h. if > 2.5 M (1) 3

(c) Neutron star density

(i) ρ = m ÷ V and 4/3 π r3 (1)

4.2 [or 4.3] × 1017 (kg m–3) (1)

ρ = m ÷ V

= 3m ÷ 4 π r3

= 3 × 6.0 × 1024 kg ÷ (4 π (150 m)3)

= 4.2 × 1017 kg m–3 2

(ii) Neutron formation


Main sequence: fusion (reaction) / (ms) p → n / beta plus decay (1)

[post ms] p + e– → n (1)

[post ms] due to gravitational collapse / implosion (1) 4

(d) Intensity of Sun

(i) Use of I = L ÷ (4π D2) (1)

597 OR 1380 (ignore 10n) (1)

597 W m–2 AND 1380 W m–2 [accept W km–2 with appropriatevalues] (1) 3

I = L ÷ (4π D2)

IM = 3.90 × 1028 W ÷ (4π (2.28 × 1011 m)2)

= 597 W m2

IE = 3.90 × 1028 W ÷ (4π (1.50 × 1011 m)2)

= 1380 W m–2

597 ÷ 1380 [ecf, accept (2.28 ÷ 1.50)2] (1)

43% (1) 2[26]

227. Property definitions

(i) Tough: absorbs energy (before breaking) (1)

by plastic deformation (1)

Strong: high(er) UTS / high(er) stress (before breaking) (1) 3

Force calculation

(ii) Attempted use of σ = F / A [accept use of r instead of A for 1/3] (1)

Use of A = π r2 (ignore 10n) (1)

3.8 (or 4) × 10–3 N (1) 3

σ = F / A = F / π r2

F = σ π r2

= 3 × 108 Pa × π (2.0 × 10–6 m)2

= 3.8 × 10–3 N

Stiffest part of curve

(iii) Initial slope indicated (1) 1

Young modulus calculation

(iv) Any attempt at gradient / stress ÷ strain (1)

Correct pair of values for linear region above stress 0.25 / Extendedgradient at start of curve (1)

5.5 GPa [5.2 – 5.6 with GPa or GNm–2] (1) 3[10]

228. (a) Why transverse waves can be polarised but not longitudinal waves[Marks can be earned in diagram or text]Transverse waves have * perpendicular to direction of ** (1)

* = vibration/displacement/oscillation/motion of particles

** = travel/propagation/motion of wave/energy transfer/wave

In a transverse wave, * can be in different planes but polarisationrestricts it to one plane (1)Longitudinal waves have * parallel to ** (1) 3

[Don’t accept “motion” for **Diagrams to earn marks must be clearly labelled, but don’t insiston a label “looking along direction of travel” in the usual diagramsto illustrate polarised and unpolarised waves]

(b) (i) Effect of Polaroid on intensityIntensity is reduced (OR halved) [not zero] (1)[Accept slightly reduced and greatly reduced]Polaroid stops (OR absorbs) vibrations (OR waves OR light) inone plane/ (1)Polaroid only lets through vibrations (OR waves OR light)in oneplane/Light has been polarised 2

(ii) Effect of rotating PolaroidNo effect (1)[ignore incorrect reasons accompanying statements of effect] 1

[6]

229. (a) Definition of SHMAcceleration (OR force) is proportional to displacement/allowdistance from point (from a fixed point) / a (OR F) = (–)constant xwith symbols defined (1)Acceleration (OR force) is in opposite direction to displacement/Acceleration is towards equilibrium point [Allow “towards a fixedpoint” if they have said the displacement is measured from thisfixed point] / Signs in equation unambiguously correct, e.g.

a (OR F) = –ω2 x

[Above scheme is the only way to earn 2 marks, but allow 1 mark formotion whose period is independent of amplitude OR motionwhose displacement/time graph is sinusoidal] 2

(b) (i) Calculation of periodUse of T = km /2π (1)Correct answer [1.1 s] (1) 2

e.g. T = 2π√(0.120 kg / 3.9 N m–1)= 1.10 s

(ii) Calculation of maximum speedUse of vmax = 2 π fxo and f = 1/T (1)

Correct answer [0.86 m s–1] (1) 2

e.g. f = 1/(1.10 s) = 0.91 Hz

vmax = 2π (0.91 Hz)(0.15 m)

= 0.86 m s–1

(iii) Calculation of maximum acceleration

Use of amax = (–)(2πf)2x0 (1)

Correct answer [4.9 ms–2] (1) 2

e.g. amax = (2π × 0.91 Hz)2(0.15 m)

= 4.9 m s–2

(iv) Calculation of mass of blockUse of T α m / Use of T = 2 km /π (1) Correct answer [0.19 kg] (1) 2

e.g. m = (0.12 kg)(1.4 s /1.1 s)2

= 0.19 kg

OR m = (3.9 N m–1)(1.4 s / 2π)2

= 0.19 kg

[Apply ecf throughout][10]

230. (a) (i) How the bow causes the wave patternEITHERBow alternately pulls and releases string (or sticks and slips) (1)Creates travelling wave (OR travelling vibration ) (on string) (1)Wave reflects at the end (OR bounces back) (1)Incident and reflected waves (OR waves travelling in opposite (1)directions) superpose (OR interfere OR combine)[Don’t accept collide] max 3

ORBow alternately pulls and releases string (or sticks and slips) (1)Produces forced oscillation/acts as a driver/exerts periodic force (1)[Don’t accept makes it vibrate] At a natural frequency of the string (1)Causing resonance (OR large amplitude oscillation) (1) max 3

(ii) Determination of wavelengthUse of node to node distance = λ/2 / recognise diagram shows 2λ] (1)Correct answer [0.4 m] (1) 2

e.g. λ = 2 × 0.2 m = 0.4 m

(iii) Differences between string wave and sound waveAny TWO points from:– String wave is transverse, sound wave is longitudinal /…can be polarised, … can’t– String wave is stationary (OR standing), sound wave is travelling(OR progressive) / … has nodes and antinodes, …doesn’t /…doesn’t transmit energy, …does…– The waves have different wavelengths– Sound wave is a vibration of the air, not the string (1)(1) 2

[Don’t accept travel in different directions / can be seen, can’t beseen / can’t be heard, can be heard / travel at different speedsThe first two marking points require statements about both waves,e.g. not just “sound waves are longitudinal”]

(b) Sketch of the waveformSinusoidal wave with T = 1 ms (1)[Zero crossings correct to within half a small squareAccept a single cycle]Amplitude 1.6 cm (1)[Correct to within half a small square] 2

[9]

231. (a) Conditions for observable interferenceAny THREE of:• Same type of wave / must overlap (OR superpose) / amplitude

large enough to detect / fringes sufficiently far apart todistinguish [Only one of these points should be credited]

• (Approximately) same amplitude (OR intensity)• Same frequency (OR wavelength)• Constant phase difference (OR coherent OR must come from

the same source) (1)(1)(1) 3

[Accept two or more points appearing on the same line

Don’t accept– must be in phase– must be monochromatic– must have same speed– no other waves present– must have similar frequencies– answers specific to a particular experimental situation, e.g.

comments on slit width or separation]

(b) (i) Experiment description[Marks may be scored on diagram or in text](Microwave) transmitter, 2 slit barrier and receiver (1)[Inclusion of a screen loses this mark, but ignore a single slit in frontof the transmitter]Barrier, metal sheets (1)[Labels indicating confusion with the light experiment, e.g. slitseparations or widths marked as less than 1 mm, lose this mark]Appropriate movement of receiver relevant to diagram [i.e. move inplane perpendicular to slits along a line parallel to the plane of theslits, or round an arc centred between them] (1) 3

(ii) Finding the wavelengthLocate position P of identified maximum/minimum 1st/2nd/3rd etc. (1)away from centreMeasure distance from each slit to P (1)Difference = λ OR λ/2 (consistent with point 1) (1) 3

[Accept use of other maxima and corresponding multiple of λ][9]

232. (a) [Treat parts (i) and (ii) together. Look for any FIVE of thefollowing points. Each point may appear and be credited in eitherpart (i) or part (ii)]

(i) • Light (OR radiation OR photons) releases electrons from cathode

• Photon energy is greater than work function / frequency oflight > threshold frequency / flight > fo / wavelength of light isshorter than threshold wavelength / λ < λ0

• PD slows down the electrons (OR opposes their motion ORcreates a potential barrier OR means they need energy to crossthe gap)

• Electrons have a range of energies / With the PD, fewer (ORnot all) have enough (kinetic) energy (OR are fast enough) tocross gap

• Fewer electrons reach anode / cross the gap

(ii) • (At or above Vs) no electrons reach the anode / cross the gap

• Electrons have a maximum kinetic energy / no electrons haveenough energy (OR are fast enough) to cross

ANY FIVE (1)(1)(1)(1)(1)

[Don’t worry about whether the candidate is describing the effect ofincreasing the reverse p.d. (as the question actually asks), or simplythe effect of having a reverse p.d.] 5

(b) Effects on the stopping potential

(i) No change (1)

(ii) Increases (1) 2

[Ignore incorrect reasons accompanying correct statements of the effect][7]

233. (a) Comment on assumption

Yes – air resistance negligible OR still close to Earth (ignore upthrust)or No – air resistance becomes significant (1) 1

(b) Explanation of why formula for cell B6 is appropriate

Recall of v = u + at (accept ∆v = a ∆t or ∆v = at for 1st mark) (1)

(v is B6), u is zero, a is 9.81 [m s–2] and t is A6 (1) 2

(c) (i) Explanation of 2

B7) B6( +

it is average speed (for that interval)

or 2

v)u ( + (1) 1

(ii) Why 2

B7) B6( + is multiplied by 0.20

because dist = ave speed × time [accept s = vt]and 0.20 is the time (1) 1

(d) Formula for D10

= D9 + C10 (1) 1

(e) Calculation to check D11

Use of appropriate equation of motion (1)

Correct answer [12.557 m] [no ue] (1) 2


s = ut + ½ at2

= 0 + ½ × 9.81 m s–2 × (1.6 s)2

= 12.557 m

N.B. use of υ2 = u2 + 2as gives answer s = 12.563 m[8]

234. (a) Circuit diagram and explanation

ammeter and voltmeter shown in series and parallel respectively (1)

current measured with ammeter and voltage / p.d. with voltmeter (1) 2

(b) Calculation of resistance

Recall of R = V/I (1)



R = V/I

R = 3.00 V ÷ 0.12 A

= 25.0 Ω

(c) Calculation of resistance

Recall of P = V2/R (1)



P = V2/R

R = (230 V)2 ÷ 1800 W

R = 29.4 Ω

[Accept calculation of I = 7.8 A (1), calculation of R = 29.4 Ω (1)]

(d) Explanation of difference in values of resistance

At higher voltage value element is at a higher temperature (1)

(resistance higher because) increased lattice ion vibrations impedecharge flow (more) (1) 2

[8]

235. (a) Explain how vapour emits light

electrons excited to higher energy levels (1)

as they fall they emit photons/electromagnetic radiation/waves/energy (1) 2

(b) (i) Meaning of spectral line

(when the light is split up) each frequency/wavelength/photon energy isseen as a separate/discrete line (of a different colour) (1) 1

(ii) Calculation of frequency


Correct answer [f = 5.1 × 1014 Hz] (1) 2


v = f λ

3.0 × 108 m s–1 = f × 589 × 10–9 m

f = 5.1 × 1014 Hz

(c) Explanation of different colours

different colours = different freq/wavelengths / photons of differentenergies (1)

photon energy/frequency/wavelength depends on difference betweenenergy levels (1)

diff atoms have diff energy levels/diff differences in levels (1) 3

(d) Explanation of transverse waves

variation in E or B-field /oscillations/vibrations/displacementat right angles/perpendicular to direction of travel/propagation

[not just motion or movement for both 1st and 3rd part] (1) 1[9]

236. (a) Mark and label W and T

W marked and labelled (1)

T marked and labelled (1) 2

(b) Calculation of horizontal component of P

Recall of trigonometrical function (1)

Correct answer [9974 N] (1) 2


horizontal component = P cos θ

= 23600 N × cos 65°

= 9974 N

(c) (i) State magnitude of horizontal component of T T = 9974 N [ecf] (1) 1

(ii) Calculate magnitude of T

Use of trigonometrical function (1)

Correct answer [13 420 N] [ecf] (1) 2


horizontal component of T = T cos 42° = 9974 N

T = 9974 N ÷ cos 42°

= 13 420 N

(d) Scale drawing

P added (1)

resultant correctly drawn (1)

magnitude of resultant = 13 400 N (± 400 N) (1)

angle = 42° (± 3°) (1) 4

(e) Describe one other force

E.g., push from wind (1) 1[12]

237. (a) Explanation of maximum or minimum

path difference = 2 × 125 × 10–9 m = 250 × 10–9 m (1)

= half wavelength /antiphase (1)

→ destructive interference / superposition (1) 3

(→ minimum intensity)

(b) Meaning of coherent

remains in phase / constant phase relationship (1) 1[4]

238. (a) (i) Type of airflow:Laminar / streamline (1) 1

(ii) Airflow diagram:At least two streamlines drawn in front of the skier (1)At least two streamlines continuous around and behind the skier (1)[Maximum of 1 mark if the streamlines cross or touch] 2

(iii) Skier’s equipment:Smooth / tight–fitting / not baggy / elastic (1) 1

(b) (i) Desirable property:Elastic or Tough (1) 1

Reason:Correct reasoning in line with property, ie.Will return to original shape (once load removed) (1)orCan withstand shock / impact (without breaking) (1) 2

(ii) Undesirable property:Plastic (1)

Reason:Will remain deformed (once load removed) (1) 2

[8]

239. (a) (i) Energy level diagram:• Arrow showing electron moving from lower level to a

higher level (1)• Arrow downwards from higher to lower level [must

show smaller energy change than upward arrow] (1) 2

(ii) Missing energy:Causes a rise in temperature of a named item (1) 1

(iii) Range of energies:

Minimum energy when λ = 400 × 10–9 m (1)Use of f = c/λ (1)Use of E = hf (1)Correct answer [3.1 eV] (1)[allow 3.0 – 3.3 eV for rounding errors] [no u.e] 4

eg. f = 3 × 108 / 400 ×10–9

= 7.5 × 1014 Hz

E = hf = 5.0 × 10–19 JE = 3.1 eV

(b) Detecting forgeries:Forgery would glow / old painting would not glow (1) 1

[8]

240. (a) (i) Critical angle calculation:Use of sin C = 1/dµa (1)Correct answer [24.4° – only acceptable answer] [no u.e] (1) 2

eg. Sin C = 1/dµa = 1/2.42C = 24.4°

(ii) Ray diagram:Small angle ray shown passing into air, away from the normal (1)Large angle ray showing T.I.R. with angles equal [by eye] (1) 2

(iii) Labelling of angles:An incident angle correctly labelled between normal and rayin diamond (1)An angle of refraction correctly labelled between normaland ray in air (1) 2

(iv) Amount of trapped light:Any 3 of the following:• The higher the refractive index the greater the amount of

trapped light• The higher the refractive index the lower the critical angle• T.I.R occurs at angles greater than the critical angle• So, if critical angle is smaller, more light is reflected (1)(1)(1) Max 3

(b) Comment on angle:Lower critical angle so more sparkle (1) 1

[10]

241. (a) Graph scale:Log scale (1) 1

(b) (i) Choice of material:Any 2 of the following:• (almost) all of the voltage is dropped across the carbon rod• gives the greatest speed variation• others need to be longer (to have same resistance as carbon)• others need to be thinner (to have same resistance as

carbon) (1)(1) Max 2

(ii) Resistance calculation:Use of R = ρL/A (1)Correct units used for all terms [all in cm or all in m] (1)Correct answer [1.9 Ω] (1) 3[allow 1.8 Ω for rounding errors – no u.e]

eg. R = 1.4 × 10–5 × 0.4 / 3.0 × 10–6

= 1.9 Ω

(iii) Available voltage:X – 12 V Y – 0 V (1) 1

(iv) Effect of connecting wires:Less voltage available for train set as some wasted across wires (1)0.5 Ω is (relatively) large % of total resistance, so effect ishigh / not negligible (1)

or

Calculation of potential difference available now (1)[9.5 V] [allow 9.5 – 9.6 V]Significant drop from 12 V (1)Vxy = (Rxy / RTotal) × Vsupply = (1.9/ (1.9 + 0.5)) × 12 = 9.5 V 2

[9]

242. (a) Graph:

(i) Line of best fit completed curving between 5.0 and 5.5 mm (1)

(ii) X marked correctly on line (by eye) between 5.0 – 5.5 mm (1) 2

(b) (i) Energy stored calculation:Energy = ½ Fx or area under graph to intercept line (1)Correct reading of x from graph (1)Correct answer from graph in Joules (1) 3

eg. Energy = ½ Fx

= ½ × 20 × 4 ×10–3

= 0.04 J

(ii) Gradient of graph:Stiffness of wire (1) 1

(c) Thicker wire:Any 2 of the following:• Steeper gradient• More force required to produce the same extension• Limit of proportionality at a larger force (1)(1) Max 2

[8]

243. (a) (i) Recall of Q = CV or W = 1/2 CV2 (1)Correct calculation of W or V or C (1)⇒ Conclusion [must be consistent] (1)

eg W = 1/2 CV2

⇒ C = 2 W/V2 = 2 × 0.045 / (30,000)2 (F)

= 1 × 10–10 (F) = 100 (pF) (⇒ NOT COMPATIBLE)

or ⇒ W = 1/2 × 10 × 10–12 × (30,000)2 (J) = 0.0045 (J) (⇒ NOT COMPATIBLE) 3

[no mark for conclusion; but ue for saying 100pF ~ 10 pF]

(ii) Sub of one appropriate value into Q = CV or W = 1/2QV (1)Correct value (1)

eg Charge = 1 × 10–10 × 30,000 (C)

= 3 × 10–6 C 2

(b) (i) Use of E = V/d [Rearranged or subbed into] (1)Correct value (1)

eg d = V/E = 30,000/3 × 106 (m)= 0.01 m 2

(ii) Use of E = F/q [Rearranged or subbed into – any charge value] (1)Correct value (1)

eg F = Eq = 3 × 106 × 1.6 × 10–19 (N)

= 4.8 × 10–13 N 2

(e) Correct use of 1 mm in W = Fd or V = Ed [ecf from(b)(ii)] (1)⇒ 3000 (V or eV)) (1)⇒ correct value (1)

eg W = Fd = 4.8 × 10–13 N × 0.001 m (= 4.8 × 10–16 J)⇒ 3000 (eV)3000/35 = 85/86/85.7

or V = Ed = (3 × 106 V/m) × 0.001 (m)⇒ 3000 (V)3000/35 = 85/86/85.7 3

[12]

244. (a) Any 2 from:momentum conserved (1)initial momentum zero (1) (Any 2)

⇒ final momentum zero (1)[opposite charges repel ⇒ xx] 2

(b) 0.140 GeV/c2 (1)

– 1.6 × 10–19 C (1)anti-u, d (1) 3

(c) Meson (1) 1

(d) [(1) for 0.14 (alone) or correct use of 109] (1)

Minimum energy = 1.4 × 108 (eV) or 0.14 × 109 (eV) (1) 2[0.14 G is (1)x]

(e) Particles have K.E. (as well as mass) (1) 1

(f) Use of ∆E = c2∆m [rearrangement OR one correct line subbed) (1) correct value (1)

eg ∆m = ∆E / c2 = 0.14 × 109 × 1.6 × 10–19 J / (3 × 108 m s–1)2

Mass loss = 2.5 × 10–28 kg 2[ecf from (d)]

[11]

245. (a) R drawn [10° to vertical] (1)D drawn [10° to horizontal] (1)drag force D = 140 – 155 N [147.6 N by calc is OK] (1) 3

(b) Resolve vertically (1)correct value (1)

eg P cos 40° = 850 N⇒ P = 1100 N 2

(c) (i) velocity not constant / direction changing (1)[NOT “if no force, goes straight”]acceleration (towards centre of circle) (1) (Any 2)F = ma (1) 2

(ii) P/push of ice (on sled) (1)horizontal component (1) 2[“additional centripetal force” = 0]

(d) Recall circular motion formula (1)resolve horizontally (1)correct value (1)[incorrect force is eop]

[also possible: W.tan40 = mv2/r (1)(1)]

eg F = P sin 40 = 713 (643) (N) [formula or value]

R = mv2/F

= 87 × 352 / 713 (643) (m)radius = 149 (166) m 3

[12]

246. (a) emf/voltage (1)induced / created / caused by flux change (1) 2

(b) Lenz (1)effect opposes change producing it (1) 2

(c) dynamo generates emf (1)lights off ⇔ no current (1)lights on ⇔ current flowing (1) Any 4If current, then force on dynamo rotor/F = BIl (1)[or field acting against field in dynamo]This force opposes rotation (1) 4

[8]

247. (a) Shape [lines not crossing] (1)arrow(s) (1) 2

(b) [reference to] changing B field/ flux cuts coil / changing flux(linkage) (1)1induces emf or current [NOT “output”] /EM induction (1)2emf α rate of change / Faraday’s law stated (1)3output is gradient of flux graph (1)4signal +ve while φ increases / –ve while φ decreases (1)5max emf for max dφ / dt / steepest gradient (1)6 (Any 5)Emf 0 when gradient = 0 (1)7[(1)5 (1)6 (1)7 can be gained by annotations on graph] 5

(c) (Binary / 1 or 0 / 210

maximum number =) 1024 (1) 1

(d) 10010 (1) 1

(e) Attempt to calculate circumference (formula or numbers) (1)

dividing by 0.83 × 10–6 (m) (1)correct value (1)

eg C = πd = π × 0.089 m (= 0.2796 m)

No of bits along circumference = C ÷ (0.83 × 10–6 m) 3

(= 3.37 × 105)

Rate = 3.37 × 105 × 120 (7200 revs/min = 120 Hz)

= 4.04 × 107 s–1

[2.43 × 109 min–1 is OK][12]

248. (a) Complete statement of Newton’s Third Law of Motion....exerts an equal force on (body) A (1)

(but) in the opposite direction (to the force that A exerts on B) (1) 2[‘exerts an equal but opposite force on body A’ would get both marks]

(b) Complete the table

1 mark for each of the three columns (1) (1) (1) 3

[Accept from earth for up. Accept towards ground or towards earth for down]

Earth Gravitational. [Not ‘gravity’. Not

gravitational field strength]

Up(wards) / ↑

Ground Down(wards)

/ ↓[5]

249. (a) Time to fall

Use of s = ut + ½ at2 or use of 2 correct equations of motion (1)

or use of mgh = ½ mv2 and other equation(s)

[allow g = 10 m s–2] 2

Answer to at least 2 sig fig [0.69 s. No ue] (1)

Example

2.3 m = 0 + ½ 9.8 m s–2 t2

t = 0.68(5) s [0.67(8) if 10 m s–2 used][Reverse argument only accept if they have shown that height is 2.4 m]

(b) Time to riseSelect 2 correct equations (1)Substitute physically correct values [not u = 0 or a + value for g] (1)

[allow g = 10 m s–2 throughout] (1) 3Answer: [ t = 0.38 s]

Example 1

0 = u2 + 2x – 9.81 m s–2 0.71 m

0 = 3.73 m s–1 + –9.81 m s–2 tt = 0.38 s

[0.376 s if 10 m s–2]

Example 2

0 = u + – 9.81 m s–2 t; u = 9.81t

0.71 m = 9.81 t.t + ½ – 9.81 m s–2 t2

t = 0.38 s

[Note. The following apparent solution will get 0/3. s = ut + ½at2;

0.71 m = 0 + ½ 9.81 m s–2 t2; t = 0.38 s, unless the candidate makes itclear they are considering the time of fall from the wicket.]

(c) Velocity u

Use of tdv (1)

[d must be 20 m, with any time value from the question eg 0.7 s]

Answer: [18.9 m s–1 or 18.2 m s–1 if 0.7 s + 0.4 s = 1.1 s is used. (1) 2ecf value for time obtained in (b).]

Example

ssmv

38.068.020

+=

= 18.86 m s–1 [18.18 m s–1 if 1.1 s used]

(d) Why horizontal velocity would not be constant

Friction/drag/air resistance/inelastic collision at bounce or impact (1) 1/ transfer or loss of ke (to thermal and sound) at bounce or impact(would continuously reduce the velocity/ kinetic energy).[also allow ‘friction between ball and surface when it bounces(will reduce velocity/kinetic energy)’].

[Any reference to gravitational force loses this mark.A specific force must be mentioned, eg resistive forces is not enough.]

[8]

250. (a) Newton’s Second Law of Motion

(The) force (acting on a body) is proportional/equal to the rate of (1)change of momentum (1)and acts in the direction of the momentum change[accept symbols if all correctly defined for the first of these marks][ignore any information that is given that is not contradictory] 2

(i) Calculate the mass

Correct calculation for volume of air reaching tree per second[Do not penalise unit error or omission of unit] (1)Correct value for mass of air to at least 3 sig fig [246 kg. No ue.] (1)[If 1.23 × 10 × 20 = 246 kg is seen give both marks.Any order for the numbers]

Example

20 ms–1 × 10m2 = 200 m3

1.23 kg m–3 × 200 m3 = 246 kg

(ii) Calculate the momentum

Answer: [ (246 kg × 20 m s–1 =) 4920 kg m s–1]

[Accept (250 kg × 20 m s–1 =) 5000 kg m s–1. Accept 4900 kg m s–1. (1) 2Ecf value for mass. Ignore signs in front of values.]

(iii) Magnitude of the forceAnswer: [F = 4920 N or 5000 N or 4900 N.][Ecf value from b(ii). Ignore signs in front of values] (1) 2

[6]

251. (i) Work done

Use of work done = force × distance (1)

Answer given to at least 3 sig fig. [2396 J, 2393 J if 9.8 m s–2 is used, (1) 2

2442 J if g = 10 m s–2 is used. No ue.]

Work done = 110 kg × 9.81 m s–2 × 2.22 m = 2395.6 J

(ii) Power exerted

Use of power = time

donework or power = F × v (1)

Answer: [799 W. 800 W if 2400 J is used and 814 W if 2442 J is 2used. Ecf value from (i)] (1)

Power = 3s

J2396

= 798.6 W

(iii) Principle of Conservation of Energy

EitherEnergy can neither be created nor destroyed (1) (1)OREnergy cannot be created/destroyed or total energy is not lost/gained (1)(merely) transformed from one form to another or in a closed/isolatedsystem. (1) 2

[Simple statement ‘Energy is conserved’ gets no marks][Information that is not contradictory ignore. ∆ Q = ∆ U +∆ W, withterms defined acceptable for 1st mark]

(iv) How principle applied to...

Lifting the bar: -Chemical energy (in the body of the weightlifter) or work done(lifting bar) = (gain in) g.p.e. (of bar) (1)[Reference to k.e. is acceptable]

The bar falling: -Transfer from g.p.e. to k.e. (1)(and that) g.p.e. lost = k.e. gained (1) 3

[‘g.p.e. converted to k.e.’ would get one mark][References to sound and thermal energy are OK, but gpe to sound orthermal energy on its own gets no marks]

(v) Speed of bar on reaching the floor

Setting ½ mv2 = m g h or ½ mv2 = work done or 2400 J (1)[ecf their value][Shown as formulae without substitution or as numbers substitutedinto formulae]Correct values substituted (1)

[allow this mark if the 110 kg omitted – substitution gives v2 = (1)

43.55(6) m2 s–2 or 44.4 m2 s–2 if g = 10 m s–2 is used]

Answer: [6.6 m s–1. 6.7 m s–1 if g = 10 m s–2 is used.]

½ 110 kg × v2 = 110 kg × 9.81 m s–2 × 2.22 m or = 2400 J / 2396 J

v = 6.6 m s–1 [6.66 m s–1 if 10 m s–2 used] (1)

OR

Selects v2 = u2 + 2as or selects 2 relevant equations (1)Correct substitution into equation (1)

Answer [6.6 m s–1] (1)

v2 = 0. + 2 × 9.81 ms–2 × 2.22m 3

v = 6.6 m s–1

[12]

252. (i) Plot a graph

Check any 2 points.[Award if these correctly plotted in appropriate square] (1)Curve of best fit. (1) 2

(ii) Half life average time required (1)for the count rate / activity / intensity to reach half the originalvalue or time taken for half of the atoms / nuclei/nuclides to decay (1)[NOT mass / particles / atom / (radio)isotope / count / sample/cells/ nuclide]

(iii) Use the graphValue of half life [Allow answers in the range 3.1 – 3.3. (1)Mark not to be awarded if a straight lined graph was plotted]Two or more sets of values used to find half life.[Could be shown 1 on graph] (1) max 3 (ii &iii)

(iv) Similar toeg (The programme) obeys an exponential law or once a cell has‘decayed’, it is not available to decay later or (the ‘decay’ is)random or it is impossible to predict which cell will ‘decay’ next. (1) 1

(v) Differenteg (Far) fewer cells available than atoms (in a sample of radioactivematerial) or it is a different ‘scenario’ eg. they are not atoms butcells on a grid generated by computer. (1) 1

[7]

253. (a) (i) Potential difference = work (done)/(unit) chargeOR Potential difference = Power/current (1) 1

(ii) J = kg m 2 s –2 (1)

C = A s or W = J s1 (1)

V = kg m2 A–1 s–3 (1) 3

(b) Converts 2 minutes to 120 seconds (1)Multiplication of VI∆ t or V∆ Q (1)Energy = 1440 J (1) 3

Example of answer:Energy = 6.0 V × 2.0 A × 120 s

= 1440 J[7]

254. (a) n = number of charge carriers per unit volume OR

n = number of charge carriers m–3 ORn = charge carrier density (1)

v = drift speed/average velocity/drift velocity (of the charge carriers) (1) 2

(b) n is greater in conductors / n less in insulators. (1)[There must be some comparison]larger current flows in a conductor. Dependant on havingreferred to n (1) 2(statement that n large in conductor and so current large max1)

(c) (In series), so same current and same n and Q (1)vB greater vA (1)vA/vB = ¼ // 0.25 (1) 3

[7]

255. (a) pd = 3.6 V (1) 1

Example of answer;p.d. = 0.24 A × 15 Ω = 3.6 V

(b) Calculation of pd across the resistor (1)[6.0 – 3.6 = 2.4 V]Recall V = IR (1)I1 calculated from their pd / 4Ω (1)

[correct answer is 0.60 A. Common ecf is 6V/4Ω gives 1.5 A] 3

Example of answer:I1 = 2.4 V / 4.0 Ω = 0.6 A

(c) Calculation of I2 from I1 – 0.24 [0.36 A] (1)[allow ecf of their I1. common value = 1.26 A]Substitution V = 3.6 V (1)R = 10 Ω (1) 3

[7]

256. (a) (i) (– gradient =) r = 1.95 – 2 Ω (1)E = 8.9 – 9 V (1) 2

(ii) I = 2.15 – 2.17 A (1) 1

(iii) Use of V = IR (1)R = 2.1 – 2.2 Ω (1) 2

(b) (i) Battery or cell with one or more resistive component (1)Correct placement of voltmeter and ammeter (1) 2

(ii) Vary R e.g. variable resistor, lamps in parallel (1)Record valid readings of current and pd (consequent mark) (1) 2

[Do not give these marks if the candidate varies the voltage as well][9]

257. H–R Diagram

(i) L and T (1)L and K (1) 2

or L and L (1), T and K (1), not W]

(ii) Any 2 correct [of 102, 1 or 10°, 10–2] (1)All 3 correct (1) 2

(iii) 20 000 and 5 000 (1) 1

Identify stars

(iv) Red giant = (B and) C (1)Low mass ms star = E [ignore X] (1) 2

Zeta Tauri Luminosity

(v) Use of L = 4 π D2 I (1)Correct substitution (1)

3.8(2) × 1030(W) (1) 3

Zeta Tauri identification (ecf)

(vi) 3.8(2) × 1030 W ÷ 3.9 × 1026 W [or 4 × 1030 W used] (1)

Correct ratio [e.g. 9700, 9800, 10300 or 104, etc.] (1)Hence A [from answer in range 9700 to 10300] (1) 3

[13]

258. (i) Fusion calculationsMass difference substitution [(2 × 5.0055) – (6.6447 + 2 × 1.6726)] (1)

2.11 × 10–29 kg [or 0.0211 × 10–27 kg] (1) 2

(ii) E = mc2 seen (1)

1.9 × 10–12J[ecf] (1) 2[4]

259. White dwarf density

(i) M ÷ 34

π r3 [allow M, m, R, r] (1) 1

(ii) Any pair of values correctly read [may be implied, ignore 106] (1)

Any correct substitution [with 2.0 × 1030 and 106, ecf on (i)] (1)

Two correct answers [in kg m–3, no statement required] (1) 3

White dwarf future

(iii) Cools / temperature decreases (1)Becomes dimmer / changes colour [not brown dwarf] (1) 2

[6]

260. Stress – strain graph

(i) Stress (1)

Pa/MPa/GPa/Nm–2 (1) 2

(ii) Use of E = σ ÷ ε / E = gradient (1)Any correct substitution [for linear region] (1)

Suitable scale: 1, 2, 3, 4, 5 and × 109 / G (1) 3

UTS and yield stress

(iii) 5GPa [ecf] (1) 1

(iv) The stress at which plastic deformation begins / beyond elasticregion [not just ‘beyond Hooke’s law’] (1) 1

(v) Y at or just beyond end of straight line on graph [0.03 < ε < 0.04] (1) 1

Second material

(vi) Lower gradient initially (1)Straight line to right-hand edge of graph (1) 2

Energy density and Work done

(vii) Any reference to area [may be implied] (1)Correct technique: rectangle (and triangle) or counting squares (1)

7.5 – 8.5 × 108 J m–3 [no ecf] (1) 3

(viii) 8 × 108 J m–3 × 3.8 × 10–7 m3 [ecf on energy density from (vii)] (1)Correct answer [300 J, ecf] (1) 2

[15]

261. β – decay equations

(i) n = udd and p = uud (1)

β – and v have no quarks / are leptons / are fundamental (1) 2

(ii) p → n (1)

β+ and ν [on RHS, allow e+] (1) 2[4]

262. Antihydrogen

(i) Antiproton [or anti–up quark, anti–down quark] and positron (1) 1

(ii) p = –1 and e+ = +1 [accept correct u , d charges for p ] (1)u u d (e+ fundamental / no quarks) [ecf from (b), credit if in (i)] (1) 2

(iii) zero / neutral (1) 1

Antimatter storage

(iv) Annihilates (1)(On contact) with matter / container / protons / HOR Not charged: not affected by magnetic fields (1) 2

[6]

263. (a) (i) l 65.0 cm to 75.0 cm and recorded to nearest mm or better (1)w 29.0 cm to 31.0 cm and recorded to nearest mm or better (1)

[Ignore l and w reversed] [Unit error once only][For precision of l and w over half of the reading must beto correct precision]

Both repeated (1)

Measurements taken perpendicular/parallel to the edge ofthe foil / eye vertically above edge of foil (1) 4

(ii) 16t recorded to 0.01 mm or better + unit and 0.15 mm to 0.30 mm

Repeat readings shown (1)

Zero error checked (1)Other precaution e.g. smoothed foil at each fold to excludeair/careful folding to avoid wrinkles 4

(iii) Sensible ∆ (16t)

=∆

∆

used. Δ if obtained becan mark 2nd

shown. 16

)Δ(16t if obtained becan mark 1

.instrument ofdivision scale halfor division scale allow reading oneonly or reading identical If values.of range halfor rangeExpect ).(16 Sensible

st

tt

t

t

Correct calculation of percentage (1) 2

(iv) Attempt at density = volumemass (1)

Correct substitution into volume formula with consistentand correct l, w, and t (1)

Value 2.3 to 3.0 g/cm3 and > 2.s.f. + unit (1)

[2.3 × 10–3 → 3.0 × 10–3 g/mm3 or 2300 → 3000 kg/m3] 3

(b) (i) Circuit set up correctly without help (2) 2

(ii) I to nearest mA or better and 40 mA to 55 mA with units (1)V repeated for same x (1)V to 0.1 mV or better with unit (1) 3

(iii) Correct substitution into R (1)Correct calculation, 2/3 s.f. + unit (1) 2

(iv) b to nearest mm or better and 8.0 mm to 12.0 mm with unit (1)b repeated (1)Correct substitution with consistent units for b, t, and x (1)Correct calculation, consistent and correct unit (1) 4

[24]

Sample results

(a) (i) l = 67.0,67.2, 66.8 cm l = 67.0 cm

w = 29.7, 29.7, 29.7 cm w = 29.7 cm

(ii) 16t = 0.15, 0.16, 0.15, 0.16 mm

t16 = 0.155 mm

t = 16

mm 0.155 = 0.0097 mm

(iii) ∆ (16t) = 0.01 (mm)

Percentage uncertainty =mm 515.0

mm 0.01 × 100%

= 6.5(%)

(iv) Mass = 5.68 g

Density = cm 0.00097 cm 29.7 cm 67.0

5.68g××

= 2.94 g/cm3

(b) (ii) I = 50 mAV = 6.5, 6.7, 6.9 mV V = 6.7 mV

(iii) R = V / I = mA 50mV 6.7

= 0.134 Ω

(iv) b = 1.0, 0.8, 1.2 cmb = 1.0 cm

ρ = xRbt

3.0

10 9.7 0.01 0.134 6-×××

= 4.3 × 10 –8 Ω m

264. (a) (i) d to 0.01 mm or better and ± 0.03 mm of Supervisor’s[unit seen somewhere]

Repeat shown (2)

Other precaution e.g. measured in 2 perpendicular directions/measured in different places/ used ratchet to avoid overtighteningmicrometer/ avoided kinks in wire (1) 3

(ii) Sensible ∆ d consistent with d (1)Correct calculation of percentage (1) 2

(iii) Sensible mass and correct formula for density (1)

Correct substitution into volume formula with consistentlength units (1)

Density to 2/3 s.f. + unit (1)

Value 8.4 → 9.8 × 10x (1)

Correct x value (1) 5

+

−

3

3

3

m kgfor 3cm gfor 0mm gfor 3 e.g.

-

-

-

(d) (i) Circuit set up correctly without help 2

(ii) I to nearest mA or better and approximately 50 mA with unit (1)

V to nearest mV or better and in range 50 mV to 90 mV with unit (1)

V repeated (1) 3

(iii) Correct calculation of R (1)

R from IV to 2/3 s.f. + unit (1) 2

(iv) I ≤ I1, to nearest mA or better + unit (1)

V = 3 × to 4 × V1 to nearest mV or better + unit (1)

Correct calculation of R ≥ 2 s.f. + unit (1) 3[Only penalise a particular unit error once in (ii), (iii) and (iv)]

(v) Correct calculation of R (1)Correct substitution for ρ with consistent length units (1)Unit of ρ (1)

Value 4.2 → 5.2 × 10–7 Ω m and ≥ 2 s.f. (1) 4[24]

Sample results

(a) (i) m = 2.74 gd = 0.31, 0.31, 0.31 mm

d = 0.31 mm

(ii) ∆ d = 0.01(mm)

Percentage uncertainty = mm 0.31mm 0.01 × 100%

= 3.2(%)

(iii)4

m 4.0 m) 10 (0.31 23 ××=-

Vπ

= 3.02 × 10–7m3

ρ = 37

3

m1002.3

kg 10 2.74−×

× -

= 9080 kg m –3

(b) (ii) I = 49 mAV = 89, 90, 88 mVV = 89 mV

(iii) R1 = mA 49mV 89

= 1.82 Ω m

(iv) I = 48 mAV = 266, 265, 267 mVV = 266 mV

R2 = IV =

mA 48mV 266 =5.54 Ω

(v) R = 5.54 Ω – 1.82 Ω = 3.72 Ω

ρ = m 0.64

m)1031.0(π72.3 23

××××Ω −

= 4.7 × 10–7 Ω m

265. Explanation

There is a resultant (or net or unbalanced) force (1)

Plus any 3 of following:–

Direction of motion is changing (1)Velocity is changing (1)Velocity change implies acceleration (1)Force produces acceleration by F = ma (or N2) (1)Force (or acceleration) is towards centre / there is a centripetal (1)force (or acceleration) / no force (or acceleration) parallel to motionNo work done, so speed is constant (1) Max 3

[4]

266. (a) Experimental verification

QOWC (1)Measure T using clock or motion sensor or video camera or digital (1)camera [Don’t accept light gates]for a range of masses (or various masses) (1)

Plot T vs m1/2 / Plot T2 vs m / Plot log T vs log m / calculate T/m1/2

or T2/m (1)

Str line through origin / Str line through origin / Str line gradient (1)0.5 / constantOne precaution (1)e.g. Use fiducial (or reference) mark

Repeat and averageNo permanent deformation of springSmall amplitude or displacementMeasure at least 10T Max 5

(b) ExplanationsNatural frequency:Freq of free vibrations / freq of unforced vibrations / freq when it (1)oscillates by itself (or of its own accord) / freq of oscillation ifmass is displaced[Don’t accept frequency at which it resonates, frequency at whichit oscillates naturally, frequency if no external forces]

Resonance:When vibration is forced (or driven) at natural frequency (1)Amplitude (or displacement or oscillation) is large(or violent or increases) (1)Amplitude is a maximum / large energy transfer (1) 4

[Accept 4 Hz for natural frequency][9]

267. (a) Solar Power

Use of P = Iπr2 [no component needed for this mark] (1)Use of cos 40 or sin 50 (with I or A) (1)2.2 [2 sf minimum. No ue] (1) 3

e.g. P=1.1 × 103 W m–2 × cos 40 × π(29 × 10–3 m)2

= 2.2 W

(b) EnergyUse of E = Pt (1)

1.8 × 104 J/2.0 × 104J (1) 2

e.g. E = 2.2W × (2.5 × 3600 s)

= 2.0 × 104 J[5]

268. (a) GraphStraight line with positive gradient (1)Starting the straight line on a labelled positive fo (1)[Curved graphs get 0/2. Straight line below axis loses mark 2unless that bit is clearly a construction line.] 2

(b) Work function

From the y intercept (1)[Accept if shown on graph]OR Given by gradient × f0 (or h × f0) [Provided that f0 is markedon their graph, or they say how to get it from the graph]OR Read f and Ek off graph and substitute into Ek = hf – φ[Curved graph can get this mark only by use of hf0 or equation methods.] 1

(c) GradientGradient equals Planck constant (1) 1

[Curved graph can’t get this mark][4]

269. (a) WavelengtheV to J (1)Use of ∆ E = hf (1)Use of c = fλ (1)

1.8 × 10–11 [2 sf minimum. No ue] (1) 4

e.g. f =

(–1.8 keV – (– 69.6 keV)) × (103 × 1.6 × 10–19 J keV–1) / 6.6 × 10–34 J s

= 1.64 × 1019 Hz

λ = 3.00 × 108 m s–1/1.64 × 1019Hz

= 1.8 × 10–11 m

(b) TypeX rays [Accept gamma rays] (1) 1

[5]

270. (a) (i) Hubble constantUse of v = Hd or gradient = H (1)Converts y to s i.e. × (365 × 24 × 60 × 60) (1)Correct × by ‘c’ (1)

[Seeing 9.46 × 1015 gets previous two marks]

1.7 to 1.8 × 10–18 (s–1) (1) 4

[No marks for a bald answer]

e.g. H = 60 × 106 m s–1 /

(3.6 × 107 1y × 365 × 24 × 3600 × 3 × 108 m 1y–1)

= 1.8 × 10–18 s–1

(ii) UncertaintyDistance / d (1) 1

(b) Age of UniverseStates that d = vt (any arrangement) (1)Combines this with restated Hubble law (any arrangement) to give

t = H1 (1) 2

(c) Recessional SpeedRed shift = 76 nm / 469 – 393 nm (1)

Use of cv=∆

λλ (1)

5.8 × 107 m s–1 (1) 3

e.g. v = 76 × 10–9 m × 3 × 108 m s–1/393 × 10–9 m

= 5.8 × 107 ms–1

(d) Average mass-energy densityClosed : high density/above critical density (1)Then gravitational pull (or force or attraction) sufficient to causeBig Crunch/pull everything back/stop expansion (1)[NOT to hold the galaxies together]OR equivalent argument for Open[Don’t accept mass for density in mark 1 or just “gravity” in mark 2] 2

[12]

271. (a) Formula in words

(The force between two charged particles is directly) proportional tothe product of their charges [plural] and (1)

inversely proportional to the square of their separation [not just ‘radius’]. (1)

OR Either equation for F*, with valid word replacements for Q1, Q2 (1)

and r or r2 symbols. One mark for numerator, one for denominator. (1)

=

4in or Fin wordsi.e.*

20

212

21

r

QQ

r

QkQ

π ε

[If equation given in symbol form, followed by a key to the symbol 2meanings, then 1/2.]

(b) Base units of constant

[Either k or (4π )ε 0, be sure which][ecf from part a if power of Q or r wrong]

F = 221

r

QkQ or F = 2

0

21

4 r

QQ

π ε [OR using k units N m2 C–2]

Q1Q2 (or C2) → A2s2 (1)

F (or N) → kg m s–2 (1)

→ (units of) k = kg m3 A–2 s–4 OR (units of) ε0 = kg–1 m–3 A2 s4 (1)

OR using ε0 units F m–1:

C = As and either F = CV–1 or V = JC–1 (1)

J = kg m2 s–2 or N = kg m s–2 (1)

→ (units of) ε0 = kg–1 m–3 A2 s4 (1) 3[5]

272. (a) Electron speed

Substitution of electronic charge and 5000V in eV (1)

Substitution of electron mass in ½ mv2 (1)

Correct answer [4.2 (4.19) × 107 (m s–1), no ue] to at least 2 sf (1) 3[Bald answer scores zero, reverse working can score 2/3 only]

Example of answer:

v2 = (2 × 1.6 × 10–19C × 5000 V)/(9.11 × 10–31 kg) = 1.76 × 1015

v = 4.19 × 107 m s–1

(b) (i) Value of E

Correct answer [2.80 × 104 V m–1/N C–1 or 2.80 × 102 V cm–1] (1) 1

Example of answer:

E = V/d = 1400 V / 5.0 × 10–2

= 28 000 V m–1

(ii) Value of force F

Correct answer [4.5 × 10–15 N, ecf for their E] (1) 1

Example of answer:

F = Ee = 2.80 × 104 V m–1 × 1.6 × 10–19 C

= 4.48 × 10–15N

(c) Calculation of h 4

See a = their F / 9.11 × 10–31 kg (1)

[→ a = 4.9 × 1015 m s–2]

See t = 12 (× 10–2) m / 4 × 107 m s–1 (or use 4.2 × 107 m s–1) (1)[t = d/v, with d = plate length; 12 cm]

[→ t = 3.0 × 10–9 s, or 2.86 × 10–9 s]

See substitution of a and t values [arrived at by above

methods] into ½ at2 (1)

Correct answer [h = 0.020 m – 0.022 m] (1)

[Full ecf for their value of F if methods for a and t correct andtheir h ≤ 5.0cm]

Example of answer:

h = ½ a t2

= ½ × 4.9 × 1015 m s–2 × (2.86 × 10–9 s)2

= 2.0 × 10–2 m

(d) (i) Path A of electron beamLess curved than original (1) 1

(ii) Path B of electron beamMore curved than original, curve starting as beam enters field[started by H of the Horizontal plate label] (1) 1

[For both curves:• ignore any curvature beyond plates after exit• new path must be same as original up to plates]

[No marks if lines not identified, OK if either one is labelled][11]

273. (a) Newton’s law

Equation route:

F = 221

R

MGM (1)

M1, M2, R defined correctly, G defined correctly or not defined (1)[Both marks can be awarded for word equation]

OR Proportion route:(force is directly) proportional to the product of the masses[plural] and

inversely proportional to the square of their separation [not just‘radius’, unless related to orbital motion] (1) 2

(b) (i) Graph

Take two pairs of values off graph (1)

A). Find gR2 for one pair [≈ 400 ( × 1012)]

Attempt to show gR2 ≈ same for second pair (1)(within uncertainty limits of data read from graph) (1)

OR B). Compare pairs of values to show that as R changes by a

factor n, g changes by a factor 1/n2. (1) (1)

OR C). Substitute into formula with one pair to give a value ofM or some other constant. (1)

Repeat with second pair to give same value OR substitute back to confirm agreement of second pair of values. (1) 3

(ii) Gravitational field strengthValid approach via routes A, B or C above. (1)

g = 0.0027 – 0.0031 N kg–1 (1) 2

Example of answer:

g × 3802 = 400 → g = 400/ 3802 = 0.00277 N kg–1

(c) EffectMaintains the Moon in orbit around the Earth / keeps Moon (1)rotating around the Earth / provides (all the) centripetal (1)force/acceleration for its circular motion / pulls Moon towards 1Earth. [not just exerts force on the Moon]

[8]

274. (a) (i) Additional force

Correct answer [3.9 × 10–3 N ] (1) 1Example of answer:

0.4 × 10–3 kg × 9.81 N kg–1 = 4 × 10–3 N

(ii) ExplanationQuality of written communication (1)(Current produces) a magnetic field around the rod (1)[Do not accept in the rod]

There is an interaction between the two magnetic fields / fieldscombine to give catapult field (1)

Fleming’s Left Hand Rule/ Fleming’s Motor Rule (1)The rod experiences an upward force (1)

Using Newton 3 → downward force on magnet Max 4

(b) (i) DiagramLower pole labelled North/N and upper pole labelled South/S (1) 1

(ii) Calculation of current in rod

Use of F = BIl. (Ignore 10x. F is their force and l is 5cm) (1)See conversions; mT to T and cm to m (1)Correct answer [2.6/2.7 A ] (1) 3

Example of answer:

I = 3.9 × 10–3 N / (30 × 10–3 T × 5 × 10–2 m) = 2.6 A

(iii) New reading on the balanceValue < 85g [not a negative value] (1)84.6 g (1) 2

[11]

(2)

(Total 5 marks)

275. (a) Circuit set up correctly without help (2) 2

(b) (i) Units for X and V [seen anywhere] (1)

5 good values (± 0.01 V of examiner’s best line) (1)[4 good values + units only (1)]

(ii) GraphPlots – accurate to ½ square (1)Lines – both thin, straight and best fit (1)

(iii) 5 good values (± 0.01 V of examiner’s best line) (2)[4 good values only (1)][Allow point(s) off end of voltage scale]

(iv) Read off correctly, 0.66 – 0.71 m + unit (1) 7

(c) Both triangle’s large [mX base ≥ 6 cm, mY base ≥ 4 cm] (1)mX correctly calculated to > 2 s.f. (1)mY correctly calculated to > 2 s.f. [No units required] (1) 3

(d) Ratio correctly calculated to > 2s.f. [ecf] (1)

Correct percentage difference with 0.360 as denominator (1)Sensible comment based on percentage difference (1)

With reference to experimental error or sensible estimate of 4experimental error e.g. error in V or gradient (1)

[16]

Sample results

(a) x = 0.500 mV = 0.39 V

(b) (i)AB x/m V/V

0.100 0.080.200 0.150.300 0.230.400 0.310.500 0.390.600 0.460.680 0.52

(ii)

1 . 2

1 . 1

1 . 0

0 . 9

0 . 8

0 . 7

0 . 6

0 . 5

0 . 4

0 . 3

0 . 2

0 . 1

0 . 1 0 . 2 0 . 3 0 . 4 0 . 5 0 . 6 0 . 7 0 . 8 0 . 9 1 . 0x m/

V / V

X

Y

××

×

×

×

×

G R A P H

P l o t s ( a c c u r a t e t o ½ s q ) L i n e s ( b o t h t h i n , s t r a i g h t a n d b e s t f i t )

( 1 )

( 1 )

(iii)BC x/m V/V

0.750 0.660.800 0.780.850 0.880.900 0.990.950 1.101.000 1.20

(iv) x = 0.68(5) m

mX = 44.000.100.078.0

−− = 0.78 V m–1

mY = 44.000.100.020.1

−− = 2.14 V m–1

(d) 14.278.0X =

Ymm

= 0.36(4)

Percentage difference = 360.0

360.0364.0 − × 100

= 1%

This is acceptable experimental error, suggesting the resultssupport the suggestion.

276. (a) (i) Geiger-Muller tube must be labelled (1)

Level and ≤ 1 cm away [between ‘g’ and ‘e’ of ‘arrangement’] (1)

Counter, ratemeter or scaler must be labelled

[Or G-M tube (1), Close (1) and Interface + computer(or datalogger) (1)] 3

(ii) Well-away from other sources/sample and NOT in several places (1)

Record for at least 3 minutes (1)b 2

(iii) Suitable time interval (5 s → 20 s)for≥ 2 minutes (1) 2

(iv) Correct arrangement into y = mx + c [need not relate to (1) 1y = mx + c]

(b) Corrected count rate & lnA to ≥ 3 s.f. (1) (1)

GraphScale at least ½ graph paper in each direction, avoiding awkward scalessuch as 3’s (1)

Axes: labelled with units, must have ln(A/min–1) (1)

Plots: accurate to ½ square (1)

Line: straight thin & best fit (1) 5

(c) Large triangle [base ≥ 8 cm] (1)

Correct calculation of λ ≥ 2 s.f. [ignore unit] (1)

69 – 73 s with unit and 2 s.f. [No ecf] (1) 3

×−×

(1)

(1)

(1) (1)

s.f. 2 tounit] [ignore calculated ence

only lives 2[

unit with s), 73 (69correctly off read lives 3

plotted v If

21

21

λH

tA

[16]

Sample results

(a) (i)

L a y e rc o n t a i n i n gd e c a y i n gp r o t a c t i n i u m

G e i g e r - M u l l e r t u b e . . . . . . . . . .( C l o s e t o P a ) . . . . . . . . . .

( 1 ) ( 1 )

C o u n t e rR a t e m e t e r . . . . .o r s c a l e r

( 1 )

(ii) Keep G-M tube well away from all radioactive sources (or Pa sample)and record count for 5 minutes to determine the average count rate.Subtract this from subsequent count rates.

(iii) Record readings from ratemeter every 15 s for about 5 minsOrDiagram interface to integrate for 3 s and find count ratefor a duration of 5 minutes.

(iv) A = A0e–λ t

lnA = –λ t + lnA0

y = mx + c

(b)

Corrected count

rate / min–1ln(A/min–1)

500 6.21420 6.04340 5.83273 5.61235 5.46189 5.24156 5.05

6 . 2

6 . 0

5 . 8

5 . 6

5 . 4

5 . 2

5 . 0

0 2 0 4 0 6 0 8 0 1 0 0 1 2 0 t / s

l n ( A / m i n )– 1

(c) λ = – gradient = 012600.522.6

−−

s–1

= 9.68 × 10–3 s–1

21t = 31068.9

69.0−×

s

= 71 s

277. (a) (i) Correct circuit [–1 per error or omission] (1) (1) (1) 3

(ii) Connect capacitor to P to charge it (1)Switch to Q and start stopwatch (1)Record current I at regular time intervals t (1) 3

(iii) Correct arrangement into y = mx + c form[need not relate to y = mx + c] (1)Correct calculation shown + unit (1) 2

(b) Correct values of ln(I/uA) to 3/4 s.f. (1)Scale: at least ½ paper in each direction, avoiding awkward (1)scales, eg 3’sAxes: labelled + unit, must be of form ln (I/uA) (1)Plots: accurate to V2 square (1) 5

(c) Large triangle [base ≥ 8 cm] (1)Correct value [ignore units] [Ignore signs] (1)4.5 mF < C < 4.7 mF from correct calculation 2/3 s.f. + unit (1) 3[must be F, mF or µ F]

[16]

Sample results

(a) (i)

A

1 0 kΩ

P Q

(iii) I = I0e–at

1nI = –at + 1nI0

y = mx + c

At t = 0, I = kΩ 10V 1.5≈

RV

≈ 0.15 mA (150 µ A)

(b)

ln(I / µ A)4.804.584.374.153.943.733.50

5 . 0

4 . 8

4 . 6

4 . 4

4 . 2

4 . 0

3 . 8

3 . 6

~~0 1 0 2 0 3 0 4 0 5 0 6 0 7 0 t / s

l n ( I A )

(c) a = – gradient = 062

68.302.5−−

= 0.0216 s–1

C = aR1 = 310100216.0

1××

= 4.6 × 10–3 F

= 46000 µ F (to 2 s.f)

278. (a) Quality of written communication (1)Protons drift/move uniformly inside tubes (1)Accelerate between the tubes/in the gaps (1)Alternating p.d. reverses while p is in tube (1)The tubes must get longer as p speeds up (1)For time inside tube to be constant or to synchronisemovement with the pd (1) Max 5

(b) (i) Multiply by 419 or 420 (1)

Multiply by 1.6 × 10–19 (1)

Correct answer to at least 2 sf (1)

[5.36/5.38/5.4 × 10–11 (J)] [no ue]

∆ m = energy ÷ (9.0 × 1016 m2 s–2) (1)

[ecf their energy or 5 × 10–11] (1)

∆ m ÷ 1.01 × 1.66 × 10–27kg [ecf their ∆ m] (1)

Correct answer (1)

[0.36 or 36%] [Use of 5 × 10–11 gives 33%] (1) 6

[Accept routes via ∆ m in u and mp in J]

(ii) Use of 1/f (1)

∴ time down linac = 420 ÷ 3.9 × 108 s–1

or 210 ÷ 3.9 × 108 s–1 (1)

[t = 1.07/1.08/1.1 × 10–6(s) or 0.54 × 10–6 (s)] 2

(c) (i) Fixed target:Large(r) number of /more collisions or more likely to get collisions[not easier to get collisions] (1)

Other particle beams produced (1)

(ii) Colliding beams:More energy available for new particles (1)p = 0 so all energy available (1) Max 2

[15]

279. (a) Mention of natural frequency (of water molecules) (1)

At f0 there is a large/increased amplitude (1)

and hence max energy transfer / max power transfer / maxefficiency / max heating (1) 3

(b) (1.2 kg)(3200 J kg–l K–1)(75 K) seen (1)⇒ 288 kJ+ 600 s to give a power in W [⇒ 480 W] (1)Efficiency 480 W e.c.f ÷ 800 W [= 60%] (1)

There will be heat/energy/power losses from the meat/to the (1) 4surroundings or water evaporation needs LHV or waterevaporation leaves fewer molecules to vibrate

[7]

280. Calculation of time for ball to travel 30 m

Recall of v = s/t (1)

Correct answer [1.2 s] (1) 2


v = s/t

t = 30 m ÷ 25 m s–1

= 1.2 s

Calculation of westward component of ball’s velocity

Recall of v = u + at (1)

Correct answer [9.6 m s–1] [ecf] (1) 2


v = u + at

v = 0 + 8 m s–2 × 1.2 s

= 9.6 m s–1

Calculation of distance ball travels to west

Use of appropriate equation of motion (1)

Correct answer [5.76 m] [ecf] (1) 2


s = ut + ½ at2

= 0 + ½ × 8 m s–2 × (1.2 s)2

= 5.76 m

Calculation of final velocity of ball

Use of Pythagoras’ theorem or scale drawing with velocity triangle (1)

Correct answer for magnitude of velocity [26.8 ms–1, accept inrange 26 – 27.5] [allow ecf] (1)

Use of trigonometrical equations with velocity triangle (1)

Correct answer for direction of velocity [21.0° West or 339.0° ±2°] (1) 4[allow ecf]

[Allow 3rd and 4th marks for displacement triangle instead of velocity]


magnitude of velocity = √((9.6 m s–1)2 +(25 m s–1)2)

= 26.8 m s–1

angle = tan–1(9.6 m s–1/25 m s–1)

= 21.0° West (or 339.0°)[10]

281. Meaning of energy level

Specific allowed energy/energies (of electron in an atom)(1) 1

Meaning of photon

Quantum/packet/particle of energy/radiation/light/electromagnetic wave (1) 1

Formula for photon energy

E2 – E1 (1) 1[Allow E1 + Ephoton = E2]

Explanation of photon wavelengths

Same energy change / same energy difference / energy the same (1) 1

Meaning of coherent

Remains in phase / constant phase relationship(1) 1

282. Show that the average deceleration is about 0.1 m s –2

Use of v2 = u2 + 2as (1)

Correct answer [0.13 m s–2] to at least 2 sig fig [no u.e.] (1) 2[ignore + or – in answer and reversal of v and u in calculation][Bald answer scores 0, reverse calculation 2/3]


v2 = u2 + 2as

0 m2 s–2 =(13 m s–1)2 + 2 × a × 640 m

a = –(13 m s–1)2 ÷ (2 × 640 m)

a = – 0.13 m s–2 OR deceleration = 0.13 m s–2

Calculation of average resultant force

Recall of F = ma (1)

Correct answer [182 N] [allow ecf] (1) 2

[Use of a = – 0.1 m s–2 gives answer of 140 N]


F = ma

= 1400 kg × 0.13 m s–2

= 182 N

Explanation of graph shape

gradient decreasing / slope of graph becoming less steep(1) 1

Using graph for speed after 15 s

Tangent touching at 15 s, not crossing curve(1)

Use of ∆s/∆t(1)

Correct answer calculated using values from curve or tangent(1) 3

[range 14.0 m s–1 to 17.0 m s–1]


Speed = gradient = 800 m / 52 s

= 15.4 m s–1

Calculation of average deceleration

Recall of a =(v – u) / t(1)

[Allow use of alternative equation of motion and values fromgraph or previous parts]

Correct answer [0.97 m s–2] [allow ecf](1) 2


a = (v – u) / t

= (15.4 m s–1 – 30 m s–1) ÷ (15 s)

= (–)0 .97 m s–2

Explanation of difference in values of deceleration

Deceleration greater when car at higher speed(1)

(because) e.g. more air resistance / greater drag(1) 2[12]

283. Description of sound

Particles/molecules/atoms oscillate/vibrate (1)

(Oscillations) parallel to/in direction of wave propagation / wavetravel / wave movement [Accept sound for wave] (1)

Rarefactions and compressions formed [Accept areas of high and low pressure] (1) 3

Meaning of frequency

Number of oscillations/cycles/waves per second / per unit time (1) 1

Calculation of wavelength

Recall v = fλ (1)

Correct answer [18 m] (1) 2

Example of calculation

v = fλ

λ = 330 m s–1 ÷ 18 Hz

= 18.3 m[6]

284. Explanation of standing waves

Waves reflected (at the end) (1)

Superposition/interference of waves travelling in opposite directions (1)

Where in phase, constructive interference/superpositionOR where antiphase, destructive interference/superpositionOR causes points of constructive and destructiveinterference/superposition [Do not penalise here if node/antinode mixed up] (1) 3

Mark node and antinode

Both marked correctly on diagram (1) 1

Label wavelength

Wavelength shown and labelled correctly on diagram (1) 1

Explain appearance of string

Any two from:

• light flashes twice during each oscillation / strobefrequency twice that of string [accept light or strobe]

• string seen twice during a cycle

• idea of persistence of vision (2) max 2

Calculate speed of waves

Use of v = √T/µ (1)

Correct answer [57 m s–1] (1) 2


v = √T/µ

= √(1.96 N / 6.0 × 10–4 kg m–1)

= 57.2 m s–1

[9]

285. Complete table:

Plastic: Not desirable (1)Reason: Would remain deformed(once load is removed) (1)

Tough: Desirable (1)Reason: To withstand dynamic loads/impacts/shocks (1)

Brittle: Not desirable (1)Reason: Would crack / shatter / snap / break with no(plastic) deformation (1) 6

Calculate stress:

Use of m × g(1)

[g = 10 m s–2 will not be penalised]

Correct answer [1.9 × 104 Pa] (1) 2

e.g.Force = 80 × 9.81 = 785 N

σ = F/A = 785/4.2 × 10–2

= 18686 = 1.9 × 104 Pa

[allow 1.8 – 2.0 × 104 Pa and allow Nm–2 as units]

Running athlete:

On one foot / part of foot (1)As less(surface) area (1)ORWhen landing/pushing off (1)As the force is greater (1) 2

[10]

286. Distance to aircraft:

Use of distance = speed × time(1)

Correct answer [7.2(km) / 7200(m) is the only acceptable answer. No u.e.] (1) 2

e.g.

Distance = speed × time = 3 × 108 × 24 × 10–6

= 7.2 km

Why pulses are used:

Any two of the following:

• Allow time for pulse to return before next pulse sent

• To prevent interference/superposition

• A continuous signal cannot be used for timing

• Can’t transmit / receive at the same time (2) max 2

Doppler shift:

Any three of the following

• Change in frequency/wavelength of the signal [allow specified change,either increase or decrease]

• Caused by (relative) movement between source and observer[accept movement of aircraft/observer]

• Size of change relates to the (relative) speed of the aircraft[Allow frequency increasing; do not allow frequency decreasingunless linked to aircraft moving away]

• Quote v/c = ∆f/f (3) max 3[7]

287. Airflow diagram:

At least two continuous lines drawn around and beyond sky diver (1) 1[Ignore turbulence around feet – lines must not touch]

Force diagrams:

Weight arrow vertically downwards [allow W, mg and gravitational force] (1)(Viscous) Drag and Upthrust arrows upwards (1) 2[Allow air resistance for drag and u for upthrust][Upthrust must be vertically upwards]

Forces relationship:

Weight = Drag + Upthrust OR in equilibrium(1) 1[Allow symbols or formulae for any of these quantities]

Why Stokes’ law not valid:

Sky diver not spherical / flow not laminar(1) 1

Upthrust:

Use of Upthrust = Weight of displaced air (1)Use of Mass = density × volume (1)

Correct answer [4.1(N) or 4.2(N) if g = 10 m s–2 is used] (1) 3

e.g.Weight = 1.2 × 0.35 × 9.81= 4.1 N

Comment on size of force:

Smaller than weight / small(1)Very little effect on terminal velocity / small decrease in terminal velocity (1) 2

Airflow:

Turbulent/slower (1) 1[11]

288. Area of wire:

Use of A = πr2 (1)

Correct answer [1.9 × 10–7 (m2). Allow 1.9 × 10–7 and 2.0 × 10–7 (m2)](1) 2[No u.e.]

e.g.

A = πr2 = π ×(2.5 × 10–4)2

= 1.96 × 10–7 m2

Table + graph:

Length / Area / × 106 m–1

0.0

0.5

1.0

1.5

2.0

2.5

3.1

3.6 (1)

4.0 – 4.1

First 2 points plotted correctly to within 1 mm (1)Rest of points in straight line with origin by eye (1) 3

Resistivity calculation:

Drawn through the origin, ignoring first 2 points (1)Recall ρ= R /(L/A) [in any form] (1)Large triangle drawn on graph OR accept the use of a pair of values (1)read from the line

[ x> 3 × 10-6 m–1)is required in both cases][x-axis allowed as bottom of triangle]

Correct answer [1.2 × 10-7 Ω m)] (1)

[allow 1.1 – 1.3 × 10-7(Ω m)] [no u.e.] 4

e.g.

0.4 / 3.4 × 106 = 1.2 × 10–7 Ω m

Anomalous results:Any two of the following:

• Higher current/lower resistance for shorter lengths/at these points

• At shorter lengths/at these points wire gets hotter

• Non-uniform area/diameter

• Cable / contact resistance

• Sensitivity of meters

• Effect on resistance of any of the above (2) max 2[11]

289. Unpolarised and plane polarised light:

Correct diagrams showing vibrations in one plane only and in all planes (1)

Vibrations/oscillations labelled on diagrams (1) 2

Telescope adaptation:

Fit polarising filter / lens [must be lens not lenses] (1)At 90° to polarisation direction to block the moonlight / rotate until 2cuts out moonlight (1)

[4]

290. Calculation of voltage

Use of E=V/d [could be rearrangement] (1)

Correct answer [1.5 × 109 V] (1) 2


V = Ed = 3 × 105 × 5000 V

V = 1.5 × 109 V

Calculation of capacitance

Recall Q = CV(1)

Correct answer [2.7 × 10–8 F](1) 2


C = Q / V

= 40 C / 1.5 × 109 V

= 2.7 × 10–8 F

Resistance

Use of RC = 20 ms, or an appropriate time (eg, 20 ms ÷ 5 = 4 ms)(1)OR attempt to find current from I = Q/t

Correct answer [7.5 × 105 Ω (1.5 × 105 Ω)](1)

Example 1:

⇒ R = 20 ms ÷ 2.7 × 10–8 F (4ms ÷ 2.7 × 10–8 F)

= 7.5 × 105 Ω (1.5 × 105 Ω)

Example 2:

I = 40 C / 20 ms = 2000 A(1)

⇒ R = A2000

V105.1 9×=I

V

⇒ 7.5 × 10 5 Ω (1) 2

[Also give credit for using “average” pd which is less than 1.5 GV

say V/2 → 3.75 × 105 Ω]

Drawing of electric field region

–

G r o u n d

5 k m

≥ 2 radial lines (1)

Arrow(s) (1) 2

Electric field

Recall E = 204 r

q

π ε OR k = 04

1

π ε = 9 × 109 m F–1) (1)

Correct answer [1.44 × 104 V m–1] (1) 2

[Use of d

V scores 0]


E = 204 r

q

π ε

212– 50001085.84(

40

×××=

π V m–1

= 1.44 × 104 V m–1

Lightening strike

Field stronger near cloud OR Greater force/acceleration on chargesOR Mention of force on charges OR Mention of ionising atoms by collision (1) 1

[11]

291. Speed of sphere

Momentum conserved [stated or implied] (1)

Correct subs L.H.S or R.H.S of conservation of momentum equation (1) 3

Correct answer [ν = 1.43(m s–1)] (1)


54 × 2.57 (+ 0) = 54 × ν + 29 × 2.12 (g m s–1)⇒ 138.78 = 54 × ν + 61.48

⇒ ν = 1.43(m s–1)

Elastic or inelastic collision

Recall K.E: 1/2mν2(1)

Correct values for both KEs [178(mJ), 120(mJ), no ue](1)

Conclusion consistent with their results for KE(1) 3

[max 1 if only words used and inelastic ≡ energy lost implied]


= ½ × 54 × 2.572 = 178 mJ

Final total K.E: ½ × 29 × 2.122 + ½ × 54 × 1.432 mJ= 65 mJ + 55 mJ= 120 mJ⇒Inelastic

Average speed of the spheres

Recall ν = 2πr / t (1)

Correct answer [2.9 m s–1] (1) 2


ν = 2πr / t = π × 0.17 × 2 m / 0.37 s

= 2.9 m s–1 or 290 cm/s

Calculation of centripetal force

Recall F = mν2 / r OR mrω2 OR m ν ω (1)Correct answer [1.43 N, ecf for their ν ] (1) 2


F = mν2 / r

= 0.029 × 2.92 / 0.17 N [watch out for 29 twice]= 1.43 N [ecf]

Tension

Weight of sphere (= mg = 0.029 × 9.81 N = )0.28 N (1)

T = F – W OR F = T + W [using their values for F and T] (1) 2


= 1.43 – 0.28 (N)⇒ T = 1.15 N

[12]

292. “The standard model”

Everyday matter/atoms: p,n, e [maybe in two places] (1)

Protons / neutrons are made from quarks (1)

p: uud and n:udd (1)

show charge of either [p: u(+2/3) u(+2/3) d(–1/3) ⇒ + 1 OR n: u(+2/3) d(–1/3) d(–1/3) ⇒ 0] (1)

All baryons have three quarks (1)

Hadrons contain quarks (1)

Electron is fundamental/leptons are fundamental (1)

Electron-neutrino created during β-decay (1) Max 6[6]

293. Calculation of voltage

Use of ∆E = c2∆m (1)

Use of eV (1)

Correct answer [4.1 × 109 (V)] [no ue] (1)


∆E = c2∆m = eV

⇒ V = c2∆m / e = 9 × 1016 × 8000 × 9.1 × 10–31 / 1.6 ×10–19 V

= 4.1 × 109 V 3

Role of magnets

Field deflects/bends/curves the path (1)

Field is at curved parts / field at AD and BC / no field on straight parts (1)

Field perpendicular to page / velocity (1)

Force perpendicular to velocity or field (1)

Force is centripetal / towards centre (1) Max 4

Calculation of field strength:

r = p / Bq rearranged to B = p/rq (1)

correct substitution of either p OR of r and q (1)

Correct answer [0.124(T), no ue] (1)


r = P / Bq ⇒ B = P / qr

= 8000 × 9.1 × 10–31 × 3.0 × 108 / 110 × 1.6 × 10–19 T= 0.124 T 7

[10]

294. Proton numbers:

55 and 94 (1) 1

Fuel for the power station:

(i) (Nuclear) fission (of 235U) (1)

(ii) Absorption of a neutron by(238)U(followed by β-decay) (1) 2[not bonding, not fusion, allow combining] [Any other particlementioned in addition to neutron loses the mark]

Calculate emission rate:Use of λ = ln 2 / t½ [allow either Cs t½] (1)

See 1.5 × 106.e–0.023 × 20 [allow ecf of λ for this mark] (1)

Correct answer [9.5 × 105(Bq m–2)] (1)

[2040(Bq m–2) scores 2/3]ORWork out number of half lives (1)Use the power equation (1)Correct answer (1) 3


λ = ln 2 / 30 = 0.023 yr–1

R = 1.5 × 106.e–0.023 × 20 Bq m–2

R = 9.5 × 105 Bq m–2

Assumption:

the only source in the ground is 137Cs / no 137Cs is washed out of(1) 1soil / no clean-up operation / no further contamination / reference toweather not changing the amount

Scattered isotopes:

(131)I and 134 Cs (1)For either isotope: many half lives have passed / half life short 2compared to time passed / short half life therefore now low emission (1)

Comment:Even the isotopes with a thirty year half life are still highlyradioactive [eg accept strontium hasn’t had a half life yet] (1)Plutonium will remain radioactive for thousands of years (as thehalf life is very large) [accept the alpha emitting isotopes forplutonium] [accept plutonium half lives much longer than 20 years] (1) 2

[11]

295. Doppler effect:2 points from

• As ambulance approaches, frequency is higher (than normal) (1)

• As ambulance recedes, frequency is lower (than normal) (1)

• At moment of passing observed frequency = siren frequency (1) max 2

Diagram and explanation:

Diagram shows planet, star and Earth in line, with planet on far side of star (1)Diagram shows planet, star and Earth in line, with planet betweenstar and Earth (1)Idea that planet attracts star towards itself (1)Hence star experiences a change in velocity towards planet (1)Max f / min λ / blue shift when planet is nearest Earth or min f /max λ / red shift when planet is furthest away from Earth [must be max 4referring to radiation from the star] [must not refer to sound] (1)[A good diagram can score all 4 marks. Information on the diagramoverrules written]

Time for orbit:Idea of measuring/identifying the time taken for planet to return tothe same position in its orbit or half an orbit (1) 1[e.g. the time between successive minima or maxima / the time betweenthe start of red shift and the start of blue shift is half an orbit]

Explanation:

F = GMm/r2 quoted [watch out for F change to g] (1)(For the Earth) m is very small [accept size is very small] (1)Hence the force exerted on the star is smaller(despite a smaller r) (1)Change in velocity / wobble produced is too small to give anobservable Doppler shift [frequency shift, red/blue shift accepted in 4place of Doppler shift] (1)

[11]

296. Definitions:An electric field is a region where charged objects experience a force (E = F/Q) (1) 1A gravitational field is a region where masses experience a force(g = F/m) (1) 1

Similarities:

• Both fields obey an inverse square lawOR inverse square equations quoted (1)

• Both fields are radial for point objects / spherical distributions (1)

• Both fields have an infinite range / field strength approacheszero a long way from source (1) max 2

Differences:

• Electric forces can be attractive or repulsive but gravitationalforces are always attractive (1)

• Electric forces are (much) stronger than gravitational forcesOR comparison of size of coupling constants in the twoforce equations (1)

• Electric forces only act on charged particles but gravitationalforces act on all matter (1)

• Electric forces can be shielded (e.g. by use of a Faradaycage) but gravitational forces cannot (1) max 3

[6]

297. Meaning of plane polarised

Oscillations/vibrations/field variations (1)

Parallel to one direction, in one plane [allow line with arrow at both ends] (1) 2

Doppler effect

Doppler (1)

If source/observer have (relative) movement [reflections offvibrating/moving atoms] (1)

Waves would be bunched/compressed/stretched or formula quoted[accept diagram] (1)

Thus frequency / wavelength changes [accept red /blue shift] (1) 4

Frequency about 3 × 10 14 Hz

Evidence of use of 1/wavelength = wavenumber (1)

laser wavenumber = 9400 or wavelength change =7.69×10–4 (1)

New wavenumber = 10700 [or 8100] or conversion of wavelength

change to m [7.69 × 10–6] (1)

New wavelength = 935 nm [or 1240 nm]

Use of frequency = c / wavelength [in any calculation] (1)

f = 3.2 × 1014 Hz [note answer of 2.8 × 1014 = 3 , 3.4 × 1014 = 4](1) 5

Model of light

Particle/photon/quantum model (1)

Photon energy must have changed / quote E = hf (1)

Energy of atoms must have changed [credit vibrating less/more/faster/slower] (1) 3[14]

298. Smoke detectorsRecognition that mass alpha = 4 (1)Idea of - 4 to find resulting nucleus mass [237] (1)

Mudaughter product = mvalpha or momentum equations in context (1)

valpha = 59(.25) udaughter product [allow ecf incorrect masses eg 4 and 241 : 60.25] (1)

Use of ½ mv2 To give ratio = 59(.25) [allow ecf as long as rounds (1)

to 60 ; must have speeds sub ; valid use of p2/2m]

Energy:

E = mc2 / energy must have come from mass (1)

Total mass after is a (little) less than before/mass loss/mass defect/binding energy (1) max 6

[6]

299. (a) Calculation of weight

Use of L × W × H (1)

Substitution into density equation with a volume and density (1)

Correct answer [49.4 (N)] to at least 3 sig fig. [No ue] (1) 3[Allow 50.4(N) for answer if 10 N/kg used for g.]

[If 5040 g rounded to 5000 g or 5 kg, do not give 3rd mark; if conversion to

kg is omitted and then answer fudged, do not give 3rd mark][Bald answer scores 0, reverse calculation 2/3]

80 cm × 50 cm × 1.8 cm = 7200 cm3

7200 cm3 × 0.70 g m–3 = 5040 g

5040 g × 10–3 × 9.81 N/kg

= 49.4 (N)

[May see :

80 cm × 50 cm × 1.8 cm × 0.7 g m–3 × 10–3 × 9.81 N/kg

= 49.4(N)]

(b) (i) Horizontal and vertical components

Horizontal component = (83 cos 37 N) = 66.3 N / 66 N (1)Vertical component = (83 sin 37 N) = 49.95 N / 50 N (1) 2[If both calculated wrongly, award 1 mark if the horizontal was identifiedas 83 cos 37 N and the vertical as 83 sin 37 N]

(ii) Add to diagram

Direction of both components correctly shown on diagram (1) 1

(iii) Horizontal force of hinge on table top

66.3 (N) or 66 (N) and correct indication of direction [no ue] (1) 1[Some examples of direction: acting from right (to left) / to the left /West / opposite direction to horizontal. May show direction by arrow.Do not accept a minus sign in front of number as direction.]

[7]

300. Expression for Ek and work done / base unit

(a) (i) Kinetic energy = ½ mu2

Work done = Fd[must give expressions in terms of the symbols given in the question] (1) 1

(ii) Base units for kinetic energy = (kg (m s–1)2 ) = kg m2 s–2 (1)

Base units for work done = kgms–2 .m = kg m2 s–2 (1)

[derivation of kg m2 s–2 essential for 2nd mark to be given] 2

[Ignore persistence of ½] [ For 2nd mark ecf mgh for work from (a)(i)]

(b) Show that the braking distance is almost 14 m

[Bald answer scores 0; Reverse calculation max 2/3]

Either

Equating work done and kinetic energy [words or equations] (1)

Correct substitution into kinetic energy equation and correct substitution (1)into work done equation

Correct answer [13.8 (m)] to at least 3 sig fig. [No ue] (1)

0.5 × m × (13.4 m s–1)2 = m × 6.5 m s–2 × d

(m)8.13ms5.6

)ms 4.13(5.02

21

=×××

−

−

m

m3

[m may be cancelled in equating formulae step and not seen subsequently]

OR

Selecting v2 = u2 + 2as OR 2 correct equations of motion (1)Correct magnitudes of values substituted (1)

[i.e. 0 = (13.4 m s–1)2 + 2((-)6.5 m–2)s]Correct calculation of answer [13.8 (m)] to at least 3 sig fig. [No ue] (1)

(c) Why braking distance has more than doubled

QOWC (1)

Either(Because speed is doubled and deceleration is unchanged) time (1)(to be brought to rest) is doubled/increased.

(Since) distance = speed x time [mark consequent on first] or s = ut + ½ at2 (1)the distance is increased by a factor of (about) 4 (1) 4

Or

Recognition that (speed)2 is the key factor (1)

Reference to v2 = u2 + 2as or rearrangement thereof or kinetic energy (1)[second mark consequent on first](Hence) distance is increased by a factor of (almost) 4 (1)

Or

Do calculation using v2 = u2 + 2as and use 26.8 m s–1 and 6.5 m s–2 (1)Some working shown to get answer 55.2 m (1)(Conclusion that) distance is increased by a factor of (almost) 4[Note : unlikely that QOWC mark would be awarded with this method] (1)

OrAccurate labelled v-t graphs for both (1)Explanation involving comparison of areas (1)Distance is increased by a factor of (almost) 4 (1)

[In all cases give 4th mark if 4 is not mentioned but candidate shows morethan doubled eg “Speed is doubled and the time increased, thereforemultiplying these gives more than double.”]

[10]

301. (a) From what height?

Use of mg∆ h and ½ mv2 (1)[ignore power of 10 errors]

mg ∆ h = ½ mv2 (1)

[shown as formulae without substitution, or as numbers substitutedinto formulae]

Answer [0.8(2) m] (1)

[It is possible to get 0.8 m by a wrong method:

• If v2 = u2 + 2as is used, award 0 marks

• If you see v2/a then apply bod and up to 2/3 marks – the 2nd and 3rd 3

marks. Note that v2/g is correct and gains the first 2 marks, with the

3rd mark if 0.8 m is calculated]

80 × (10–3) kg × 9.81 N kg–1 × ∆ h = ½ × 80 (× 10–3) kg × (4 m s–1)2

h = 13

213

kg N 18.9kg1080

)ms4(kg10805.0−−

−−

×××××

= 0.8(2) m

(b) (i) Law of conservation of linear momentum

Provided no external [other/resultant/outside] force acts (1)The total momentum (of a system) does not change / total momentum (1)before(collision) = total momentum after (collision) 2[Total seen at least once] [Ignore all references to elastic and inelastic][Do not credit simple statement that momentum is conserved]

(ii) Speed of trucks after collision

Any correct calculation of momentum (1)

Use of conservation of momentum leading to the answer 1.3(3) m s–1 (1)

80 × (10–3) kg × 4 m s–1 = 240 × (10–3) kg × u, giving u = 1.3(3) m s–1 2

(c) Time for trucks to stop

[Do not penalise candidates for using a total frictional force of 0.36 N.3/3 possible]

Either

Correct use of power = f × v and ½ mv2 (1)[Do not penalise power of 10 errors or not dividing by 2 in f × V equation]

Use of energy divided by power (1)

Answer in range 2.6 s to 2.7 s (1)[ecf their value for u]

P = 0.12 N × 233.1 m s–1 = 0.08W

KE = ½ × (3) × 80 × (10–3) kg × (1.33 m s–1)2 = 0.21 J

W0.08J 21.0

powerEnergy =

t = 2.6(5) s

[accept 2.6 or 2.7 as rounding] 3

OR

Use of F = ma (1)

Use of either v = u + at i.e or a = tv∆ (1)

Answer in range 2.6 s to 2.7 s (1)

(–)0.12 N = (3) × 80 × (10–3) kg × a (a = (–)0.5 m s–2)

0 = 1.33 m s–1 – 0.5 m s–2 × t or (–)0.5 m s–2 = t

–1s m )1.33(−

t = 2.6(6) s

OR

Select Ft = ∆ p (1)

Substitution (–)0.12t = (–3) × 80 × (10–3) kg × 1.33 m s–1 (1)Allow omission of any bracketed value]Answer in range 2.6 s to 2.7 s (1)

[10]

302. (a) (i) Newton’s First law of Motion

An object will remain (at rest or) uniform/constant velocity/speed/motionin a straight line unless (an external/impressed) force acts upon it /provided resultant force is zero. (1) 1

(ii) Everyday situation

Reference to air resistance / friction / drag etc. (1) 1

(iii) Equilibrium

The resultant force is zero / no net force /sum of forces is zero /forces are balanced / acceleration is zero (1) 1[Accept moments in place of force]

(b) (i) Identify the other force

Earth (1)Gravitational [consequent on first mark] [Do not credit gravity.] (1) 2

(ii) Why normal contact forces are not a Newton’s third law pair

Do not act along the same (straight) line / do not act from the same point (1)They act on the same body (1)They act in the same direction / they are not opposite forces (1)They are of different magnitudes (1)

max 2[7]

303. (a) Sources of background radiation

2 from:Cosmic rays, rocks, soil, food, nuclear power/industry[buried waste asalternative], atmosphere, building material, medical uses, nuclear weaponstesting (in the 60 s), Sun, radon gas 2

[Do not credit more than 1 example in each category e.g. coffee and Brazil nutsis 1 mark not 2]

(b) (i) Measurement of background count rate

• Use GM tube or stop watch/ratemeter/datalogger (1)

• All sources must be in their (lead) containers / placed awayfrom the (1) experiment / place thick lead around tube

• Measure count over measured period of time (1)(and divide count by time)

• Repeat and average / measure the count for at least 5 minutes (1)

• Subtract background (count rate) from readings (1)max 4

(ii) Why it might be unnecessary to measure background count rate

Count rate for the radioactive material is much greater than thebackground count rate. (1) 1[Comparison required with count rate of radioactive material]

[7]

304. (a) (i) Stable ?

Will not: decay / disintegrate / be radioactive / emit radiation / emit (1) 1particles / break down[Do not accept will not emit energy]

(ii) Complete equation

Y1

1 (1) 1

(iii) Identify particles

X = neutron (1)Y = proton (1) 2

(b) (i) Decay Constant

Use of λ = 2/1

69.0t i.e. =

s102.3556869.0

7×× (1)

[Do not penalise incorrect time conversion]

Correct answer [ 3.87 × 10–12 (s–1)] to at least 2 sig fig. [No ue] (1) 2[Bald answer scores 0]

(ii) Number of nuclei

Use of A = λN eg 6016 = (–) 4 × 10–12 N (1)

[Ecf their value of λ] [Do not penalise incorrect time conversion]

Answer in range 6.6 × 1010 to 7.0 × 1010 (1) 2[8]

305. (a) (i) Lamp brightness

Lamp A (1)

Larger current through it (at 9.0 V)/greater power (1) 2(at 9.0 V)/smaller resistance (at 9.0 V)

(ii) Battery current

Addition of currents (1)

Current = 1.88 – 1.92 A (1) 2

(iii) Total resistance

R = 9 V/1.9 A or use of parallel formula (1)

R = 4.6 – 4.9 Ω (1) 2[full ecf for their current]

(b) Lamps in seriesCurrent same in both lamps/current in A reduced from original value (1)Pd across A less than pd across B (1)Lamp A has a lower resistance than lamp B (1)

P = VI or P = RI2 (1) Any 2

Lamp A will be dimmer than B [conditional on scoring ONE of (1) 1the above marks]

[9]

306. (a) (i) Resistance

Use of V/I [ignore 10x] (1)

3800 Ω (3784 Ω ) (1) 2

(ii) Resistance of thermistor

Use THTH

R

RR

VV

= OR 9V/.74mA – R OR (1)

6.2 V = 0.74 mA × RTH

8400 Ω [8378 Ω ] [substituting 4000 Ω gives 8857 Ω ie 8900 Ω ] (1)[method 2 substituting 3800 Ω gives 8362 Ω : substituting 4000 Ωgives8162 Ω ] 2

(b) Suggestion and Explanation

The milliammeter reading increases (1)

Thermistor resistance ‘becomes zero’ /Short circuit (1)

Since supply voltage is constant / I = 9.0 V/R (1)ORCircuit resistance reduced 3

[7]

307. (a) Definition of E.M.F.

Energy (conversion) or work done (1)

Per unit charge (1)

ORE = W/Q (1)Symbols defined (1)[E = 1J/C scores 1]

ORE = P/I (1)Symbols defined (1)

[terminal pd when no current drawn or open circuit scores max 1] 2

(b) Voltmeter calculation

Any attempt to find any current (1)

Attempt to calculate pd across 10 Ω resistor (1)

5.77 V 2

ORPotential divider method; ratio of resistors with 10.4 Ω on the bottom (1)

Multiplied by 6.0 V (1)

5.77 V (1) 3[For either method, an answer of 0.23 V scores max 1]

(c) Second battery added

Voltmeter reading increased (1)

Any two of:

EMF unchanged

Total resistance reduced

current increases or “lost volts” decreases (2) 3[8]

308. (a) Diagram of apparatus

• Trapped gas/fixed mass of gas with fixed volume (1)

• Pressure gauge/U-tube or mercury/Pressure sensor (1)

• Water bath completely surrounding gas (1)

• Thermometer in water bath or gas /Temperature sensor (1)[Boyle’s law apparatus 0/4] 4

(b) Method

Record pressure and temperature (1)for a range of temperatures/ every x K deg C or min, due to heating (1)

Processing results

Plot graph of p against T (1)for temp in Kelvin straight line through origin (1)ORCalculate p/T average (1)and show it is constant for Kelvin temperatures (1)

QOWC (1) 5

(c) Precaution

• Stir water (1)

• Remove energy and await steady temperatures (1)

• Wide range of readings/extend range by use of ice bath (1)

• Eye level with mercury meniscus (1)

• Short/thin tube between gauge and sensor (1) max 1[10]

309. Background wavelength

Use of λ max T =2.90 × 10–3 m K (1)


1.06 (or 1.1) × 10–3m (1) 3

Part of spectrum

Microwave or infra-red (1) (1)[3]

310. Main sequence star definition

(Fusion of) hydrogen (nuclei) / protons to helium (nuclei) (1)

stably / in equilibrium / in core (1) 2

Hertzsprung-Russell diagram

Diagonal falling line (1)

Correct curvature above 20 000 K and below 5000 K (1) 2

X on line and level with 100 (to ± 1 mm) [must be clearly indicated] (1) 1

Dwarfs and Giants

(i) bottom left quadrant (1) 1

(ii) top right quadrant (1) 1[no region indicated max. (1) x]

T consistent with diagram at centre of region and 2500 < T/K < 10000 (1) 3[10]

311. More MS stars

(i) γ Cas (1)

(ii) α Cen B (1) 2

Diameter of Sirius A

26 × 3.9 × 1026 (1)

1.0 × 1028 W (ue) (1) 2

L = σ T4A (or implied by substitution) (1)

A = 2.46 (or 2.43 or 2.45) × 1019 (m2) (1) 2

Use of π d2/4π r2/1.4 × 109 m (1)

2.8 × 109 m[no ecf] (1) 2[8]

312. (a) Neutron Capture Equation

UnU 23992

10

23892 →+ (1) (1)

Beta minus decay

v++−

→ β1

0

93

239

92

239 NpU

U → Np + β – (1)

Hence all six numbers correct (1)

antineutrino (1) 3

(b) (i) Binding energy per nucleon graph

Nucleon number / mass number (1) (1)

(ii) Nuclei on graph

H at start of curve (< 3 MeV), Fe at peak of curve (at 56),U at end of curve (at 235) [to ± 1mm]

any two (1)

all three (1)

(iii) Fe (ecf) (1) 3

(iv) Binding energy of U

7.5/7.6(ecf) (1)

× 235 (1)

1.8 (GeV) [allow 1.76 – 1.80, no e.c.f] (1) 3[9]

313. Conservation laws

Baryon (1)

– 1 (1)

Q: (–1) + (+1) = (0) + (+1) + (X) (1)

B: (0) + (+1) = (0) + (0) + (X) (1) 4

Quark content

uud (1)

su (1) 2[6]

314. (a) (i) t 1.8 to 2.1 mm [when rounded to 2 significant figures] or (1)

better with unit to 0.1 mm

tn 14.4 to 16.8 mm with unit to 0.1 mm or better (1)[unit penalty once only. Ecf unit]

Repeat of both [recorded] (1)

Sensible n between 6 and 10 [integer value] from correct calculation (1) 4[ecf wrong values]

(ii) Correct arrangement including centre (of mass of rule) and sensible (1)marked distances

Centre of mass found to nearest mm or better in range 490 to 510 mm (1)with unit [can be seen in calculation]

x and y (shown) to centre of coins and recorded to the nearest mm (1)

Method for x and y e.g. readings at side of coins (1)

Correct calculation from correct method giving n 7.5 → 8.4

Hence n = 8 (1) 6

(iii) Error in method (i): Variable thickness of coin/coins

Rim at edgeThickness of overlapping tape Any one (1)

Error in method (ii): x is a very short distancemass of tape Any one (1) 2

(b) (i) Circuit set up correctly without help (2) 2

ε and V recorded to 0.01 V or better, with V < ε, and correctorder of magnitude, with unit seen on at least one value and (1)V ≥ 1.0 V

Correct calculation with unit and 2/3 significant figures [not negative] (1) 2

(ii) R0 0.5 → 5.0Ω with unit to a precision of 0.1 Ω without help on (2) 2range by Supervisor

[If help given by Supervisor only 1 mark out of 2 for correct R0]

(iii) Circuit set up correctly without help (1)

I to 0.01 A or better with unit and in the range 0.12 A to 0.18 A (1)[No penalty for mA]

Correct calculation [candidate V ÷ candidate I] with unit and2/3 significant figures [No penalty for mA] (1)

Comparison with R and sensible conclusion (1) 4

(iv) Correct calculation of ratio with no unit and > 2 significant figures (1)

T read off correctly within ½ square with unit (1)

OR

A sensible comment on ratio provided R0 is in range (2) 2[24]

Sample results

(a) (i) t = 2.0, 1.9 mm t = 1.95 mm

nt = 15.9, 15.9 mm tn = 15.9 mm

n = 95.19.15 = 8.15

∴ n = 8 coins

(ii)

4 2 4m m

4 5 0m m

4 9 6 m m9 7 4 m m

1 0 0 0 m m

x y

C e n t r e o f g r a v i t y a t 4 9 6 m m m a r k

x = 496 – 437 = 59 mm

y = 987 – 496 = 491 mm

n = 59491 = 8.3 = 8

(b) (i) Error in method (i): Variable thickness of coin/coinsRim at edgeThickness of overlapping tape

Error in method (ii): x is a very short distance mass of tape

(i) Circuit set up correctly without help

(ii) ε = 1.48 V

V = 1.44 V

r =

−1

44.148.1 10

= 0.28 Ω

R0 = 1.1 Ω

(iii) Circuit set up correctly without help

I = 0.162 A

RT = 162.044.1 = 8.89 Ω

It is reasonable to use V = 1.44 V because the resistance of the lamp is approximately equal to R in the first circuit

(iv) 1.189.8

R 0

T =R

= 8.1

= 1585 K = 1590 K (to 3 significant figures)

315. (a) (i&ii) Units shown in table or on graph axes (1)

Temperatures taken every 0.5 mins or more frequently (1)

At least one attempt at better than 1 °C in each run (1)

Overall fall in temperature greater for 250 ml beaker (1)

6 good points for θ1 (1)

6 good points for θ2 (1)

[Good ± 0.5 °C of examiners curve] 6

(b) Sensible scales [Allow 2 cm = 60 s] (1)

Axes and curves labelled (1)

Plots [check the plot furthest from each curve] (1)

At least one best fit smooth curve (1) 4

(c) Good tangent at common temperature with at least 2 points eitherside of tangent. (1)

∆ x∆ y ≥ 64 cm2 or max possible (1)

Correct reading of sides of triangle (1)

Hence correct calculation to 2/3 significant figures (1) 4

(d) (i) Use more than 2 beakers (1)

Of different diameter (1)

Which are lagged/insulated (1)

Use same volume of water in each beaker (1)

Measure temperature as a function of time (1)

Measure diameter of beakers to find area OR describe suitable (1)technique for finding diameter

Same starting temperature (1)

(ii) Plot θ against t for each beaker (1)

Measure gradient at same temperature for all beakers used

Plot gradient against area OR calculate area

gradient (1)

(iii) Graph should be a straight line through origin OR area

gradient

= constant (1)

[Must have more than 2 beakers] 10[24]

Sample results

(a) (i&ii)

Time t / mins Temperature θ1 ofthe water in

the 250 ml beaker / °C

Temperature θ2 ofthe water in the

100 ml beaker/°C

0.0 82.0 88.00.5 77.0 85.01.0 74.5 82.51.5 72.0 80.52.0 69.5 78.52.5 67.8 76.53.0 65.5 74.53.5 64.0 73.04.0 62.5 71.04.5 60.8 69.55.0 59.5 68.5

Note: When the experiment was being trialled the 100 ml beaker was used first.

(b)

8 8

8 4

8 0

7 6

7 2

6 8

6 4

6 0

5 60 1 2 3 4 5

° C

tm i n s

1 0 0 m l b e a k e r( B )

2 5 0 m l b e a k e r( A )

(c) 72 °C chosen because at least 3 points either side of this on both curves

Gradient = 95.4

561.79 − = 4.67 °C/min

(d) (i) Use more than 2 beakers

Of different diameter

Which are lagged/insulated

Use same volume of water in each beaker

Measure temperature as a function of time

Measure diameter of beakers to find area

Same starting temperature

(ii) Plot θ against t for each beaker

Measure gradient at same temperatures for all beakers used

Plot gradient against area / calculate area

gradient

(iii) Graph should be a straight line through origin / area

gradient

area constant [Must have more than 2 beakers]

316. (a) Units

s–1 / km s–1 kpc–1 /km s–1 Mpc–1 (1) 1

(b) Estimate

See d = vt or rearrangement (1)

Substitution in v = Hd for v to give t = 1/H (1)

[Substitute value of H to obtain t]

Assumption

Since the Big Bang/start of time (1)

(All) galaxies/galaxy is/are travelling at constant speed /no (1)gravitational attractive forces / Universe expands at a constant rate

[H is constant scores max 1 for Assumption. Allow credit for the 4marking points anywhere within (b)] 4

[5]

317. Frequency

(a) (i) 1.0(3) × 1010 Hz (1) 1

Electromagnetic Spectrum

(ii) IR, microwave & radio in correct order above visible (1)UV with either X rays / Gamma rays / both in correct order belowvisible (1)

(iii) Wavelength at boundary 1 × 10–8 m / 1 × 10–9 m (1) 3

Plane polarised

(b) (i) Vibrations/oscillations (of electric field/vector) (1)In one direction/plane (of oscillation) (1) 2

Description

(ii) Diagram showing generator labelled transmitter/generator/source/emitter (1)And suitable detector eg shows how signal is observed by using (1)(micro)ammeter/cro/loudspeaker/computer with interface[Ignore anything drawn between generator and detector but for eachmark do not give credit if a grille etc is attached]

To detect max and min (1)(Rotate through) 90° between max and min (1) 4

[10]

318. (a) Explanation

QOWC (1)

UV/red photon (1) 2

EUV > ER / fuv (1)EUV > Φ / fuv > fTH (so electron can break free) (1)One photon absorbed by one electron (1)Both metal plate and electron are negative or repel (each other) (1) max 2

(b) (i) Intensity red light increased

nothing / no discharge (1)

(ii) Intensity of UV increased

(Coulombmeter) discharges quicker (1) 2

(c) Max KE

Use of E = hc/λ (1)

conversion of eV to J or vice versa i.e. appropriate use of 1.6 × 10–19 (1)Subtraction hc/λ – Φ [must use same units] or use of full equation (1)

max KE = 2.2 × 10–19J (1) 4

[Candidates may convert photon energy to eV leading to max KE = 1.4 eV][10]

319. (a) Definition of longitudinal waveOscillations OR particles (of medium) move (1)Parallel to direction of wave propagation/travel / energy transfer (1) 2

2nd mark consequent on 1st]

(b) Collapsing towerResonance (1)Frequency of quake = natural frequency of tower (1)[Allow resonant frequency for natural frequency]Max energy transfer (1)Very large increase in amplitude of oscillation or maximum amplitude (1) max 3

[5]

320. (a) Radius of circular path

Correct use of v = T

rπ2 (allow substitution of their T) (1)

Radius = 70 – 80 m (74.48 m) (1) 2

(b) Resultant force

rmvF

2

= [seen or used] (1)

Force = 0.08 N (0.077 N) [Allow ecf of their radius.] (1)

Towards the centre of the circular path / towards hub. (1) 3

(c) Forces on the man

(i) Force P : Normal contact/reaction force / EM force / push of (1)capsule or floor on man

Force Q : Pull of Earth on man / weight / gravitational pull (1) 2

(ii) Resultant force (to centre) (1)(at A provided by) friction (1) 2

(iii) at B resultant provided (by force Q being greater than P) (1) 1[10]

321. Explain deduction that legs in contact for 0.001 s

Acceleration changes / discontinuity / vertical velocity max / onlyaccelerates during this time [acceleration related] (1)

After this upward force ceases / stops when legs no longer pushing leaf [forces related] (1) 2

Show that acceleration is about 3000 m s –2

Acceleration = gradient of graph (1)

= 2.8 m s–1 ÷ 0.0010 s

= 2800 m s–2 [No ue] (1) 2

Calculation of force exerted by leg muscles

F = ma (1)

= 1.2 × 10–5 kg × 2800 m s–2

= 3.4 × 10–2N (1) 2

Show that froghopper rises about 0.001 m

Distance = area under graph [or implied by shading etc] (1)

= ½ × 2.8 m s–1 × 0.001 s

= 0.0014 m [No ue](0.0013 to 0.0015 m for square counting) (1) 2

Calculation of work done by leg muscles

W = Fx (1)

= 0.034 N × 1.4 × 10–3m [allow e.c.f.]

= 4.7 × 10–5J (1) 2

Calculation of power developed by leg muscles

Power =W ÷ t (1)

= 4.7 × 10–5 J ÷ 0.001 s

= 0.047 W (1) 2[12]

322. Explanation increase of resistance with temperature

Temperature increase leads to increased lattice vibrations (1)

scattering flowing electrons / increased collisions of electrons. (1) 2

Calculation of resistance at 200 °C

R = V/I [stated or implied] (1)

= 7.4 V ÷ 0.19 A

= 39 Ω (1) 2

Discuss whether results support hypothesis

No. Resistance is not increasing with temperature. (1) 1

Calculation of mains voltage

P = V2 ÷ R (1)

V2 = PR

= 1200 W × 41 Ω [Mark for rearrangement OR substitution] (1)

[Accept 39 – 41 Ω ] [ecf]

V = 220 V (1)

[Allow P = I2R (1), 3

calculate I = 5.4 A and use in 1200 W = 5.4 A × V (1),V= 220 V (1)][8]

323. Show that expected speed is about 35 m s –1

Ek = ½ mv2 and Ep = mg∆ h (1)

½mv2 = mg∆ h (1)

v = gh2

= ) . ( m 64kg N8192 1 ×× −

= 35.4 ms–1 [No ue] (1) 3

[For v2 = u2 + 2as mark u = 0 (1), rest of substitution 1), evaluation (1)]

Assumption

No resistive force, all gpe → ke, constant accn (1) 1

[Do not accept g = 9.81 m s–2]

Reason for lower speed

Work done against resistive force/frictional forces oppose motion/ (1)some g.p.e. → heat/sound ...

reduces maximum kinetic energy / acceleration is reduced/less than (1) 2

9.8 m s–2

Calculate efficiency

Efficiency = (actual max k.e. ÷ theoretical max k.e.) × 100%

OR efficiency = (actual max k.e. ÷ initial p.e.) × 100% (1)

= (½mυ act2 ÷ ½ mυth

2) × 100% OR = (½ m υ act2) ÷ (mgh) × 100%

21–

2–1

)s m 435(

)s m532(

.

.× 100% =

648.9)s m532( ½ 2–1

× . × 100% (1)

= 84.2% (1) 3

Reason why speed greater than expected

e.g. motor assisted / initial speed > 0 / run up before drop (1) 1[10]

324. Calculation of energy provided by people in one second

E = Pt= 27 × 80 W × 1 s

= 2160 J [Accept J or W or J s–1] (1) 1

Show that temperature might rise by about 7 °C

Heat generated = Pt = 2160 J s–1 × 40 × 60 s = 5 184 000 J [allow e.c.f.] (1)

∆ Q = mc∆ θ

5 184 000 J = 740 kg × 960 J kg–1 °C–1 × ∆ θ [Substitution] (1)

∆ θ = 5 184 000 J ÷ (740 kg × 960 J kg–1 °C–1) (1) 3

= 7.3°C [Noue] (1)

Explanation of whether actual difference is more or less than answer

Less due to heat loss ... (1)

... through windows /draughts / walls / to contents ... (1) 2[Allow more due to heat gained (1) from valid source (1)]

That air must be exchanged at about 0.3 kg s –1

m = ∆ Q ÷ c∆ θ

= 2160 J ÷ (960 J kg–1 °C–1 × 8 °C) [Substitution] (1)

= 0.28 kg s–1 [No ue] (1) 2

Calculation of time to exchange air

740 kg ÷ 0.28 kg s–1

= 2640 s OR 44 minutes (1) 1[9]

325. Explanation of ‘excited’

Electrons/atoms gain energy (1)

and electrons move to higher (energy) levels (1) 2

[Credit may be gained for diagrams in this and the next 3 parts]

Explanation of how radiation emitted by mercury atoms

Electrons (lose energy as they) drop to lower levels (1)

Emit photons / electromagnetic radiation (1) 2

Explanation of why only certain wavelengths are emitted

Wavelength (of photon) depends one energy (1)

Photon energy depends on difference in energy levels (1)

Levels discrete / only certain differences / photon energies possible (1) 3(and therefore certain wavelengths)

Why phosphor emits different wavelengths to mercury

Different energy levels / different differences in energy levels (1) 1

Calculation of charge

Q = It (1)

= 0.15 A × 20 × 60s

= 180 C (1) 2[10]

326. Why microwaves are reflected

Wave is reflected when passing from one medium to another / when density changes / when speed changes (1) 1

Varying amplitude

Any two of the following:

Varying differences in density of the two mediums produce different intensities of signal (1)

Different distances travelled give different amplitudes (1)

Following a reflection there is less energy available (1) Max 2

Varying time

Different thicknesses of medium (1) 1

What is meant by Doppler shift

Change in frequency/wavelength (1)

Caused by movement of a source (1) 2

Changes due to Doppler shift

Wavelength increases (1)

Frequency decreases (1)

[Allow e.c.f. from incorrect wavelength]

Any one of the following:

• Each wave has further to travel than the one before to reach the heart

• The waves are reflected from the heart at a slower rate (1) 3[9]

327. Add forces to diagram

Downward arrow labelled weight / mg (1)

Upward arrow labelled (viscous) drag (1) 2

Expression for upthrust

Upthrust = weight of displaced fluid (1)

= volume × density × g (1) 2

Relationship between forces

Upthrust + (viscous) drag = weight OR F + U = W (1) 1[ecf. From diagram – must be 3 forces]

Expression for velocity

F = W – U

6π ηrv = 4/3πr3 ρsg – 4/3 πr3ρwg (ecf) (1)

6ηrv = 4/3 r3g (ps – pw)

v = η ν

ρρ9

)(2 ws2 −gr

(1) 2

Velocity change with temperature

Velocity will increase (1)

As viscosity will decrease with temperature/as velocity increases with decreasing viscosity / as density of wine decreases (1) 2

Explanation of what is meant by laminar

Diagram showing at least 3 reasonably parallel and straight lines (1)

No abrupt change in direction/no whorls/no eddies (1) 2[Both marks may be awarded from the diagram.]

[11]

328. Resistance calculation

Use of R = ρL/A (1)

Substitution R = 1.6 × 10–4 × 0.02/(5 × (10–3) × 0.02 × (10–3)) (1)

= 32 Ω (1) 3

Total resistance

Either Section 2 = ½ × R1 (16 Ω ) OR Section 3 = 31 × R1 (10.7 Ω ) (1)

Use of RTotal =R1 + R2 + R3 (1)

RTotal = 58.7 Ω [55 Ω if 30 Ω used as starting point] (1) 3

[ecf if section 3 calculated as ¼ × R1 = 56 Ω OR 52.5 Ω if 30 Ω usedas starting point]

Why thermochromic ink becomes warm

Current produces heat / reference to I2RORThermal conduction from conductive ink (1) 1

[Mark for identifying that the heating effect originates in the conductive ink]

Why only thin section transparent

Thinner / section 1 has more resistance (1)

So even a small current will heat it/Power (heating effect) given by

I2R / current will heat it more (1) 2

[OR opposite argument explaining why thicker section is harder to heat][9]

329. Example of light behaving as a wave

Any one of:

• diffraction

• refraction

• interference

• polarisation (1) 1

What is meant by monochromatic

Single colour / wavelength / frequency (1) 1

Completion of graph

Points plotted correctly [–1 for each incorrect point] (1) (1)

Line of best fit added across graph grid (1) 3

What eV s tells us

Maximum (1)

Kinetic energy of the electrons / ½ mv2 of electrons (1) 2

Threshold frequency for sodium

Correct reading from graph: 4.3 × 1014 Hz (1) 1

[Accept 4.1 × 1014 – 4.7 × 1014Hz]

Work function

f = hf0 = 6.63 × 10–34 J s × 4.3 × 1014 Hz (1)

= 2.9 × 10–19J [Allow ecf] (1) 2

Why threshold frequency is needed

• Electron requires certain amount of energy to escape from surface (1)

• This energy comes from one photon of light (1)

• E = hf (1) Max 2[12]

330. Adding angles to diagram

Critical angle C correctly labelled (1) 1

Calculation of critical angle

Use of µ = 1/sin C (1)

Sin C = 1/1.09

C = 66.6° (1) 2

Why black mark not always seen

At (incident) angles greater than the critical angle (1)

t.i.r. takes place (so black mark not visible) (1)

light does not reach X / X only seen at angles less than C (1) 3

[OR opposite argument for why it is seen at angles less than C]

Comparison of sugar concentration

Lower µ means greater density (1)

Greater density means more sugar (1) 2[8]

331. Material property

Tough (1) 1

Crosses on graph

P at end of straight line section [allow 2.6 – 2.8 m extension] (1)

E: accept between P and maximum force value (1) 2If P not shown allow 2.6 – maximum force value]

Stiffness of rope

k = F/x; k = gradient of graph (1)

= 7 kN ÷ 2.25 m = 3.1 kNm–1 (1) 2

Breaking strain

From graph, maximum extension = 3.4 m (1)

Breaking strain = 3.4/50 = 0.07 (0.068 or 6.8%) (1) 2[allow ecf. from extension]

Energy stored in the rope

Force = 90 kg × 9.81 =883 N (900 N) (1)

Extension for this force from graph = 0.28 m (1)(allow 0.25 → 0.3 m)

Energy = ½ Fx [OR area under graph] (1)

= 124 J (allow 110 → 132 J) (1)[112 → 135 J if 900 N is used]

OR alternative method:

Force = 90 kg × 9.81 =883 N [900 N] (1)

Extension found by substitution: x = F/k (1)

Energy = ½ Fx OR ½ F × F/k (1)

– 125 J [128 J if 900 N is used] (1) 4[allow ecf for value of k used]

[11]

332. Conserved quantities

Momentum, charge, (mass-)energy, lepton number (1) (1) 2

[2 right gets 1 mark; all 3 right get 2 marks][Do not credit kinetic energy]

Charge of the pentaquark

31

312

322 +−×++×

= (+) 1(e) (1) 1

Charge on X

Positive since pentaquark was positive, neutron neutral [ecf] (1) 1

[Reasoning needed]

Possible quark composition for X with explanation

u s (1)

Left behind (after removing neutron/udd) (1) 2

Mass of pentaquark

Conversion from GeV to J or substitution of c2 (1)

answer [no ue] (1)

1.54 × 109 × 1.6 × 10–19/(3 × 108)2 2

= 2.7 × 10–27kg[8]

333. How electron gun creates beam of electrons

Any four from:

1. hot filament (1)

2. thermionic emission / electrons have enough energy to leave (1)

3. anode and cathode / ± electrodes [identified] (1)

4. E–field OR force direction OR cause of acceleration (1)

5. collimation [eg gap in anode identified as causing beam] (1)

6. need for vacuum (1) Max 4

Speed of electrons

(eV =) ½ mv2 (1)

Use of eV [ie substituted or rearranged] (1)

Answer [1.09 × 107 m s–1] (1)

1.6 × 10–19 × 340 (J) = ½ × 9.11 × 10–31 (kg) × v2

v =1.09 × 107 m s–1 3

Definition of term electric field

Region/area/space in which charge experiences force (1) 1ertical acceleration of electrons due to field[Bald answer =0]

Use of equation E = V/d (1)

E = V/d = 2500 V ÷ 0.09 m = 28 (kV m–1)

Rearranged equation E = F/q or substitution into it (1)

F = Eq = 28 000 × 1.6 × 10–19 (N) 4.4 × 10–15 (N)

Equation F = ma seen or substitution into it (1)

A = F/m = )(1011.9

)(104.431

15

hg

N−

−

××

= 4.9 × 1015 (m s–2) (1) 4

[at least 2 sig fig needed] [No u.e.] [Reverse calculation max 3][12]

334. Direction

Force → centre/perpendicular to velocity/motion (1) 1[Accept sideways/inwards]

Why force required

2 of:

Changing direction /charging velocity (1)

Acceleration (1)

Reference to or explanation in terms of NI or NII (1) Max 2

What provides force

Friction between tyres and road surface (1) 1

Maximum speed

F = mv2/r [Accept F = mrω2] (1)

⇒ v = mFr OR

16014470 × (1)

= 6.4 m s–1 (1) 3

Why skid occurs

Smaller r (1)

⇒ F(mv2/r) (required) increases / use of F = mv2/r to deduce v decreases (1)

Only 470 N (available) / the force is the same (1) 3

Explanation

(m increased ⇒) F (needed) increases (1)

EITHER only 470 N available or the force is the same ⇒ NO OR friction increases as mass increases( ⇒YES) (1) 2

[12]

335. Approximate energy of alpha particle in MeV

1. r = 0.09 (m) [accept in range 0.07 – 0.12] (1)[must have unit if given in cm]

2. q = 2 × 1.6 × 10–19 (C) (1)

3. m = 4 × 1.7 × 10–27(kg) (1)

4. r = p/Bq ⇒ p = rBq or v = rBq/m (1) [see equation or substitution]

[p = 0.09 × 3.7 × 3.2 × 10–19 N s]

5. =1.07 × 10–19 (Ns) OR v = 1.6 × 107 (m s–1) (1)

6. E = p2/2m or use of ½ mv2 (1)

[E = (1.07 × 10–19)2 / (2 × 4 × 1.67 × 10–27)J]

7. 8.6 × 10”13J (1)

(5.4 MeV/5.4 × 106 eV) 7[7]

336. (a) Advantage of avoiding metal contacts

Any one from:

• makes possible a sealed unit

• avoids electrocution

• stops corrosion (by water)

• water cannot enter/short contacts (1) 1

(b) How arrangement is able to charge the battery

Any six from:

1. current (in X) produces magnetic field

2. field links second coil

3. metal = iron

4. metal core increases field

5. field changes/alternates

6. changing φ /B or dφ /dt or Faraday induces/causes V

7. V causes I

8. diode needed (or a.c. so won’t charge)

9. field penetrates plastic

10. like a transformer / X is a primary and Y is a secondary

11. electromagnetic induction Max 6[7]

337. (a) Free body force diagram for magnet

(Electro)magnetic / (force of) repulsion / push (1)

Weight / W / mg / pull (of Earth) / gravitational (attractiveforce) / attraction (of Earth) (1) 2[NOT gravity][An additional incorrect force cancels 1 mark awarded]

(b) Newton’s third law pairs 6

Force Body on which correspondingforce acts

Direction of thecorresponding force

Contact (Wooden) stand/baseDownwards /down / ↓

(1) (1)

Magnetic (Magnet) M1 Upwards / up / ↑ (1) (1)

Weight Earth / Earth’s surface Upwards / up / ↑ (1) (1)

[8]

338. (a) Explanation

Vb has a horizontal component equal to Va (1)Vb has a vertical component (1) 2

[Vb has two components of velocity is 1 mark][Vb cos 45 = Va is 2 marks]

(b) ExplanationEITHERQoWC (1)The average speed / velocity of A is greater (than B) / converse (1)(because) A continually accelerates whereas B slows down / (1)decelerates (initially)

[description of both A and B necessary for this 2nd physics mark]ORQoWC (1)Va = horizontal component of Vb and they travel the same (1)horizontal distanceVertical component of projectile’s motion does not affect (1)horizontal motion 3

[5]

339. (a) Energy change

Both parts correct [NB 1 mark only] (1) 1Gravitational potential (energy) to kinetic / movement (energy) /work done

(b) Principal of conservation of energy

EITHER (1) (1)Energy can be neither created nor destroyedOREnergy cannot be created/destroyed / total energy is not (1)lost/gainedmerely transformed from one form to another / in a closed/isolated system (1) 2

(c) Speed of waterCorrect substitution into correct formula (1)Correct value with correct unit (1) 2Power = force × velocity

1.7 × 109 (W) = 3.5 × 108 (N) × V

V = 4.86 m s–1

–

(d) Explanation

Not all the energy of the falling water is transferred to the outputpower OR system is not 100% efficient OR water is not brought (1) 1to rest OR friction OR some of the energy is transferred toheat/sound/surroundings.

(e) TimeCorrect value with correct unit. (1) 1

Time = ( )1–3

36

sm390

)m(107 × = 17 949 s (= 299 min) (= 5 h)

(f) Work done

Correct substitution into correct formula to find mass of water (1)

Identifying“work done = force x distance moved in direction of force” (1)

Correct value with correct unit (1)

Mass of water = volume × density 3

= 7 × 106 (m3) × 103 (kg m–3) (= 6.9 × 109 kg)

Work done = force × distance

Work done = 6.9 × 109 (kg) x 9.81 (ms–2) x 500 (m)

= 3.43 × 1013 J[10]

340. (a) (i) Complete equation

Correct identification of 42 for α (1)

Correct substitution (1) 22713 OR correct values which balance the candidate’s equation

(ii) Completion of 2 nd equation

01 (1)

Correct identification of positron / positive (+ ve) electron / β+ / (1) 2antielectron

[If incorrectly given 01− allow electron / β– ie 1 mark]

[Correct spelling only]

(b) Half-life

Average (1)

Time taken for the activity/intensity/count rate to drop by halfOR time taken for half the atoms/nuclei to decay (1)

[NOT mass, count, particles, radioisotope, sample]

Isotope

Same: proton number / atomic number (1)[Not same chemical properties]

Different: neutron number / nucleon number / mass number (1) Max 3

[Not different physical properties/density]

(c) γ-ray emission 1

EITHER

(The loss of a heliumnucleus/electron has left theremaining) nucleus in anexcited state/with a surplusof energy

OR

The nucleus emits its surplusenergy (in the form of a quantumof γ -radiation) (1)

[8]

341. Circuits

Base unit: ampere OR amperes OR amp OR amps (1)Derived quantity: charge OR resistance (1)Derived unit: volt OR volts OR ohm OR ohms (1)Base quantity: current (1) 4

[If two answers are given to any of the above, both must be correct to gain the mark][4]

342. (a) Io and Jupiter: Time taken for electrons to reach Jupiter

t = s/υ = (4.2 × 108 m)/(2.9 × 107 m s–1) = 14.48 s

Correct substitution in υ= s/t (ignore powers of ten) (1)

Answer: 14.48 s, 14.5 s [no ue] (1) 2

(b) Estimate of number of electrons

Q = ne = It

n = It/e

n = (3.0 × 106 A) (1s)/(1.6 × 10–19 C)

Use of ne = It (1)

(1.8 – 2.0) × 1025 (1) 2

(c) Current direction

From Jupiter (to Io) / to Io / to the moon (1) 1[5]

343. (a) p.d. across 4 Ω resistor

1.5 (A) × 4 (Ω)

= 6 V (1) 1

(b) Resistance R2

Current through R2 = 0.5 A (1)

R2 = 0.5(A)

(V) 6

R2 = 12 Ω (1) 2

[allow ecf their pd across 4 Ω]

(c) Resistance R1

p.d. across R1 = 12 − 6 − 4

= 2 V (1)

Current through R1 = 2 A (1)

R1 = 2(A)

V)(2 = 1Ω (1)

[allow ecf of pd from (a) if less than 12 V]

Alternative method

Parallel combination = 3Ω (1)

Circuit resistance = 12(V)/2 (A) = 6Ω (1)

R1 = 6 – (3 + 2) = 1 Ω (1) 3

[allow ecf of pd from (a) and R from (b)][6]

344. (a) Current in filament lamp

P = VI or correct rearrangement (1)

2 A (1) 2

(b) (i) Sketch graph

Correct shape for their axes (1)

−I−V quadrant showing fair rotational symmetry (1) 2

I

V

(ii) Explanation of shape

(As the voltage/p.d. increases), current also increases (1)

(As the current increases), temperature of lamp increases (1)

(This leads to) an increase in resistance of lamp (1)

so equal increases in V lead to smaller increases in I OR rate ofincrease in current decreases OR correct reference to their correct (1) 4gradient

[8]

[If a straight line graph was drawn though the origin then (1)(0)(0)(1) forthe following:

V is proportional to Rtherefore the graph has a constant gradient]

345. (a) (i) Graph

Attempt to find gradient at start of graph ie over 11°C rise or less (1)

Value calculated with units in K s–1 / K min–1 / °C s–1 / °C min–1

Range 0.07 – 0.18 K s–1 or 4.4 – 11.0 K min–1 (1) 2

(ii) Power of heater

Formula ∆Q/∆t = mc∆T/∆t used (1)

Converts g to kg (1)

Value for rate within acceptable range 18 – 50 W (1) 3

or 1100 – 3000 J min–1

[no ecf from gradient]

(b) Heating process

(rate of) energy lost to the surroundings OR due to evaporation[do (1)not credit boiling] (1) 2approaches (rate of ) energy supply OR increases with temperature difference.

(c) Graph

(i) Curve of reducing gradient starting at 20 °C, 0 s (1)

initially below given graph (consequential mark) (1) 2

(ii) Explanation

Reference of need to heat mug (1)

Hence reduced rate of temperature rise [consequential mark] (1)

Reference to insulating properties of mug (1) Max 2[11]

346. (a) (i) Replacement

V1 (1) 1

(ii) Explanation

[ONE pair of marks]Resistance: resistance of V1 [not just the voltmeter] is much largerthan 100 Ω OR combined resistance of parallel combination is (1)approximately 100 Ω

Voltage: p.d. across V1 is much greater than p.d. across 100 Ω OR (1)all 9 V is across V1

OR

Current: no current is flowing in the circuit / very small current (1)Resistance: because V1 has infinite/very large resistance (1)

OR

(Correct current calculation 0.9 x 10 –6 A and) correct pd calculation

90 x 10 –6 A (1)This is a very small/negligible pd (1) 2

(b) Circuit diagram

(i) or equivalent resistor symbol labelled 10 MΩ (1)

or equivalent resistor symbol labelled 10 MΩ (1) 2

[They must be shown in a correct arrangement with R]

(ii) Value of R

6 (V): 3 (V) = 10 (MΩ): 5 (MΩ) / Rtotal of parallel combination is 5 (1)MΩ

1/5 (MΩ) = 1/10 (MΩ) + 1/R OR some equivalent correct (1)substitution to show working

R = 10 MΩ (1) 3[8]

347. Base units of intensity

(i) W = J s–1 / N m s–1 or P = E / t or P = F v (1)

J = kg m2 s–2 or kg m s–2 m (1)

Algebra to kg s–3 shown (e.g. kg m2 s–2 s–1 m–2) (1) 3

Luminosity calculation

(ii) Correct substitution (1)

3.82 or 3.8 [ignore 10n] (1)

hence 3.8(2) × 1026 W [ue] [allow 3.9 or 4] (1) 3[6]

348. Forces within star

(i) 1. Fusion forces [allow ‘pressure from nuclear reactions’ or (1)‘hydrogen burning’] or radiation / photon pressure

2. Gravitational / Weight (not just gravity) (1)

(ii) Equal (1) 3

(iii) White dwarf & red giant differences

Any three from:

• Temperature: Twd (6000K – 30000K) > Trg (2000K – 5000K)• Volume: Vrg > Vwd – allow A / d / r / bigger• Mass: e.g. Mwd < 1.4 M AND (0.4M <) Mrg < 8m• Fusion (of He / heavier elements) in rg / no fusion in wd

• Luminosity: Lrg [102 – 106] > Lwd [10–2 – 10–4] in terms of L• Wd is (core) remnant of rg / rg before wd stage• Density: ρwd > ρrg (1) (1) (1) Max 3

[no numerical values for any property – max 2/3]

(iv) Neutron star

Core remnants’ mass (1)

Must be > 1.4 M or < 2.5 M (1) 2[8]

349. When Sun was formed

(i) Attempted use of L = 1.4 L (1)

2.8 × 1026 W (1)

(ii) 1.062 used (1)

5.5 × 1018 m2 / 5.5 × 1012 km2 (1) 4

Show temperature change

(iii) L = σT4A (or implied) (1)

Correct substitution [ecf] (1)

Hence 5500 (K) [no ecf] (1)

Hence 5800 − 5500 [or 330, 308, 310] (1) 4

Wien’s law

(iv) Use of λ(max)T = 2.90 × 10–3 m K (1)530 nm or 500 nm [no ue] (1)∆λ = 30 nm (when rounded to 1 s.f.) (1) 3

[11]

350. Base units of energy density

(i) J m–3 or N m–2 (1)

J = kg m2 s–2 or N = kg m s–2 (1)

Algebra to kg m–1 s–2 shown (i.e. kg m2 s–2 m–3 or kg m s–2 m–2) (1) 3

(ii) Energy density calculation

200 × 106 used (1)

Energy density = ½ σ ε (or implied) (1)

Correct substitution to 95 000 [no ue] (1) 3[6]

351. Rubber band graph

(i) Clear labels (or arrows up & down) (1) 1

(ii) Hysteresis (1) 1

Maximum stress

(iii) Use of F/A with 12 (N) (1)

2 × 106 Pa / N m–2 [ue, no ecf] (1) 2

(iv) Internal energy gain

Any attempt at area / 0.5 F x (1)

Correct values approximated [ignore 10n] [allow counting squares] (1)[ecf]

(½ ×) 12 N × 500 × 10–3 or counted squares conversion to energy (1)

(1cm2 : 0.2 J)3 J [when rounded to 1sf, ue, no ecf] (1) 4

(v) Hence show loop area

Attempt at loop area / attempt at area under unloading line (1)

Hence working to show 1 J (1) 2

Mechanism

(vi) Creep (1) 1Hooke’s law

(vii) (Loading) force is proportional to extensionOR may be F = k∆x with symbols defined] (1) 1Force–extension apparatus

(viii) Valid diagram (1)Clamp and rubber band, both labelled (1)Ruler and masses/weights, both labelled (1)Accuracy technique (eye-level, clamp ruler, use set-square) (1) 4

[16]

352. Base units of eV

(i) Reference to joule (1)

Useful energy equation / units shown [e.g. ½mv2, mgh, mc2, Fd, not (1)QV or Pt]

Algebra to J = kg m2 s–2 shown (e.g. kg (m s–1)2 or kg m s–2 m) (1) 3

(ii) Energy released

146 shown or used (1)∆m calculation [1.9415, ecf] (1)

Multiply by 930 [allow E = mc2 with mass in kg] (1)1800 MeV [no ue] (1) 4

[7]

353. (i) Decay numbers

11 p and 1

0 n (1)

01 β + and

00 ν (1) 2

(ii) Tick the boxes

Proton: baryon and hadron only (1)

neutron: baryon and hadron only (1)

β+: lepton and antimatter only (1)

ν: lepton only (1) 4

[only penalise once for including meson] [if both baryon correct but no hadrons 1 mark out of 2 and vice versa]

[6]

354. (a) Resultant force required

The direction of speed OR velocity is changing (1)There is an acceleration/rate of change in momentum (1) 2

(b) (i) Angular speed

Use of an angle divided by a time (1)

7.3 × 10–5 rad s–1 OR 0.26 rad h–1 OR 4.2 × 10–3 o s–1 OR 15o h–1 (1) 2

(ii) Resultant force on student

Use of F = mrω2 OR v = rω with F = r

mv 2

(1)

2.0 N (1) 2

(iii) Scale reading

Evidence of contact force = mg − resultant force (1)Weight of girl = 588 (N) OR 589 (N) OR 60 × 9.81 (N) (1)Scale reading = 586 N OR 587 N [ecf their mg − their F] (1) 3

[9]

355. Table 6

Wavelength of light in range 390 nm – 700 nm (1)

Wavelength of gamma ≤ 10–11 m (1)

Source (unstable) nuclei (1)

Type of radiation radio (waves) (1)

Type of radiation infra red (1)

Source Warm objects / hot objects /above 0 K

(1)

[6]

356. (a) AmplitudeMaximum distance/displacementFrom the mean position / mid point / zero displacement line / (1) 1equilibrium point[If shown on a diagram, at least one full wavelength must be shown,the displacement must be labelled “a” or “amplitude” and the zerodisplacement line must be labelled with one of the terms above.]

(b) Progressive wave

Displacement at A: 2.0 (cm) [accept 2] (1)Displacement at B: 2.5 (cm) to 2.7 (cm) (1)Displacement at C: 1.5 to 1.7 (cm) (1) 3

Diagram

[Minimum] one complete sinusoidal wavelength drawn (1)

Peak between A and B [accept on B but not on A] (1)

y = 0 (cm) at x = +2.6 cm with EITHER x = +6.2 cm OR x = − 1.0 (1)cm 3

[7]

357. (a) Transverse wave(Line along which) particles/em field vectors oscillate/vibrate (1)Perpendicular to (1)Direction of travel or of propagation or of energy flow or velocity (1) 3

(b) Differences

Any two:

Standing waves Progressive waves

1. store energy 1. transfer energy (1)2. only AN points have max 2. all have the max ampl/displ (1) ampl/displ3. constant (relative) phase 3. variable (relative) phaserelationship relationship (1) Max 2

(c) (i) DropletsFormed at nodes / no net displacement at these points (1) 1

(ii) Speed

Use of υ= fλ (1)Evidence that wavelength is twice node–node distance (1)Wavelength = 1.2 (cm) (1)

Frequency = 8.0 [8.2 / 8.16] Hz or s–1 only (1) 4[10]

358. (i) Diagram

Component (mgcosθ) correctly drawn – good alignment and (1) 1approximately same length

(ii) Diagram

Component (mgsinθ) correctly drawn, reasonably perpendicular (1) 1to T to the left

(iii) Acceleration

Use of mgsinθ= ma [must see 9.8(1) (m s–2) not 10 for this mark] (1)

a = 0.68 m s–2 [for this mark allow 0.69 m s–2 ie 10 m s–2 for g] (1) 2

(iv) Direction

Directed to O along arc/in same direction as mgsinθ/tangential to (1) 1arc

[5]

359. (a) Electromagnetic Doppler effect

Change in the frequency/wavelength (of the light/radiation from a source) (1)because of relative motion between source and observer (1) 2[If giving specific examples must cover both possibilities of changein frequency and relative motion eg describe red shift and blue shift]

(b) Hubble’s conclusions

Any two from:

• (Recession) velocity ∝ galaxy distance [NOT stars]• Red shift due to a galaxy moving away from Earth/observer• Deduction of the expanding Universe [not the Big Bang] (1) (1) 2

[only penalise lack of galaxy once]

(c) Minimum velocity

∆λ= 660 (nm) − 390 (nm) = 270 (nm) (1)

Their ∆λ/ their short λ = v/c (1)

Correct substitution of c = 3 × 108 (m s–1) (1)

Maximum velocity = 2.1 × 108 (m s–1) (1) 4

(d) Critical mean density

Density is large enough to prevent Universe expanding for ever (1)but not too big to cause a collapse/contraction of the Universe (1) 2

[10]

360. Photoelectric effect

(a) Explanation:

Particle theory: one photon (interacts with) one electron (1)

Wave theory allows energy to ‘build up’, i.e. time delay (1) 2

(b) Explanation:

Particle theory: f too low then not enough energy (is released byphoton to knock out an electron) (1)

Wave theory: Any frequency beam will produce enough energy (to release an electron, i.e. should emit whatever the frequency) (1) 2

[4]

361. (a) (i) Definition of law

EITHEREquation given and all symbols defined (1) (1)[For each symbol incorrectly defined −1 mark; 3 incorrect getzero, not −1]ORForce proportional to product of masses (1)Force inversely proportional to square of distance between themasses (1) 2

(ii) Derivation

Set mg = 2r

GMm hence g = 2r

GM (1) 1

(iii) Graph

Starting point [R, g] (1)

(R, g) and (2R, 4

g) plotted (1)

(3R, 9

g) and (4R,

16

g) ~ plotted (1) 3

[Ignore the line joining origin to (R, g)]

(b) (i) Equipotential surface

Surface containing all points at the same (gravitational) potential(energy) (1) 1

(b) (ii) Drawing of equipotential surfacesThree concentric circles drawn (1)Increasing separation (1) 2

(c) ExplanationThe weight / g must remain constant OR uniform gravitational (1)field(For this to be true) changes in height must be small (1) 2

[11]

362. (a) Direction of field lines

Downwards (1) 1

(b) (i) Calculation of force

Use of V/d i.e. 250 V/0.05 m [if 5 used mark still awarded] (1)

Use of d

V e [Mark is for correct use of 1.6 × 10–19 C] (1)

= 8.0 × 10–16 N (1) 3

(ii) Direction and explanation

(Vertically) upwards / towards AB (1)

No (component of ) force in the horizontal direction OR because (1) 2(the force) does no work in the horizontal direction

(c) Calculation of p.d.

Use of ∆EK = ½ mυ2 / ½ 9.11 × 10–31 (kg) × (1.3 × 107)2 (1)

Use of Ve / V × 1.6 × 10–19 (C) (1)

= 480 V (1) 3

(d) Beam of electrons

Diagram showing:

Spreading out from one point (1)fastest electrons labelled (1)

2[11]

363. (a) Sensible use of rule in contact with block or clamp (1)and set square or pin (1)Readings on both sides of mass or b + ½ diameter or uses slot (1)

Sensible l to 1 mm i.e. ‹ 880 mm or centre value (1)± 0.02 s of Supervisor’s from ≥ 30 T (1) (1)

≥±≥±

(1)

(1)

T

T

30from s03.0

10 from s02.0

Repeat readings shown (1)Fiducial marker (pin) at centre of oscillations [must be at centre] (1) 8[Do not allow “eye level”]

(b) w ± 0.03 cm of Supervisor’s from ≥ 2 readings (1) (1)

≥±

±(1)

(1)

readings 2 from cm 0.05

reading 1 from cm 03.0

t ± 0.03 cm of Supervisor’s from ≥ 2 readings (1) (1)

≥±

±(1)

(1)

readings 2 from cm 0.05

reading 1 from cm 03.0

[Readings to 0.1 mm, otherwise max 2/4]

(c) Consistent (for l, w, t) substitution [allow ecf] (1)Correct calculation [dependent mark] (1)Unit consistent with substitution (1)

Value in range (0.5 → 2.5) × 1010 kg m–1s–2and 2 / 3 (1)significant figures [No ue] 4

[16]

Sample results

W o o d e n b l o c k s

B e n c h

M e t r e r u l e

l

a b c

Second metre rule and set square used as showna = 90.0 cmb = 5.5 cmc = 0.2 cm

l = 90.0 −

+

2

2.05.5 cm

= 87.2 cm20 T / s = 13.13, 13.00, 13.16average T = 0.655 sRepeat readingsFiducial marker (pin) at centre of oscillations

(b) w / cm = 2.78, 2.82, 2.86average w = 2.82 cmt / cm = 0.64, 0.67, 0.66average t = 0.66 cm

(c) E = 32

3216

wtT

Mlπ

= ( ) ( )32–2–2

32

1066.01082.2655.0

)872.0(600.016

××××

××π

= 1.8 × 1010 kg m–1 s–2 [Pa]

364. (a) (i) QOWC (1)Link track to bubbles (1)Which reflects light / are illuminated (1)(produced as) the electron / it ionises liquid / particles / (1)H2 / air 4

(ii) Mention of B–field/F = Bqυ/ F = Beυ/ FLHR (1)B is perpendicular to υ/ direction of motion / in or out (1)of pageElectron loses energy/slows down (1)Colliding with / interacting with / ionising liquid particles / H2 (1) 4

(b) (i) & (ii)

r/m r/mm p/kg m s–1 m/kg

P 62 – 67 × 10–3 62 – 67 1.2 – 1.3 × 10–20 4.0 – 4.3 × 10–29

Q 43 – 48 × 10–3 43 – 48 0.83 – 0.92 × 10–20 2.8 – 3.1 × 10–29

R 28 – 33 ×10–3 28 – 33 0.54 – 0.63 × 10–20 1.8 – 2.1 × 10–29

Values for r in range above [ignore 10n and units] (1)

p = Ber ⇒ any one correct p [ignore 10n but must have (1)unit] [ecf] (1) 3

All ps correct numerically [no ue] (1)p = mυ ⇒ m = p/υ (1)

Any one correct m [ignore 10n but must have unit]

EITHER

Comment [e.c.f.]: any reference to 9 × 10–31 kg/rest (1)mass (of electron) / electron massBecause electron is moving close to / at the speed oflightOR(effective) mass (of electrons) is decreasing (1)

reference to E = mc2 / ∆E = c2∆m / mass–energy (1)conservation 4

[15]

365. (i) Lead shot loses g.p.e. (which becomes k.e.) (1)(which becomes/lost to/transfers to) internal (1) 2energy/heat

(ii) Use of 60 mg∆h [allow between 0.70 m and 0.80 m] (1)Use of mc∆θ/ mc∆T (1)= 3.6 K [⇒ 3.2 K] / 3.6 °C (1) 3

(iii) Expect ∆T to be less (1)Any 2 of: Tube/plastic warms up; cork/air warms up;because lead falls < 80 cm; energy lost to surroundings/tube/cork/air ; poor thermal contact withthermocouple (1) (1) 3

(iv) As m cancels / mass does not matter (1)but as c is higher (1)∆T will be lower (1) 3

[11]

366. (i) Its chemical composition / surface temperature (1)(not velocity) 1

(ii) Use of ∆λ/λ = υ/c [some substitution or rearrange] (1)see λ = 440 or 400 (1)

= 1.36 × 107 m s–1 (1)

[if bald answer: 1.43 × 107 (1)xx; 1.4 × 107 (1)xx ;

1.50 × 107 (1) (1)x; 1.5 × 107 (1) (1)x]towards the Earth / us (1) 4

[5]

367. Show that vertical component of velocity is about 70 m s –1 down

Use of ∆ξ/∆t (1)

Tangent - touching at 300 s, not crossing curve (1)

(65 000 m – 86 500 m) ÷ 290 s

= [–]74.1 m s–1 [no ue][accept answers in range 68 m s–1 to 79 m s–1] (1) 3

Calculation of average vertical acceleration

a = (v – u)/t (1)

= (−38.0 m s–1 – (− 74.1 m s–1 )) ÷ 100 s [ecf]

= [+] 0.36 m s–2 (1)

Upwards (1) 3

Calculation the weight of the shuttle

W = mg

= 2.0 × 106 kg × 9.6 N kg–1

= 1.9 × 107 N (1) 1

Calculation of average upward vertical force

[Resultant] force = ma (1)

= 2.0 × 106 kg × 0.35 m s–2 [ecf]

= 0.7 × 106 N (1)

Upward force = resultant force + weight [consistent with second part]

= 2.0 × 107 N (1) 3[10]

368. State energy extracted in one second

35 000 J or 35 kJ [accept J s–1 or kJ s–1] (1) 1

Calculate mass of water which flows through system in one second

In one second, ∆ Q = mc∆θ

35 000 J = m × 4200 J kg–1 K–1 × 5 K – substitution [ecf] (1)

m = 35 000 J ÷ (4200 J kg–1 K–1 × 5 K) – rearrangement (1)

m = 1.7 kg [accept kg s–1] (1) 3

Give reason for use of solar cells

e.g. cooling needed most during the day or at sunny times of year/reduces use of non-renewable resources/uses renewable source/reduces production of greenhouse gases/low power/doesn’t needmuch energycost arguments must explain long term savings [high start up costsbut low running costs][do not accept “environmentally friendly” unqualified] (1) 1

[5]

369. Description of photon

Packet/quantum/particle of energy [accept E = hf for energy] (1) (1)

[allow packet/quantum/particle of light/e-m radiation/e-m wave etc for (1) X] 2[zero marks if error of physics such as particle of light with negative charge]

Show that energy to move electron is about 8 × 10 –20 J

W = QV (1)

= 1.6 × 10–19 C × 0.48 V

= 7.7 × 10–20 J [no ue] (1) 2

Calculate efficiency of photon energy conversion

Efficiency = (7.7 × 10–20 J ÷ 4.0 × 10–19 J) [ecf] (1)

= 0.19 or 19 % (1) 2[6]

370. Explanation of pressure nodes or antinodes

Pressure constant (1)

Node as a result (1) 2

Relationship between length and wavelength

l = λ/2 or λ = 2l (1) 1

Calculation of fundamental frequency

λ = 2 × 0.28 m = 0.56 m [ecf for relationship above] (1)

v = f λ (1)

f = v/λ = 330 m s–1 ÷ 0.56 m

= 590 Hz (1) 3

Calculation of time period

T = 1/f (1)

T = 1 ÷ 590 Hz [ecf]

= 0.0017 s (1) 2

State another frequency and explain choice

e.g. 590 Hz × 2 = 1180 Hz (or other multiple) (1)

multiple of f0 or correct reference to changed wavelength (1)

diagram or description, e.g. N A N A N, of new pattern [ecf for A & N] (1) 3[11]

371. Calculate kinetic energy

Ek = ½ m υ2 (1)

Ek = ½ × 1800 kg × (53 m s–1)2

= 2.53 × 106 J (1) 2

Show that max height would be about 140 m

Ep = mg∆h (1)

Initial Ek = final Ep/½ mυ 2 = mg∆h/2.53 × 106 J = mg∆h (1)

∆h = 2.53 × 106 J/(1800 kg × 9.81 N kg–1)

∆h = 143 m [no ue] (1)

OR

υ 2 = u2 + 2as

0 m2 s–2 = (53 m s–1)2 + 2 × (– 9.81 m s–2) × s [subst] (1)

s = (53 m s–1)2 ÷ (2 × 9.81 m s–2) [rearrangement] (1)

s = 143 m [no ue] (1) 3

Show that energy loss is about 3 × 10 5 J

Ep = 1800 kg × 9.81 N kg–1 × 126 m = 2.22 × 106 J (1)

Ek – Ep = 2.53 × 106 J – 2.22 × 106 J

= 3.1 × 105 J [no ue] (1)

OR

For 143 m

Ep = 1800 kg × 9.81 N kg–1 × 143 m = 2.53 × 106 J

For 126 m

Ep = 1800 kg × 9.81 N kg–1 × 126 m = 2.22 × 106 J (1)

Energy lost =2.53 × 106 J – 2.22 × 106 J

= 3.1 × 105 J [no ue] (1)

OR

Energy lost = 1800 kg × 9.81 N kg–1 × (143 m – 126 m) (1)

= 3.1 × 105 J [no ue] (1) 2

Calculation of average resistive force

Work = force × distance (1)

Force = work ÷ distance

= 3.1 × 105 J ÷ 126 m

= 2500 N (1) 2

Calculation of time for climb

s = ½ (u + v) × t (1)

t = 2s ÷ (u + v)

= 2 × 126 m ÷ 53 m s–1

= 4.8 s (1)

[Use of g = 9.81 m s–2 in equations of motion to get a consistent valueof t [υ = u + at → t = 5.4 s] → 1 mark]

Assumption: eg assume uniform acceleration/constant resistive force/constant frictional force (1) 3

[12]

372. Explain zeroing of meter

No resistance when leads touched together/short circuit/calibration forzero error (1) 1

Show that resistance is about 70 Ω

R = V ÷ I (1)

= 0.54 V ÷ 0.0081 A

= 67 Ω [no ue] (1) 2

Explain section from passage

Other currents/voltages/resistances present (1)

change in current changes reading for resistance (1) 2

Explain changes in meter reading with temperature increase

Increased lattice vibrations/vibration of atoms/molecules (1)

scattering flowing electrons/more collisions (1)

increased resistance/increase meter reading (1) 3[8]

373. Name process of deviation

Refraction (1) 1

Completion of ray diagram

B – no deviation of ray (1)

A and C – refraction of ray away from normal on entering hot air region (1)

A and C – refraction of ray towards normal on leaving hot air region/ (1) 3

Show positions of tree trunks

B the same (1)

[consistent with ray diagram]

A and C closer to B (1) 2

Explanation of wobbly appearance

Hot air layers rise/density varies/layers uneven (1)

Change in the amount of refraction [accept refractive index]/changein direction light comes from (1) 2

[8]

374. Why gamma radiation used

γ is the most/more penetrating (1) 1

(OR α/β less penetrating)

Factors controlling amount of radiation

Any 2 from:

• Strength/type of radiation source/half-life/age of source• speed of conveyor belt/exposure time• shape/size of food packages/surface area• distance from radiation source (1) (1) Max 2

Suitable material for wall

Concrete/lead (1)

Suitable thickness

30 cm – 1 m/1 – 10 cm (1) 2[thickness mark dependant on named material]

Source of natural radiation

Rocks, soil, cosmic rays, named radioactive element, sun, space, air (1) 1[6]

375. Circuit diagram

Ammeter and power source in series (1)

Voltmeter in parallel with electrodes (1) 2

[Allow both marks if diagram shows an ohmmeter without a powerpack –1 if power pack]

Calculation of resistance

Use of area = πr2 (1)

R = 2.7 × 10–3 Ω m × 5.0 × 10–4 m/A (1)

= 172 Ω (171.9 Ω) (1) 3

Plotting graph

Axis drawn with R on y-axis and labelled with units (1)

Points plotted correctly [−1 for each incorrect] (1)

Sensible scale (1)

Curve added passing through a minimum of 4 points (1) 4

Diameter of hole

Correct reading from graph = 0.23 mm [Allow 0.22 – 0.26 mm] (1) 1[10]

376. Stress on thread

(Weight) = 1.0 × 10–3 kg × 9.81 m s–2 (1)

Use of A = πr2 OR 7.9 × 10–11 m2 (1)

Stress = 1.2 × 108 Pa (1) 3

Maximum stress

Use of σ = ε × E (1)

ε = 0.001 (if incorrect no further marks for this section) (1)

σ = 0.001 × 2 × 1011 Pa = 2 × 108 Pa (1) 3

Weight

F = σ × A = 2 × 108 Pa × 3 × 10–4 m2 (1)

= 6 × 104 N (allow ecf) (1) 2

Comparison with spider silk

Larger (1)

Discussion of figures in table, eg E ~ 3 times smaller, but εmax ~ 300

times larger/calculation of breaking force on silk (5.4 × 106 N allow

ecf on ε value)/comparison of stress values ( σsilk = 1.8 × 1010 Paallow ecf on ε value) (1) 2

Assumption

Young modulus still holds at breaking point/elastic limit notreached/elastic behaviour/obeys Hooke’s law (1) 1

[11]

377. Unpolarised and plane polarised light

Minimum of 2, double-headed arrows indicating more than 1 planeand 1 double-headed arrow indicating 1 plane labelled unpolarisedand polarised (1)

Vibrations/oscillations labelled (1) 2

Appearance of screen

Screen would look white/bright/no dark bits/light [not dark = 0] (1)

Explanation

As no planes of light prevented from leaving screen/all light getsthrough/all polarised light gets through (1) 2

Observations when head is tilted

Screen goes between being bright/no image to image/dark bits (1)

Every 90°/as the polarising film on the glasses becomes parallel/perpendicular to the plane of polarisation of the light (1) 2

Comment on suggestion

Image is clear in one eye and not the other (1)

If plane of polarisation is horizontal/vertical (1)

ORImage is readable in both eyes (1)

As the plane of polarisation is not horizontal or vertical (1) 2[8]

378. Speed of car 2

υ 2 − u2 = 2as

⇒ (−)u2 = (−) 2 × 3.43 × 23.9 [substitution or rearrange] (m s–1) (1)

⇒ u = 12.8 (m s–1) [value] (1) 2

Magnitude of the momentum of car 2

p = mυ (1)

= 1430 × 12.8 (13) (kg m s–1)

= 18 300 (18 600) N s or kg m s–1 (1) 2

Calculation of easterly component of momentum

Component = momentum × cos θ (1)

Car 1: 23 800 Ns × cos 45

= 16 800 N s (1)

Car 2: 18 300 (18 600) N s × cos 30 = 15 800 (16 100) N s (1) 3

Whether car 1 was speeding before accident

(Sum of two easterly components) ~ 33 000 N s [ecf] (1)

(÷ mass of car 1) ⇒ ~ 16.8 m s–1 [ecf] (1)

Conclusion related to speed limit (17.8 m s–1) (1) 3

Explanation of how investigator could use conservation law

Any two from:

• Momentum conservation• After collision there is significant northerly momentum• Before collision car 1 had no northerly momentum/only car 2

had northerly momentum (1) (1) 2[12]

379. Explanation

energy gained by electron accelerated through 1 V/W = QV (1)

W = 1.6 × 10−19 C × 1 V = 1.6 × 10−19 J (1) 2

Unit of mass

∆E = c2∆m so ∆m = ∆E/c2 (1)

GeV is energy ⇒ GeV/c2 is mass (1) 2

Mass of Higgs boson

m = 115 × [109] × 1.6 × 10–19/(3 × 108)2 (1)

= 2.04 × 10–25 kg (1) 2

Antiparticle

Same mass and opposite charge (1)[Accept Particle and its antiparticle annihilate (→ photons)] 1

Explanation of need for a magnetic field and why it can be small

Force deflects particles/force produces circular motion (1)

Force is perpendicular to motion/force provides centripetal force (1)

r is large or curvature is small/gentle (1)

reference to B = p/rQ to show why small B is needed (1) 4[11]

380. Minimum charge on balloon

0 . 5 m

1 . 8 m

M g

F

T

any 2 forces correct (1)

T 3rd force correct (1) 2

F = kq1q2/r2/F = kq2/r2 (1)

mg = T cos θ/T = 1.8 × 10−2 N (1)

F = T sin θ/F = mg tan θ/F = 4.6 × 10−4 N (1)

r = 0.5 + 2 × 1.8 sin 1.5° (= 0.594 m) (1)

⇒ q2 = Fr2/k = Fr2 × 4πε0 (1) Max 3

= 0.0018 × 9.81 × tan 1.5° × 0.5942 × 4π ε0

⇒ q = 1.36 × 10–7 C (1)

[NB 2 for diagram, maximum 3 for intermediate parts, and final 1for the answer] 1

[6]

381. Number of protons and neutrons in isotope of cadmium

Protons: 48 (1)Neutrons: (122 – 48) = 74 (1) 2

Process which occurs in a fission reaction

3 points, e.g.• starts with large/heavy nucleus/atom• nucleus captures/absorbs neutron• (becomes) unstable• splits into (two) smaller nuclei (and more neutrons)• with emission of energy (from mass defect)• increased binding energy (per nucleon) (1) (1) (1) 3

Calculation of change in mass

Use of ∆E = c2∆m (1)

∆m = ∆E/c2

∆m = 3.2 × 10–11 J/(3 × 108 m s–1)2

= 3.6 × 10–28 kg (1) 2

Calculation of number of fissions required each second

Correct use of efficiency (1)e.g. fission energy required = 660 MW × 100/30 = 2200 MW

No. of fissions per second = 2200 × 106 J/3.2 × 10–11 J = 6.9 × 1019 (1) 2

Why it is necessary to have a lot more fuel in nuclear reactor

2 points, e.g.• reactor needs to run for longer than 1 second• need sustained chain reaction• fuel must last for years• only a small proportion of U-235 undergoes fission• only small proportion of uranium is U-235• greater certainty/frequency of neutron collisions (1) (1) 2

[11]

382. Age of part of the stalagmite

λ = ln2/t1/2 = 1.2 × 10–4 years–1 (= 3.8 × 10–12 s–1) (1)

Use of N = N0e–λt (1)

1 = 256 e–1.2 × 10 – 4t

[allow 255 instead of 1 for this mark but do not carry forward]

t = 46 000 years ( = 1.45 × 1012 s) (1) 3

[OR recognise 1/256 (1)8 half-lives (1)45 800 years (1)]

Carbon-14 concentration

Carbon-14 measurement would be greater (1) 1

Validity of radio-carbon dating

3 points, e.g.• not valid• twice original concentration gives greater proportion measured now• object seems younger than it actually is• older parts could have more carbon-14 than younger parts• technique relies on constant levels, therefore unreliable• mixture of old and young carbon-14 in 1 stalagmite makes dating

impossible (1) (1) (1) 3[7]

383. Velocity of jumper

ω = 2π/T = 2π/5.0 s (= 1.26 s–1) (1)

υ max = Aω

= 4.0 m × 2π/5.0 s

=5.0(3) (m s–1) (1) 2

Why tension in rope and jumper’s weight must be balanced

When υ is maximum, acceleration = 0 (1)

so net force = 0 (1) 2

[OR: If forces not in equilibrium, he would accelerate/decel. (1)So velocity cannot be maximum (1)]

Calculation of force constant for rope

Use of T = 2π km / (1)

Hence k = 4π2m/T2 = 4π2 × 70 kg/(5.0 s)2

= 109 – 111 N m–1 [kg s–2] (1) 2

Verification that rope is never slack during oscillations

F = mg = 70 kg × 9.81 N kg–1 = 687 N (1)

At centre of oscillation, when forces in equilibrium,

x = F/k

= 687 N/110 N m–1 (allow e.c.f. from previous part) (1)

= 6.2 m which is larger than amplitude (1) 3

OR

Calculation of amax (= – ω2A) [6.32 m s–2] (1)

Comparison with g 9.81 m s–1 (1)Deduction (1)

Likewise for forces approach.

Motion of jumper

Any 1 from:• motion is damped shm• so amplitude decreases• but period stays (approximately) the same (1) 1

[10]

384. How sound from speakers can reduce intensity of sound heard by driver

Any 6 from:

• graphs of 2 waveforms, one the inverse of the other

• graph of sum showing reduced signal

• noise detected by microphone

• waveform inverted (electronically)

• and fed through speaker

• with (approximately) same amplitude as original noise

• causing cancellation/destructive superposition

• error microphone adjusts amplification 6[6]

385. Momentum of neutron

Use of p = mυ (1)

p = 5.03 × 10–20 N s/kg m s–1 (1) 2

Speed of nucleus

Total mass attempted to be found (1)Conservation of momentum used (1)

υ = 2.01 × 106 m s–1 [ecf from p above only] (1) 3

Whether collision was elastic

Use of k.e. = ½ mυ 2 (1)

ke = 7.45 × 10–13 (J) / 5.06 × 10–14 (J) (ecf) (1)

A correct comment based on their two values of ke. (1) 3[8]

386. Maximum velocity

Area = 100 m (1)

Attempt to find area of trapezium by correct method (1)

υ = 10 m s–1 (1) 3

Sketch graph

Horizontal line parallel to x axis

Some indication that acceleration becomes 0 m s–2

The initial acceleration labelled to be υ max ÷ 2 [ initial a = 5 (m s–2) (1)(ecf)]

t = 2 (s) where graph shape changes (1) 4[7]

387. Decay constant

λ = 0.69/432 (yr–1) (1)

λ = 5.1 × 10–11 (s–1) [At least 2 significant figures] (1) 2

Number of nuclei

3.0 × 1013 (1) 1

Activity calculation

Use of A = λ N (1)

A = 1.5 × 103 Bq / s–1 [ecf ] (1) 2

Explanation

Range few cm in air / short range (1)

Alpha would produce enough ions (to cross gap) OR ionisesdensely/strongly/highly (1) 2

Features of americium sample

Half–life long enough to emit over a few years (1)

Count well above background (1)

Suitable as safe as range very low / shielded (1) 3[10]

388. Criticism of statement

Not a Newton third law pair (1)

Forces in equilibrium but not for reason stated (1)

N3 pairs act upon different bodies (1)

N3 pairs same type (1)

Line of action different / rotation (1) Max 3

Table

Gravitational (1)

Earth (1)

Upwards and downwards [both must be correct] (1)

Table (1) 4

Force Type of force Direction of Newton

3rd law ‘pair’ force

Body ‘pair’ forceacts upon

Weight Gravitational Upwards Earth

Push of table Electro-magnetic Downwards Table

[7]

389. Principal energy transformation

Kinetic energy to internal energy/heat/work done against friction (1) 1

Explanation of braking distance

Q0WC (1)

Car is (also) losing gpe (1)

Total work done against friction is greater OR more energy to be converted to heat (in the brakes) (1)

Since force is same, distance must be greater [consequent] (1) 4[5]

390. Temperature calculation

Current = 4.5 × 10–3 A (1)

p.d. across thermistor is 4.2 V (1)

Rthermistor = 930 Ω ecf their current and pd subtraction error (1)

Temperature = 32 °C − 34 °C [Allow ecf for accurate reading] (1) 4

Supply doubled

Any two from:

• Current would increase / thermistor warms up / temperature increases

• Resistance of thermistor would decrease (1) (1)

• Ratio of p.d.s would change

No OR voltmeter reading / pd across R more than doubles (1) 3

[This mark only awarded if one of the previous two is also given][7]

391. Diagram

Labelled wire and a supply (1)

Ammeter in series and voltmeter in parallel (1)

OR

Labelled wire with no supply (1)

Ohmmeter across wire (1) 2

Readings

Current and potential difference OR resistance ( consistent with diagram) (1)

Length of wire (1)

Diameter of wire (1) 3

Use of readings

R = V/I OR ρ = RA/l (1)

Awareness that A is cross–sectional area (may be seen above and credited here) (1)

Repetition of calculation OR graphical method (1) 3

Precaution

Any two from:

• Readings of diameter at various places /different orientations

• Contact errors

• Zeroing instruments

• Wire straight when measuring length

• Wire not heating up / temperature kept constant (1) (1) 2[10]

392. Conductor resistance

R = ρ l/A (1)

Correct substitution of data (1)

R = 4.3 × 10–2 Ω (1) 3

Manufacturer’s recommendation

Larger A has a lower R (1)

Energy loss depends on I2R / reduces overheating in wires (1) 2[5]

393. Car battery

Voltmeter reading: 12.2 (V) (1) 1

Equation

Terminal p.d. = 12 V + (5.0 A × 0.04 Ω )

See 12V (1)

See 5.0 A × 0.04 Ω (1)

Addition of terms (1) 3

Wasted power

See 0.04 Ω + 0.56 Ω OR 2.8 V + 0.2 V OR 5 x (15 – 12) W (1)

Power = 15 W (1) 2

Efficiency

(same current) 12 V / 15 V OR POUT/PIN = 60 W/75 W (1)

Efficiency = 0.8/80% Efficiency = 0.8/80% (1) 2

Explanation

Any two from:

• Starter motor / to start car needs (very) large current

• I = rR

E

+

• (E and R fixed) rmin ⇒ Imax (1) (1) (1) 2[10]

394. (a) Graph

Falling concave curve (1)

Not intercepting x–axis or y–axis (1) 2

Two reasons

• light is scattered by dust (or air molecules)/refraction (1)[allow twinkling]

• some wavelengths are absorbed by atmosphere (1) 2

(b) Two spectra

β car is bluish; β And is reddish [not just different colours] (1) 1

Read off λ max ≈ (760 − 770) nm [Beware 680 nm] (1)

Use of Wien’s law (1)

Answer T = 3800 K [allow 3600 K to 4000 K] (1) 3

Calculation

Use L = σ AT4 (1)

A = 2.0 × 1028 W ÷ (9300 K4 × 5.67 × 10–8 W m–2 K–4) (1)

= 4.7 × 1019 m2 (1) 3

Estimate

Attempt at areas giving ~ 7 × [(× 5 – × 8) allowed] (1)

= 1.4 × 1029 W [(1.0 – 1.6) × 1029 W allowed] (1) 2[13]

395. What is mean by MS star

One burning/fusing (H as fuel) (1) 1

Outline

Quality of written communication (1) 1

• fuses He/other elements AND becomes red giant (1)

[OR ≤0.4 M⊙ not red giant/becomes white dwarf]

• ceases fusion AND becomes white dwarf (1)

• white dwarf fades to cold lump/becomes (specified colour)dwarf/no longer visible (1) Max 3

What determines whether ms star becomes a white dwarf

The mass of the star (1)

if mass star < 8 M⊙ OR if mass core remnant < 1.4 M⊙ (1) 2[7]

396. Determine period of pulsation of star A

2.4 to 2.7 days (1)

with evidence of averaging (1) 2

Addition to graph

B is approximately the same height (1)B has a longer period (1)Shape – more steeply up than down (1) 3

Description

Measure period and hence work out luminosity (1)

Measure intensity (1)

Use I = L/4π D2 (1) 3[8]

397. (a) Show that energy stored can be written as in formula

W = ½ Fx [allow x or ∆ x] (1)

F = kx (1) 2

Graph

Rising curve [either shape] (1)

starts at origin and correct shape, i.e. gets steeper (1) 2

(b) Young modulus of copper

Read valid pair off straight line region (1)

1.3 [when rounded to 2 s.f.] (1)

correct power of 10, i.e. × 1011 (1)

Correct unit: Pa OR N m–2 4

Copper wire

Obtain reasonable extension/reduce uncertainty (1) 1

Calculation of stress

Use of π r2 (1)

Substitution in F/A i.e. allow 280 N/r [OR their A] (1)

1.8 × 108 Pa [No e.c.f.] (1) 3

Point P

Point P marked on graph [e.c.f.] 1

Justification

Behaviour is elastic since on straight line region [e.c.f.] 1[14]

398. (a) Explanation of binding energy

Energy required to separate a nucleus (1)

into nucleons (1)

What this tells about an iron nucleus

Iron is the most stable nucleus (1) 3

(b) Nuclear equation for decay

ν++→ −− e/βNC 01

01

147

146

Symbols [C → N + β ] (1)Numbers [14, 6, 14, 7, 0, –1] (1)

Antineutrino / ν / eν (1) 3

Estimate of age of a fossil

3 half–lives (1)

giving 17 000 years to 18 000 years (1) 2

399. Particle X

Positive (1) 1

Is a baryon (1) 1

Quark compositions

Proton uud; neutron udd BOTH (1) 1

Explanation and deduction of identity of X


Strong / not weak interaction (1)

One strange quark on each side / no flavour change (1)

X is a proton (1) 4[7]

400. (a) (i) Table with units (1)Temperature to better than 1 °C (1)Readings every 0.5 minutes or less up to 5 minutes (1)Cooling curve established (1)8 points ± 0.5 °C from your best curve (1) 5

(ii) Data must occupy more than ½ page in both directionsScale [Allow scale of 1 cm ≡ 30 s] (1)Axes labelled + unit (1)Plots (1)Line (1) 4

(iii) ΔxΔy ≥ 64 cm2 or as large as possible (1)Correct calculation ≥ 2 significant figures (1)[Tangent at the correct point and sides read correctly][Ignore sign and unit]Correct calculation [allow J/min] [allow e.c.f. from gradient] (1)2/3 significant figures + unit from calculation (1) 4

(b) (i)

I n n e r c u p

A i r g a p

O u t e r c u p

C o l l a r

Correct arrangement with collar (1)Collar and air gap labelled (1) 2

(ii) Same volume of water (1)Same place on bench (1)Same starting temperature (1)Stir water (1)Extend the time range/same temperature range (1)Compare temperature falls after 5 minutes/compare times toreach same temperature (1) Max 3

(iii) Correctly labelled and double cup curve above single cup curve (1)Concave curves (1)Single curve always steeper than double curve (1) 3

8 0

6 0

0 5

/ ° C

t / m i n

(iv) Take gradient of the curves at the same temperature (1) (1)[OR Time for same temperature fall measured / temperature fall for thesame time measured (1)]EITHERGradient of the double cup should be half (or less) than the single cup (1)OR For same starting temperature [stated or seen] the time for a givenΔθ will be twice as big for the double cup (1) 3

[24]

Sample results

(a) (i)θ/°C t/min80.0 0.077.7 0.575.7 1.074.0 1.572.0 2.070.5 2.569.3 3.067.8 3.566.5 4.065.2 4.564.0 5.0

(ii) Graph

8 0 . 0

7 8 . 0

7 6 . 0

7 4 . 0

7 2 . 0

7 0 . 0

6 8 . 0

6 6 . 0

6 4 . 0

6 2 . 00 1 2 3 4 5 t

m i n

θ° C

(iii)t∆

∆θ =

0–48.5

9.77–0.62

= –2.90°C min–1

t

Q

∆∆

= mC t∆

∆θ

= 100 × 4.2 × 2.9

= 1220 J min–1

= 20.3 W

(b) (i)

I n n e r c u p

A i r g a p

O u t e r c u p

C o l l a r

(ii) Same volume of waterSame place on benchSame starting temperatureStir waterExtend the time range/same temperature rangeCompare temperature falls after 5 minutes/compare times to reach same temperature

(iii)

8 0

6 0

0 5

/ ° C

(iv) Take gradient of the curves at the same temperature[OR Time for same temperature fall measured / temperature fall for thesame time measured]EITHERGradient of the double cup should be half (or less) than the single cupOR For same starting temperature [stated or seen] the time for a givenΔθ will be twice as big for the double cup

401. Resultant force

Direction of travel changing (1)

Velocity changing/accelerating (1)

Force is towards centre of circle (1) 3

Why no sharp bends

Relate sharpness of bend to r (1)

Relate values of υ , r and F (1) 2

[e.g. if r large, υ can be large without force being too large/if r small,υ must be small to prevent force being too large]

Bobsleigh

Ncosθ = mg (1)

Nsinθ (1)

= mυ 2/r or ma(1)

Proof successfully completed [consequent on using correct formula] (1) 4

Calculation of angle

77 − 78° (1) 1[10]

402. Wavelength

0.30 m (1) 1

Letter A on graph

A at an antinode (1) 1

Wavespeed

Use of υ = fλ (1)

11(10.8) m s–1 (1) 2

[allow ecf λ = 0.15 m ie υ = 5.4 m s–1]

Phase relationship

In phase (1) 1

Amplitude

2.5 mm (1) 1[6]

403. Diagram

One arrow straight down (from −3.84 to − 5.02) (1)Two arrows down (from −3.84 to −4.53, then − 4.53 to − 5.02) (1) 2

Transition T

T from − 5.02 to − 1.85 upwards (1) 1

Kinetic energy values and explanation of what has happened to lithium atomin each case

0.92 eV (1)Atom stays in −5.02 (eV) level/nothing happens to it (1)

0.43 eV (1)

Atom excited to − 4.53 (eV) level (1) 4

Full credit is given to candidates who take the k.e. of the electron to be 0.92 Jafter collision. Any TWO correct energies with correct statement.

[7]

404. (a) Hubble constant

Attempt to find gradient (1)

1.9 × 10–18 s–1 (1) 2

Distance of this galaxy from Earth

∆ λ = 37.3 or see (410 – 372.7) (1)

Use of ∆ λ /λ = υ /c (1)

Use ofυ = Hd [υ = 3.0 × 107 m s–1] (1)

1.6 × 1025 m (1) 4

[full ecf H = 2 × 10–18 s–1 → 1.5 × 1025 m]

(b) Balloon – position of three dots

P, Q, R further apart on larger balloon (1)

Approximately similar triangles, i.e. approx. isosceles with baseapproximately ½ of long sides (1) 2

How balloon can be used to model expansion of Universe


Dots represent galaxies (1)

Balloon inflation represents expanding universe (1)

Dots further apart move apart faster, (as with galaxies) (1) 4[12]

405. Calculation of weight of shuttle

W= mg

= 2 × 106 kg × 9.81 N kg–1

= 1.96 × 107 N (1) 1

Labelling of forces acting on shuttle

Upward force = 3 × 107 N (1)

Downward force = 1.96 × 107 N [Allow ecf] (1) 2

Show that initial acceleration is about 5 m s –2

F = ma [Stated or implied] (1)

a = (3 × 107 N − 1.96 × 107 N) ÷ 2 × 106 kg [Allow ecf]

= 5.2 m s–2 [no ue] (1) 2

Use of graph to find height of shuttle after 120 s

Use of area [stated or implied e.g. by shading etc] (1)

At least to level of ½ vt approximation [44000 m to 63 000 m] (1)

Evidence of improved method, e.g. counting squares (1)

Answer within range 49 000 m to 53 000 m (1) 4

Suggestion and explanation of increased acceleration

Mass decreases [as fuel is burnt] (1)

Decrease in mass causes increase in acceleration / a α 1/m (1)

OR

Weight decreased because fuel used (1)

Resultant force increases (1) 2

[Accept Resultant force increased because of decrease in airresistance or in gravitational field strength for (1)×]

Reason for reduction in thrust

e.g. to go into orbit/stop going up/danger for craft/crew ifacceleration too large/required speed already reached (1) 1

[12]

406. Value of wavelength

λ = 13.9 cm − 0.5 cm (using interpolated sine curve) (1)

= 13.4 cm [accept 13.2 to 13.6 cm] (1) 2

[12.3 to 12.5 cm for distance using rods (1)× ]

Value of amplitude

Peak to peak = 4.5 cm [Accept 4.3 cm to 4.7 cm] (1)

Amplitude = ½ × peak to peak

= 2.25 cm [Accept 2.15 cm to 2.35 cm] [Allow ecf for 2nd mark if (1) 2

first part shown]

Calculation of frequency

f = 1/T

= 1 ÷ 2 s

= 0.5 Hz (1) 1

Explanation of why waves are transverse

Oscillation/vibration/displacement/disturbance at right angle (1)

to direction of propagation/travel of wave (1) 2

[Oscillation not in direction of wave (1)×]

Description of use of machine to illustrate sound wave

Sound is longitudinal/not transverse (1)

with oscillation along the direction of propagation / compressions and rarefactions (1)

so model not helpful (1) 3[10]

407. Show that initial vertical component of velocity is about 50 m s –1

Identify v = −u

OR

Time to top, with v = 0, is 5 s in v = u + at

OR

s = 0 with t = 10 s in s = ut + ½ at2 (1)

0 = u + (−9.81 m s–2 × 5 s)

u = 49 m s–1 [no ue] (1) 2

Calculation of velocity at which arrows left the bow

Vertical component = vsinθ [Stated or implied] (1)

v = 49.1 m s–1 ÷ sin 50°

= 64.1 m s–1 [Allow ecf] (1) 2

Show that range is about 400 m

Horizontal component = 64.1 m s–1 × cos 50°] [ecf] (1)

= 41.2 m s–1

Horizontal distance = υ × t [stated or implied] (1)

= 41.2 m s–1 × 10 s

= 410 m [no ue] (1) 3

Explanation of difference between recorded and calculated ranges

For example: air resistance [has been ignored in calculations]/wind against arrows (1)

[air resistance] will cause deceleration / decelerating force [reducing range] (1) 2[9]

U ø408. Circuit diagram

Variable voltage (1)

Includes ammeter and voltmeter (1)

…. in series and parallel respectively (1) 3

[No penalty for LED bias]

Description of current variation in LEDs

Initially, increasing voltage still gives zero current

OR

Current doesn’t flow until a specific minimum voltage (1)

Current then increases… (1)

….with an increasing rate of increase (1) 3

Discussion of whether LEDs obey Ohm’s law

No (1)

I not proportional to V

OR

R not constant / V/I not constant / R decreases (1) 2

Calculation of resistance of green LED at 1.9 V

R = V/I [Stated or implied] (1)

= 1.9 V ÷ 1.46 × 10–3 A

= 1300 Ω (1) 2

Calculation of power dissipated by red LED at 1.7 V

P = IV [Stated or implied] (1)

= 3.89 × 10–3 A × 1.7 V [do not penalise mA twice]

= 6.6 × 10–3 W (1) 2[12]

409. Process at A

Refraction [Accept dispersion] (1) 1

Ray diagram

Diagram shows refraction away from normal (1) 1

Explanation of condition to stop emergence of red light at B

Angle greater than critical angle (1)

Correctly identified as angle of incidence [in water] (1) 2

Calculation of wavelength of red light in water

c = fλ [stated or implied] (1)

λ = 2.2 × 108 m s–1 ÷ 4.2 × 1014 Hz

= 5.24 × 10–7 m (1) 2[6]

410. Forces on sphere (diagram)

Weight [downwards] W/mg (1)

Upthrust [upwards] U (1)

(Viscous) drag [upwards]/D/R (1) 3

Comparison

Weight is greater than upthrust and (viscous) drag (1) 1

Volume

4/3π r3 = 7.2 × 10–6 m3 (1) 1

Mass

Volume × density

= 7.2 × 10–6 × 1020

= 7.3. × 10–3 kg [e.c.f.] (1) 1

Upthrust

Weight of liquid displaced (1)

= 7.6 × 10–3 × 9.81

= 7.2 × 10–2 N [ no e.c.f.] (1) 2

Explanation of where sphere found

Overall force upwards U > W (1)

so sphere accelerates upwards / higher (1) 2

[OR stays at the top]

Properties

Any two from:

• Viscosity / temperature

• density (1) (1) 2[12]

411. (i) Tough

Can withstand dynamic loads / shock / impact / repeated deformation (1)

(ii) Material

Polythene / Kevlar / rubber / carbon fibre / steel (1) 2

Meaning of elastic

Return to original shape after force removed (1) 1

Calculation

(i) Force:

Use of F = kx (1)

= 1250 × 0.090

= 113 N (1)

(ii) Energy:

Use of E = ½ Fx [OR ½ kx2] (1)

= ½ × 113 × 0.090

= 5.1 J (1) 4

Calculation and discussion

Weight of child = 30 × 9.81 = 294 N (1)

Any two from:

• Child will fully compress the spring if landing on handle bar or 113 N implied

• time of impact longer

• impact force felt by child less

• some energy absorbed by spring (1) (1) 3

Why incorrect

Any two from:

• spring and wire are different

• length of “spring” is not equal to length of metal/material used

• area of “spring” is not the cross sectional area of the material used

• YM = F × 1/Ae

• as 1 is longer, A is smaller value of YM above would be underestimate (1) (1) 2[12]

412. Why γ rays are dangerous

For example:

Penetrates (skin) (1)

Can cause ionisation / cell damage / mutation (1) 2[not kill cells]

Material for shielding

Lead (1)

Several centimetres (0.5 → 5 cm) (1) 2

Why α radiation not used

Not (sufficiently) penetrating (not absorbed by luggage) (1) 1

Increased background radiation

Exposed to more cosmic radiation (1)

Less atmosphere above them for shielding (1) 2[7]

413. Difference between polarised and unpolarised light

Polarised: vibrations in one plane (at right angles to direction of travel) (1)

Unpolarised: vibrations in all planes [NOT 2 planes] (1) 2

OR

Correct drawing (1)

Vibrations labelled (1)

Meaning of advertisement

(Light vibrations are) in one plane (1) 1

Evidence that glare comprises polarised light

Glare is eliminated, so must be polarised light (1) 1

Sunglasses turned through 90°

Glare would be seen through glasses (1)

since they now transmit the reflected polarised light (1) 2[6]

414. Results of experiments and conclusions

Most pass straight through/undeflected (1)

A few deflect/reflect (at large angles) (1)

Small nucleus/mostly empty space (1)

Concentrated mass and/or positive charge (1) 4

How to determine x graphically

Plot log N v. log (sin θ / 2) [OR ln on both sides] [Any base] (1)

Gradient = x (1) 2

Meaning of numbers in the symbol for the gold nucleus

Bottom number: 79 protons (1)

Top number: 197 ns + ps )

OR )

197 nucleons ) (1)

OR )

197 − 79 = 118 ns ) 2

Mass of alpha particle

Mass of alpha particle ≈ 4 × mp

= 4 × 1.67 × 10–27 = 6.7 [or 6.68] × 10–27 kg (1) 1

Calculation of electric force

F = kq1q2/r2 OR q1q2/4π ε o r2 (1)

q1 = 79 × 1.6 × 10–19 C and q2 = 2 × 1.6 × 10–19 C (1)

[stated or subbed]

→ F = 14.56 N (1) 3[12]

415. Situation to which equation refers

F = force on particle (1)

υB = ( m a g n e t i c ) f l u x d e n s i t y / f i e l d s t r e n g t hυ = v e l o c i t y / s p e e d o f p a r t i c l eq = c h a r g e o f p a r t i c l e

= a n g l e b e t w e e n B a n d υ / m o t i o n/ c u r r e n t

( 1 )

( 1 )

( 1 )

F is perpendicular to B and υ (1)

[Some of these may be shown by diagram] Max 4

Description of situation modelled by equation

Curved/circular motion of particle (1)

p = momentum (1) 2

Why path of a particle is curved

Charged particles (1)

with (component of) motion perpendicular to field (1)

Force perpendicular to motion/ Fleming’s L.H. rule (1) Max 2

Why spiralling path decreases as it nears North Pole

Nearer pole → field stronger (1)

Reference to r = p(mυ )/Bq OR r ∝ 1/B

OR B increasing → centripetal/inward F increases

Alternative: υ ↓ due to resistive force (1)

Reference to r = p(mυ )/Bq OR r ∝ p/υ 2[10]

416. Discussion of type of collision

Inelastic (1)

Energy → heat at impact/plastic deformation (1) 2

Momentum vector diagram

Diagram [right–angle triangle with arrows on two perpendicular sides] (1)

Labelling (1) 2

M a s s : 5 . 8 × 1 0 k g × 3 5 0 0 0 m s ( = 2 . 0 × 1 0 N s )

C a s t a l i a1 . 2 × 1 0 k g × 2 5 0 0 0 m s( = 3 × 1 0 N s )

– 1 2

1 6

– 1

6 – 1

1 1

Calculation of change in direction

θ /sinθ /tan θ = 00025102.1

00035108.512

6

××××

= 6.77 × 10–6 (1)

θ = 6.8 × 10–6 rad OR 3.9 × 10–4 degrees OR 1.4 seconds (1) 2

Formula for net force in terms of momentum

F = d / d t (mυ ) OR words (1) 1

Calculation of number of rockets required

N × F × t = 2 × 1011 OR N = 2 × 1011 / (7 × 106 × 130) (1)

N = 220 [must be a whole number] (1) 2[9]

417. How properties of particles and antiparticles compare

Same mass/properties, opposite charge (1) 1

Energy

E = mc2 = 1.67 × 10–27 × (3 × 108)2 J [m or c subbed correctly] (1)

= 1.503 × 10–10 J [u.e. if comparison made here]

= 1.503 × 10–10/109 × 1.6 × 10–19 GeV (1)

= 0.94 GeV (1) 3

[jump to “≈ 1 GeV” omitting last line scores (1)(1)× ]

Survival of anti-atom

Anti–proton meets proton OR positron meets electron OR (anti-atom)meets atom (1)

(leads to) annihilation (1) 2

Table 2

Meson Baryon Lepton

proton (1)

antiproton

electron (1)

positron

Quark structure

Antiproton: 2 × −2/3 (anti u) + 1 × + 1/3 (anti d) (1)

= −1 (e not needed) (1)

[3 × d ⇒ –1 scores × × ] 2[10]

418. Explanation of why resultant flux in iron core is zero

Same current (in both coils) OR same turns (1)

Wound opposite ways (1) 2

OR leading to cancelling of magnetic effects

Explanation of how RCCB breaks circuit

Any five from:

Different currents give different (noncancelling) effects (1)

∴ net B OR φ /B ≠ 0 (1)

Faraday/changing φ /B (1)

⇒ V induced in third coil [“I induced” is 4th (1) only] (1)

⇒ I in third coil/relay coil (1)

⇒ relay coil magnetized (1)

⇒ relay contact opens (1) Max 5[7]

419. What happens in circuit after switch closed then opened again

Any seven from:

S closed → C charges (1)

up to VS (1)

Instantly/very quickly (1)

S open: discharge starts (1)

Exponential discharge (1)

(Vc = Vs e−t/RC )

¾ Vs = Vs e−t/RC (1)

⇒ ln ¾ = −t/RC (1)

⇒ t = 29.7 s OR RC = 103 s [if no other calculation] (1)

Buzzer sounds for 29.7 s [ecf] (1) Max 7

[Marks 1-5 and mark 9 are available via appropriate graph. For mark 5graph must have axes labelled with a V/Q/I and same t, and a recognisableexponential curve.]

420. Table

Type of force Example

Gravitational Weight/attraction between two masses (1)

Electromagnetic Normal reaction/friction/drag/tension/forcebetween two charges or magnets/ motoreffect/ elastic strain forces/contact forces

(1)

Nuclear Strong/Weak/force keeping protons (and/orneutrons) together/beta decay/forces withinnucleus

(1)

3

Forces

Any three from:

• same type (1)

• same magnitude/equal (1)

• act on different bodies/exerted by different bodies (1)

• opposite direction (1)

• same line of action (1)

• acts for same time (1) Max 3[6]

421. Deceleration of trolley

Select υ 2 = u 2 + 2ax /both appropriate formulas (1)

Correct substitutions (1)

0.309 [2 significant figures minimum](1) 3

Frictional force

Use of F = ma (1)

8.7 / 8.6 N [8.4 if 0.3 used] (1) 2

Power

Use of P = Fυ (1)

9.6 / 9.5 W [9.2 if 0.3 used] (1) 2

Force

Use of a = (υ – u)/t (1)

Add 8.6 /8.7 N to resultant force [8.4 if 0.3 used] (1)

42.8 N [42.6 if 0.3 used] [Accept 42.2 N] (1) 3[10]

422. Explanation

Some energy converted to internal energy [or heat or sound] / work done againstfriction [or air resistance] (1) 1

Experiment

Measure υ at the bottom (1)

Suitable apparatus, e.g. motion sensor and data logger/light gate(s) and timeror computer (1)

Detail of technique, e.g. sensor sends pulses at regular time intervals and timeto return is measured/gate measures time for card of known length to pass/tickertapemeasures length between dots made at regular time intervals (1)

Measure mass of trolley with balance (1)

Calculate kinetic energy from mυ 2/2 (1)

Measure vertical drop with ruler (1)

Calculate (gravitational) potential energy from mgh (1)

Calculate gpe

ke × 100 Max 6

[7]

423. Isotopes

same different

Number of protonsAtomic numberElementProton number

Number of neutronsNeutron numberNucleon numberAtomic massMass number (1) 1

Polonium decay

Po at (84, 210) with label (1)

2 steps west (1)

4 steps south (1) 3

Experimental check

Use of GM tube (1)

Inserting sheet of paper/aluminium foil/very thin aluminium/a few cm ofair stops the count (1)

Measure background, and look for count rate dropping to background (1) 3

NB Award points 2 and 3 for correct converse argument.[7]

424. Meanings ot terms

Range: distance travelled (before being stopped) (1)

Ionises: removes electron(s) from atoms (1) 2

Explanation

More strongly ionising means shorter range (1)

ionising means energy lost (1) 2

Mass

Use of m = ρ V (1)

8.1 kg (1) 2

T’hickness of lead sheet

Use of 8.1 kg/ρ OR t proportional to 1/ρ (1)

0.7 mm (1) 2[8]

425. Rutherford scattering experiment

Most went (nearly) straight through (1)

A small proportion deflected through large angles (1) 2

Arrows to diagram

Two arrows directed away from N (1) 1

Sketch graph

Speeds equal at A and B (1)

A non-zero minimum at P (1) 2

Shape of graph

A to P: Force (component) against velocity so decelerates (1)

P to B: Force (component) in direction of velocity so accelerates (1) 2

Add to diagram

Same initial path but deflected through larger angle (1) 1

Observations

More alpha particles deflected/ alphas deflected through largerangles/fewer pass straight through (1) 1

[9]

426. Charge

Charge is the current × time (1) 1

Potential difference

Work done per unit charge [flowing] (1) 1

Energy

9 V × 20 C (1)

= 180 J (1) 2[4]

427. Number of electrons

(–64 × 10–9 C) / (–1.6 × 10–19 C) = 4.0 × 1011 electronsUse of n = Q/e (1)

Seeing 1.6 × 10–19 C (1)

Answer of 4.0 × 1011 (electrons) (1) 3

[Use of a unit is a ue]

[–ve answer: 2/3]

Rate of flow

(6.4 × 10–8 C)/3.8 s = 16.8/17 [nC s–1] OR 16.8/17 × 10–9 [C s–1]

(6.4) / 3.8 s i.e. use of I = Q/t [Ignore powers of 10] (1)

Correct answer [No e.c.f.] [1.7 or 1.68 x 10–8 or 1.6 × 10–8] (1) 2

Unit

Amp(ere)/A (1) 1[6]

428. Explanation of observation

Any two from:

• LED on reverse bias/R in LED infinite/ LED wrong way round

• so current is zero /LED does not conduct / very small reversebias current

• since V = IR

• V = 0 × 1K = 0 V (1) (1) 2

Explanation of dimness

RV very large / RV much greater than RLED (1)Current very low / pd across LED very small (not zero) (1) 2

Circuit diagramLED in forward bias (1)Variation of pd across LED (1)Voltmeter in parallel with LED alone (1) 3[LED in series with voltmeter 0/3]

[7]

429. Circuit diagramAmmeter in series with cell and variable resistor (correct symbol) (1)Voltmeter in parallel with cell OR variable resistor (1) 2

Power output at point XPower = voltage × current (1)= 0.45 V × 0.6 A= 0.27 W (1) 2

Description of power output

Any three from:

• Current zero; power output zero/small/low

• As current increases power output also increases

• Then (after X ) power decreases

• Maximum current; power output zero (1) (1) (1) 3

[Accept reverse order]

e.m.f. of cell

0.58 V (1) 1

Internal resistance

Attempt to use current

"lost volts" OR ε = V+ IR (1)

= A6.0

0.45V–V58.0

= 0.217 / 0.2 Ω (1) 2

[ecf an emf greater than 0.45 V][10]

430. Statement 1

Statement is false (1)

Wires in series have same current (1)

Use of I = nAeυ with n and e constant (1) 3

[The latter two marks are independent]

Statement 2

Statement is true (1)

Resistors in parallel have same p.d. (1)

Use of Power = V2/R leading to R ↑, power ↓ (1) 3

OR as R ↑, I ↓ leading to a lower value of VI 3rd mark consequenton second

[6]

431. Ways in which white dwarf star differs from main seguence

Lower mass/volume/radius/(surface) area (1)

Fusion (burning) finished [Not luminosity] (1) 2

Fate of star after it has become a white dwarfCools (gradually)until no longer visible/becomes dimmer/changes colour (1) 2

[Allow brown dwarf]

HR diagram

Any two from:

• temperature scale in reverse direction

• at least two reasonable T values shown/ eg 40 000,10 000,4000)

• indication of log scale (1) (1) Max 2

(Single) star selected at L/L = 1 [≈ ± 2mm vertically by eye] (1) 1(Region) W clearly below MS [No ecf on T] (1) 1M to include top of MS (1) 1

Explanation

Quality of written communication (1)Greater mass means greater luminosity or greater temperature or greatergravitational forces (1)

Burns hydrogen/fuel much faster (1)Runs out of fuel quicker (1) 4

[13]

432. Temperatures of the two stars

They are similar/same (1)because of Wien’s law/same λ /f (1) 2

Greater radius

Deneb (1)More luminous (1)

Refer to σ AT4 and state T similar (1) 3

Which star will appear brighter?Vega (1)

Use of I = L/4π D2 (1)

One correct value: D = 8.8 OR V = 29 (× 10–9 W m–2) (1)Second correct value with unit (1) 4

[9]

433. Area under graph

It represents energy (stored) per unit volume/energy density (1) 1

Volume of seat belt

1.8 × 10–4 (m3) (1) 1

Kinetic energy

Attempt to use ½ mυ 2 (1)= 15.8 (kJ)/15 800 (J) (1) 2

Energy per unit volume

8 or 9 × 107 J m–3 (88 MJ m–3)OR

2nd answer divided by 1st with correct unit (1) 1

Belt satisfactory

Attempt to find area under graph / ½ ε σ used (1)

Value ≥ 9.6 × 107 (J m–3) (so total area is greater than above) (1) 2

Design change

Wider or thicker or harness shaped belt/more straps (1)

Need greater volume/need to reduce pressure on driver/needto absorb more kinetic energy [Not faster] (1) 2

[9]

434. Energy conversions

GPE to KE (1)to EPE or internal energy/strain/elastic (1) 2

Three properties

Strength, toughness, elasticity

Any TWO correct (1)

Third property correct 2

[–1 per incorrect answer if ≥ three circles]

Calculation of theoretical extension of rope

Correct substitution in A = π r2 / 9.5 × 10–5 m2 (1)

Sensible use of any TWO from:

• Stress = F ÷ A [Ignore 10n, ecf A]

• E = stress ÷ strain

• Strain = ∆ l ÷ l (1) (1)

Answer: 2.0 m [No ecf] [ue] (1) 4

Suggested reason why rope should be replaced

May have exceeded its elastic limit or may have deformed plastically ormay have been damaged on sharp rock/ fibres may be broken (1) 1[NOT rope has broken]

[9]

435. Emission - written above arrows

α β – β – α αAll five correct [Allow e–, 4He 2+] (1) (1) 2[For each error –1][α β β α α gets 1/2]Number of alpha particles emittedFive (1) 1

[3]

436. Comparison between antiparticle and its particle pair

Similarity: same mass as its particle pair (magnitude of charge) (1)

Difference: opposite charge/baryon number/(Iepton number / spin) (1) 2

Quark composition

ū ū –

d [OR anti-down etc] (1) 1

Baryon number

–1 (1) 1

Why difficult to store antiprotons

As soon as they contact protons/matter (1)

they annihilate (1) 2

Maximum possible mass

×2 (1)

÷ 0.93 or equivalent [OR by using E = mc2 to 1.6 × 10–25 kg] (1)96 (u) OR 97 (u) (1) 3[48u x (1) (1)]

Two reasons why interaction cannot take place

Q/charge not conserved (1)B/baryon number not conserved (1) 2

[11]

437. (a) All temperatures recorded with units (1)∆ θ ≈ (5 → 15) K [Need not be calculated] (1)Some attempt at temperatures better than 1°C (1)Precautions (Max 2): (1) (1) 5

• stirred water

• took θ i just before water added

• recorded θ f immediately after rapid fall

• equilibrium between water and thermometer

(b) Correct re-arrangement and substitution (1)Correct calculation, 2/3 significant figures and no unit (1)Value 0.15 to 0.30 or centre value if > 0.30 (1) 3

(c) Discard 20 cm3 into the waste beaker (1)

Add further 20 cm3 into the hot/warm water (1)Temperatures recorded, with units and θ i < θ f from previous value (1)

Correct calculation ≥ 2 significant figures and no unit (1)Value ± 0.04 of previous k (1) 5[Dependent mark]

(d) Correct calculation with average as denominator (1)Sensible comparison with 10% (1)Sensible conclusion [Dependent mark] (1) 3

(e) (i) Description:

• Keep volume of hot water constant at 100 ml (1)

• Volume of added water constant at 20 ml (1)

• Discard (20 ml) each time so that initial volume remains constant (1)

• (Measure ∆ θ ) by measuring θ i and θ f (1)

• for several values of θ i (1)

Any two sensible precautions (eg lagging/lid/largervolumes/check θ 0 (1) (1) (Max 5)

(ii) Axes labelled correctly (1)Straight line through origin (1)ORif ∆ θ against θ f straight line negative intercept (1)

(iii) k = gradient [Dependent mark] (1) Max 8[24]

Sample results

(a) θ o = 18.5 °C

θ i = 85.0 °C

θ f = 75.0 °C

∆ θ = 10.0°C

(b) k = 0– θθ

θf

∆ =

5.18–0.75

0.10

= 0.177

(c) Discard 20 cm3 into the wastebeaker

Add further 20 cm3 into the hot/warm waterθ i = 57.5 °C

θ f = 51.5 °C

∆ θ = 6.0 °C

k = 5.18–5.51

6= 0.182

(d) % difference = ( ) %100177.0182.02/1

177.0–182.0 ×+

=3.6%

Difference is less than 10% ∴ relationship is supported

(e) (i) • Keep volume of hot water constant at 100 ml

• Volume of added water constant at 20 ml

• Discard (20 ml) each time so that initial volume remains constant

• (Measure ∆ θ ) by measuring θ I and θ

• for several values of θ I

Any two sensible precautions: eg lagging/lid/larger volumes/check θ 0

(ii)

∆ θ

θ θ–f 0

(iii) k = gradient

438. Velocity of galaxyCalculation of 7 or 11nm, (1)Consistent values substituted in ∆ λ /λ ∆ λ must be 7 or 11 (1)

[Ignore 10X errors]

5.0 or 5.12 ×106 ms–1 ( consequent mark) (1)Moving away from the Earth/Milky Way/us/observer (1) 4

Estimation of distance of galaxy from Earth

Use of υ = Hd (1)

d = 2.8-2.9 × 1024 m [Allow e.c.f their υ above] (1) 2[6]

439. Explanation

Diffraction (1)Molecular/atomic separation ≅ 1nm/de Broglie wavelength (1) 2

Kinetic energy

Use of λ = h/mv (1)

Use of k.e. = 1/2mv2 OR p2/2m (1)

k.e. = 9.1-9.2 × 10–23 J [no ecf] (1) 3[5]

440. Description + diagram

Diagram to show:Microwave source/transmitter and detector (not microphone) (1)Transmitter pointing at metal plate/second transmitter from same source (1)Written work to include:Move detector perpendicular to plate/to and fro between /accept ruler on diagram (1)Maxima and minima detected/nodes and antinodes detected (1) 4[Experiments with sound or light or double slit 0/4]

Observation

In phase/constructive interference → maximum/antinode (1)Cancel out/out of phase/Antiphase/destructive interference → minimum /node (1) 2

How to measure wavelength of microwavesDistance between adjacent maxima/antinodes = λ /2 (1)Measure over a large number of antinodes or nodes (1) 2

[8]

441. Incident photon energies

Use of E = hf (1)

Use of c = f λ [ignore × 10X errors] (1)÷ e (1)For 320 nm E = 3.9 (eV) and 640 nm E = 1.9 (eV) (1) 4

Photocurrent readings

Work function of Al > 3.9 / energies of the incident photonsOR threshold frequency is greater than incident frequencies (1)

For Li (ϕ = 2.3 eV / f = 5.6 × 1014 Hz / λ = 540 nm hence) a photocurrentat 320 nm but not 640 nm (1)If intensity × 5 then photocurrent × 5 (1) 3

Stopping PotentialKEmax = 4.00/3.88 –2.30 = 1.7/1.58 [ignore anything with only e] (1)Vs = 1.7/1.58 V (1) 2

[9]

442. Gradient of graph

Gradient = 2.5 (1)

Unit s–2 or negative sign (1)Frequency

(2π f)2 = 2.5 [or above value] (1)f= 0.25 Hz [ecf ONLY for gradient error] (1) 4

Period

T = 4 s ecf their f (1) 1

Acceleration against time graph

Any sinusoidal curve over at least two cycles (1)

Negative sine curve (1)

y axis scale showing a = 20 (mm s–2) OR x axis scale showingT = 4(s) / their T 3

[8]

443. Speed of rim of drum

v = rω or v = 2π r/T [either used] (1)

ω = s60

min rev 8002 –1×π OR T = 1–min rev 800

s60 (1)

= 18.4 m s–1 [3 sf min.] (no ue) (1) 3

Acceleration

Use of a = rω 2 OR a = v2 /r (1)

1.5 × 103 ms–2 (1) 2

Addition of arrow and explanation

Arrow labelled A towards centre of drum (1)Push of drum on clothing/normal contact exerted by drum on clothing (1) 2[Normal reaction accepted]

Arrow of path

Arrow labelled B tangential to drum, from P, in anticlockwise direction (1) 1[8]

444. Wavelength and wave speed calculation

λ = 0.96 m (1)

seeing f = 2 their λ (f = 2.1 Hz) (1) 2

Qualitative description

(Coil) oscillates / vibrates (1)

With SHM / same frequency as wave (their value) (1)

Parallel to spring / direction of wave (1) 3[5]

445. Meaning of uniform magnetic field

Magnetic flux density constant / magnetic field lines parallel / (1)magnetic field strength is constant/ does not vary 1

Sizes and directions of forces on LM and NO

Force on LM: 2.4 × 10–4 N/ 0.24 mN (1)Direction:Downwards/into (paper) (1)

Force on NO: )

2.4 × 10–4 N / 0.24 mN [No unit penalty] ) Must have both (1) 3

Direction: )Upwards / out of (paper) )

Why no forces on MN and OL

Wires/current and B field directions are parallel [allow ‘samedirection’] / field due to current and B field of magnet areperpendicular to each other (1) 1

The effect on the square

A (turning) moment will be applied / it will (begin) to turn / spin / rotate (1) 1

Moving the pole pieces further apart

Reduces the size of the forces, (1)Because the flux density is reduced/ magnetic field (strength)reduced / B (field) reduced (1) 2

[8]

446. Charge on capacitor

220 µ F × 5 V [use of CV ignore powers of 10] (1)= 1100 µ C (1) 2

Energy on capacitor

2

220µ F × (5 V)2 /

2

1100µ C ×5 V /

Fµµ

2202

C 1100 22

× [ignore powers of 10] (1)

= 2750 µ J (2.8 × 10–3 J) (1) 2

Experiment

Method 1 (constant current method):

• Circuit (1)

• For a given V record time to charge capacitor at a constant rate (1)

• for a range of values of V (1)

• Use Q = It to calculate Q (1)

• Plot Q → V– straight line graph through origin / sketch graph /dive Q/V and obtain constant value (1)

Method 2:

• Circuit (1)

• For a given value V measure I and t (1)

• Plot I → t find area under graph Q (1)

• Repeat for a range of values of V (1)

• Plot Q → V for straight line graph through origin/ sketch graph /dive Q/V and obtain constant value (1)

Method 3 (joulemeter method):

• Circuit (1)

• Record V and energy stored (1)

• For range of V (1)

• Determine Q from ½ QV or C

Q

2

2

(1)

• Plot Q → V– straight line graph through origin / sketch graph /divide Q/V and obtain constant value (1) 5

[Coulombmeter (will not work with this value of capacitor)circuit (1) ; record charge Q on colombmeter (1); for a range of values of V (1); PlotQ → V for straight line through origin (1) – Max 3]

[9]

447. Lenz’s law

The direction of an induced current/emf/voltage is such as (1)to oppose the change (in flux) that produces it (1) 2

Polarity at top of coil

North (1)Direction of current

Only ONE arrow required (1) 2

Graph

Magnet is moving faster / accelerating (under gravity) (1)

(Rate of) change/ cutting of flux is greater (1)

Induced emf is greater (1) Max 2[6]

448. (a) (i) Straight wire, power supply, resistor and ammeter in series (1)Flat face of probe in same plane as wire/statement that probeis perpendicular to field line (1)Distance r shown correctly to centre of sensor (1)Probe alongside wire and approximately at mid-point (1)Meter connected to probe in some way (1)Half-metre rule perpendicular to wire (1) Max 4

(ii) Resistor limits current (prevents short circuit) (1)[Not reduces current]and so prevents damage/overheating (1) 2

(iii) Between 5 A and 20 A (1) 1

(iv) Several parallel wires (or long length of wire wound into a largerectangular coil, with the probe near vertical side of coil) (1) 1

[NOT coils or solenoids unless clear that only one limb is usedand the other limbs are far enough away not to have an effect.]

(b) Correct 1/r values to 2/3 sf and unit [here or on graph] (1)[Only penalise wrong or no unit for 1/r once]Graph:Scale: at least half paper in each direction, avoiding awkwardscales, e.g. 3s (1)Axes: labelled with units (1)Plots: accurate to ½ square (1)Line: thin, straight, best fit (1) 5

(c) Large ∆ (∆ x∆ y ≥ 100 cm2) (1)Correct calculation of gradient to ≥ 2 sf [ignore unit] (1)Correct calculation of I to 2/3 sf + unit (1) 3

[16]

Sample results

(a) (i)

A

2 . 2 Ω

+

–

0 – 1 2 VP o w e rs u p p l y

V

H a l ls e n s o r

H a l lp r o b ep o w e rs u p p l y

m e t e r t om e a s u r eo u t p u t o fH a l l p r o b e

W i r e

r

(ii) Resistor limits current (prevents short circuit)and so prevents damage/overheating

(iii) Between 5 A and 20 A

(iv) Several parallel wires [or long length of wire wound into alarge rectangular coil, with the probe near vertical side of coil]

(b) Plot B against 1/r

B/µ T r/m

r

1/m–1

93 0.020 50.074 0.025 40.054 0.035 28.638 0.050 20.019 0.100 10.0

(c) Gradient = 0.0–0.50

0.1–5.92 = 1.83 µ T m

Gradient = π

µ2

0 I → I =

0

gradient2

µπ ×

I = 7–

6–

104

1083.12

×××

ππ

= 9.2 A

Graph:

1 0 0

9 0

8 0

7 0

6 0

5 0

4 0

3 0

2 0

1 0

0 1 0 2 0 3 0 4 0 5 0 1r m – 1

B / Tµ

G R A P HS c a l e :

A x e s :P l o t s :L i n e :

a t l e a s t h a l f p a p e r i n e a c h d i r e c t i o n ,a v o i d i n g a m b i g u o u s s c a l e s , e g . 3 ’ s .

l a b e l l e d , w i t h u n i t s a c c u r a t e t o ½ s q u a r e

t h i n , s t r a i g h t , b e s t f i t

( 1 )( 1 )( 1 )( 1 )

449. (a) (i) Solenoid, power supply and ammeter in series (1)Hall probe sensor shown at centre of solenoid (1)Flat face of probe perpendicular to axis of solenoid (1)[Details of Hall probe circuit are not required]Length of solenoid (1)Number of turns (1) 5

(ii) Sensor shown near mid-axis of magnet (1)Correct orientation of flat face of sensor [ecf] (1)d shown correctly from axis of magnet to centre of sensor (1) 3

(b) Correct expansion [Need not be related to y = mx + c] (1)Correct ln values to 2 or 3 decimal places (1)Units [here OR on graph] (1)Graph:Scale – at least ½ paper in each direction, avoiding awkward scales,e.g. 3s, AND axes – labelled [ignore units here] (1)Plots – accurate to ½ square (1)Line – thin, straight, best fit (1) 6

(c) Large ∆ (∆ x∆ y ≥ 80 cm2) [Allow if a tangent drawn to a curve] (1)Correct calculation giving n in range –2.05 to –2.11 (1) 2

[16]

Sample results

(a) (i)

A x i s o fs o l e n o i d

V a r i a b l ed . c . p o w e rs u p p l y

A+ –

Length of solenoid

Number of turns

(ii)

N S

d

S e n s o r

M a g n e t

(b) lnB = nlnd + lnky = mx + c

B/mT d/mm ln(b/mT) ln(d/mm)

5.26 20 1.66 0.721 3.00 1.30

3.31 25 1.20 0.520 3.22 1.40

2.27 30 0.82 0.356 3.40 1.48

1.67 35 0.51 0.223 3.56 1.54

1.24 40 0.22 0.093 3.69 1.60

1.04 44 0.04 0.017 3.78 1.64

Graph:

l n ( / m T )B

1 . 8

1 . 6

1 . 4

1 . 2

1 . 0

0 . 8

0 . 6

0 . 4

0 . 2

03 . 1 3 . 2 3 . 3 3 . 4 3 . 5 3 . 6 3 . 7 3 . 8

l n ( / m m )d

(c) Gradient = – 00.3–80.3

00.0–67.1

n = –2.08

450. Show that average daily capacity provides about 2 × 10 13 J

Ep = mgh (1)

= (28 × 106 m3 × 103 kg m–3) × 9.81 N kg–1 × 64.5 m

= 1.8 × 1013 J [no up] (1) 2

Calculation of efficiency over one year

Efficiency = (useful energy output/total energy input) × 100%

6.1 × 1015 J (1)

÷ 365 × 1.77 × 1013 J (1)× 100%

= 94.4 % [Accept fractional answers. Allow use of 2 × 1013 J, whichgives 83.6%, or ecf, but check nos.] (1) 3

Calculation of average power output over year

P = W/t (1)

= 6.1 × 1015 J ÷ 3.16 × 107 s

= 1.9 × 108 W (1) 2

Reason for difference from max power output

Any sensible reason, e.g., river flow varies over the year / variationsin rainfall [Accept answers related to demand] (1) 1

[8]

451. Explanation of emission of radiation by hydrogen atoms

Electrons excited to higher energy levels (1)

as they fall they emit photons / radiation (1) 2

[Accept 21 cm line arises from ground state electron changing spinorientation (1) / relative to proton (1)]

Why radiation is at specific frequencies

Photon frequency related to energy / E = hf (1)

Energy of photon = energy difference between levels / hf = E1 – E2 (1)

Energy levels discrete/quantised / only certain energy differences possible (1) 3

Show that hydrogen frequency corresponds to λ = 21 cm

f = 4.4623 × 109 ÷ π= 1.42 × 109 Hz (1)

c = f λ

λ = 3 × 108 ÷ (1.42 × 109 Hz) (1)

λ = 0.211 m or 21.1 cm [no up] (1) 3[8]

452. Explanation of assumption that voltmeter does not affect values

Voltmeter has very high resistance/takes very small current (1) 1

Current through X

4.8 A ÷ 6 = 0.8 A

OR 48 V ÷ 60 Ω = 0.8A (1) 1

Value missing from E7

P =IV

P = 4.4 A × 53 V = 233 W (1) 1

Description of appearance of lamp X as lamps switched on

Gets dimmer

from table, voltage decreasing / current in X decreasing / power per lamp decreasing (1)

So P decreases (1) 3

Formula for cell C6

I = ε / Rtot (1)

I = 120 / (15 + B6) (1) 2

Effect of internal resistance on power

Power has a maximum value (1)

when external resistance = internal resistance (1) 2[10]

453. Show that lift is about 14 700 N

Lift = weight = mg

= 1500 kg × 9.81 N kg–1

= 14 700 N (1) 1

Explanation of why vertical component equals weight

No vertical acceleration / resultant vertical force = zero / verticalforces balanced (1) 1

Show that horizontal component is about 4500 N

Horizontal component = Fsinθ

OR

= 15 400 N × sin 17° OR 15 400 N × cos (90° – 17°) (1)= 4503 N [no up] (1) 2

Calculation of forward acceleration

a=F / m (1)

= 4503 N ÷ 1500 kg

= 3.0 m s–2 (1) 2

Calculation of distance travelled after 10 s

s = ut + ½ at2

= 0 + ½ × 3.0 m s–2 × (10 s)2 [e.c.f.] (1)

= 150 m (1) 2

Explanation of whether likely to be actual distance

Distance likely to be less (1)

Air resistance / drag will decrease resultant force / acceleration (1) 2[10]

454. Fundamental frequency of note

440 Hz (1) 1

Frequencies of first three overtones

880 Hz

1320 Hz

1760 Hz

Two correct frequencies (1)

Third correct frequency (1) 2

Comment on the pattern

Any 2 from the following:

[Allow ecf]

880 Hz = 2 × 440 Hz

1320 Hz = 3 × 440 Hz

1760 Hz = 4 × 440 Hz

1760 Hz = 2 × 880 Hz (1) (1) 2

[OR They are multiples (1) of the fundamental (or similar qualification) (1)]

[Allow 1 mark for amplitude decreasing with frequency]

Measurement of period

Example: 7 cycles takes (0.841 – 0.825) s [at least 5 cycles] (1)

Period = 0.016 s ÷ 7

= 2.3 × 10–3 s [in range 2.2 × 10–3 s to 2.4 × 10–3 s] (1) 2

Calculation of frequency

f = 1/T (1)

= 1 ÷ 2.2 × 10–3 s [Allow ecf]= 454 Hz (1) 2

[9]

455. Mark on diagram

Correctly drawn normal (1)

Correctly labelled angles to candidate’s normal (1) 2

Show that refractive index of water is about 1.3

Angles correctly measured:

i = 53 (± 2)°

r = 39 (± 2)° (1)

µ = sin i / sin r = sin 53° / 39°

= 1.27 [Allow ecf] [Should be to 2 d.p. min] (1) 2

Critical angle

µ = 1/sinC (1)

so sin C = 1/1.27 so C = 52° [ecf] (1) 2[use of 1.3 gives 50°]

Explanation of reflection of ray

Internal angle of incidence = 39° ± 1° (1)

Compare i with critical angle (1)

Valid conclusion as to internal reflection being total/partial (1) 3

Refractive index

It varies with colour (1) 1[10]

456. Weight

mg

= 70 × 9.81

= 690 N (1) 1

Meaning of upthrust

There is an upward force (1)

in a fluid / equal to weight of air displaced (1) 2

Upthrust in newtons

Upthrust = mass of air displaced × g

= volume of air displaced × density of air × g (1)

= V × 1.29 × 9.81

= 12.65V (1) 2

Weight of helium

Volume × density × g

= 0. 18 V g (= 1.77 V) (1) 1

Total volume of balloons

Upthrust = weight of man + weight of helium (1)

12.65V= 690 + 0.18Vg (1)

10.88V= 690

V= 63 m3 [Allow ecf] (1) 3

Why viscous force can be ignored

Any two from:

• Quote of 6π η rυ

• υ is small

η is small (1) (1) 2[11]

457. Explanations

(i) Refraction:e.g. bending of wave when travelling from one medium to another [OR change of speed] (1)

(ii) Diffraction:e.g. spreading of wave when it goes through a gap (1) 2

Diagram of wavefronts near beach

Gradual bend in wavefronts (1)

Smaller wavelengths (1)

Waves bending upwards as they approach shore (1) 3

Diagram of wavefronts in bay

Constant wavelength (1)

Waves curve (1) 2

Explanation

Refraction/diffraction causes waves to bend towards the beach (1) 1[8]

458. Nuclear radiation which is around us

Background (1) 1

Source of radiation

e.g. Sun / rock (eg granite) / cosmic rays [not space] / nuclear power stations (1) 1

Why exposure greater today

Nuclear power stations/nuclear bomb tests/X–rays/Radon from building materials (1) 1

Beta radiation

(i) Any two from:

• γ more difficult to shield

• β lower range (than γ )

• β more ionising (than γ ) (1) (1)

(ii) α stopped by a few cm of air or has a short range/muchlower range (than β ) / β radiation has a long range (1) 3

Why gamma radiation is suitable

Any two from:

· γ will pass through (metal of) wing / α and β cannot pass through the wing

· but passes more easily through cracks

· hence crack shows as darker mark on photo or increased counton detector (1) (1) 2

[8]

459. Meaning of terms

(i) HardDifficult to scratch or dent (1)

(ii) Brittle:Breaks without plastic deformation / shatters / cracks (1)

(iii) Tough:Can withstand deformation/dynamic loads/shock/impact (1)Plastic (1) 4

How electrons can be used to examine arrangement of atoms

Any three from:

• electrons have wave properties

• beam of electrons directed at the specimen

• electrons diffract by spaces between atoms

• superposing constructively / series of dots seen

• pattern can be used to determine arrangement of atoms

• no (or irregular) pattern indicates an amorphous structure

• pattern seen on a fluorescent screen as electrons hit it (1) (1) (1) 3[7]

460. Measurement needed

Any three from:

• Resistance

• Distance between probes

• Effective area/cross sectional area

• R = ρ A

L (1) (1) (1) 3

Equation of line A

Intercept = –3.5 (Ω m) (+/– 0.3) (1)

Gradient = 1.5 (Ω mm–1) (+/– 0.05) (1)

So equation is ρ = 1.5 d – 3.5 [Or equivalent, e.c.f. allowed] (1) 3

Addition of line

Points correctly plotted (–1 for each error, allow ½ square tolerance) (1) (1)

Line of best fit drawn (1) 3

Best distance

Between 1.90 and 1.99 km (1) 1[10]

461. Ultrasound

Ultrasound is very high frequency sound (1)

How ultrasound can be used

Any three from:

• gel between probe and body

• ultrasound reflects

• from boundaries between different density materials

• time taken to reflect gives depth of boundary

• probe moved around to give extended picture

• size of reflection gives information on density different (1) (1) (1) 3

How reflected ultrasound provides information about heart

Any two from:

• Doppler effect

• frequency changes

• when reflected from a moving surface

• gives speed of heart wall

• gives heart rate (1) (1) 2[6]

462. Explanation of what has happened in circuit

Charging process (1)

Plates oppositely charged OR charge moves from one plate to another (1)

Charge flows anticlockwise OR electrons flow clockwise OR leftplate becomes positive OR right plate becomes negative (1)

Build up of Q/V reduces flow rate (1) Max 3

Explanation of what would have been seen

Same as ammeter 1 (1)

Reason: Same I everywhere OR series circuit OR same I/Q in eachcomponent (1) 2

Estimate of charge

Attempt to find area under correct region of graph (1)

= 52 µ C (1) 2

[Allow 45 – 65 µ C]

Estimate of capacitance

p.d. across resistor at t = 10 s = 100 × 103 Ω × 3 × 10–6 A = 0.3 V (1)

(hence p.d. across capacitor = 1.5 V – 0.3 V = 1.2 V)

C = V

Q =

V2.1

C105 5–× (equation or sub) [ecf] (1)

C = 42 µ F [If 1.5 V is used to obtain C = 33 µ F, then 2/3] (1) 3

Alternative method using e –t/RC

Correct answer appropriate to set of values (1)

Correct ln line (1)

Correct answer (40–44uF) (1)

Alternative method using T = RC

Using T = RC (1)

Appropriate T value (1)

⇒ correct answer (1)

Observations

Same picture as before (1)

since same ∆ V (1) 2

[OR C now carries twice the previous charge][12]

463. Meaning of E

Voltage/e.m.f. (1)

Induced/caused/created (when magnetic field/flux changes) (1) 2

Definition of φ

φ = BA (1)

(Magnetic) flux OR magnetic field lines (1) 2

Additions to diagram of paths of currents

Joined up, and one each side and wholly on disk (1) 1

Explanations

(i) Disc cuts B/φ [relative motion implied] (⇒ V/I induced) (1)

(ii) B + I [or two interacting magnetic fields] ⇒ force (1)

Lenz or LH rule ( ⇒ opposing force) OR energy argument (1) 3

Explanation of reasoning

Mention of F = BIl (1)

B ∝ Ib (1)

I (in disc) ∝ B/φ (1) 3[11]

464. Addition to diagram

2 lines or more, vertical (1)

Arrow downwards (1) 2

N i o b i u m s p h e r e

T o p p l a t e

B o t t o m p l a t e

Electric field strength

E = V/d OR 2000 ÷ (0.8 × 10–2) (1)

= 250 000 V m–1 OR N C–1 [OR 2500 V cm–1] (1) 2

(2.5 × 105)

Magnitude of charge

F = ma OR 1.8 × 10–7 × 3.0 × 10–7 (N) OR 5.4 × 10–14 (N) (1)

(F = Eq ⇒) q = F/E OR 5.4 × 10–14/2.5 × 105 (C) (1)

= – 2.16 (2, 2.2) × 10–19 C (1) (1) 4

[NB 1 mark for –, 1 mark for rest of answer]

Why a vacuum

Air/gas/molecules would alter acceleration OR provide anotherforce OR collide with niobium sphere (1) 1

[9]

465. Comparison of positron with electron

Same mass (1)

Opposite charge (1) 2

Minimum energy

Use of E = mc2 (1)

= 2 × 9.11 × 10–31 × (3 × 108)2 J

= 1.6398 × 10–13 J (1)

[Factor 2 omitted: lose second tick]

= 1.6398 × 10–13/1.6 × 10–19 (× 106) MeV= 1.02 MeV (1) 3

How process releases energy

Annihilation (1) 1

Any two from:

• ⇒ em radiation/photon(s)

• 2 photons

• 0.51 MeV each (1) (1) Max 2[8]

466. Conditions for simple harmonic motion

Acceleration OR restoring force ∝ displacement (1)

in opposite direction / towards equilibrium position (1) 2

Why child’s motion only approximately simple harmonic

Any one from:

• damped / friction opposes motion / air resistance

• swing’s path is arc of circle, not straight line

• angle too large (1) 1

Calculations

(i) Period of the motion:

T= 20 s/6 = 3.3 s (1)

f = 1/T= 0.30 Hz (allow e.c.f. for T) (1)

(ii) Value of ωUse of ω = 2π f

= 2π × 0.30

= 1.9 rad s–1 (allow e.c.f. for f, no repeat unit error) (1)

(iii) Child’s acceleration:

amax = –ω 2 A (1)

= –1.92 × 1.2 (allow e.c.f for ω ) (1)

= (–) 4.3 ms–2

Swing an example of resonance

Push (driver) at same frequency as swing (driven) (1)

causes increase of amplitude / energy transfer (1) 2[10]

467. Formula

F = GMm/r2 (1) 1

Show that tan θ = MR 2 / M e r 2

Horizontally Tsinθ = Fmountain and vertically Tcosθ = mg (1)

[OR vector diagram showing forces and θ ]

Dividing equations [OR from vector diag.]: tanθ = Fmountain ÷ mg (1)

and Fmountain = GMm ÷ r2 and mg = GMem ÷ R 2 (1)

so tan θ = GMm/r2 ÷ GMem/R2 (1) 4

Value for gravitational constant, G

Volume = 4/3π R3 = 4/3π (6.4 × 106 m)3 = 1.1 × 1021 (m3) (1)

Me = Vρ = 1.1 × 1021 m3 × 4.5 × 103 kg m–3 = 4.9 × 1024 (kg) (1)

G = eM

gR 2

= ρπ 3

2

4

3

R

gR = ρπR

g

4

3 = 3–36

2–

mkg105.4m104.64

sm8.93

×××××

π

= 8.1 × 10–11 N m2 kg –2 (1) 3

Reason for inaccuracy

Any one from:

• Maskelyne’s density is incorrect

• Earth/mountain not uniform density

• (centre of) mass of mountain not known

• mountain is not spherical

• difficult to determine vertical / measure very small θ

• Earth not a perfect sphere / point mass (1) 1

Earth’s core

Much denser than mountain (1) 1[10]

468. Plutonium-238

238 protons + neutrons [OR nucleons] in the (nucleus of the) atom (1) 1

Why plutonium source caused concern

If accident at launch, radioactive Pu would be spread around Earth (1) 1

Activity of plutonium source

λ = ln2/88 × 3.16 × 107 s = 2.5 × 10–10 (s–1) (1)

Use of dN/dt = –λ N (1)

= 2.5 × 10–10 s–1 × 7.2 × 1025 = 1.8 × 1016 (Bq) (1) 3

Power delivered by plutonium

Use of power = activity × energy per decay (1)

= 1.79 × 1016 Bq × 5.6 × 106 × 1.6 × 10–19 s

[conversion of MeV to J] (1)

= 1.6 × 104 (W) (1) 3

[2 × 1016 Bq gives 1.79 × 104 (W)]

Whether power can be relied upon

Large number of nuclei present, so decay rate (almost) constant (1) 1

Percentage of power still available after 10 years

Percentage = N/N0 × 100 = 100 e–λ t (1)

= 100 e –10 × ln2/88 = 92% (1) 2

[After 10 y, N = N0 e–λ t = 7.2 × 1025 × 0.92 = 6.65 × 1025 (1)]

Why plutonium was chosen for Cassini mission

Examples:

• long (enough) half–life for duration of mission

• Power constant / no orientation problems compared with solar

• α -emitting, so energy from particles easily transferred

• availability (1) 1[12]

469. Physics principles

Requires 9 V battery:

Battery required for electronic circuitry / microphone / speaker (1)

Rubberized foam ear cups:

Air filled material / material has large surface area (1)

Air molecules collide frequently with material (1)

Foam deforms plastically/collisions are inelastic (1)

Sound converted to heat in material (1)

Active noise attenuation:

Noise picked up by microphone (1)

Feedback signal inverted / 180° out of phase with noise / antiphase (1)

Amplified [OR amplitude adjusted] and fed to earphones / speaker (1)

Sound generated cancels/superimposes/minimum noise (1)

Diagrams of superposing waves showing (approx.) cancellation (1)

Amplifier gain automatically adjusted if noise remains (1)

Device only works over frequency range 20 – 800 Hz (1) Max 6

Where does the energy go?

Some places will have constructive interference (1)

More intense noise (1)

Some noise dissipated as heat in air / foam (1)

increased kinetic energy of air [OR foam] molecules (1) Max 2[8]

470. Explanation of processes involved

• induced / induction (e.m.f).

• Faraday’s law or E = (–) d (Nφ )/t

• I0 is cutting flux

OR

• moving electrons (in I0)

• Force on them due to magnetic field / F – Bqυ

• Pushes them to one side (1) (1) (1)

• (Gases conduct current) if there are charged particles

• (Massive current) if low resistance/large e.m.f.

• E/V = IR (1) (1) (1)

• inelastic (collisions)

• electrons in atoms/molecules excited/change energy levels/diagram

• emit photons/hf

• different ∆ E = different f/colour (1) (1) (1) (1) Max 7[7]

471. (i) Distance travelled

Attempt to find area under curve/use of suitable equations (1)

Distance = 300 m (1)

(ii) Averape speed

Use of total distance/20 (1)

Average speed = 15 m s–1 [e.c.f. distance above] (1)[4]

472. Average deceleration

Select υ 2 = u2 + 2ax, ½ m υ 2 = Fx and F = ma OR equations of motion (1)

Correct substitutions of 40 m and 25 m s–1 (1)

a = 7.8 m s–2 [If a = –7.8 m s–2 → 2/3] (1) 3

Depth of sand and stopping distance

More sand ⇒ shorter stopping distance/stops more quickly/slowsdown faster Because lorry sinks further/ bigger resistingforce / bigger friction force (1) 1

[4]

473. Alpha particle scattering experiment


Most alpha went straight through/deflected very little (1)

A tiny minority were deflected through large angles / > 90° (1)

Atom had a dense/massive nucleus (1)

Most of the atom was empty space/small nucleus (1)[5]

474. Resultant force

4 N to the right / 4 N with correct arrow (1) 1

Motion of object

(i) Constant velocity / a = 0 / constant speed (1)

(ii) Accelerates upwards (1)

(iii) Slows down (1) 3

Student’s argument

The forces act on different bodies (1)

Therefore cannot cancel out / there is only one force acting on thebody [consequent] 2

[6]

475. Graph

Sensible scale + point (0, 192) plotted (1)

Rest of points [ –1 mark for each misplot] (1) (1) 3

[(1,96); (2, 48); (4, 12)]

[Accept bar chart]

Random process

Cannot predict which nuclei will decay/when a particular nucleuswill decay (1) 1

Model

Cannot predict which children will flip a head/which coins will beheads/when a particular coin /child will flip a head (1) 1

Half-life

Time taken for activity/count rate to drop by half/time taken for halfthe atoms/nuclei to decay (1) 1

How model illustrates half-life

Yes, if children were told to flip coin at regular time intervalORYes, because about half of the children flipped a head each timeORNo, because time is not part of the experiment (1) 1

[7]

476. Mass of head of malletSelecting density x volume (1)Correct substitutions (1)Mass = 1.15 (kg) [3 significant figures, minimum] (1) 3

Momentum change

p = mυ used (1)

∆ p = 1.15 or 1.2 kg (4.20 + 0.58) m s–1 (1)

= 5.50 / 5.74 kg m s–1/N s (1) 3

Average force

Their above / 0.012 s (1)

F = 458/478 N [e.c.f. ∆ p above] (1) 2

Value for force

Handle mass/weight/ head weight/force exerted by user (handle)neglected (1) 1

Effectiveness of mallet with rubber head

∆ t goes up/∆ p goes up (1)

⇒ less force, less effective/more force, more effective [consequent] (1) 2[11]

477. Vehicle movement

mgh and ½ mυ 2 [Both required] / mgh and mgh / ½ mυ 2 and ½ mυ 2 (1) 1

Expression for speed

Kinetic energy gained = gravitational potential energy lost /

mgh = ½ mυ 2 (1)

υ = ( )gh2 (1) 2

Assumption

No friction/air resistance/rolling (1) 1

Explanation

Yes, because C is lower than A / potential energy is lower at C thanat A (1)Yes so it will still have some kinetic energy at C (1)No because:Frictional forces do act to slow the vehicle (1)even though C is lower than A the vehicle has insufficient kineticenergy to reach C (1) 2

[6]

478. Resistance calculationsEvidence of 20 Ω for one arm (1)

20

1

20

11 +=R

(1)

R = 10 Ω (1) 3

Comment

This combination used instead of a single 10 Ω resistor [or samevalue as before] (1)

because a smaller current flows through each resistor/reduce heatingin any one resistor/average out errors in individual resistors (1) 2

[5]

479. Graphs

Diode:

RH quadrant: any curve through origin (1)Graph correct relative to labelled axes (1)LH side: any horizontal line close to axes (1) 3

I

V

L i n e o n o r c l o s e t ov o l t a g e a x i s

Filament lamp

I

V

RH quadrant:Any curve through origin (1)Curve correct relative to axes (1)LH quadrant:Curve correct relative to RH quadrant (1) 3[Ohmic conductor scores 0/3]

[6]

480. CircuitAmmeters and two resistors in series (1) 1[1 mark circuit penalty for line through cell or resistor]Cell e.m.f

E= 150 x 10–6 (A) x 40 x 103 (Ω ) total R (1)Powers of 10 (1) 2E = 6.0 (V)

New circuit

Voltmeter in parallel with 25 (kΩ ) resistor (1) 1Resistance of voltmeter

(Total resistance) = A)(10 170

V)(66–×

= (35.3 kΩ )

(Resistance of ll combination) = 35 – 15 kΩ= (20 Ω ) [e.c.f. their total resistance]

VR

1

25

1

20

1 +=

100

4–51 =VR

RV = 100 kΩ [108 kΩ if RT calculated correctly]

(1)

(1)

(1)

Alternative route 1: 3

p.d. across 15 kΩ = 2.55 V(∴ p.d. across ll combination = 3.45 V)resistance combination = 20 kΩ→ RV = 100 kΩ

(1)

(1)

(1)

[7]

Alternative route 2: 3

p.d. across parallel combination = 3.45 VI through 25 kΩ = 138 µ A→ RV = 100 kΩ

(1)

(1)

(1)

481. Resistance of strain gauge

State R = A

lρ (1)

Use of formula (1)x 6 (1)R = 0.13 Ω [ecf their l] (1) 4

Ω=Ω×=

××××Ω×==

13.0

106.129

m101.1

6m104.2m109.9ρ

3–

27–

2–8–

R

A

lR

Change in resistance

∆ R = 0.13 Ω × 0.001

∆ R = 1.3 × 10–4 (Ω ) [no e.c.f.]OR∆ R = 0.02 × 0.001

∆ R = 2.0 × 10–5 Ω

0.1% → 0.001 (1)Correct number for ∆ R (1) 2

Drift velocity

Stretching causes R to increase (1)Any two from:• Current will decrease• I = nAυ Q• Drift velocity υ decreases• nAe constant (1) (1) 3

[9]

[For R decreasing, max 1:Any one from:• I will increase• I = nAυ Q• υ will increase• nAe constant]

482. Relationship

Interpretation of line passing through origin i.e. direct proportionality ORinverse proportionality (1)between appropriate quantities (1) 2

[Comment that: as p↑, V↓ / as p↑, 1/ V↓ / p times V is constant,scores 1 mark only]

Sketch graph

A line with negative gradient (1)

Concave curve [must not touch either axis or continue parallel to axes] (1) 2

Units of pV

Nm–2 × m3 [or correct more complex version] (1)Nm = J (1) 2

Explanation

Quality of written communication (1) 1Any four from the following:• molecules collide with walls of container• molecules undergo a change of direction/momentum• force is rate of change of momentum

• pressure = area

force (1) (1) (1)

• large number of molecules hence pressure same/constant (1) Max 4[11]

483. DiagramTo include

• lagging

• clock or top pan balance

• variable supply/rheostat + supply OR joulemeter + supply

• V and A correct OR joulemeter parallel to supply (1) (1) (1) Max 3

Measurements

Mass of aluminium (m) (1)

Initial and final temperature (θ 1, θ 2) (1) 2

EITHER

Current (I) and p.d. (V) (1)Time (t) that current flows (1)

OR

Initial joulemeter reading (1)

Final joulemeter reading (1) 2

Use of measurements

EITHER

Find temperature rise

Rearranged equation C = θ∆m

VIt OR equivalent (1)

OR

Plot graph θ → t (1)

C = gradient m

VI (1)

Any one assumption (1)

Assume no heat losses to surroundings OR heater completely within block

OR heater 100% efficient

OR Good thermal contact between heater and block OR temperature of block uniform throughout OR stop/start time of clock and heater are the same

[10]

484. (a) Stefan-Boltzmann lawT = absolute or Kelvin temperature/K (1)[Not °K or k]of surface (1)Unit luminosity: watt/W (1) 3

(b) GraphState Wien’s law OR see evidence of λ max × T (1)

at two different points to give same product [Ignore 10–6] (1)

More than two points and allow 2.8 – 3.0 × 10–3 (mK) (1) 3[No ue]

Surface temperature of star

Temperature = 7300 [7000 to 7500] [No ue] (1) 1Luminosity calculation

Use of T4 [7300 K4] [e.c.f] (1)

A = 4π r2 substitution correct (1)

(5.67 ×10–8 W m–2 K–4) σ T4A used [ecf any A,T] (1)

= (1.4 → 1.8) ×1025 (W) [No ecf] (1) 4(7000 K) (7500K) [No ue]Matter consumed

∆ E = c2 ∆ m (1)

Substitute their L and 9 × 1016 m2 s–2 [Beware ∆ E = c∆ m used] (1)

Mass = 1.8 × 108 (kg s–1) [No ue] [ecf] [Range: 1.5 – 2.0 ×108 allowed] (1) 3[14]

485. (a) Select wordsCool (1)High (1)Surface area (1)Off the main sequence (1) 4

Hertzsprung-Russell diagram

(i) Temperature scale “←“ (1)

Values in range (20 000 K to 60 000 K) and (2000 Kto 4000 K) and non-linear showing at least 3 values (1)[e.c.f. scale in wrong direction]

(ii) Level with 10° marked Xs [Check with ruler if unsure] (1)

(iii) Region below MS marked W (1)

(iv) Region above MS marked R (1) 5

(b) Recognition of supernova

Extremely bright star [not “explosion”] (1)Suddenly appearing/time reference (1) 2

How supernova is formed

Any three from:

• quality of written communication

• when star collapses/implodes

• shock wave / explosion blows outer layers away [both needed]

• (H) fusion ceases/other fusion begins

• protons combine with electrons to form neutrons (1) (1) (1) Max 3[14]

486. Expression

Energy density = joule/m3 (1)

Stress = N/m2 (1)

Strain = m/m OR no unit stated (1)

J = N m / kg m2 s–2 (1) 4[4]

487. Hooke’s law

Tension/force proportional to extensionOR formula with symbols defined (1)

Up to a certain limit/limit of proportionality (1) 2

[Accept elastic limit]

Calculation of Young, modulus of brass

Stress = 34 N/1.5 × 10–7 m2 OR E= lA

Fl

∆ used (1)

Strain = 5.3 × 10–3 m/2.8 m [ie substitution]

[ignore 10n] (1)

Young modulus = 1.2 × 1011 [No ecf] (1)

Pa / N m–2 [Not kg m–1 s–2] (1) 4

Graph

[Mark alongside]

Straight line from origin to 46 N (1)

going through 34 N, 5.3 mm (1) 2

Energy stored

By finding area/area shaded/ ½ Fx up to 24 N (1) 1

Second wire

Less energy stored (1)

Less extension (1)

Smaller area under graph OR smaller ½ Fx (1) 3[12]

488. Homogeneity

p = mass × velocity (1)

p units N s or kg m s–1 [This alone implies above mark] (1)

E unit (J) N m or kg m 2 s–2 (1)

c unit m s–1 (1) 4[4]

489. Classification of particles

Ξ – is a baryon (1)Λ is a baryon (1)

π – is a meson (1) 3[Allow bbm]

Charge of strange quark

Show that –1 = –1/3(d) + –1/3(s) + –1/3 (s) (1) 1Λ particle

Λ is neutral (1)+2/3 + –1/3 + –1/3 = 0 and udsOR charge conservation (–1) = 0 + (–1) (1) 2

[6]

490. (a) (i) Initial level of water below height of plasticeneORSensible trial and error shownORBeaker used to collect correct volume of waterORPlasticene removed with pencil explained (1)Plasticene fully immersed (1)Correct volume by difference method, with unit (1)

[Ignore incorrect conversion to m3]

[Accept readings to 0.5 cm3]Two precautions given (1) (1)Correct calculation, 2/3 significant figures + unit (1)

Value 1.6 – 2.0 g cm–3 (1) 7[OR ± 0.2 on centre value]

(ii) Sensible ∆ V (1)

[1 cm 3 or 2 cm3 or ½ range or range of values]Correct calculation of percentage from sensible ∆ V (1)Correct calculation of percentage difference with 1800 (1.8)as denominator (1)Sensible comment based on their percentage difference andthe manufacturer’s specification [±10%] (1) 4

(b) (i) Circuit set up correctly without help (1) (1) 2[Ignore polarity errors]

(ii) VAC ≈ 3 V to 0.1 V or better, with unit (1)VBC < VAC to 0.1 V or better, with unit (1)[Use unit penalty once only]Correct difference calculated with unit (1)Correct calculation to more than 2 significant figures + unit (1)Correct calculation to more than 2 significant figures + unit (1) 5

(iii) VBC > VBC in part (ii) and > 2 V and to 0.1 V with unit (1)

[Allow VBC ≈ VAC]VBC increases because (1)resistance of LDR increases (1)as light intensity reduced/when LDR is covered (1)

[dependent on previous mark because Ω=

k

R

V

V LDR

AB

BC

1 (1)]

so share of voltage increases OR circuit current reduces sop.d. (1)across 1 kΩ reduces ∴ p.d. across LDR increases (1) 5

Correct calculation of RLDR with unit, using ratios OR

current method and ≥ 2 significant figures (1) 1[24]

Sample results:

(a) (i) m = 75.0 g

A l l v o l u m e s m u s t b e m e a s u r e d b y m e a s u r i n g c y l i n d e r ,e l s e 0 / 2

Volume of water and Plasticene = 95 cm3

Volume of water = 53 cm3

Volume of Plasticene = 42 cm3

Any two precautions from:Tilt the measuring cylinder/lower the Plasticene carefully to avoid splashingEye level with meniscus/when taking surface of water readingsRepeat readings seenCarefully exclude air when shapingTap the measuring cylinder to remove air bubbles

Density = 42

75 = 1.79 g cm–3

(ii) ∆ V = 1 cm3

Percentage uncertainty = 1÷ 42 ×100 = 2.4%

Percentage difference = 8.1

01.0 × 100

= 0.6%Percentage difference is smaller than 10% ∴ well within manufacturer’s specification

(b) (i) Ignore polarity errors

(ii) VAC = 3.08 VVBC = 0.95 VVAB = 3.08 – 0.95 = 2.13 V

I = R

V =

1000

13.2 = 2.13 10–3 A

R = I

V = 3–1013.2

95.0

× = 446Ω

OR

AB

BC

V

V =

1000

R

I = 10001 +

=Ω

=R

V

k

V

R

V ACABBC

(iii) VBC = 2.68 VVBC increases because:resistance of LDR increases as light intensity reduces/when the LDRis covered, so either share of voltage increases or circuit current reducesso p.d. across 1 kΩ reduces, therefore p.d. across LDR increases.

AC

BC

V

V =

LDR

LDR

R

R

+0.1

LDR

LDR

R

R

+==

0.187.0

08.3

68.2

∴ 0.87 + 0.87 RLDR = RLDR

0.13 RLDR = 0.87

RLDR = 6.7 kΩ

491. ExplanationClarity of written communication (1)Wave reflects off bench (1)(Incident and reflected) waves superpose/stationary wave is formed (1)Maxima or antinodes where waves in phase or constructiveinterference occurs (1)Minima or nodes where waves exactly out of phase or destructiveinterference occurs (1) 5

Speed of soundSee a value between 5.0 and 5.6 (cm) (1)Use of υ = fλ (1)λ = 2 × spacing (1)

320 m s–1 to 360 m s–1 (1) 4

Explanation of contrastAs height increases, incident wave gets stronger, reflected wave weaker (1)So cancellation is less effective [consequent mark] (1) 2

[11]

492. Magnitude of F

F = mυ 2 /r (1)Towards the centre (1) 2

Calculation

(i) 9.07 × 10 3 N (1)

(ii) R = mg – mυ 2 / r (1)Substitutions (1)

5.37 × 103 N

[Calculation of mυ 2 / r max 1] (1) 4

Explanation

Required centripetal force > mg (so cannot be provided) (1) 1

Critical speed

Use of (m)g = (m)υ 2 / r (1)

15.7 m s–1 (1) 2

Apparently weightlessThis means no force exerted on/by surroundings OR R = 0 OR onlyforce acting is weight (1)When car takes off it is in free fall [consequent] (1) 2

[11]

493. Simple harmonic motion

Acceleration proportional to displacement (from equilibrium position/ point) (1)and in opposite direction/directed towards equilibrium position / point) (1) 2

OR accept fully defined equation

Oscillations

x0 = 0.036 m (1)

Period = 7.60 s/20 = 0.380 s (1)f = 2.63 Hz (1) 3

Displacement when t = 1.00 s

x = (–)0.026 m (1) 1

How and why motion differs from prediction

Motion is damped/amplitude decreases with time (1)(Because of) air resistance (1) 2

[8]

494. Oscillating, system

Diagram: suitable oscillator (1)method of applying periodic force of variable frequency (1) 2

Natural frequency:

(With no periodic force) displace oscillator and let it oscillate(freely) (1)

Frequency of this motion is natural frequency (1)

Forced oscillation:

Their system is being forced to oscillate/vibrate at driver’s frequency (1)

Resonance:

Vary the frequency (1)

Oscillator has large amplitude at / near natural frequency (1) 5[7]

495. Ionisation energy

(10.4 eV) × (1.6 × 10 –19 J eV–1) (1)

(–) 1.66 × 10–18(J) (1) 2

Kinetic energy

0.4 (eV) (1) 1

Transition

Use of E = hc/λ (1)3.9 (eV) (1)Transition is from (–)1.6 eV to (–)5.5 eV 3

[6]

496. Deductions about incident radiations

(i) Radiations have same frequency/same wavelength/ same photon energy (1)

(ii) Intensity is greater in (a) than in (b) (1) 2

Sketch graph (c)

Line of similar shape, starting nearer the origin on negative V axis (1) 1

Maximum speed

Use of E = hf (1)

Subtract 7.2 × 10–19 (J) (1)

Equate to ½ mυ 2 (1)

3.1 × 106 ms–1 (1) 4[7]

497. F proportional to I Quality of written communication (1)Any two from:[In words or on diagram]

• Method of producing and measuring a varying direct current

• Wire perpendicular to B field

• Method of measuring/detennining forces, e.g. moments / acceleration (1) (1)

Graph of F - I straight line through origin for F = added weight (1) 4

[OR correct straight line if F is total weight OR I

F constant]

Calculation of Initial acceleration

F = BIl

= 0.20 T × 4.5 A × 5.0 × 10–2 m

= 4.5 ×10–2 Na = F/m

= kg1050

N104.53–

–2

××

= 0.90 m s–2

Recall/state/use F – ma and F = BIl (1)

Use 5.0 × (10–2 m) for length (1)

Conversion of g to kg 50 × 10–3 (1)

a = 0.90 m s–2 (1) 4[8]

498. E.m.f.Motion of magnet (1)produces changing magnetic field over the coil (1) 2ORField lines (of magnet) cut across coilORProduces changes in flux linkage between coil and magnetsDiagramX at both ends of path (1)X in middle of path (1) 2

Rate of change of flux

500

))(10(3 3– V

t

×=∆∆φ

(1)

= 6.0 ×10–6 wb s–1/V/T m2 s –1 (1) 2

Changes to apparatusAny three from:

• more coils• stronger magnet [Accept ‘more powerful’]• decrease length of suspension ) [Not just ‘increase’• larger amplitude ) speed of magnet]• larger cross sectional area of coil• iron core within coil (1) (1) (1) 3

[9]

499. Diagram

Electric pattern:

Straight, parallel, reasonably perpendicular to plates and equispaced[Minimum 3 lines] (1)Correct direction labelled on one line [Downwards arrow] (1) 2Equipotential lines:Any two correct equipotentials with any labelling to identifypotentials (rather than field lines) (1) 1[Arrows on electric field lines – none on equipotential beingsufficient labelling]

Force

E = ( )m1025

30003–×

V [Correct substitution] (1)

Use of F = Ee even if value of “e” is incorrect (1)

F = 120 ×(103) V m–1 × 1.6 × 10–19 C

= 1.9 (2) × 10–14 (N) (1) 3

Graph

Straight horizontal line [Even if extending beyond 25 mm] (1)Value of F marked [e.c.f. their value] provided graph begins onforce axis and is marked at this point (1) 2

SpeedUse (1)

eV = ½ mv2

v2 = 2 eV/mv2 = 2

m

Fs

Fd = ½ mv2

v2 = 2Fd/m

Substitution (1)

V2 = ( ) ( )

kg109.11

V3000C101.6231–

–19

××××

= 2 kg109.11

N)10(1.9231–

–14

××

× 25 10–3 m

= kg109.11

m1025N101.92231–

–3–14

×××××

Answer: V = 3.2 × 107 ms–1 (1) 3

[If F= 2 × 10–14 N, then V= 3.3 × 107 ms–1][11]

500. (a) (i) Pitch (frequency) is higher (1) 1[Accept greater or more]

(ii) Resonance occurs and (1)energy is transferred from the loudspeaker to the vibrating airORamplitude of vibration is a maximum (1) 2

(iii) Fill flask with water and find volume (with measuring cylinder) (1)

Empty out some water and measure its volume; this is the volume of air (1)

Repeat, removing more water each time to increase the volume of air (1)

[OR empty out all water to find volume of air initially;add water to reduce volume of air]

Plot Inf against ln V (1) 4

[Allow lnf = n lnV + lnK or if correct graph plotted]

(b) Correct ln (or lg) values to 2 or 3 d.p. (1) 1Graph:Suitable scale (at least ½ paper used for each axis, avoiding scales of 3, etc) (1)Axes labelled with units (or units in table) (1)Plots (1)Line (1) 4[f v V can get all 4 marks]

(c) Large triangle [or maximum] (∆ y∆ x ≥ 80 cm2) (1)Correct calculation with n in range 0.49 – 0.51, , given to 2/3 s.f.and no unit (1)Negative sign (1)Negative sign indicates that f increases as V decreases, as wasobserved in (a)(i) (1) 4

[16]

Sample results

(a) (i) Pitch (frequency) is higher

(ii) Resonance occurs andenergy is transferred from the loudspeaker to the vibrating airORamplitude of vibration is a maximum

(iii) Fill flask with water and find volume (with measuring cylinder)Empty out some water and measure its volume; this is the volume of airRepeat, removing more water each time to increase the volume of air[OR empty out all water to find volume of air initially; add water to reduce volume of air]Plot lnf against ln V

(b) Table:

V/cm3 f/Hz ln(V/cm3) ln(f/Hz) log(V/cm3) log(f/Hz)

554 219 6.32 5.39 2.74 2.34454 242 6.12 5.49 2.66 2.38354 274 5.87 5.61 2.55 2.44254 324 5.54 5.78 2.40 2.51204 361 5.32 5.89 2.31 2.56154 415 5.04 6.03 2.19 2.62

Graph

6 . 1

6 . 0

5 . 9

5 . 8

5 . 7

5 . 6

5 . 5

5 . 4

5 . 35 . 0 5 . 2 5 . 4 5 . 6 5 . 8 6 . 0 6 . 2 6 . 4 6 . 6

l n V ( c m )3

l n ( / H z )f

(c) n = gradient = –00.5–50.6

30.5–05.6

= –50.1

75.0 = –0.50

Negative sign indicates that f increases as V decreases, as wasobserved in part (a) (i).

501. (a) More than four radial lines/four symmetric lines (1)Arrows inwards (1) 2

(b) Reference to speed of (gas) molecules [e.g. > 10 m s–1] (1)Greater than escape speed/υ e (1) 2

(c) (i) mA in kg and rA in m (1)

N (in G) in kg m s–2

(ii) mA = rAυ e 2/2G (1)

⇒ mA = 5.8 × 1015 (kg) (1)

Use of ρ = m/V and V = 4/3 π r3 (1) 5

⇒ ρ = 2900 kg m–3 / 2940 kg m–3 [3018 kg m–3 from 6 × 1015 kg]

(d) (i) Size or volume of Universe/distance between galaxiesagainst t/time (1)Line rising which does not level off (1)Big Bang/t = 0 labelled (1)

0

S o r S

t

(ii) Average (1)Density/mass-density (of Universe) (1) Max 4

(e) (i) Use of ∆ p/m∆ υ = F∆ t (1)

∆ υ = (2 × 106N) (7000 s) ÷ 5.8/6 × 1015 kg (1)

= 2/2.3 × 10–6 m s–1 (1)

(ii) Will/will not alter asteroid’s course [No mark]Justification: refer to ∆ s = t∆ υ /s = υ t (1) 4

[17]

502. Initial speed in x direction

Speed= distance ÷ time (1)= 1.2 m/0.2 s

= 6.0 m s–1 (1) 2

Initial speed in y direction

Speed = distance ÷ time= 1.9 m/0.2 s

= 9.5 m s–1 [No u.e.] (1) 1

Why answers are estimates

Speed not constant / some deceleration / acceleration ignored / (1) 1speed an average over 2.0 s

Initial velocity

υ 2 = (6.0 ms–1)2 + (9.5 ms–1)2 [e.c.f] (1)

υ = 11.2 m s–1

tan θ = 9.5 m s–1 ÷ 6.0 m s–1

θ = 58° [No u.e.] (1) 2

Kinetic energy

k.e. = ½ mυ 2 (1)

½ × 0.0052 kg × (11.2 m s–1)2 = 0.33 J (1) 2

Gravitational potential energy

Ep = mg∆ h (1)

= 0.0052 kg × 9.81 m s–2 × 5.3 m = 0.27 J (1) 2

Why answers not the same

Horizontal component of motion (1)shuttle still has ke (1)OREnergy converted to heat/work done against/energy lost because of (1)air resistance (1) 2

[12]

503. Wavelength

Distance between two points in phase (1)

Distance between successive points in phase (1) 2

[May get both marks from suitable diagram]

Sunburn more likely from UV

UV (photons) have more energy than visible light (photons) (1)

Since shorter wavelength / higher frequency (1) 2

What happens to atoms

Move up energy levels/excitation/ionization (1)

Correctly related to electron energy levels (1) 2[6]

504. Weight

750 N (1) 1

Mass

W=mg (1)

= 750 N ÷ 9.81 m s–2

= 76 kg (1) 2

Figure 2 completion

W arrow, labelled 750 N (1)R arrow, labelled 250 N (1) 2

[No u.e. in either case]

[Allow 1 mark for downward arrow longer than upward arrow]

Acceleration

F=ma (1)

= 750 N – 250 N = 500 N (1)

a = 500 N ÷ 76 kg

= 6.6 m s–2 (1) 3

Description of motion for t = 1.3 s to 1.5 s

Constant/steady velocity (1)

No acceleration (1) 2

Description of motion for t = 1.5 s to 2.0 s

Deceleration down/accelerating up (1)

Until υ = 0 / to rest / v decreasing (1) 2[12]

505. Revision Notes: Radiation

One suitable source, e.g. cosmic radiation, rocks, soil, medicalequipment, power stations. (1) 1

Nuclear radiation properties

Alpha Beta Gamma

Ionising ability (Very) strong Medium Weak

Penetrationpower(stopped by)

Thin paper or 3-10 cm air

Few mmaluminium or few × 10 cmair

Many cm leadof m of concrete

1

Correct materials for both alpha and beta (1)

Correct thickness for one correct material (1) 3[4]

506. Emitted pulse

Greater amplitude/pulse is larger/taller (1) 1

Depth of rail

2d = vt = 5100 m s–1 × 4.8 × 10–5 s

= 0.24 m

Hence d = 0.12 m

Reading from graph [4.8 or 48 only] (1)

Calculation of 2d [their reading × timebase × 5 100] (1)

Halving their distance (1) 3

Description of trace

A reflected peak closer to emitted/now 3 pulses (1)

Exact position e.g. 1.6 cm from emitted (1) 2

Diagram

Shadow region (1)

Waves curving round crack (1) 2[8]

507. Resistance in darkness

In the dark R = 4 kΩ (1)

so resistance per mm = 4000 Ω /40 mm = 100 Ω (mm–1) (1) 2

Resistance of 8 mm length

In the light R = 200 Ω (1)so resistance of 8 mm strip = (8 mm/40 mm) × 200 Ω [= 40 Ω ] (1) 2

Calculations

Resistance of remainder = 32 mm × 100 Ω mm–1 = 3200 Ω (1) 1

(i) Total resistance = 3240 Ω (1)

Current = V/R = 1.2 V/3240 Ω = 3.7 × 10–4 A (1)

(ii) Across 8 mm, p.d. = IR = 3.7 × 10–4 A × 40 Ω (1)= 0.015 V (1) 4

Explanation of why current decreases

Any two points from:

• more of strip is now in the dark

• greater total resistance

• I = V/R where V is constant (1) (1) Max 2[11]

508. Error in circuit diagram

Cell needs to be reversed (1)Any one point from:

• electrons released from the magnesium

• copper wire needs to be positive to attract electrons (1) 2

Completion of sentence

UV is made up of particles called photons (1) 1

UV and visible light

(i) UV has shorter wavelength/higher frequency/higher photonenergy (1)

(ii) Both electromagnetic radiation/both transverse waves/samespeed (in vacuum) (1) 2

Explanation of why low intensity UV light produces a current

Any three points from:

• reference to photons or E = hf

• frequency > threshold frequency

• electron must have sufficient energy to be released

• UV photons have more energy

• electron is released by ONE photon

• brighter light just means more photons (1) (1) (1) Max 3

Why current stopped

Glass prevents UV reaching magnesium (1) 1[9]

509. Total internal reflection


• from a more dense medium to a less dense medium/high to low refractive index

• incident angle greater than the critical angle

• light is reflected not refracted/no light emerges (1) (1) Max 2

Critical angle

Sin i / sin r = µ ; gives sin 90°/sin C = µ (1)

C = 42° (1) 2

Diagram

Reflection (TIR) at top surface (air gap) (1)

Reflection (TIR) at bottom surface and all angles equal by eye (1) 2

Path of ray A

Passing approximately straight through plastic into glass (1)

Emerging at glass–air surface (1)

Refraction away from normal (1) 3

Why there are bright and dark patches on image

Bright where refracted/reference to a correct ray A in lower diagram (1)

Dark where air gap (produces TIR)/reference to correct top diagram (1) 2[11]

510. Polarisation

The (wave) oscillations (1)occur only in one plane (1) 2[OR shown with a suitable diagram]

How to measure angle of rotation

Any four points from:

• Polaroid filter at one/both ends

• with no sugar solution, crossed Polaroids (top and bottom oftube) block out light

• sugar solution introduced between Polaroids

• one Polaroid rotated to give new dark view

• difference in angle between two positions read from scale (1) (1) (1) (1) Max 4

Graph

Points plotted correctly [–1 for each incorrect; minimum mark 0] (1) (1)

Good best fit line to enable concentration at 38° to be found (1) 3

Concentration

0.57 (± 0.01) kg l–1 1[10]

511. Formula for magnitude of force

F = Eq (1) 1

Direction

Down page (1) 1

Calculation

Eq = Bqυ (1)→υ = E/B (*)

= 1.2 × 104/0.4 ms –1 (*)(*) [Equation or substitution] (1)

υ = 3 × 104 ms–1 (1) 3

Explanation

mg << eE and/or Bqυ (1) (1)

[OR gravity force << E and/or B force 2 marks

OR m very small 1 mark only

OR gravity is a weak force 1 mark only 2

OR ion moving fast 1 mark only][7]

512. Corrected errors

line 3 Mesons are made from q and antiq (1) (1)OR leptons are fundamental/not made from smaller etc.

line 4 as line 3 [only one (1) for same correction made twice]OR quarks, leptons, neutrinos, and others (1)

line 6 Neutron is made from 3 q s (1) (1)OR meson is made from q and antiq [with restriction as in line 4]

line 10........... energy .... [instead of momentum] (1) (1) Max 6[6]

513. Direction of centripetal acceleration

Towards centre/downwards/inwards (1) 1

Explanation

F=maa and F in same direction (1) 1

Resultant force

υ = T

rπ2 =

5.4

82 ×π [Equation OR substitution] (1)

= 11.2 m s –1 (1)

F =r

mv2

= 8

2.1160 2× [Equation OR substitution] (1)

= 936 N (1)

OR w = T

π2 =

5.4

2π [Equation OR substitution] (1)

= 1.396 rad s–1 (1)

F = mrw2 (*)

= 60 × 8 × 1.42 (*) [(*) Equation OR substitution] (1)= 936 N (1) 4

Calculation of weight

W = mg= 60 × 9.81= 589 N (1) 1

Calculation of magnitude of push

Fnet = W + PP = F – W= 936 – 589= 347 N (1) 1

Diagram

W ↓ (1)

P ↑ (1) 2[10]

514. Deceleration of Earth

F = m∆ υ / ∆ t = 7 × 3 × 104 N [Equation or substitution] (1)

=2.1 × 105 N (1)

a = F/m = 2.1 × 105/6.0 × 1024 [Equation or substitution] (1)

= 3.5 × 10–20 m s–2 (1) 4

OR

in 1 second: ∆ υ for debris = 3 × 104 m s–1

⇒ a = 3 × 104 m s–2

F = ma = 7 kg × 3 × 104 ms–2

= 2.1 × 105 N (1) (1)

F on Earth = F on debris

⇒ aearth = F/m = kg106

N101.224

5

××

= 3.5 × 10–20 m s–2 (1) (1)OR Conservation of momentum (1)

or m1u1 + m2υ 2 = m3υ 3 etc

6 × 1024 × 3 × 104 (+ 7 × 0) = (6 × 1024 + 7) × υ 3 (1)

OR An answer (3.5 × 10–20 m s–2) calculated from any specific (1) (1) (1)arbitrary time] (1)Comment: negligible (1) 1[OR energy loss negligible]

[5]

515. Oscillations

Correct ticks/cross (1)Reasons (1) (1) (1) 4

Oscillations SHM

Reason

Mass on end of spring Force ∝ displacement[OR acceleration ∝displacement]ORForce always towards the equilibrium position

Child jumping up anddown

Force constant when child in

the airORPeriod/frequency notindependent of amplitute

Vibrating guitar string Force ∝ displacement[OR acceleration ∝ displace-ment]

ORFrequency not dependent onamplitute

[4]

516. Explanations of observations

Speed of light is much greater than speed of sound (1)

Speed of sound in soil is greater than speed of sound in air (1) 2[2]

517. What I represents

Surface temperature of star (1) 1

Sun on diagram

S correctly marked at L / Lsun = 1 (1) 1

Flux calculation

L = 10–2 × 3.9 × 1026 W = 3.9 × 1024 W (1)

Use of F = L/4π d2 (1)

= 3.9 × 1024 W/4π × (500 × 3.09 × 1016 m)2

= 1.3 × 10–15 W m–2 (1) 3

Force which holds Sun together

Gravity (1) 1

How nuclear processes release energy

Any three from:

• Nuclear fusion

• When hydrogen nuclei combine to form helium

• There is a loss of mass [OR mass defect]

• This mass loss is converted to energy (∆ E = c2 ∆ m) (1) (1) (1) 3

Sun becoming red giant

(i) Recall of F = Gm1m2/r2 [OR g = GM/r2] (1)Gravitational force does not change, since this depends onthe mass of the Sun and distance to (centre of) Sun, whichhave not changed (1)

(ii) Arrow from S pointing right and upwards (1) 3[12]

518. Dinosaur

Weight on each leg = 2.5 × 105 N (1)

Area = 3.14 × 0.152 = 0.071 (1)(Use of stress = F/A)

= 3.5 × 106 N m–2 OR consistent units (1)

Can stand up OK (1) 4

Compression of leg bone

Use of YM = stress/strain, e.g. 1 × 1010 = 3.5 × 106/strain [Allowe.c.f] (1)

Strain = 0.00035 (1)

Compression = 1.4 mm (1) 3

Physics principles underlying claim

Would place all weight onto two legs [Fewer than 4]: this halvesstresses (1)

Correct use of F = mv/t OR ma (1)As t small,then v/t or a is (much) larger than 9.8 or F larger (than weight) (1) 3

Reduction of compressive forces

Upthrust/upward force acts (1)

Equal to weight of water displaced (1)

Reduces resultant downward force (1)

Upthrust will be significant compared with weight (1)

Swampy muddy water – larger density than water – larger U (1)

Decreases stresses calculated earlier (1) Max 3[13]

519. Millennium bridge

Resonance (1)

People walking/swaying in step/periodic (1)

Forcing frequency = natural frequency/implied by sketch graph (1)

Frequency – number of sways/swings/oscillations per second (1)

Stifffness/k is force per unit extension/deflection/F = kx (1)

Welding steel gives less deflection for a given force/increase k reduces strain (1)

Damp – remove energy from oscillations/reduce amplitude (1)

Oscillate – to and fro repetitive motion/time period/displacement time graph (1)

Out of phase – move in opposite directions /opposite displacements /graphical representation (1)

Cancels out – destructive interference/produces a minimum/node/graphical sketch (1) Max 7

[7]

520. Magnitude of resultant force

4 cm line S / 1.7 cm line N 1

8 cm line NE / 8N resolved into two perp. components (5.7E & 11.7N or 5.7N)

Correct construction for vector sum 1

5.7-6.1N 1

Name of physical quantities

Vectors 1

Two other examples

Any two named vectors other than force 1(if>2, must all be vectors)

[6]

521. Calculation of average velocity

Use of υ = s/t 1

υ = 1.86 m s–1 / 1.9 m s–1 1

Acceleration of trolley

Selecting υ 2 = u2 + 2as 1

Correct substitutions 1

2.87 m s–2 / 2.9 m s–2 / 3.0 m s–2 1

Tension in string

Use of F = ma 1

2.73 N / 2.76 N / 2.85 N 1

Assuming no friction/no other horizontal force/table smooth/light 1string/inextensible string

Explanation

Suspended mass/system is accelerating 1

Idea of resultant force on the 0.4 kg mass 1[10]

522. Acceleration of free fall

Advantage:

So time/distance can be measured more precisely/accurately 1[Allow reaction time less important]

Disadvantage:

Air resistance becomes important [NOT air resistance acting for longer time ]/may reach terminal velocity/harder to hit trap door 1

Experimental method

Diagram:

Labelled start mechanism (any part) 1Labelled stop mechanism (any part) 1

Releasing ball starts timer 1Ball opening trap door/switch stops timer 1

OR

Diagram:

Ticker timer 1Tape from sphere through timer [at least one labelled] 1

Timer makes dots at known rate 1Time = number of spaces × time interval between dots 1

OR

Diagram:

Camera 1Strobe lamp 1

Lamp flashes at known frequency 1Time = number of spaces between images × time interval between flashes 1

OR

Diagram

Light gate joined to timer 1Second light gate also joined to timer [one labelled] 1Ball passing gate starts timer 1Ball passing second gate stops timer 1

OR

Diagram

Labelled stopwatch -one mark only out of 4 1

OR

Diagram

Motion sensor labelled 1At top or bottom 1

Produces distance time graph/pulses emitted at known time intervals 1Time read off from graph 1

OR

Diagram

Video camera 1At side 1

Frames at known frequency/time interval 1Time = no. of frames × time interval 1

Statement

Weight = mass × g (allow ‘W = mg’) 1

g is the same (for all objects) [NOT ‘gravity’ is constant] 1[8]

523. Gravitational potential energy

Use of mgh 1

Vertical drop per second = (8.4 m) sin (3°) 1

3.9 × 102 J/Js–1/W 1

What happens to this lost gpe

Becomes internal energy/used to do work against friction and/ordrag/heat/thermal energy. [mention of KE loses the mark] 1

Estimate of rate at which cyclist does work

Rate of working = 2. × 3.9 × 102 W 1

=7.8 × 102 W 1

[3.9 × 102 W earns 1 out of 2][6]

524. Momentum and its unit

Momentum = mass × velocity 1

kg m s–1 or N s 1

Momentum of thorium nucleus before the decay

Zero 1

Speed of alpha particle/radium nucleus and directions of travel

Alpha particle because its mass is smaller/lighter 1

So higher speed for the same (magnitude of) momentum OR N3 argument 1

Opposite directions/along a line 1[6]

525. Nuclear equation

βe/Snln 01–

01–

11550

11549 +→

Correct symbol and numbers for tin OR beta 1

Correct symbols and numbers for the other two 1

Decay constant

Use of λ = 0.69/t1/2 1

1.57 × 10–15 y–1 OR 4.99 × 10–23 s–1 1

Activity of source and comparison with normal background count rate

Use of A = λ N. 1

0.11/0.12 (Bq) 1

Lower (than background) [Allow ecf- assume background = 0.3 to 0.5] 1[7]

526. Radiation tests

Alpha:

Test 2 or 2 and 1 1

Count drops when alphas have been stopped by the air / alphas have a definite range / (only) alpha have a short range (in air) 1

Beta:

Test 3/3 and 1 , because 1 mm aluminium stops (some) beta/does not stop any gamma rays 1

Gamma:

Test 4 or 4 and 1, because 5 mm aluminium will stop all the betas, (so there must be gamma too )/gamma can penetrate 5 mm of aluminium 1

[4]

527. Resistance of lamps

P = R

V 2

OR I= 60/12 = (5 A) 1

R = W60

V12V12 × R = V/I 1

R = 2.4 Ω 1

Resistance variation

Lamp A: resistance of A decreases with current increase 1

Lamp B: resistance of B increases with current increase 1

Dim filament

Lamps are dim because p.d. across each bulb is less than 12 V 1

Why filament of lamp A is brighter

Bulbs have the same current 1

p.d. across A > p.d. across B/resistance A> Resistance B 1

OR

power in A > power in B 2[8]

528. Current in heating element

p = VI

I=V230

W500

I = 2.2 A

p = R

V 2

R = )(8.105/500

2302

Ω

I = 2.2 A

1

1

1

Drift velocity

Drift velocity greater in the thinner wire / toaster filament 1

Explanation

Quality of written communication 1

See I = nAQυ 1

I is the same (at all points ) 1

(probably) n (and Q) is the same in both wires 1[8]

529. Resistance of films

R = A

lρ1

R = t

l

ωρ

or A = ω t [consequent on first mark] 1

[i.e. product = ω t]

Resistance calculation

R = m)10(0.001m)10(3

)10(8Ωm)10(6.03–3–

–3–5

××××××

OR

R = )m100.1)(mm0.3(

)mm8()m100.6(6–

5–

××Ω×

R = 160 Ω

Correct substitution except powers of 10 1

Correct powers of 10 1

Answer 1

Resistance of square film

l = ω 1

R = t

ρ1

[7]

530. Definition of specific heat capacity

Energy (needed) 1

(per) unit mass/kg ) 1(per) unit temperature change/ °C / K )

OR

Correct formula [does not need to be rearranged] 1

with correctly defined symbols 1

Circuit diagrams

A

A

V V

O R

Accept voltmeter across heater and ammeter as well as voltmeter across heater only

Means of varying p.d./current 1

Voltmeter in parallel with a resistor symbol 1

Ammeter in series with any representation of heater 1

Other apparatus

• (Top pan) balance / scales 1

• Stopwatch / timer / clock 1

Explanation

Energy/heat loss to surroundings/air/bench

OR

mc∆ θ +∆ Q = Vlt or equivalent in words (e.g. student ignoresenergy loss in calculations)

OR

mc∆ T +∆ Q = Vlt or equivalent words 1

Modifications

Any two from

• Use of insulation around block

• Ensure all of heater is within block

• Grease heater/thermometer 2[10]

531. Specific heat capacity calculation

c = C)22–750(kg4.1

J10860

θ

3

°××=

∆∆m

Q

c = 844 (J kg–1 K–1/°C–1)

Conversion to joule × 103 1

Subtraction of temperature 1

Answer 1[3]

532. Definition of e.m.f. of a cell

Work/energy (conversion) per unit charge 1

for the whole circuit / refer to total (energy) 1

OR

Work/energy per unit charge 1converted from chemical to electrical (energy) 1

OR

E = Q

W for whole circuit 1

All symbols defined 1

OR

E = I

P for whole circuit 1

All symbols defined 1

[Terminal p.d. when no current drawn scores 1 mark only]

Circuit diagram

A AR R

V

R 1 R (can be variable) 1 2A in series 1 A and V correct 1

V as shownOr across R + AOr across battery

[2nd mark is consequent on R(fixed, variable) or lamp]

Sketch graph

R V

1 / I I

Graph correctly drawn with axes appropriately labelled andconsistent with circuit drawn 1

Intercept on R axes Gradient ≡ (–)r [Gradient mark consequent 1≡ (–)r on graph mark]

[Gradient may be indicated on graph][6]

533. Topic A – Astrophysics

Red giant

Quality of written communication [ needs > 2 processes] 1

Any three from:

• Hydrogen burning ceases in core

• Core collapses/star collapses

• Star swells up/star expands

• other fusion processes occur in core/hydrogen burning takesplace in ‘shell’ /outer layers Max 3

Wien’s law

Star cooler/T less 1

Hence Wien’ s law means λ max greater 1

[e.c.f. T increases, hence λ max decreases]

Stefan’s law

Law states L = σ AT4 OR L ∝ AT4 1

Larger (surface) area A/radius/diameter 1

Increase in A > decrease in T4 1

OR huge/massive increase in A/r/d makes up for/compensates for decrease in T[NOT just A increases more than T decreases]

[9]

534. Topic B -Solid Materials

Stress-strain curves for two materials

(i) Tougher: B because it has larger area/greater energy density 1

(ii) Stiffer: B because steeper slope/greater Young modulus 1

(iii) More ductile: B because greater strain in plastic region 1

Line added to graph for material C

[Mark alongside graph]

Straight with sudden loop/straight line with sudden stop 1

Smallest gradient 1

Greatest stress 1[6]

535. Topic C – Particles

Binding energy

Quality of written communication 1

Energy required/ put IN to separate/break up a nucleus 1

Into protons + neutrons OR nucleons 1

OR in terms of energy given OUT when making a nucleusOR mass defect between nucleus and separate nucleons

Graph

Shape [not bell-shaped; steeper rise than fall; start near origin; fall 1less than half max height] Peak at around 50 [40 – 70] 1

Isotopes

BE/nucleon [any reference] 1See working out, e.g. 7.72: 7.44 [no u.e.] 1Hence O-16 ..I 1 1

[O-16 because O-17 is radioactive gets 1/3][8]

536. Table

( i )p a r t i c l e

( i i )q u a r k c o n t e n t

( i i i )a n t i p a r t i c l e

( i v )q u a r k c o n t e n t

p r o t o n

π

K

u u d

d u

d s

p

π

K

u u d ( 2 )

u d ( 1 )

s d ( 2 )

–

– – – –

–

–

– +

0 0

Shaded boxes show answers: circled terms count as one.

Proton is uud 1

antiproton or p is uud [allow −

p or p-bar ] 1

π + 1

Anti K0 is 0

K [ allow K0-bar] 1

Quark composition is dsanddu 1[5]

537. Equation

XCnN 11

146

10

147 +→+

14/7 and 1/0 11/1 [no e.c.f.] 1Hence X is H atom/H nucleus/proton/H/hydrogen 1

Estimation of age

Down to 1.9 cpm needs 3 half-lives 13 × 5730 1

17 000/17244 years/5.4 × 1011s 1

Suggested problem in measuring

Background count mentioned/randomness significant 1

[OR need larger mass than one gram][7]

538. (a) (i) 10 spheres used, in channel between rules 1

Close packed/touching 1

Correct use of set squares at end or correct use of set squaresto measure single diameter 1

Scale readings shown, giving l/d to nearest mm 1

d ± 0.03 cm of Supervisor’s value to ≥ 3 s.f. + unit 1

(ii) ∆ l between 0.5 mm and 2 mm with unit

OR

correct ∆ l from range or 1/2 range 1

Correct calculated percentage 1

(iii) Correct calculated to more than 2 s.f. + unit 1

[Allow e.c.f. ]

(iv) Use of 10 marbles [Allow 10 single values] 1

± 0.10 g of Supervisor’s value 1

Correct calculation of ρ , 2/3 significant figures + unit 1

Value 2.3 → 2.7 g cm–3 [Rounded to 2 s.f. ] 1

(b) (i) Measure height above bench at two places 1

Use vertical rule 1

Vertical rule checked with set square OR set square againststand to show string horizontal 1

Reasonable value to 0.1 N with unit 1

[Between 4.0 N and 5.5.N]

(ii) Three forces in correct orientation 1

[May be mirror Image]

Arrows on all forces correct and lines meeting at a point 1

[Dependent on first mark ]

Correct labels or numerical values 1

(iii) Correct values shown 1

[Allow 4 N or 0.4 g and e.c.f.][Must be evidence of a unit]

Scale diagram or trigonometry or use of Pythagoras 1

Correct calculation1 1

2/3 s.f. + unit 1

Value 1.5 → 3.5 N 1[24]

Sample results

(a) (i) 10 spheres used, in channel between rulesClose packed and touching

l = 41.2 – 26.0 = 15.2 cm

d = 10

2.15 = 1.52 cm

(ii) ∆ l = 0.2 cm

Percentage uncertainty = %3.11002.15

2.0 =×

(iii)6

52.1

6

32 ×== ππ dV

= 1.84 cm3

(iv) Mass of dish = 15.0 g

Mass of 10 marbles + dish = 60.6 gMass of 10 marbles = 45.6 gMass of 1 marble = 4.56 g

Density = 84.1

56.4

= 2.48 g cm–3

(b) (i) Measure height above bench at two places; use vertical rule

Vertical rule checked with set square OR set square against stand to show string horizontal

R = 4.7, 4.9 N

R = 4.8 N

(ii)

W e i g h t = 3 . 9 2 N

P u l l o f s t r i n g o n B = 4 . 8 NP u l l o f s t r i n g o n B

B

(iii)4 . 8 N 3 . 9 2 N

x N

x2 = 4.82 – 3.922 = 7.67

x = 2.77 N

539. (a) Measure height above bench at two places 1

Use vertical rule 1

Vertical rule checked with set square 1

[OR set square used between stand and string]

Reasonable value to 0.1 N (4.0 – 5.5 N) with unit 1

(b) h and 1 recorded to nearest mm or better 1

Scale readings shown 1

Measured height above the bench at Band C 1

with vertical rule 1

Correct calculation of cosθ to 2/3 s.f. 1

Hence value 0.70 → 0.98 1

(c) [Using g = 9.8]

Correct calculation of W/R 1

2/3 s.f. and NO unit [dependent mark] 1

W/R and cos θ within 0.05 [no e.c.f.] 1

Correct calculation of percentage uncertainty 1

Correct calculation of percentage difference [OR range of values] 1

Sensible comment [Must be quantitative] 1

(d) (i)

Method 1

• Change M

• Adjust height of newton-meter until AB horizontal

• Record R and h

• Evaluate cosθ and W/R

Method 2

• Keep M constant

• Adjust Separation of clampsuntil AB horizontal

• Record R and h

• Evaluate cosθ and I/R

1

1

1+1

1

[OR inferred from graph]

(ii) Correct graph with axes labelled 1

Straight line through origin expected ) 1) Dependent mark

Gradient ) 1[24]

Sample results

(a) Measure height above bench at two placesUse vertical ruleVertical rule checked with set square [ or set square used betweenstand and string]R = 4.7, 4.9 N Average R = 4.8 N

(b) 1 = 30.0 cm

h = 44.5 -19.0 = 25.5 cm

Measured height above the bench and B and C with vertical rule

cosθ = l

h =

30

5.25 =0.85

(c) 82.08.4

81.94.0 =×=R

W

Percentage uncertainty in R = 8.4

2.0×100% = 4.2%

Percentage difference between values =

%6.3%100)82.085.0(2

182.0–85.0 =×

+

Difference is less than percentage uncertainty, hence goodagreement between cosθ and W/R.

(d) (i)

Method 1

• Change M

• Adjust height of newton-meter until AB horizontal

• Record R and h

• Evaluate cosθ and W/R

Method 2

• Keep M constant

• Adjust Separation of clampsuntil AB horizontal

• Record R and h

• Evaluate cosθ and 1/R

(ii)

c o s θ c o s θ

W R/ 1 / R

G r a d i e n t = 1 G r a d i e n t = W

540. Angular speed

Use of ω = 2π /T 1

ω = 1.2 × 10–3 [min 2 significant figures) [No ue as units given] 1

Free-body force diagram

Pull of Earth/Weight/mg/Gravitational Pull 1

Why satellite is accelerating

Resultant/Net/Unbalanced force on satellite must have an acceleration OR Σ F = ma. 1

Magnitude of acceleration

Use of a = ω 2 r OR υ 2 ÷ r 1

a = 9.36-9.42 OR 6.5 m s–2 1

[Depends on which ω value used][6]

541. Frequency of spectral line for calcium

Use of c = fλ 1

f= 7.63 × 1014 Hz 1

Ultra violet 1

Line spectrum

(A series of) lines on a dark /white background 1

Wavelength of calcium line

Use of ∆ λ = υ /c × 393 nm 1

393 ± 18 – 19 (nm) 1

λ = 411–2 nm 1

Hubble constant

See 365 × 24 × 60 × 60/3.2 × 107 1

Use of 3 × 108 [d= 9.5 × 1024] 1

Use of υ = Hd 1

H = 1.50 × 10–18 [no ue as unit given] 1

Recessional velocity

υ = 5.72 × 107 (ms–l) [No u.e.] 1[12]

542. Wavefront

Line/surface joining points in phase 1

Addition to diagrams

Wavefront spacing ≈ as for incident waves (min. 3 for each) 1

1st diagram: wavefronts nearly semicircular 1

2nd diagram: much less diffraction 1

Reception

L W has longer wavelength 1

so is more diffracted around mountains [consequent] 1[6]

543. Alpha particle: diagram

Curving path between plates 1

Towards 0 V plate 1

Emerging from plates and carrying on straight 1

Calculation

Electric field = m)10(10

V20003–×

Substitution 1

Force = EQ

×=

3–1010

2000Vm–1 × (2) × 1.6 × 10–19 C

Substitution [ecf their E] 1

= 6.4 × 10–14 N

Correct answer 1[6]

544. Horizontal component

4.8 × 10–5 T × cos 66°

= 1.95 [2.0] × 10–5 T

Use cos 66°/sin 24° 1

Answer 1

Calculation of induced voltage

Speed after 2 seconds = 9.81 m s–2 × 2 s 1

[ecf their B]

Induced e.m.f. = 1.95 × 10–5 T × 2.5 m × [9.81 m s–2 × 2 s] 1

= 9.6 × 10–4V 1

North-south rod

Induced emf = 0 (V) 1

Rod does not cut magnetic field lines/no flux cutting/no change in flux 1[7]

545. Forces

(i) F = GMEm/R2 1

(ii) F = GMMm/r2 1

Distance R

=

=

=

r

R

M

M

r

R

M

M

r

mGM

R

mGM

m

E

m

E

mE

21

2

2

22

OR

OR

27

2

)m109.3(1

81

×= R

R = 3.5 × 108 m

Evidence that equating forces has occurred 1

Correct substitution 1

Correct answer 1[5]

546. (a) T = 0.4 ms 1

f = 2.5 kHz [e.c.f. from T] 1

(b) (i) Correct set up 2

[–1 for each error or omission. c.r.o and signal generator mustbe labelled][mic between LS and board otherwise –2]

(ii) Distance between (nodes/antinodes/max/min) measured 1

is λ /2 1

Measure across several [implied] 1

(c) 1/f values all correct, to 3 s.f. with unit [here or on graph] 1

[Allow kHz–1 OR 103 Hz–l]

Graph:

0 . 1 8

0 . 1 6

0 . 1 4

0 . 1 2

0 . 1 0

0 . 0 8

0 . 0 6

0 . 0 4

0 . 0 2

0

λ / m

0 . 1 0 . 2 0 . 3 0 . 4 0 . 5 1 / f / m s

S c a l e - a tl e a s t p a p e ri n b o t h d i r e c t i o n sa v o i d i n g 3 ' se t c .

A x e s -l a b e l l e dw i t h u n i t sf o r o rc o r r e c t w a y r o u n d

P l o t s -a c c u r a t e t o s q a r e

L i n e -g o o d , t h i ns t r a i g h tl i n e

λ

( 1 )

( 1 )

( 1 )

( 1 )

12

1 2/

G R A P H

(d) ∆ xy > 80 cm2 1

Gradient calculated correctly and 330 – 350 m S–1 1

= c, 2/3 s.f. and unit 1

Eliminates a significant systematic error OR graph averages a number of values 1

[16]

Sample results

(a) λ = 4 divisions = 4 × 0.1 ms = 0.4 ms

f = kHz5.2104.0

11 3– =×=T

(b) (i)S i g n a lg e n e r a t o r

L o u d s p e a k e r

M i c r o p h o n e

B o a r d

c . r . o .

(ii) Distance between (nodes/antinodes/max/min) measured is λ /2

Measure across several

(c) ms/1

f

0.5000.4000.3330.2860.2500.200

(d) c = gradient = ms000.0500.0

m010.0–180.0

− = 340 m s–1

Graphical method eliminates a significant systematic error OR graphaverages a number of values

547. (a) (i) Correct circuit with ammeter in series with R and C and voltmeter across R and C 1, 1

[–1 for each error or omission ]

87.6 Ω or 88 Ω with unit 1

[Allow V A–l; do not allow V mA–l]

(ii) Correct substitution 1

Hence correct calculation [> 2 s.f. + unit] 1

[Do not penalise unit if already penalised in (i)]

(iii) Replace a.c. supply with a signal generator/variable frequency supply 1

Measure V and I 1

for different frequencies of f 1

1

4 π C

1f

Z R = × + [ c l e a r l y s h o w n ]22 2 2

2

y = m x + C1

(b) Z2 and 1/f2 values all correct and 1

> 2 s.f. with units for both 1

Graph

Scale -at least half paper in each direction avoiding 3s etc and axes 1labelled [Ignore units ]

Plots- all accurate to 1 mm [ 21

square] 1

Line -thin, straight line of best fit 1

(c) Gradient (1.29 – 1.35) 107 from large ∆ [∆ x∆ y > 64 cm2] 1

[Ignore units ] [As an alternative to the gradient candidates may use a point on the lineand the intercept. This is allowed provided the point is not a data point]

Correct calculation of C with unit [Must be F or µ F] [Ignore s.f. ] 1[16]

Sample results

(a) (i)

~

A

V

4 7 Ω 4 7 Fµ

Z = Ω= 6.87mA17

V49.1

(ii) Z2 = (4π 2 × (47 × 10–6)2 × 502)-1 + 472

Z2 = 4587 + 2209 = 6796

Z = 82.4 Ω

(iii) Replace a.c. supply with a signal generator/variable frequencysupply

Measure V and I for different frequencies of f

1

4 π C

1f

Z R = × + [ c l e a r l y s h o w n ]22 2 2

2

y = m x + C

(b) Data:

Z2/Ω 2 – × 103 1/f2/s2 × 10–4 (Hz–2)

10.68.97.76.76.05.14.0

6.254.944.003.312.782.041.23

Graph:

1 1

1 0

9

8

7

6

5

4

3

2

0 1 2 3 4 5 6 7

Z / × 1 0Ω

1 / f / s × 1 0

2

2

2

2

3

– 4

(c) Gradient = 2–274–

3

s1032.110)0.0–0.6(

10)4.2–3.10( Ω×=××

C2 = (4π 2 × gradient)–1 = (4π 2 × 1.32 × 107)–1 F

C = 44 µ F

548. (i) Protons are positively charged / like current 1

refer to Fleming or motor rule / Rev / Bqv / perpendicular F and 1

υ

[not right hand rule ]

(ii) υυ2

Ber

m = mrω 2 = Beυ 2

[accept q for e ]

t

r

T

r π/

π2υ =

tT

π/

π2ω = 1

(iii) Quality of written communication 1

Each time it crosses gap/between dees it accelerated / is attracted / is given E 1

Idea that p.d. between the dees reverses while the proton completes half a revolution / c.e.p. 1

As energy becomes large the mass/inertia of the proton increases 1[not protons hit edge ]

so it cannot exceed the speed of light [i.e. ref to c ]/synchronousproperty breaks down/formula no longer gives constant f 1

(iv) ∆ E = (1.6 × 10–19 C) (12000 V) [allow × 12] 1

= 1.9/1.92 x 10–15 (J) [no e.c.f.] 1

(v) r2 = 2m × k.e. ÷ B2e2r = √same

Substitute 1.66 / 1.7 × 10–27 kg /1860 me /2000 me and

1.6 × 10–19 C 1

Use of k.e. = (1.9 × 10–15J) × 850 1

[e.c.f. for 1.9 × 10–15 J e.g. 2 × 10–15J ⇒ 1.7 × 10–12J]

⇒ r = 0.575 m /57.5 cm 1

[2 × 10–15J ⇒ 0.59 m]

[9.1 × 10–31 kg ⇒ 0.0137 m e.o.p. max 1/3][15]

549. (a) (i) a is acceleration 1

f is frequency 1

x is displacement from equilibrium/centre/mean position 1

[beware amplitude ]

Either minus sign means that a is always directed tocentre/equilibrium position/mean position OR a and x in opposite directions / a opposite to displacement 1

(ii) Attempt to use υ m = 2π fa 1

υ m = 2π (50 Hz) (8.0 × 10–6 m) [a = 4 × 10–6 m e.o.p] 1

= 2.5 × 10–3 m s–1/ 2.5 mm s–1/2500 µ m s–1 1

(b) (i) ρ in kg m–3 and g in m s–2 / N kg–1 1

Aρ g/m with units leading to s–2 1

(ii) At least three peaks of y/m measured [ignore numbers] 1

EITHER

(5.4/5/6 3.7/6/8 2.4 1.6/71.24.4 2.8/9 2.0/1.9 1.4)

Attempt to calculate successive ratios [ e.g. 0.67 or 1.5] 1Logical statement re ratios and exponential change 1

OR

Read 3 values of x and y [ignore numbers] 1Two half lives between 0.75 s and 0.95 s 1Logical statement re half-lives and exponential change 1

[beware ∆ y not constant / use of y = Yoe–kt][12]

550. Average speed of the car

Speed = s/t [stated or implied] (1)

= 15 m/0.7 s [allow 14.5 m to 15.5 m]

= 21.4 m s-1 (1) 2

Deceleration

Identify u = 24.0 m s-1 [Can show by correct substitution] (1)

s = ut + 22

1 at

12.6 m = (24.0 m s-1 × 0.7 s) + 21 × a × (0.7 s)2 + 2

1 × a × (0.7 s)2

The rest of the substitution

a = 2(12.6 m – (24.0 m s-1 × 0.7 s)) ÷ (0.7 s)2

Rearrangement

a= (–)17.1 m s-2 / deceleration = 17.1 m s-2 [No u.e] (1) 4

[If using speed limit: identify u (1);

speed limit = 18 m s-1 → v = 12 m s-1 (1);substitute in or rearrange v = u + at or v2 = u2 + 2as (1),a = (–)17.1 m s-2 (1)]

Calculation of braking force

F = ma (1)

= 1400 kg × 17.1 m s-2

= 2.4 × 104 N (1) 2[8]

551. Shape of graph

(Temperature) rises more rapidly at first then less rapidly (1)

Heat is lost to the surroundings more quickly as temperature rises (1) 2

Initial rate of temperature rise

Draw tangent at t = 0 s (1)

Use a gradient starting at 0 s, 16 to 16.25 °C (1)

Rate = (30 °C – 16 °C) ÷ 1300 s

= 0.011°Cs-l [No u.e.] (1) 3

Maximum thermal energy gained in one second

AQ = mc∆ T

= 0.13 kg × 4200 J kg-1 °C-1 × 0.011°C (1)

Energy gained per second = 6.0 J [Accept J, W or J s-1] (1) 2

Efficiency calculation

Initial energy input per second = 0.015 m2 × 500 W m-2

= 7.5 J s-1 (1)

Efficiency = useful energy (or power) output ÷ energy (or power)input × 100%

= (6.0 J s-1 ÷ 7.5 J s-1) × 100% (1)

= 80% OR 0.8 OR 4/5 (1) 3

Effect of water spillage

Lower mass or heat energy gained by water over-calculated (1)

Calculated efficiency higher than actual efficiency (1)

OR

Air in hose insulates, so less heat absorbed than with no air (1)

Calculated efficiency less than if no water spillage (1) 2[12]

552. Rate of energy transfer

Ep = mg∆ h (1)

For one person: Ep = 90 kg × 9.81 m s-2 × 420 m = 370 800 J (1)

For 2800 people: Ep = 2800 x 370 800 J = 1.04 × 109 J]

Rate = 1.04 × 109 J ÷ 3600 s

= 288 000 W [No u.e.] (1) 3

Total kinetic energy

k.e. = 21 mυ 2 (1)

21 × 2800 × 90 kg × (5.0 ms-1)2

= 3 150 000 J ÷ [No u.e.] (1) 2

Rate of energy conversion

3 150 000 J ÷ (60 × 60 s) = 875 W [No u.e.] (1) 1

Discussion of student's answers

k.e.:

Skiers gain ke, increases total energy used (1)

but not significant / 875 W << (364 000 – 288 000) W (1)

Heat:

Heat loss indicated (1)

Identified mechanism, e.g. friction electrical in motor (1) 4[10]

553. Total e.m.f of cells in series

e.m.f. in series add up / 6000 × 40 × 10-3 V (1)

= 240 V (1) 2

Internal resistance of cells in series

6000 × 0.70 Ω = 4200 Ω (1) 1

Calculation of current

I= V ÷ R (1)

= 240 V ÷ 4200 Ω = 0.057 A (1) 2

Calculation of total current

20 × 0.057 A = 1.1 A (1) 1

Explanation of voltmeter reading

Since V = IR (1)

and R = 0 Ω (1)

V = 0 V ≠ e.m.f.

OR

V= E – Ir (1)

r ≠ 0 Ω (1)

so V < E

OR

Lost volts (1)

across internal resistance (1) 2

Voltmeter reading

0 V [No u.e.] (1) 1[9]

554. Path difference

2 × 1.11 × 10-7 m = 2.22 × 10-7 m (1) 1

Explanation of why light appears dim

Path difference = ½ × wavelength (1)

so waves in antiphase/destructive interference/superposition (1) 2

Reason for increase in film thickness

Because of gravity/soap runs down (1) 1

Explanation of whether film further down appears bright or dark

Path difference = wavelength (1)

Waves in phase/constructive interference (so appears bright) (1) 2

Explain bright and dark stripes

Different positions have different thicknesses/path differences (1)

So some points in phase, some in antiphase/some points have constructive interference, some destructive (1) 2

Movement of bright and dark stripes

Soap flows down/thickness profile changes (1)

so positions of destructive/constructive interference changing (1) 2

Alternative path added to diagram

One or more extra reflections at each internal soap surface (1) 1[11]

555. Diagram

(i) Any angle of incidence marked and labelled I (1)

(ii) Any angle of refraction marked and labelled R (1)

(iii) Angle of incidence/reflection at lower surface marked and labelled G (1) 3

Refraction of light

Velocity of light is lower in glass (1) 1

Velocity of light in hot air

Velocity is greater (1) 1

Property of air

(Optical) density / refractive index (1) 1[6]

556. Graph

Suitable readings from graph (1)

Gradient = 9.5 (no u.e) (1) 2

Equation

Use of y = mx + C or υ = u + at (1)

leading to υ = 9.5t + 2 (1) 2

Weight of ball

W = mg = 0.25 x 9.81 = 2.5 N [2.4N] (1) 1

Validity of statement

(F = 6π η rυ ) = 6π × 0.040 × 1.71 × 10-5 × 32 (1)

= 4.1 × 10-4 (N) [No u.e.] (1)

[OR

υ = F /I 6π η r = 2.5 / 6π × 0.040 × 1.71 × 10-5 = 1.9 × 105 m s-1 (1)from graph υ 32 m s-1 (1)]

Therefore, viscous drag is not equal to the actual weight (1) 3

Completion of diagram

At least two streamlines drawn below ball (1)

At least one eddy drawn above ball (1) 2[10]

557. Resistor

(i) A = π r2 = π × (4.0 × 10-4)2 (1)

= 5.03 × 10-7 m2 (no u.e) (1) 2

(ii) Recall of R = ρ l/A (1)

Length l = RA/ρ

= 0.12 × 5.0 × 10-7 / 1.8 × 10-8 [substitutions]

= 3.3 m (1) 3

Advantage of using iron wire of same diameter

Shorter piece of wire needed (if iron chosen) (1) 1[6]

558. Description

Electron (near surface of cathode) absorbs photon and gains energy (1)

Work function is energy needed for electron to escape from surface (1)

Electrons released in this way are called photoelectrons (1) 3

Lowest frequency of radiation

f0 = E/h (1)

= 2.90 × 10-19 J/6.63 × 10-34 J s (1)

= 4.37 × 1014 Hz (1) 3

Suitability of potassium

λ = 3 × 108 m s-1 / 4.37 × 1014 Hz [use of lowest frequency] (1)6.86 × 10-7 m [with suitable comment] (1)

OR

f = 3 × 108 m s-1 / 4.0 ×10-7 and f = 3 × 108 m s-1 / 7.0 × 10-7 [usesrange of λ ] (1)f = 7.5 × 1014 Hz to 4.3 × 1014 Hz [with suitable comment] (1) 2

[Suitable comment – e.g. this is within range of visible light/almostall of the visible light photons will emit photoelectrons]

Maximum kinetic energy

Use of E = hc/λ AND minimum wavelength (1)

Max photon energy = hc/λ = 6.63 × 10-34 J s × 3 × 108 m s-1/(400 ×10-9m)

= 4.97 × 10-19 J [no u.e]

Max k.e. = max photon energy – work function [or use equation]

= 4.97 × 10-19 J – 2.90 × 10-19 J

= 2.07 × 10-19 J [allow ecf if wrong wavelength used] [no u.e] (1) 3

Why some photoelectrons will have less than this k.e.

One point from:

• photon energy might be transferred to electron below surface• so some energy transferred to atoms on the way to surface• hence electron leaves surface with less energy than max• max is for electron from the surface• lower energy photon responsible for emission (1) 1

[12]

559. Flow behaviour

Viscous (1) 1

Definitions

Any two from:

• elastic – is elastic - it returns to original length after stretching• brittle – not brittle - it extends a lot before breaking• hard – not hard - does not scratch other materials• durable – is durable – can be repeatedly loaded/unloaded

without change in properties• stiff – not stiff – force vs extension graph has a low gradient (2) 2

Graph

Realisation that this is area under (top) graph (1)

29 – 32 squares (1)

× 5 × 10–3 N m (1)

= 0.15 – 0.16 N m (J) [allow ecf from wrong scaling factor / no.of squares] (1) 4

Work done by rubber band

e.g. less energy given out than stored, OR it can do less work thanthat required to stretch it, OR less than the work done stretching it (1) 1

Why tyres become hot

This energy difference is stored as internal energy (heat) in the tyre (1) 1[9]

560. Direction of travel of electron

Anticlockwise (1) 1

Why track is curved and spiral

Force perpendicular v (1)

Force perpendicular B (1)

electron loses energy or v (1) 3

Measurements from photograph

rmax = 108 (100 - 120) mm (1)

onsubstitutiORequation)108(108.0106.14 19

×××=

=⇒−

BQrp (1)

= 6.9 (6.4 – 7.7) × 10-20 (kg ms-1) [no u.e.] (1) 3

Consistency of tracks

Any two from:

• photon leaves no track

• opposite/different directions of curvatures/spiral [NOT oppositedirections without reference to curve/spiral]

• similar/same curvatures/radii/shape

• no evidence of any other particles/two tracks only (2) 2

[Symmetrical scores once under bullet point 2 or 3]

How event obeys two conservation laws

Naming two laws e.g. charge and momentum (1)

Any one from:

• Charge O → +1 + – 1• Momentum ( ) γ initial → electron + positron• Momentum ↔ electron = positron• Energy of photon = mass + energy of electron and positron (1) Max 2

[11]

561. Why person moving in a circle must have an acceleration

Acceleration due to changing directionORIf not it would continue in straight line (1) 1

Centripetal acceleration

a= 22

OR rwr

v (1)

×=

=××××π==

−−

−

)srad(103.7

OR

)ms(465606024

104.622

15

16

w

T

πrv

(1)

⇒ a = 0.034 (m s-2) [ no u.e.] (1) 3

Which force is the larger

mg is larger than R/R is smaller than mg (1)

mg – R / centripetal/accelerating/resultant force acts towards centre (1) 2

Differing apparent field strength

(0.034 ÷ 9.81) × 100%

= 0.35%

OR (0.03 ÷ 9.81) × 100% = 0.31 % [NOT 0.3%] (1) 1[6]

562. Energy stored in a capacitor

Justify area: W = QVORwork/area of thin strip = V × ∆ Q (1)

Area under graph (1) 2

Energy stored when capacitor charged to 5000 V

W= 21 QV= 2

1 × 0.35 × 5000 J

= 875 J (1) 1

Time constant for circuit

5000/e or 3 = 1840/1667 V (1)

⇒ T.C = 3.3 m s [3.1 – 3.6 m s] (1)

OR

Initial tangent → t-axis (1)

Accept between 3.5 and 4.0 m s (1) 2

[Also allow use of exponential formula with appropriate substitutionof correct V and t, e.g. 2000 and 3 ms]

Capacitance

C =R

T or as numbers (1)

3.3 m s → 7.0 × 10-5 F [Allow e.c.fs.]

4.0 m s → 8.5 × 10-5 F (1) 2

[OR using graph: C = Q/V (1)= 0.35/5000 = 7.0 × 10-5 F (1)]

Energy left in capacitor

At 2 ms, V = 2700 V [2600 – 2800] (1)

⇒ E = 21 CV2 OR 2

1 QV

= 255 J [e.c.f, depends on method] (1) 2

Energy setting

Energy leaving capacitor = (875 – 255) J

= 620 J [e.c.f ] (1)

Energy delivered = 620 × 60/100 J

= 372 J

⇒ 380 J setting [Allow e.c.f] (1) 2[11]

563. Electric field

)m(10300

)V(1006−×

(1)

= 3.3 × 105 V m-1 (1) 2

Force

F = Eq = 3.3 × 1.6 × 10-19 (N)

= 5.3 × 10-14 N [Allow e.c.f] (1) 2

Why force has this direction

Vertical line ↑ (1)

directionfieldoftermsinOR

platepositivetoAttracted (1) 2

How much energy hole gains

W = F × d = 5.3 × 10-14 × 2.8 × 10-10 (J) (1)

= 1.5 × 10-23 J [Allow e.c.f] (1) 2[8]

564. How torch works and factors which affect brightness

At least two lines leaving N and S pole, diverging and crossingwires (1)

Arrow leaving N pole/towards S pole (1)

Field/flux lines cut wires (1)

→ changing B/φ OR td

dφ OR Faraday's law causes V or I (1)

If causes V followed by causes I (1)

Any two of rotation, field strength, number of coils (1)

Appropriate direction e.g. faster rotation brighter/more V/I (1)

R ↓ ⇒ brighter (1)[Max 6]

565. Simple harmonic motion – conditions

Acceleration (or force) proportional to displacement[OR a ∝ – x] (1)

Acceleration (or force) directed towards equilibrium position (1) 2

Graph

Horizontal line at Ep = 19 J (1) 1

Calculations

(i) Use of EK = Er-Ep (1)= 19J – 5J= 14 J (1)

(ii) Use of Ep = 21 kx2 (1)

with readings from graph (1)e.g. k = 2 × 5 J / (0.02 m)2 = 2.5 × N m-1 (1)

[8]

566. Expression for gravitational force

F = GMm/r2 (1) 1

Derived expression

Reasoning step must be clear, e,g, mg = GMm/r2 (1)so g = GM/r2 (1) 2

Sun's gravitational field strength

g = 6.67 × 10–11 N m2 kg-2 × 1.99 × 1030 kg / (1.50 × 1011 m)2 (1)

= 5.9 × 10-3 (N kg-1) [no u.e.] (1) 2

Diagram

(i) Jupiter marked closest to Earth (1)

(ii) (Labelled) arrows towards Jupiter and Sun (radially) (1) 2

Maximum percentage change

3.2 × 10-7 / 5.9 × 10-3 × 100% = 0.005% (1) 1

Maximum value of the ratio

Use of g ∝ M/r2 (1)

Hence gVenus/gjupiter = 152/400 = 0.56 [OR 9/16] (1) 2

Comment

E.g. reference to % change caused by Jupiter (combined with effectcaused by Venus)OR g of (all) planets is very small compared with g of Sun (1) 1

[11]

567. Binding energy

Energy released when separate nucleons combine to form a nucleus (1) 1

[OR energy required to split nucleus into separate nucleons]

Binding energy of a nucleus of uranium-235

Reading from graph: 7.2 ± 0.2 MeV (1)

Hence binding energy = 7.2 × 106 × 235 = 1.7 × 109 eV (1) 2

[Second mark is for multiplying graph reading by 235]

Energy released

Half-sized nucleus has binding energy/nucleon = 8.2 (± 0.2) MeV (1)

so released energy/nucleon = (8.2 – 7.2) × 106 × 1.6 × 10-19

(=1.6 × 10-13 J) (1)

Total energy released = 235 × 1.6 ×10-13 J =3.8 × 10-11 J (1) 3

Energy released by fission

(2).6 × 1024 × 4.0 × 10-11J = 1.0 × 1014 J

[Allow e.c.f from candidate's numbers, within range] (1) 1[7]

568. Explanation

Any two from:

• signal attenuated through interaction with particles/asteroid• signal diffracted on leaving transmitter• not all of radio signal will be received by asteroid• (reflected) signal spreads out (varying as 1/r2) (2) 2

Explanation of how reflected signal can be used to calculate speed

E.g.frequency/wavelength of reflected signal changed (1)Doppler Effect specified / Doppler equation given (with λ or f) (1)all symbols in equation υ = c ∆ f / femitted explained (1) 3

Calculations

(i) Distance = speed × time

= 3.0 × 108 × 120

= 3.6 × 1010 m (for return journey)

Asteroid distance is 1.8 × 1010 m (1)

(ii) Time to reach Earth = distance ÷ speed

= 1.8 × 1010/4000 [Allow e.c.f from (i)]

= 4.5 × 106 s

(= 52 days) (1)

Assuming that speed does not change (due to gravitationalattraction of Earth) (1) 3

Explanation of why asteroid unlikely to collide with Earth

E.g. Earth will have moved around Sun[Apply e.c.f from (ii)] (1) 1

[9]

569. Phenomenon of resonance in the context outlined etc

Any five from:

• spheres can oscillate• resonance when forcing frequency = natural frequency• sound provides forcing frequency• low frequency due to mass/density of lead spheres

At resonance, there is:

• large amplitude of oscillation (of spheres)• maximum energy transfer to spheres• energy transfer to thermal in the rubber• minimum energy transfer to neighbours 5

[5]

570. Explanation of Pelton wheel

Quote of F = ∆ mυ /t [or in words] (1)

Negative moment/velocity after (1)

Increased (twice) momentum / velocity change/mυ - mu comparedwith falling off plane paddle (1)

Idea of doubled (1) Max 3

Percentage efficiency of station

Energy available = mgh (1)

Power input = 270 × 9.8 × 250 (1)

Efficiency = 500 000 × 100 / 661 500 = 76 (%) (1) 3

Other desirable properties

For example:

Hard – does not wear/scratch/dent (2)

Tough – can withstand dynamic loads/plastic deformation (2)

Strong – high breaking stress/force (2)

Smooth – low friction surface (2)

Durable – properties do not worsen with time (2) Max 4

Time to repay initial investment

Each hour worth 500 × 0.0474 = £23.70 OR total no of kW hrequired 1.000000/0.0474 = 2.1 × 107

No. of hours to repay = 1 000 000 /23.70 OR 2.1 × 107 /500

24 × 365 = 4.8 years to repay debt (1)

Assumption

Constant power production / no interest charges / no repair costsno wages for employees (1) 4

[14]

571. Deduction

F = mυ 2 /r / mrω 2 or accel/field strength = υ 2/r/rω 2 (1)

GMm/r2 = mυ 2/r or mrω 2 / GM/r = υ 2 / υ 2α 1/r/rr

Gm 2

2

υ= (1)

So radius smaller than Earth/Sun or accompanied by “as T smaller”/ “larger gravitational force” / “larger field strength” (1)

But rPLANET = 0.18rearth (1) Max 3

Discussion

Blue green – colours absorbed by (atmosphere/surface) of planet (1)

The spectrum of light given out by this star is different to ours (1)

Periodic bright/dull – planet orbiting Bootes implied (goes in front and behind) (1)

Dimmer when it goes in front/behind/argument based on reflection (1)

Varying wavelength Doppler shifted (1)

Relative movement produces change in frequency/wavelength (1) Max 4[Max 6]

572. Gradient

Use a gradient or use of υ = u + at (1)

10 (either no unit or m s–2) (1)

[A bare answer of 9.8 gets no marks; A bare answer of 10 gets 2 marks]

Significance

It is the acceleration (due to gravity) or close to g (1) 3

Ball at point A

It hit the floor/bounces/(idea of collision with floor) (1) 1

Calculation of height of window above ground

An area / quote an equation of motion (1)

Put in relevant numbers for large triangle / correct substitution[ecf from first part, or use of 9.8] (1)

45 m [accept 44 to 46] (1) 3[7]

573. Free-body force diagrams

Tension/T/pull of string/NOT pull of ceilingReaction force from string/Contact force from string (1)

W e i g h t / W / m g / p u l l o f E a r t h /g r a v i t a t i o n a l p u l l / N O T g r a v i t y

F g

S i t u a t i o n 1 S i t u a t i o n 2

G r a v i t a t i o n a l f o r c e /p u l l o f S u n / w e i g h t /F g / m g

[Cancel a mark for every extra force (within each diagram); correct line ofaction required; penalise “gravity” once only.]

Force Newton’s third law pair, noting its direction and the body on which it acts

Weight On Earth......Upwards (1)

Tension On string......Downwards [N.B. allow ecf from ceiling pull in previous part]

(1)

Gravitational force

On Sun......Towards Earth/to the left (1)

[No other ecfs from incorrect forces: “in opposite direction” penalise once only.] 3[6]

574. Mass approximately 4 kg

Use of volume = π r2 × h (1)

Use of mass = their volume above × density (1)

Mass = 3.75 (i.e. ≈ 4) [no u.e.] [Must be calculated to 2 significantfigures at least] (1) 3

Calculation of change in g.p.e

Use of ∆ g.p.e = mg∆ h (ecf from above) (1)

39 – 44 J (positive or negative) (1) 2

Calculation of average power output

Use of Power = energy/time or use of P = Fυ (υ = 1.8 × 10–6 m s–1) (1)

Correct conversion of time into seconds (604 800 s) (1)

6.4 – 7.3 × 10–5 W [e.c.f. gpe above] (1) 3

[Answer in J/day, J/week, J/hour – can get 2 marks, i.e.1st and 3rd marks][8]

575. (a) Newton’s second law of motion

Rate of change of momentum ∝ (OR =) force/Force = mass × acceleration / F ∝ (OR =) ma with symbols defined/a ∝ F and a ∝ 1/m with symbols defined (1)

Acceleration or (rate of) change of momentum in direction of force (1) 2

Description of demonstration that acceleration is proportional toresultant force

Technique for reducing/compensation for friction (1)

e.g. Air track/friction compensated runway/low friction wheels ortrack/slight slope drawn or mentioned [not if slope used to vary theforce] / smooth runway

Correct technique for applying a constant “known” force (1)

e.g. Forcemeter/elastics of constant length/slope whereF = mgsinθ / mass on string and pulley if masses are small or movedfrom pulley to trolley

Apparatus for measuring acceleration (1)

e.g. Ticker timer/ light gate plus double interrupt/ two light gatesplus one interrupt/ motion sensor/ strobe camera/ video

Principle behind the measurement (1)

e.g. gives position at known time intervals or times for knowndistances

Vary F (1)

Graph of a versus F should be a straight line through the origin (1) Max 5OR values of F/a are constant

(b) Explanation of observations

Quality of written communication (1) 1

Pencil is accelerating/increasing momentum (1)

This requires a forward force (1)

Back edge of shelf pushes forwards (1)

Converse argument for deceleration (1)

OR

Pencil travels at constant velocity / constant momentum littleacceleration / stays at rest (1)

In line with Newton I/due to its inertia/because little or no force onit (1)

If car accelerates, it “catches up” with pencil (1)

Converse argument for deceleration (1) 4[12]

576. Amount of work done by each of the forces

(Each of the forces does)zero (1)

Forces perpendicular to motion [consequent] (1) 2

[No marks if imply that work = 0 because forces cancel]

Determination of force F

Use of gradient seen/implied (1)

F = 2.7 – 2.9 N (1) 2

Graph

Straight line finishing at (1.8, 0) (+ or – 1 small square) (1)

Starting at (0, 5) (+ or – 1 small square) (1) 2

Calculation of speed

Use of k.e. = ½ mυ 2 / use of F = ma and equation of motion (1)

υ = 3.5 ms–1 (ecf) (1) 2

Sketch of graph

Ascending line whose gradient decreases as d increases (1) 1

Shape of graph

Force greater at higher speed/gradient is the force/force decreases withdistance (1) 1

[10]

577. Number of neutrons

8 (1)

Decay constant

Use of λ = 0.69/t1/2 (1)

λ = 1.2 × 10–4 yr–1 OR 3.9 × 10–12 s–1 (1) 3

Number of nuclei

3.0 × 1014 (1)

Calculation of activity

Their N × their λ (1)

= 1170 Bq [No e.c.f. if no conversion to seconds] (1) 3

Nuclear equation

C146 → N14

7 + e01− (1) (1) 2

[8]

[1 mark for N147 , 1 mark for e0

1− as β−01 ]

[Must be on correct side of arrow]

578. Outline of evidence from Geiger’s and Marsden’s scattering experiment

Most alpha particles went (almost) straight through (1)

Some or a few deflected at larger angles/>90°/rebounded (1)

A tiny minority [e.g. 1 in 8000] were deflected at angles > 90° OR rebounded (1) 3

Suggestion

No large deflections/all go (almost) straight through (1)

Explanation

No concentrated charge/mass OR no massive object (to hit) no denseobject to hit [consequent] (1) 2

[5]

579. Unit of current

Amps/ampere (1) 1

Base units of p.d.

For V = IR method

Any three from:

• V = J C–l

• C = A s• J = N m

• N = kg m s–2

[kg m2 s–3 A–1]

[See J = kg, m2 s–2 (1) (1)]

OR

For P = VI method

• Watt is J s-1 / J/s

• V = J s–1 A–1

• J = Nm

• N = kg m2 s–2 (1) (1)] (1) (1) (1)

[See kg m2 s–2 (1) (1)] 3[4]

580. Show that resistance is approximately 45 Ω

A

lR

ρ=

7

5

100.8

m65.0m105.5R

−

−

××Ω×=

= 44.7 Ω [No u.e.] (1)

[Must see this value and not 45]

Table

Switch X Switch Y Resistance of heater/ΩOpen Closed 22.5/22.35 (1)

Closed Open 45/44.7 (1)Closed Closed 15/14.9 (1)

[No u.e.] 3

Calculation of maximum power

R

VP

2

= Use of equation with 15 Ω OR their minimum value (1)

= 3526 W,3500 W [full ecf] (1) 2

Explanation of power output fall

increases)metalsof(cetanresishot/hottergetsitasOR

increasesheatertheofetemperaturtheAs

Since V is constant P = R

V 2

OR P = VI and V = IR

[Then P ↓ as R ↑] (1) 2

OR P ∝ R

1 [so P↓ as R↑]

[10]

581. Explanation of greater drift velocity

(Electrons have greater drift velocity) in the thinner wire (1)

Any two from:

• Same current in both wires• Reference to I = nAQυ• nQ same in both wires (1) (1) 3

Explanation of higher dissipation of power

(Higher power is dissipated) by the smaller(er)/ low resistor (1)

Any two from:

• Resistors have same p.d. across them• The small resistor has the largest current [or reverse]

• Power = voltage × current, OR voltage2 ÷ resistance [NOT I2R] (1) (1) 3[6]


Resistor with another variable resistor/potential divider/variable powerpack (1)

Ammeter reading current through resistor (1)

Voltmeter in parallel with resistor (1) 3

Graph labels

labelledBothlamp–Curve

resistor–lineStraight

(1) 1

Potential difference

At 0.5 A p.d.= 3.5 V / 3.4 V + 7.8 V / idea of adding p.d. [for same current] (1)

= 11.2 V/11.3 V (1) 2

[Accept 11.0 –11.5 V]

Resistance of lamp

A5.0

V5.3 [OR their value of p.d. across lamp ÷ 0.5 A] (1)

= 7.0 Ω (1) 2

[e.c.f. their value][8]

583. Graph illustrates

p1V1 = p2V2 / pV = constant / p ∝ V

1 / p =

V

ttancons (1) 1

Pressure of trapped air

(i) 1.20 × 105 Pa (1)

(ii) 0.80 × 105 Pa (1) 2

Length of column

(0.80 × 105 Pa × 24.0 cm) = (1.2 × 105 Pa × L)

L = 16.0cm [Accept 0.16m]

See one of their pressures × L/24 (1)

Their upright pressure × L = their inverted pressure × 24 (1)

Answer L = 16 cm (1) 3

Assumption about the dry air

Constant temperature (1) 1[7]

584. Luminosity of Sun

Attempt to use I = L/4π D2

L = 4π × (1.5 × 1011m)2 × 1.4 × 103 W m–2 [ Ignore 103 error] (1)

= 3.96 [OR 4.0] × 1026 W [at least 2 s.f.] 3

Why intensity at top of atmosphere used

Atmosphere absorbs/scatters (some radiation)/reflects/filtersout/equivalent (1) 1

Estimate of energy released for each helium nucleus created

∆ m = [4(1.67) –6.64] × 10–27 kg [Ignore 10n error] (1)

= 0.04 × 10–27 kg (1)

∆ E = ∆ mc2 (1)

= 4 (OR 3.6) × 10–12 j [e.c.f. ∆ m] (1) 4

[If only ONE hydrogen atom in ∆ m: ecf to possible 3/4)]

Show number of nuclei

Number = )101(104

104 3812

26

×=××

− (1)

[ecf energy above even though ≠ 1×1038]

Mass of hydrogen

Mass = 1×1038 × 4 × 1.67 × 10–27 kg (1)

OR 1 × 1038 × 6.64 × 10–27 kg

= 7 × 1011 kg (6.6) (1) 2

[Again ecf ONE hydrogen atom → possible 1/2. If use theirvalue for number of nuclei → possible 1/2]

[10]

585. Meaning of term binary system

Two stars/masses move in (circular) orbits about commoncentre/each other (1) 1

Diagrams

At A and E small star within shaded area (1)

At C small star behind labelled or implied (1) 2

Explain why dip in curve at A is smaller than dip at C

At A (or E) small star is in front of/blocks part of large starAt C small star is behind or is blocked by large star (1)

Small star is a lot brighter (than large star) hence dip sizes (1) 2

Orbital period of binary

T = 22 hours (1) 1

How long it takes small star to cross disc of larger one

Value in range 0.7 – 1 hour (1) 1

Addition to light curve

Continuous horizontal line (1)

at same level as precious line OR very close to (1) 2[9]

586. Comparison of two metals A and B (stress-strain curves)

A is stiffer since steeper /bigger gradient/large Young modulus (1) 1

(i) Stronger: A since UTS/it breaks at 300 Mpa (± 20)OR since B breaks at 190 (± 30) MPa (1)

(ii) More ductile: B since strain 0.25 OR since A strain = 0.15 (1) 2

Identify A and B

A is mild steel; B is copper (1) 1

Estimate of work done in stretching A to breaking point

Attempt at area [NB not just a triangle] (1)

300 ×106 Pa × 0.15 [Ignore 10n error] (same stress ranges) (1)

4 → + 5 × 107 J m – 3 (1) 3

Quench hardening

Heat and cool (1)

Rapid cool/plunge into water (1) 2[9]

587. Diagram of apparatus to determine Young modulus of copper

Were firmly fixed to ceiling/beam/end of bench (1)

Load and ruler/scale (1)

Means of reading small extensions e.g. pointer againstscale/vernier (1) 3

Length of wire being tested

Appropriate length ≥ 2 m [Less if vernier used] (1)

Cross-sectional area of wire

Micrometer (1)

Diameter in several places (1) 3

Unit of k in Hooke’s law

N m –1/kg, s –2 (1) 1

Show that

E = F/A ÷ e / l (1)

= Fl/Ae (1)

but F/e = k/substitute F = ke (1) 3[10]

588. Calculation of binding energy

∆ m = 8 (1.007276 + 1.008665) u – 15.990527 u (1)

∆ E = ∆ m × 930 [Their ∆ m OR 0.137001 u) (1)

= 127.4 MeV [130]

[Answer in joule, max 2 out of 3] 3

Binding energy per nucleon

= answer above/16 [ecf joule also]

= 7.9 → 8.1 MeV [ecf] (1) 1

Graph

Steep rise and less steep fall; starts close to 0,0 and falls nomore than ½ way to axis (1)

peaking in region 25 – 75 (1) 2

Positions on Graph

[for full marks. must be placed on or close to drawn line]

(i) O labelled at approximately 16 (½ way to 50) (1)(ii) Fe labelled at peak, wherever it is (1)(iii) U labelled at just short of 250 (1) 3

[9]

589. Fundamental particle

A particle which cannot be further divided/which has no “parts”inside it/one of the 12 particles of which all matter is made (1)

[Not “one which cannot decay to another particle”]

Circled fundamental particles in list (2)

Positron and muon

[If more than two circled, –1 for each extra one] 3

Explanation

Any three from:


Mesons are composed of a q and an q (1)

These have charges ± 2/3 and ± 1/3 (1)

Shows all possibilities (+1, 0, – 1) OR other convincingarithmetic to show max +1 (1) Max 3

[6]

590. (a) (i) Use of set square against side of foil to ensure rule isperpendicular to side

OR

Repeats at different point shown on the diagram (1)

Both recorded to nearest mm with unit, and sensible (1)

Both repeated and correctly averaged (1) 3

(ii) Minimum 3 readings with unit and to at least 0.01 mm (1)

t ± 0.002 mm [of Supervisor’s value] (1)

∆ (16t) found [from range or ½ range], hence correctcalculation of percentage[but allow 0.01 mm if single readings or all readings thesame] (1)

Advantage:

Measuring, larger thickness/average of several thicknesses/enable centre thickness to be measured (1)

So that percentage uncertaintly is reduced (1)

Disadvantage:

May be trapped air between the layer/not compressedenough (1) 6

(iii) Correct calculation [ecf 16t for t or volume of folded foil]

≥ 2 significant figures + unit (1)

Correct calculation with unit [ecf] [rounded to 2 significantfigures] (1)

Value 1.8 – 2.8 g cm–3 and 2/3 significant figures (1) 3

(b) (i) Circuit set up correctly without help. [Ignore polarity errors] (1) (1) 2

(ii) V ≈ 3 mV to 0.1 mV with unit (1)

I ≈ 6 mA to 0.1 mA with unit (1)

Correct calculation with unit [Allow ecf] (1)

Two significant figures (1) 4

(iii) Greater than the resistance of the foil because of resistanceat the connections between the plugs and the foil (2 or 0) (1) (1) 2

[OR Greater because of resistance of connecting wires, 1 outof 2]

(iv) Correct calculation > 2 significant figures with unit[ecf 16t for t or wrong t] (1)

Correct substitution (1)[If the candidates has attempted to convert to a consistent setof units, allow the mark.]

Correct calculation + unit (1)

Value 2.0 → 5.0 × 10 – 8 (Ω m) (1) 4[24]

Sample results:

(a) (i) l = 30.0, 30.0 cml = 30.0 cmw = 20.0, 20.0 cmw = 20.0 cm

(ii) 16t = 0.24, 0.28, 0.25, 0.24, 0.19, 0.25 mm

1 6 t = 0.24 mmt = 0.015 mmRange = 0.09 mm in 16t

Percentage uncertainty = %

.

.19100

240

0902

1

=××

Advantage

Measuring larger thickness / average of several thicknesses/ enable centrethickness to be measured, so that percentage uncertainty is reduced.

Disadvantage:

May be trapped air between the layer/not compressed enough

(iii) V = l w t = 20×30×10

0150.

= 0.90 cm3

m = 1.72 g

Density = 900

721

.

.

= 1.9 g cm–3

(b) (ii) V = 2.1 mV

I = 6.37 mA

R = V/I = 2.1/6.37

= 0.33 Ω

(iii) Greater than the resistance of the foil because of resistance atthe connections between the plus and the foil

(iv) A = wt = 5 × 10–3 × 0.015 × 10–3

= 7.5 × 10–8 m2

ρ = 80

1057330 8

.

..

l

RA −××=

= 3.1 ×10–8 Ω m

591. (a) t ≈ 3 s from ≥ 3 readings (1) (1)

[...≥ 2 readings, (1)]

[Misread stopwatch 0/2Nearest second – 1Unit omitted – 1Incorrect average – 1]

Correct calculation, more than two significant figures, + unit (1) 3

(b) Vertical rule checked with set square (1)

Correct h shown (1)

[Ball must be shown]

Difference method shown with consistent heights, 1 m apart (1)

Correct calculation of Ep (1)

Correct calculation Of EK [ecf incorrect m] (1)

[Units seen at least once In parts (b) and (c), otherwise – 1]

k: 1.5 → 3.0 (1) 6

[ecf systematic error in Ep or EK]

(c) t < t from (a) and from ≥ 3 readings (1) (1)

[≥ 2 readings, (1)]

[Allow ecf for stopwatchNearest second – 1Units omitted – 1Incorrect average – 1]

Correct calculation of Ep (1)Correct calculation of EK (1)

[ecf for mass only]

Value: 1.5 →.2.5(1) 5

[Ignore units on K. ecf is a systematic error in Ep or EK]

(d) Correct calculation with average as denominator (1)

Sensible conclusion related to experimental uncertainties (1) 2

(i) Use blocks of different height or lab jacks/use balls ofdifferent mass

(ii) Several values of h or m indicated [ ≥ 5 if numerical] (1)

Measure t (1)

to travel a measured distance s (1)

Calculate EP and EK (1)

Use light gates to find υ (1)

υ = 2 υ (1)

(iii) Ep against EK OR EK against Ep stated or sketched (1)

Then straight line through origin shown with axes labelled (1)

Correct. hence gradient (1) Max 8[24]

Sample results

(a) t = 2.44, 2.56, 2.43, 2.40

t = 2.46 s

υ = 462

0012

.

.×

= 0.814 m s–1

(b)

1 . 0 0 m

h

h = 85 – 15 = 70 mm

m = 56 g

Ep = mgh = 0.056 × 9.81 × 0.070= 0.0385 J

Ek = ½ mυ 2

= ½ × 0.056 × (0.814)2 = 0.0186 J

k = K

P

E

E = 072

01860

03850.

.

. =

(c) t = 2.04, 2.04, 1.88, 1.94 s

t = 1.98 s

υ = t

s2 =

98.1

2 = 1.01 ms –1

h = 117 –15 = 102 mm

Ep = 56 × 10–3 × 9.81 ×0.102= 0.0560 J

EK = ½ × 56 ×10–3 × (1.01)2

= 0.0287 J

k = 95.10187.0

0560.0 =

(d) Percentage difference %100)95.107.2(2/1

95.1–07.2×

+

= 6.0%

Difference is less than 20% (or 40%), therefore k is likely to be constant

(e) (i) Use blocks of different height or lab jacks / use balls of different mass

(ii) Several values of h or m indicated (≥ 5 if numerical)Measure tto travel a measured distance sRepeat values for each h or mCalculate Ep, or EK

(iii)

E

E

p

K

(iv) Gradient = k


Acceleration/force is proportional to displacement (1)

but in the opposite direction / towards equilibrium point /mean point (1) 2

Graph

Sine curve (1)

-ve [consequent] (1)

A and B / (i) and (ii) / a & x [beware a v x] (1) 3[5]

593. Angular speed

Conversion of 91 into seconds – here or in a calculation (1)

Use of T = 2π /ω allow T= 360/ω

ω = 1.15 – 1.20 × 10–3 rad s–1 /6.9 × 10–2 rad min–1 /0.066 deg s–1 (1) 3

Acceleration

Use of a = rω 2 / υ 2 / r (1)

Adding 6370 (km) to 210 (km)/ 6580 (km) (1)

a = 8.5 to 9.5 m s–2 [No e.c.f. for 210 missed but allow for ω in rad s–1] (1) 3

Resultant force

Recall/Use of F = ma (1)

F = 35 –39 N [Allow e.c.f their a above only] (1)

Towards (centre of the) Earth (1) 3[9]

594. Table

Radio waves Sound wavesTransverse LongitudinalTravel much faster than sound Travel more slowly(Can) travel in a vacuum Cannot travel in a vacuumCan be polarised Not polarisedElectromagnetic Pressure/Mechanical wave

Any three of the above Max 3

Assumption

Attempt to calculate area (1)

Intensity = 0.02 kW m–2 OR 20 W m–2 (1)

Efficiency at collector is 100%/beam perpendicular to collector

Power

Use of I P/4π r2 (1)

Power = 3.3 × 1017 W [ecf their I]

No energy “lost” due to atmosphere (not surroundings) OR Inversesquare applies to this situation (1)

More efficient method

Use a laser (maser) / reference to beaming/ray (1) 1[10]

595. Ionisation energy

Use of × 1.6 × 10–19

2.2 × 10–18[No u.e.] (1) 2

Addition to diagram

(i) From 4 to 3 labelled R / (i) (1)

(ii) From 1 to 4 labelled A / (ii) (1) 2

Emission spectrum

Hydrogen ‘excited’ in a discharge/thin tube/lamp [not bulb] (1)

Viewed through a diffraction grating/prism/spectrometer (1)

Appearance of emission spectrum

A series of lines / colours on a dark background [accept bands] (1) 3

Region of spectrum

Radio/microwave (1) 1

Speed of galaxy and deduction

∆ λ = 8 (mm) / 211 - 203 (mm) (1)

Use of 3 × 108 (1)

υ = 1. 1(4) × 107 ms–1 [No e.c.f.] (1)

Moving towards Earth (us) (1) 4[12]

596. Photoelectric effect

Any two features and explanation from the following:

Feature: Experiments show k.e(max) ∝ f, OR not intensity[Accept depends upon] (1)

Explanation: Photon energy ∝ f (1)[Consequent]

k.e(max) ∝ intensity is a wave theory (1)

Feature: Emission of photoelectrons immediate (1)

Explanation: One photon releases one electron particle theory (1)[Consequent] Wave theory allows energy to “build up” (1)

Feature: (Light) below a threshold frequency cannot releaseelectrons (1)

Explanation: Particle theory-f too low as not enough energy is released[Consequent] by photon to knock out an electron (1)

Wave theory- if leave a low frequency beam on longenough, it will produce enough energy to release anelectron (1)

[Max 5]

597. How stationary waves could be produced on a string

Diagram showing:

String and arrangement to apply tension (1)

Vibration generator and signal generator (1) 3

Vary f / tension / length until wave appears (1)

Determination of speed of travelling waves

QOWC (1)

Determine node-node spacing; double to obtain λ (1)

Read f off signal generator / cro / use a calibrated strobe (1)

Use υ = f λ for υ (1) 4[7]

598. Cathode Ray Tube

Electron emission

• Heating effect (due to current) (1)• (Surface) electrons (break free) because of energy gain (1) 2

[Thermionic emission scores both marks]

Electron motion towards anode

The electrons are attracted to/accelerated by the positive anode (1) 1

Energy

Electron energy = (10 × 103 V) (1.6 × 10–19 C)

=1.6 × 10–15 J

Correct use of 1.6 × 10–19 OR use of 10 × 103 (1)Answer (1) 2

Number of electrons per second

Number each second = J106.1

A105.119

3

−

−

××

9.4×1015s–1

Correct conversion mA →AAnswer (1) 2

Rate

Energy each second = (9.4 × 1015 s–1) (1.6×10–15 J) (1)

= 15 Js–1 (W) / 14.4 Js–1 (1) 2

[ecf their energy][9]

599. (a) Any four points from:

• B is set into forced oscillations by A• and its amplitude of oscillation increases• Energy is transferred from A to B• Amplitude then reduces• Energy is transferred back to A• Process then repeats• Same natural frequency (1) (1) (1)• A is driver/A and B are coupled (1) 4

(b) ± 0.03 s of Supervisor from ≥ 40T (1) (1)

[± 0.03 from > 20T (1)]

± 0.05 from > 40T (1)]

TB > TA (1)

from ≥ 40TB (1)

[If no units at all –1]

Precautions (1) (1) 6

(c) T measured and repeated (1)

Correct calculation of 1/r and 1/TA, – 1/TB to ≥ 3 significant figures (1)

[Ignore unit]

Percentage difference found [Either value or average as denominator] (1)

≤ 5% (1) (1)

Hence sensible comment (1)

[10% acceptable experimental error] 6[16]

Sample results

(a) • B is set into forced oscillations by A• and its amplitude of oscillation increases• Energy is transferred from A to B• Amplitude then reduces• Energy is transferred back to A• Process then repeats• Same natural frequency

(b) 20TA = 15.94, 15.87 s

TA = 0.795 s

20TB = 17.32, 17.28 s

TB = 0.865 s

Fiducial marker in centre of oscillation.Eye level with markerRepeats shown

(c) 5T = 46.66, 47.54 sT = 9.42 s

11060429

11 −== s..T

865.0

1–

795.0

11–

1 =BA TT

= 1.258 – 1.156

= 0.102 s–1

Percentage difference = %.%.

.83100

1040

0040 =×

Acceptable experimental error therefore supported

600. (a) Potential divider (1) (1)

Ammeter in series with lamp (1)

Voltmeter in parallel (1)

Ammeter: 100 mA, 150 mA, 200 mA (1)Voltmeter: 10 V, 15 V, 20 V (1)

[ORBoth values >values for lamp but not standard (1)]

Heating effect correctly related to resistance change (1)

Sensible argument (1) 8

(b) lnI = nlnV + lnk

y = mx + c [OR log10] (1) 1

(c) Table with correct values to ≥ 2 decimal places (1)

Graph:

Suitable scale (1)Axes labelled with units (1)Plots (1)Line (1) 5

(d) ∆ x∆ y ≥ 64 cm2 (1)

Correct calculation, no unit, 2/3 significant figures and 0.50 and0.55 [when rounded to 2 significant figures] (1) 2

[16]

Sample results

(a) Diagram

Ammeter: 100 mA, 150 mA, 200 mAVoltmeter: 10 V, 15 V, 20 V

Heating effect correctly related to resistance change

Sensible argument

(b) lnI = nlnV + lnk

y = mx + c [OR log10]

(c)

V/V I/mA ln(V/V) log10 (V/V) ln(I/mA) log10(I/mA)0.87 20 –0.14 –0.060 3.00 1.301.33 25 0.29 0.124 3.22 1.401.81 30 0.59 0.258 3.40 1.482.51 35 0.92 0.400 3.56 1.543.08 40 1.12 0.489 3.69 1.604.02 45 1.39 0.604 3.81 1.654.66 50 1.54 0.668 3.91 1.705.66 55 1.73 0.753 4.01 1.746.79 60 1.92 0.832 4.09 1.78

(d) n = gradient = )2.0(––0.2

0.3–16.4= 0.53

601. (i) Tracks (of alphas) are the same length/alphas travel same orequal distance (1)

(ii) H/p + Li → 2α /2He (1)

Heandp 42

11 correctly labelled (1)

Li73 (1) 4

(iii) Mass defect = 0.01865u (1)

Either Or

Use of × 1.66 × 10–27 Use of × 930 (1)

Use of × 9.0 × 1016 Use of × 1.6 ×10–13 (1)

⇒ 2.79 ×10–12 J ⇒ 2.78 × 10–12 J (1)

Assume: proton has zero/very little k.e. (1) Max 4[8]

602. Show that deceleration = 7 m s –2

a = (υ – u) / t (1)

a = (0 m s–1 – 11.5 m s–1)/1.68 s (1)

a = 6.85 m s–2 [No u.e.] (1) 3

Calculation of frictional force

F = ma (1)

= 1400 kg × 6.85m s–2 [Allow ecf]

= 9590 N (1) 2

[F = 9800 N if 7 m s–2 used]

[Alternative: Force = kinetic energy ÷ distance (1) F = 9590 N (1)]

Why driver should not be charged

υ 2 = u2 + 2as

0 m2 s–2 = u2 + 2 × 6.85 m s–2 × 20.0 m (1)

u = m0.20ms85.62 2 ×× −

u = 16.6 m s–1 (1)

16.6 m s–1 < 18 m s–1, so not speeding (1) 3

[u = 16.7 m s–1 if 7 m s –2 used]

[OR substitute: 0 m2 s–2 = (18 m s–1)2 + 2 × (– 6.85 m s–2) × s (1)

s = 23.6 m (1), 23.6m > 20.0m, so not speeding (1)]

Assumption

e.g same deceleration as test car. OR same mass and braking force as test car/F÷ m same as for test car, OR not wet / icy weather OR speed zero at end ofskid/doesn’t collide before stopped OR skid starts when brakes applied/stops (1)when comes to rest OR same conditions 1

[9]

603. Explanation of superposition

When 2 (or more) waves meet / cross / coincide /interfere...(1)

Reference to combined effect of waves, e.g. add displacement / amplitude - maybe a diagram [constructive/destructive interference not sufficient withoutimplication of addition] (1) 2

Calculation of thickness of fat layer

Thickness = half of path difference

= 0.5 × 3.8 × 10–7 m

= 1.9 × 10–7 m (1) 1

Explanation of constructive superposition

Path difference of 3.8 × 10–7 m same as a wavelength of green light (1)

Waves are in phase / phase difference 2 π or 360° (1) 2

Explanation of what happens to other wavelengths

Path difference greater than/less than/not one wavelength/ waves not in phase / out of phase (1)

Constructive interference will not take placeOR (1)These colours will not appear bright (1) 2

Explanation of why colours are seen at other places

Thickness of fat variesORLight seen at a different angle to the meat surface (1)

Other wavelengths may undergo constructive interference/be in phaseOR (1)Path difference will vary 2

[9]

604. Meaning of m

× 10–3 (1) 1

Calculation of resistance for reading 3

R = V/I OR R = 74 × 103 V ÷ 150 × 10–9 A [ecf for milli] (1)

R = 4.9 × 105 Ω (1) 2

Calculation of power for reading 4

P = I × V OR P = R

V 2

OR P = I 2R (1)

= 210 × 10–9 A × 57 × 10–3 (1) 2

= 1.2 × 10–8 W

Plotting points on graph

Two correct points (1)Third correct point (1)Best fit straight line for points as they appear on student’s graph (1) 3

Predicting short-circuit current

Correct from graph, e.g 450 nA (1) 1

Suggested e.m.f

Correct from graph, or table, 110 mV (1) 1

Explanation of why voltage falls

Cell has internal resistance/ “lost volts” (1)

“Lost volts” = Ir, so lost volts increase as current increasesORV = E – Ir, so V decreases as I increases (1) 2

[12]

605. Show that value in cell B3 is correct

Vertical component = υ sin θ OR vertical component G2*sin(F2) (1)

Vertical component = 21 m s–1 × sin 36° (1)

Vertical component = 12.3 m s–1 [No ue] (1) 3

Suitable formula for B4

υ = u + at OR B4 = B3 – (I2*A4) (1) 1

Explanation of constant horizontal velocity

Any 2 from the following: (1) (1)

• Horizontal and vertical motion independent• No air resistance/horizontal force components• so no acceleration 2

Show that value in cell D7 is correct

Distance = speed × timeORD7 = C7*A7ORD7 = 17 m s–1 × 2s (1)D7 = 34m [No ue] (1) 2

Explanation of why flight time of 2.5 s suggested

Vertical component equal in magnitude and opposite in directionto initial vertical component (1) 1

Discussion of distance travelled for real discus

Air resistance [will decrease horizontal component of velocity] (1)so it will travel a smaller distance (1)ORAir flow may produce an upward force [decreasing, vertical acceleration] (1)increasing [the time of flight and therefore] the horizontal distance (1)ORDiscus launched above ground level (1)so it will travel a greater distance (1) 2

[11]

606. Diameters of dark ring

Diameter in frame 1 = 9 mm (± 1 mm)Diameter in frame 2 = 19 mm (± 1 mm) [No ue] (1) 1

Show that ripple travels about 25 Mm

Difference between diameters = 19 mm – 9 mm = 10 mmDistance travelled by one part = 10 mm ÷ 2 = 5 mm (1)

Scale: 200 Mm = 40 mm (39 mm to 41 mm)Distance = 5 mm × 200 Mm ÷ 40 mm = 25.0 Mm [No ue] (1) 2

Calculation of speed of ripple

Speed = distance ÷ time (1)

= 25.0 ×106 m ÷ (10 × 60) s (1)

= 41 600 m s–1 [no ue] (1) 3

How to check speed constant

Use third frame to calculate speed in this time intervalOR plot diameter (or radius) against time to get a straight lineOR compare distance travelled between frames 3 and 2 with distance travelledbetween frames 2 and 1 (1) 1

Cross-section of wave

Wavelength (1)Amplitude (1) 2

Calculation of frequency of waves

Wavespeed = frequency × wavelength (1)

Frequency = wavespeed ÷ wavelength = 41 700 m s–1 ÷ 1.4 × 107m (1) = 3.0 × 10–3 Hz (1) 2

[11]

607. Calculation of energy to heat water

∆ E = mc∆ θ = 0.2 kg × 4200 J kg.–1 °C–1 (1)

× (75 °C – 22 °C) (1)

[i.e. subst mc (1) subst ∆ θ (1)]

= 44 500 J (1) 3

Calculation of maximum thermal energy from heater

∆ E = P∆ tOR ∆ E = 2500 W × 6 s (1)

15 000 J (1) 2

Explanation of which suggestion) most likely to be correct

Reservoir, as heater supplies insufficient energy in 6 s [ecf] (1) 1

Effect of heat losses

More energy would be required (1) 1[7]

608. Preferred airflow

Streamlined/laminar flow (1) 1

Diagrams

At least one continuous curve drawn above body of cyclist (1)

Turbulence shown behind cyclist (1) 2

Advantage

Less drag on cyclist behind (1)

OR airflow above bodies more streamlined

OR less work needs to be done by following cyclists

Material of bodysuit and explanation

Elastic (1)

e.g. stretch around body in use/nothing loose to cause turbulent flow (1) 2

Material of helmet and explanation

Tough (1)

e.g deforms plastically before breaking (1) 2

Deformation

(In crash energy, deform s/absorbed by helmet rather than causing injury (1) 1[9]

609. Movement of water molecules

Molecules oscillate/vibrate (1)

Movement parallel to energy flow (1) 2

Pulses

To prevent interference between transmitted and reflected signals (1) 1

OR allow time for reflection before next pulse transmitted

Calculation

Time for pulse to travel to fish and back again = distance ÷ speed

υ∆=∆ x

t

= 1ms1500

m3002−

× (1)

= 0.4 s (1) 2

[0.2 s = 1 mark]

Effect used in method

Doppler effect (1)

Any two from:

• a change in frequency of the signal• caused by relative movement between the source and the observer• size and sign of change relate to the relative speed and direction of the

movement between shoal and transmitter• frequency increase - moving towards• frequency decrease - moving away (1) (1) 3

[8]

610. Equation to define resistivity

l

RA=ρ (1)

All symbols defined (resistivity, resistance, length, cross-sectional area) (1) (1)

[3 symbols only defined (1)] 3

Resistance meter

Any two from:

• the resistance between the two probes is measured, not the resistivity• because you cannot measure the cross-sectional area of skin between the probes• A and 1 both vary; cannot calculate resistivity (1) (1) 2

Whether results support claims

Yes (1)

Any two from:

• resistance chances with programme content• least resistance with political programme• sweat reduces resistance / is a better conductor (1) (1) 3

[8]

611. Weight of submarine

Weight = mg = 2110 kg × 9.81 m s–2 = 20 700 N (1) 1

Submarine at rest

(i) 20 700 N [ecf from previous part] (1)

(ii) Forces in equilibrium, since submarine at rest (1) 2

Adjustment of weight of submarine

(i) Expel some water from/ add air to buoyancy tanks (1)

(ii) Use of F = 6 π η υ r (1)

= 6 π × 1.2 ×10–3 kg m–1 s–1 × 0.5 m s–1 × 0.8 m

= 0.0090 N (1)

(iii) Flow not streamlined [or equivalent] (1) 4

Strain calculation

Use of Strain = stress ÷ E (1)

Strain = 1.1 ×106 Pa/3.0 × 109 Pa

= 3.7 × 10–4 (1) 2[9]

612. How diagram confirms pion is negatively charged

Any two from:

• bends opposite way to proton• reference to magnetic interaction/Fleming’s left-hand rule• proton + ⇒ pion – (1) (1) 2

Charge carried by lambda particle

Neutral (1)

because charge conserved OR +1 – 1 ⇒ 0 OR λ not ionising/no track (1) 2

Deduction

r pion < r proton / straighter / less curved (1)

⇒ since r = p / BQ (P pion < P proton) (1) 2

Scale drawing

2 straight lines lpr > lpi (1)Orientation of lines (49°) joined correct way (1)Answer 10 ± 1 kg m s–1 (1) 3

Classification of particles

baryon meson

pion (1)

lambda (1) 2

Charge of a down quark

–1 / 3 (e) (1) 1[12]

613. Reason

Even a very small resistance (in series) with 2000 A through it would generatemuch heat

[Answer must refer to where heating effect occurs]

OR

Such a big current would need thick wires in the meter design

OR

Have to break circuit to insert meter (1) 1

Explanation

Any four from:

• I1causes flux (in iron ring and coil)

• (flux) in iron ring/coil• changing I1 ⇒ changing flux

• ∴ V in coil induced / V = N dt

dφ /cutting flux /Faraday’s law in context/

I induced (1) (1)• V in coil I ⇒ I in coil (1) (1) 4

[5]

614. How ions are accelerated

Electric field exists between +, – electrodes (1)

⇒ force on ions / force → acceleration (1) 2

Speed of xenon atom

eV = ½ mυ 2/eV = E k (1)

⇒ υ = m/eV2 (1)

= 125

19

ms102.2

)2251060(106.12 −−

−

×+×××

(1)

= 4.3 ×104 m s–1 [No u.e.] (1) 4

Thrust on space probe

Force = rate of change of momentum (1)

= 2.1 × 10–6 × 43 000 N (1)

= 0.090 N (1) 3

[Using 4 × 104 m s–1 gives F = 0.084 N]

Reason for reduced thrust

Xenon ions attracted back OR similar (1) 1

Why ion drives maybe preferable

Any two from:

• less fuel required in total• for example, 66 kg for a year• thrust provided for longer/fuel lasts longer/accelerates for longer• lower payload for initial launch/ion drive lighter (1) (1) 2

[12]

615. Diagrams showing forces:

D i a g r a m 1 D i a g r a m 2

(1) (1) 2

[Each diagram, one arrow only]

Discussion re artificial gravity

Any four from:

• centripetal force, in context• forces felt by astronaut both “upward”• so feels like weight/gravity• a υ 2/r / F = mυ 2 / r / a = ω 2r / F = mω 2r / F = m υ ω

• = 2

2

2

22

10

210442 )/(

t

rr/

t

r ×π=π=

π

[Any one] or calculate υ or ω

• = 2.0 m s–2 [No u.e.]• artificial field varies with / α radius (1) (1) (1) (1) 4

≡ 1.6 N kg–1 [to one significant figure] OR other justification [e.c.f] (1) 1[7]

616. Observations of circuits

Any six from:

• capacitor is charged• energy stored in C/goes to lamps...• → heat and light in lamp• as I/Q passes through lamp/discharges through lamp• E = ½ CV2

• ⇒ V × 2 ⇒ E × 4• (hence) 4 lots of energy/4 lamps lit similarly• • 5 V across 1 lamp ≡ same Q through each lamp as before• discussion of T = RC• R same for both circuits• flash is bright and dies exponentially (1) (1) (1)• V × 2 → Q × 2 (1) (1) (1)• → same Q or I as before alone each parallel branch

[6]

617. Data for speaker and equation

Equation for shm: x = A cos ω t

A amplitude = 1.0 mm or 1.0 × 10–3 m

ω = 2π f = 6.28 × 102 (rad s–1) – no unit penalty for ω (1) 2

Calculations

(i) A =A ω 2

= 1.0 × 10–3 m × (6.28 × 102 rad s–1)2 = 394 m s–2(1)

(ii) υ = A ω (1)

= 1.0 × 10–3 m × 6.28 × 102 rad s–1 = 0.63 m s–1 (1) 3

Acceleration - time graph

Two cycles of sinusoidally shaped graph (1)

Period = 10 m s (1)

Amplitude = 394 m s–2 [e.c.f from (i)] (1) 3

Explanation

Resonance (stated or implied by explanation) (1)

Increased amplitude at resonant frequency (1) 2[10]

618. Minimum mass for comet

Volume = 4/3π (9 × 103/ 2)3 m3 (= 3.8 × 1011 m3) (1)

Use of mass = density × volume

= 500 kg m–3 × 4/3 π (9 × 103 / 2)3 m3 (1)

1.91 × 1014 kg (1) 3

Jupiter’s gravitational field strength

g = GM/r2 (1)

= 6.6720 × 10–11 N m–2 kg,–2 × 1.8987 ×1027 kg / (96 009 × 103)2 (1)

= 13.7432 N kg–1 (1) 3

Explanation

Any two from:

• Jupiter’s force/field strength different on the two sides of the comet (1)• difference sufficient to pull comet apart (1)• Jupiter’s force larger than (cohesive) force between particles of comet (1) Max 2

Difference and similarity between gravitational and electric fields

Difference: e.g. gravitational fields only attractive, electrical can be attractive orrepulsive, gravitational fields due to mass, electrical due to charge, cannot shield (1)g-field, can shield E field

Similarity: e.g. follow equivalent mathematical formulae,both obey inverse square law (1) 2

[10]

619. Volume of bubbles in squashed cup

Recall of pV = nRT (1)

OR pV constant

∴ Vnew = pold × Vold / pnew (1)

1 × 105 Pa × 7 ×10–5 m3 / 3.5 × 107 Pa

= 2.0 × 10–7 m3 (1) 3

Assumption

Example of valid assumption: temperature remains constant, air is not an idealgas, no air escapes (so n constant) (1) 1

Percentage change in average kinetic energy

T changes from 298 K to 283 K (1)

Mean KE α T

so percentage change in mean KE = (298 – 283) × 100/298 5% (1) 3[7]

620. Calculation of age of the Moon

Any six from:

λ = ln 2 / half-life (1)

= ln 2 / 1.3 × 109 y

5.3 × 10–10 y–1 (1)

Original mass of 40K = 0.10 + 0.840 = 94 µ g (1)

Use of N = N0e– λ t (1)

So 0.10 = 0.94 e– λ t (1)

So ln(0.10/0.94) = – λ t (1)

So t = 4.2 × 109 y (1)

[A valid assumption may be given a mark][Max 6]

621. Deceleration of carsAcceleration = gradient / suitable eqn. of motion. (1)Correct substitutions [ 0.9 for t is wrong] (1)

6.1 – 6.3 m s–2 [-ve value –1] [no ecf] (1) 3

Area under velocity-time graphDistance/displacement (1) 1

Shaded area6.9 – 7.5 (1)m (1) 2

[Allow 1 mark for 5.5 – 6.1 cm2.]

Minimum value of the initial separationSame as above [ecf] (1)Area is the extra distance car B travels/how much closer they get (1) 2

GraphBoth sloping lines continued down to time axis [by eye] (1) 1

ExplanationArea between graphs is larger/B travels faster for longer/B still moving when A stops (1)Extra distance B goes is larger/ > ‘7.2’ (1)Initial separation must be greater (1) Max 2

[11]

622. Experiment2 light gates (1)Gate gives time trolley takes to pass [ not just ‘the time’] (1)Speed = length of ‘interrupter’/time taken (1)

OR

2 ticker timers (1)Dots at known time intervals (1)Speed = length of tape section/time taken (1) 3[ruler + clock could obtain third mark only, specifying a length/time]

Total momentum of trolleysZero (1)It was zero initially or momentum is conserved [consequent] (1) 2

Speed v of A Use of momentum = mass × velocity (1)Use of mass × speed (A) = mass × speed (B) (1)

1.8 m s−1 [ignore -ve signs] (1) 3

[8]

623. Explanation of why kicking a door is more effectiveQuality of written communication (1) 1Foot decelerates/ loses momentum (1)This takes place rapidly giving a large force by Newton 2 or equation versions [consequent] (1)Door is providing this force [consequent on mark 1] (1)Door acts on foot; by ‘Newton 3’ foot acts on door’ (1) Max 3

[4]

624. Free-body force diagramNormal reaction/contact force [or Nor R or push of table] upwards (1)E-M/Magnetic force [or magnetic attraction or pull of magnet] to right (1)Weight [or W or mg or gravitational force or gravitational attraction orpull of Earth] downwards (1) 3[Ignore labelled forces of fiction. or drag] [if unlabelled −1 each force]

ForcesPull on earth (1)Upwards [consequent] (1)

OR

Push/contact force/force on table (1)Downwards [consequent] (1)

OR

Force on magnet X (1)To left [consequent] [allow ecf] (1) 2

[5]

625. PrecautionsMeasure background radiation //shield apparatus (1)Subtract it off/ because it may vary//to eliminate background (1)Repeat the count and average (1)Because count (or emission) is random/varying (1)Source the same distance from GM on both occasions (1)Because count rate varies with distance (1) Max 3[NB Marks must come from any TWO precautions.]

Ratio0.88 or 1.1 [min. 2 sfi [not %] (1) 1

Count for year 311 994 (1) 1

GraphSuitable axes and scales [don’t award if factors 3, 7 used] [not Bq] (1)Correct plotting of points (1)Use of curve and halving count rate (1)5.3 to 5.4 yr (1) 4

[9]

626. Name of nucleiIsotopes [not radioisotopes] (1)

Nuclear equationInβ)(oreSn 111

4901

11150 +→

Electron numbers correct anywhere (1)Correctly balanced (1)

Densest materialSn-115 (1)

[4]

627.

Word Equation Quantity Defined

Voltage ÷ Current Resistance (1)

Voltage × Current Power (1)

Charge ÷ Time Current (1)

Work done ÷ Charge Voltage/p.d./e.m.f (1)

[4]

628. Charge calculation

Q = 20 000 × 4.0 × 10−4 s [substitution]

Q = 8.0 C/A s 2

Resistance calculation

R = Alρ

= )m100.1(

)m50)(107.1(23–

8–

×Ω×

R = 8.5 × 10−4 Ω

Formula (1)

Correct substitution (1)Answer (1) 3

Potential difference calculationV = IR

= (20 000 A) × (85 × 10−5 Ω ) [or their value] (1)

= 17 V [Allow full e.c.f] (1) 2

ExplanationFor the tree: R or p is larger (1) 1

[8]

629. NetworksFirst network: 2.5(Ω ) (1)Second network: 25 (Ω ) (1)Third network: 10 (Ω ) (1) 3

Meter readingsAmmeter: 25 (mA) (1)Voltmeter V1: 25 × 10 OR 50 × 5 [ignore powers of 10] (1)= 0.25 V (1)Voltmeter V2: 50 × 25 [ignore powers of 10] (1)= 1.25 V (1) 5

[Allow full e.c.f. for their resistance for 2nd network OR their V1 answer][8]

630. Potential difference across resistors

2.0 MΩ : 6.0 V 5.99998 V (1) OR

4.0 Ω : 0V 1.2 × 10−5 V (1) 2

Second potential divider circuit

p.d. across 45 Ω :

( 5045

× 6.0 V) = 5.4 V (1)

p.d. across diode:

(6.0 V – 5.4 V) = 0.6 V (1) 2

[Allow e.c.f. for 2nd mark if candidate uses

455 × 6.0 V = 0.7 V for diode

then

6.0 V – 0.7 V5.3 V for 45 Ω ]

Graph

I / A

0 V / Vm u s t h a v e t h i si n i t i a l s h a p e

B e l e n i e n t e h e r e p r o v i d e dg e n e r a l l y r i g h t s h a p e

(1) 1[5]

631. Hertzsprung-Russell diagram

xA, xB and xC all marked (1) 1

α Ori – red giant (1)Procy B – white dwarf (1)β Per – main sequence (1) 3[e.c.f. wrong x positions]

[If no xs on diagram, 3/3 for identifying still possible]

Calculations

Luminosity α Ori = 6 × 104 × 3.8 × 1026 W (1)

= 5.67 × 10−8 Wm

−2 K−4 × A m2 × (3500)4K4 (1)

A = 2.7 × 1024 m2 (1)

A = 4 7π r2 (1)

R = 4.6 × 1011 m [e.c.f. A] (1) 5[9]

632. Light yearIs the distance travelled by light in one year (1) 1

Show that ly is equivalent toDistance = speed × time [no mark]

(365 × 24 × 60 × 60) s OR 3.15 × 107 s (1)

× 3 × 108 m s−1 = 9.5 × 1015 (m) OR 9.4 (1) 2

[Accept “working backwards”, i.e. distance ÷ speed → time ≈ 3.15 × 107 s]

ExplanationA distant star does not seem to move (1)against the background/with respect to a distant star (1)

OR

Angle between star and distant star (1)does not change/difference in angles, too small to measure (1)

As the Earth moves over 6 months/June & January (1) Max 3

[Do not accept “angle too small to measure”; do not award 1 mark fromtop and 1 mark from bottom pair]

[6]

633. Supernova and description

• Quality of written communication (1)

• A star which suddenly becomes (1)

• very bright/luminous (1)

• Fusion/hydrogen burning ceases/runs out of hydrogen (1)

• Collapse of core/collapse of star/implosion (1)

• Outer layers bounce off core/shock wave/explosion (1)

• Blowing away outer layers (1)

• Nuclear reactions take place in outer layers/ejected material (1) Max 5

Fates for the central core remnantNeutron star (1)Black holes (1) 2

[7]

634. Topic B – Solid Materials

Calculation

Stress = 8.0 N / 1.5 × 10−7 m2 (1)

= 5.3 × 107 Pa/N m−2 (1) 2

Graph:Extension = 0.67 mm [Accept 0.66 to 0.68] 1

Strain = 0.67 mm/2.6 × 103 mm [e.c.f extension from above]

[Ignore, 10n error] (1)

= 2.6 × 10−4 [Do not penalise presence of unit]

Substitute in Young modulus = stress/strain [e.c.f stress from above.

e.c.f. strain, their value OR 3 × 10−4] (1) 4

[= 2 × 1011 Pa/N m−2 (200 Gpa)] (1)

Calculation of work done

Find area of triangle OR use ½ kx2 (1)

Substitute correct pair of values off line [ignore 10n errors] 4.8 N/4.7 N, 0.4 mm

OR

Determine k = gradient (1)

= 9.6 × 10−4 J [± 0.2] [Accept N m] (1) 3

[Allow e.c.f. ONLY for grid error 4.4 N – 8.8 × 10−4 J gets 2/3]

Force-extension graph for wire of twice length:Add line ½ as steep to graph [by eye] (2)[Less steep, but not approx ½, 1 out of 2] 2

[11]

635. Topic C – Nuclear and Particle Physics

SimilarlySame mass

DifferenceCharge OR baryon number OR uud quarks → duu (1) 2

Any two lepton pairs from the following:

e– e+ (

µ – µ + ( NOT e.g. muon and antimuon/µ μ

τ -τ +

ve v e vµ v µ 0R just v v (2) 2

vτ v τ

CollisionParticle and antiparticle annihilate/produce a burst of energy/of photons/of gamma rays (1) 1

[5]

636. (a) (i) ± 1 of Supervisor’s value (1)l ± 0.3 mm of Supervisor’s value, with unit (1)[to 0. 1 mm precision or better]Repeat (1)d calculated correctly [e.c.f]2/3 significant figures + unit (1) 4[Unit penalty once only]

(ii) Table with quantities and units for m and x (1)2 good values < M (1)2 good values > M [Supervisor’s M] (1)[0/2 if systematic error, e.g. l and not x][“Good” is ±2 mm from best fit line]M read off graph correctly, with unit[x must be shown in table or graph] (1)±10 g of Supervisor’s value (1)

Graph:Scale: occupying at least ½ grid each way, avoiding 3s etc[Can ignore last point if off scale] (1)

Plots: penalise a serious mis-plot or two or more inaccurate (1)

Line: Good, sharp [Accept best straight line through points if not forced through origin] (1) 8

No marks for calculation of mass at this point

θ 1 and θ f recorded with unit for both and θ∆ 5 °C – 20 °C

Stirred water and took highest steady temperature

Any one of the following:

• both temperatures better than 1 °C

• thermometer not touching beaker

• eye level for thermometer or measuring cylinder (1)

Correct calculation with corresponding unit [allow J min–1] (1)2 s.f. [OR 1 s.f. if < 10 W] (1) 6

Masses all recorded to 0.0 1 g, including ∆ m (1)∆ m found correctly and sensible unit [∼ 0.1 – 0.5 g] (1)Suitable working that would lead to correct time (1)Hence, correct calculation of time + unit (1)Sensible comparison of power [5 W – 60 W for small lamp] (1)and time (1) 6

[24]

Sample results:

(a) (i) N = 33l = 23.7, 23.9, average 23.8 mm.

d = 33

mm23.8= 0.72 mm

(ii)m/g Scale rdg/mm x/mm

0 46 050 50 4

100 68 22150 88 42200 108 62250 126 80300 144 98

Block 92 46

From graph M = 160 g

(b) (i) m = 18.58 – 3.86 = 14.72gθ 1 = 20.9 °C

θ f = 32.7 °Cθ∆ = 11.8 K

Stirred water and took highest steady temperature.Both temperatures better than 1 °C/thermometer not touchingbeaker/eye level for thermometer or measuring cylinder.

P = s 605

J 11.8 500××

= 20W

(ii) m′ = 18.37 – 3.86 = 14.51 g∆ m = 14.72 – 14.51 = 0.21 g

mmt

∆=

5

t = 5 ×21.072.14 min,

= 350 min

= 5 h 50 min

20 W is comparable to the power of a small lamp, but the experiment suggests the candle would burn for less than 6 h, which is not “all night”.

637. (a) (i) d to. 0.01 cm precison or better and ± 0.03 cm of Supervisor’s value,with unit (1)Repeat or zero error check (1)Correct V, with unit ≥ 2 s.f (1)Correct calculation of ρ with unit (1)

1.00 → 1.30 g cm–3 and 3 s.f. (1) 5

(ii) Correct calculation of V′ [Ignore missing unit]Hence volume of “space” + unitHow answer could be checked experimentally (3 points) (3) 5[If a displacement method is not used 0/3]

(b) (i) Circuit correctly set up without help (3) 3[Help with potentiometer, –2; help with meter location –1;diode reversed –1 [Ignore meter polarity corrections]

(ii) Table with units [Quantity and unit] (1)3 sensible values for I< 10 mA [2 values → 1 mark] (2)5 values for I ≥ 10 mA [ 4 values → 1 mark (2) 5[No values for I < 80 mA, – 1 from last two marks][Ignore any I value > 100 mA]

(iii) Graph:Good smooth curve (2) 2[–1 if serious misplot; –1 if curve slightly “wobbly” or thickish; –1 if plots < 0.4 V not shown] [Ignore plots which are off the grid]

(iii) I read off correctly @ 0.70 V [Ignore missing unit] (1)Correct R + unit [e.c.f.] 1Correct I by plotting “R = 5 0 Ω “ line on graph, with unit. (2) 4[By trial and error can get 2 marks.][Allow one mark if any V value between 0.7 V and 0.8 V isdivided by 50 Ω to give correct I with unit.]

[24]

Sample results:

(a) (i) d/cm = 4.28, 4.28 [no zero error]d = 4.28 cm

V = 6

)28.4( 3×π 41.1 cm3

m = 45.3 g

ρ = 3cm 41.1

g 45.3 = 1.10 g cm–3

(ii) V′ = 3d × d × d = 3d3

= 3 × 4.283 = 235 cm3

∆ V= 235 – 3 × 41.1

112 cm3

d

3 d

Either

• Record volume of water in measuring cylinder

• Add water until packet brimful

• Deduce volume of water added (= volume of “space”)

Or

• Weigh with balls in packet

• Fill to brim with water

• Re-weigh. Hence mass (and volume) of water added (= “space”)

(b) (ii)V/V I/mA I/mA V/V0.00 0.0 10.0 0.6680.10 0.0 20.0 0.7020.20 0.1 30.0 0.7190.30 0.1 40.0 0.7320.40 0.2 50.0 0.7400.50 0.2 60.0 0.7450.55 1.1 70.0 0.7530.60 2.8 80.0 0.7570.65 7.3 90.0 0.760

100.0 0.763

(iii) Graph:

(iv) When V = 0.70 V, I = 20.0 mA

R = A10 20

V 0.703-×

= 35 Ω

From graph I = 13.4 mA[Check: V = 0.68 V

R = 0.0134

0.68 0.0134 = 50.7 Ω ]

638. Wavelength0.80 m (1)

Out of phaseEither X as in diagram below (1)

At restY at crest or trough as in diagram below (1)

Y

Y

YA

X X

Direction of movementArrow at C up the page (1) 4

Time calculation (1)Use of t = λ /υ (1)0.25 s [ecf λ ] 2

[6]


Force/Acceleration proportional to displacement/ a = ω 2 x (1)[define a and x but not ω ]

In opposite direction to displacement/ towards a fixed point/towards equilibrium position/minus sign (1) 2

Oscillation of massClarity of written communication (1)Down: T > W [W varying 0/3] (1)Up: T < W (1)T > W gives resultant force/acceleration UP (1)[or equivalent argument for displaced up] 4

Velocity-time graphEither Cosine graph [ zigzag –1 here] (1)Starting at positive maximum [if very poorly synchronised 0/2] (1) 2

Maximum velocity of massPeriod = 0.50 or amplitude = 0.07 (1)Use of f = I/T / Use of x = x0 sin or cos (ω t) (1)

Use of υ max = 2π fx0 (1)

0.88 m s–1 (1)

OR 4

υ max = gradient at 0.50 s (or equivalent) (1)Correct method for gradient (1)

Answer in range 0.8 to 1.0 m s–1 (1)[Max 3 for gradient method]

[12]

640. Electromagnetic waves experiment

EITHER

‘Lamp’, 1 polaroid // LASER (1)

2nd polaroid, suitable detector [e.g. eye, screen, LDR] (1)Rotate one polaroid [ consequent on 2 polaroids] [one if LASER] (1)Varies [consequent] (1)

OR

Microwave transmitter (and grille) [not polaroid or grating] (1)Receiver (or and grille) (1)Rotate ANY [if 2 grilles; must rotate a grille] (1)Varies [consequent] (1) 4

Nature of wavesransverse (1) 1

[5]

641. Speed of electronSelection of λ = h/p and p = mυ (1)

m = 9.11 × 10–31 (1)

7.2 – 7.3 × 106 m s–1 (1)

Kinetic energy

Use of Ek = 1/2 mυ 2 (1)147 – 152 [ecf] (1) 5

High energy electronNucleus tiny/a lot smaller so λ very small (1)υ or p very large [consequent] (1) 2

[7]

642. Planck constantRealise that h is the gradientCorrect attempt to find gradient [but ignore unit errors here]

h = (6.3 to 6.9) × 10–34 J s [No bald answers] 3

Work functionUse of hf0 / use intercept on T axis/use of φ = hf – T (1)

φ = (3.4 to 3.9) × 10–19 J [ -1 if –ve] [2.1 to 2.4 eV] (1) 2[5]

643. Base units of G: kg–1 m3 s–2 (1)Equation homogeneity:

Correct substitution of units of G, r3 (m3), M (kg) 1

leading to S2 and linked to T2 (1)[Allow e.c.f. of their base unit answer into substitution mark] 3

Use of relationship to find mass of the Earth

Any two from:Adding 20 000 + 6400 (1)Converting km to m (1)h to s (× 43 200) (1) 2

Answer M = 5.8 (4) × 1024 kg (1) 1[6]

644. Faraday’s law of electromagnetic inductionThe induced e.m.f. (1)in a conductor is equal to/proportional tothe rate of change of magnetic flux linkage (1)

OR

E = ( )t

φ

d

d− or E ∝ t

N

Δ

Δφ [Accept φ∆ or dφ ]

E – induced voltage

dϕ – change of magnetic flux (1)dt – time

[All symbols defined] 2

Conversion of sound waves into electrical signals

Any four from:

• quality of language (1)

• sound waves make the diaphragm/coil vibrate/oscillate (1)

• coil: change in flux linkage/coil cuts field lines (1)

• induced voltage across coil (1)

• frequency of sound wave is frequency of induced voltage/current/electrical wave (1) Max 4

[6]

645. Calculation of charge

6000 V × 20 × 10–6 F (1)= 0.12 C (1) 2

Energy stored in capacitor

2

2CV

2

V)6000(C1002 26 ×× −

(1)

= 360 J (1) 2

Resistance

A40

V6000= 150 Ω (1) 1

Time to discharge capacitor

Time = A40

C12.0 /their Q (1)

= 0.0030 s / 3.0 × 10–3 s [e.c.f.] (1) 2

Reason

Time is longer because the rate of discharge decreases/ current decreases with time (1) 1

[8]

646. Free-body force diagram

Tension/pull of thread (1)F/push of charged sphere/electric force/electrostatic force (1)Weight/W/pull of Earth [Not mg, unless W = mg stated] (1) 3

T e n s i o n / p u l l o f t h r e a d

F / p u s h o f c h a r g e d s p h e r e / e l e c t r i c f o r c e / e l e c t r o s t a t i c f o r c e

W e i g h t / / p u l l o f E a r t h [ N o t g , u n l e s s = g s t a t e d ]W m W m

Force equation

W = T cos θ ( ( (1)F = T sin θ (

Processing mark, e.g. F = θcos

W sin θ OR tan θ = θ

θcos

sin (1)

OR

F, T, W labelled (1)both angles labelled (1) 2

T

TW

θ

Table

Distance r = 36 × 10–3 mF = 35.5/36 [No u.e.] (1)

Distance r = 27 × 10–3 m

Using any pair of values (1)Seeing correct constant for their pair of values (1)→ F = 63.1 [n.o u.e.] (1)

OR

Valid simple ratio calculation using a pair of values (1)stating produce Q1Q or kQQ2 constant (1)

→ F = 63.1 [no u.e.] (1) 4

Measurements taken quickly because

Leakage/discharge of charge [Allow dissipation or description of process] (1) 1[10]

647. (a) LDR connected to ohmmeter (1)Light shield around LDR (1)Two polaroid filters between light source and LDR (1)θ shown (1)Measure θ with protractor [shown or stated] (1)A correct technique showing overlapping filters (1)Starting point when maximum (or minimum) LDR resistance (1)

θ = 90o [or θ = 0o] (1) 8

(b) I = k (cos θ )n

ln I = n ln(cos θ ) + ln k (1)y = mx + c 1

(c) ln values and ≥ 2 decimal places (1)Graph:Scale (1)Axes labelled with unit (1)Plots (1)Line (1)[Allow reversed negative axis][Mark any graph which is plotted using this scheme] 5

(d) ∆ x ∆ y ≥ 64 cm2 [or limits of graph] (1)Value 1.45 → 1.55 with no unit (1) 2

[16]

Sample results

(a)

Ω

O h m m e t e r

L i g h t s h i e l d

L D R

S m a l l d i s t a n c e

* L i g h t s o u r c e

P o l a r o i d f i l t e r s

θ

Starting point when maximum (or minimum) LDR resistance.

θ = 90o (or θ = 0o)

(b) I = k (cos θ )n–

ln I = n ln (cos θ ) + lnky = mx + c

(c)

θ /o I/lux Cos θ ln(I/lux) log10

(I/lux)ln (cos

θ )log10

(cos θ )10 481 0.985 6.18 2.68 –0.02 –0.0120 450 0.940 6.11 2.65 –0.06 –0.0330 398 0.866 5.99 2.60 –0.14 –0.0640 330 0.766 5.80 2.52 –0.27 –0.1250 256 0.643 5.55 2.41 –0.44 –0.1960 172 0.500 5.15 2.24 –0.69 –0.3070 98 0.342 4.58 1.99 –1.07 –0.4780 40 0.174 3.69 1.60 –.75 –0.76

Graph:

(d) n = )76.1(0

6.322.6

−−−

= 1.49

648. (a) (i) T1 and T2 reasonable and T2 > T1 from ≥ 10 oscillations (2)

[from ≥ 5 gets (1)]Both repeated (1)T2: 4.0 − 6.0 s (1) 4Correct use of set squares against rule (2)D1 between 63.0 − 68.0 mm and D2 between 31.0 − 35.0 mm to nearest mm or better (1)D1 and D2 repeated (1)ks calculated [ignore units] (1)Quantitative comparison (1)ORComment that results agree/disagree within experimental error [where nominal 10% for experimental error] 6

(b) Difference method used when filling can (1)Volume 335 – 360 ml with unit (1)L: 113 – 117 mm (1)Correct calculation with unit and 2/3 s.f. [e.c.f.] (1)Suitable suggestion why V differs from volume stated on can (1)Suitable suggestion why V ′ differs from V (1) 6

[16]

Sample results:

(a) (i) 10T1 = 32.95, 33.18 s

∴ T1 = 3.31 s5T2 = 23.99, 23.56 sT2 = 4.76 s

(ii)

D1 = 65. 65 mm D 1 = 65 mmD2 = 33, 33 mm D 2 = 33 mm

k1 = T12 D1 = 3.312 × 65

= 711 s2 mm

k2 = T22 D2 = 4.762 × 33

= 746 s2 mm

% difference = 100729

35 ×

= 4.8%which is acceptable experimental error

(b) Measuring cylinder filled and poured into can.Filled again and poured into can until fullV = 250 + (250 − 160) = 340 mlL = 115, 115 mm, L = 115 mm

V ′ = π × 6.52 × 4

5.11

= 382 cm3

Contents do not usually completely fill canCan narrows at top and bottom OR V ′ is external rather than internal volume.

649. (a) Value ± 2 cm of Supervisor’s value (2)[± 3 cm – (1)]Repeat readings etc (1)Approach resonance from both directions (1)Correct calculation from sensible ∆ l (1)[Range, ½ range or 0.1 – 1.0 cm if only one reading] 5

(b) Value 0.21 – 0.26 mm [unless alternative wire used] (1)Repeat measurements or zero error [dependent mark] (1)Correct % from range, ½ range or precision [allow 0.005 mm] (1) 3

(c) Correct SI substitutionCorrect calculation ≥ 2 s.f. + unit [dependent mark] [−1 if g = 10] (1)

Value: 7500 → 10 000 kg m–3 [no e.c.f.] [when rounded to 2 s.f.] (1) 3

(d) 2 × % uncertainty in d and l (1)Added [Not a dependent mark] (1)Calculation of % difference with 8880 as denominator (1)Compared to % uncertainty (1)Hence sensible conclusion (1)ORAdds/subtracts their % uncertainty from their value (1)Checks whether in range (1)Hence sensible conclusion (1) 5

[16]

Sample results:

(a) l = 71.5, 72.0, 72.5 cml = 72.0 cm

Repeat readings shown Or eye level with wire/use of care to observe resonanceApproach resonance from both directions

Percentage uncertainty = 1000.72

5.0 ×

= 0.7%

(b) d = 0.23, 0.23 mmd = 0.23 mm

Percentage uncertainty = 10023.0

01.0 ×

= 4.3%

(c) ρ = 2223 72.050)1023.0(

81.92.0

×××××

−π= 9110 kg m–3

(d) Percentage uncertainty in density= 2 × 4.3 + 2 × 0.7= 10.0%

Percentage difference

= %1008880

88809110 ×−

= 2.6%

Much less than total % uncertainty ∴ wire could be made from constantan.

650. (a) Light shield around LDR (1)LDR connected to ohmmeter (1)Lamp connected to power supply (1)[0/2 if all in series circuit]d measured to front of LDR (1)and filament (1)Appropriate method for measuring d (1)Precautions (2) 8

(b) Equation (1) 1

(c) ln values as shown and ≥ 2 decimal places (1)Graph:Must occupy at least ½ grid, in both directions and avoiding 3s, 7s etc. (1)ScaleAxes labelled and units [Must be in ln(I/lux) etc] (1)Plots [to a precision of 1 mm or better] (1)Line [well drawn line of best fit] (1)[Allow e.c.f. if wrong graph plotted] 5

(d) ∆ x∆ y ≥ 64 cm2 [or to limits of axes] (1)Value −2.50 → −2.70 with no unit [must have negative sign] (1) 2

[16]

Sample results:

(a) Diagram:

Ω

L i g h t s h i e l d

L D R

d

From face of LDR to filament of lampUse apparatus in a darkened roomORTake a reading of the background light level and subtract it frommeasured light levels

(b) lnI = lnk + nlndy = c + mx

(c)

d/mm I/lux ln(d/mm) ln(I/lux) lg(d/mm) lg(1/lux)40 8.20 4.94 2.10 2.15 0.9160 6.05 5.08 1.80 2.20 0.782

200 3.35 5.30 1.21 2.30 0.525250 1.80 5.52 0.59 2.40 0.255300 1.15 5.70 0.14 2.48 0.061350 0.79 5.86 –0.24 2.54 –0.102

Graph:

(d) n = 855.40.6

4.26.0

−−−

= −2.62

651. (i) Wc = ½ CV2 = ½ (0.0047 F) (25 V)2 [Ignore 10n] (1) = 1.5 J / 1.47 J [no e.c.f.] (1) 2

Quality of written communicationWc is (very) small (1)Even at 50 V it is only 6 J (1)Any ∆ T is difficult to measure/wire spread out/ (1)something like a thermocouple is needed (1)Wire (might) melt/fuse (1)Heat/energy loss to air/surroundings [not to connecting wires] (1) Max 4

(ii) Exponential (decay) (1)Radioactive decay/radioactivity [independent] (1)Use of one of five approved methods [Name it] (1)

Data off graph appropriate to method [ignore 10n] (1)Use of RC/use of R = V/I (1)R = 7.2 Ω − 8.5 Ω [no e.c.f.] (1)[7200 Ω – 8500 gets 3/4] 6

[12]

Methods:M1 RC = time to Q0 ÷ e [35 – 39 ms]M2 RC ln2 = t1/2 [24 – 28 ms]M3 RC = where initial tangent hits t axis [32 – 40 ms]

M4 Use of RC in Q = Q0e–t/RC with numbers

[≈ correct]M5 Calculation of T0 initial current from gradient [2.7 – 3.0 A]

652. Calculation of current

P = IV

I = P/V (stated or implied) (1)

= 0.78 W ÷ 6 V

= 0.13 A (1) 2

Calculation of resistance

P = V2/R

OR R = V/I

OR R = (6 V)2 ÷ 0.78 W

OR R = 6 V ÷0.13 A [ecf] (1)

= 46 Ω (1) 2

Explanation of operation from mains

In series (1)

240 V ÷40 lamps = 6 V per lamp (1) 2

Explanation of constant brightness of lamps

Current equal – justified, e.g. in series or same V/R or same P/V 1

Statement and explanation of different resistance with ohmmeter

Lower resistance with ohmmeter (1)

Identify lower temperature with ohmmeter [may be implied] (1)

(Lattice) ions’/atoms’ vibrations impede electrons/current (1)

[Require interaction]

[Allow converse argument] 3[10]

653. Show that vertical component of velocity is about 14 m s –1

Vertical component = υ sin θ

OR

Vertical component = 22.5 m s–1 × sin 38° (1)

= 13.9 m s–1 (1) 2

Show that time of flight is about 3 s

s = ut + ½ at2

Identify s = 0 (can show by correct substitution) (1)

0 = 13.9 m s–1 × t – ½ × (9.81 m s–2) × t2

Manipulation so t on one side only (e.g. 13.9 = 4.9t) (1)

t = 2.8 s (1)

OR

υ = u + at

Time to top assumes υ = 0 (can show by correct substitution) (1)

0 = 13.9 m s–1 – (9.81 m s–2)t

t = 1.4 s (1)

Time of flight = 2 × 1.4 s

t = 2.8 s (1) 3

Calculation of range

Horizontal component = 22.5 m s–1 × cos 38° (1)

= 17.7 m s–1

Horizontal distance = υ ×t [or any speed in the question × time] (1)

= 17.7 m s–1 × 2.8 s

= 49.6 m (1) 3

Effect of work done on range

Work done = force × distance in direction of force (1)

Any two from:

• assuming force constant or relevant discussion of size of force

• increases distance (moved by force) → more work done

• more work done →more k.e. gained

• more k.e. gained → greater initial speed

• greater initial speed → greater range 3[11]

654. Explanation of amp hours

Charge = current × time or Q = It (1)

Amp →current and hour → time (so amp × hour → charge) (1) 2

Show that charge about 5000 C

Charge = 1.5 A × 1 × 60 × 60 s

= 5400 C 1

Calculation of energy stored

W = QV OR W = I v t (1)

= 5400 C × 3 V [may use 5000 C]

= 16 200 J [up] (1) 2

Show that energy is about 20 000 J

Energy = Ivt (1)

= 0.3 A × 3.1 V × 6 × 60 × 60 s

= 20 100 J (20 088 J) (1) 2

Calculation of efficiency

Efficiency = (stored energy/input energy) × 100%

= 16 200 J [allow ecf from 3rd part] (1)

÷ 20 088 J [allow 20 000 J from 4th part] (1)

× 100%

= 80.6% [Accept fractional answers. Allow ecf, but check nos.] (1) 3[10]

655. Draw and label forces

Weight, W, mg (not “gravity”) (1)

Air resistance/drag/friction (1)

–1 for each extra force, ignore upthrust, ignore line of action 2

A i r r e s i s t a n c e / d r a g / f r i c t i o n

W e i g h t / /W m g

Discussion of forces

(Constant velocity) → zero acceleration / resultant force = zero /forces in equilibrium / sum of forces = 0 / forces balanced (1)

Forces equal (and opposite) / weight = drag (1) 2

Show that mass is about 70 kg

m = W/g

= 690 N ÷ 9.81 m s–2

= 70.3 kg 1

Calculation of gravitational potential energy

∆ Egrav = mg∆ h

= 690 N × 2000 m [e.c.f.] (1)

= 1.4 × 106 J (1) 2

Comments on suggestion of gravitational potential energy to kinetic energy

Any two from:

• No gain in Ek here

• Air resistance ignored/should be taken into account

• Should be Egrav lost = Ek gained + work done against air resistance/drag

• At this stage work done against air resistance = Egrav lost 2[9]

656. Reason for non – destructive testing

Sensible reason e.g.

• destroyed rails would require replacement

• trains continuously using tracks, so removing them would cause greater disruption

• saves money 1

Description of sound wave

Particles oscillate / vibrate (not move)

… in direction of wave propagation/longitudinal

causes rarefactions and compressions

[Marks may be gained from suitable diagram] 3

Show that wavelength about 1.5 × 10 –3 m

Wavespeed = frequency × wavelength, v = fλ , any correct arrangt (1)

Wavelength = wavespeed ÷ frequency

= 5900 m s–1 ÷ 4 000 000 Hz

= 1.48 × 10–3 m (1) 2

Meanings

Frequency:

Number of oscillations/waves per second/unit time (may be 4 000 000 oscillations per second) (1)

Wavelength: [may be from diagram]

Distance between 2 points in phase/2 compressions/2 rarefactions (1)

Distance between successive points in phase etc. (1) 3

Calculation of length of track

Length of track = area under graph (or sign of finding area, e.g. shading) or 3 calculated distances using const acceleration formulae (1)

Use of 18 m s–1 as a speed x a time in a calculation (1)

E.g., distance = 0.5 × (116 s + 96 s) × 18 m s–1

= 1908 m (1) 3[12]

657. Explanation of superposition

When 2 (or more) waves meet / cross / coincide … (1)

Reference to combined effect of waves, e.g. add displacement /amplitude – may be a diagram [constructive/destructive interferencenot sufficient without implication of addition] (1) 2

Explanation of cancellation effect


• path/phase difference between direct and reflected waves

• destructive interference/superposition

• path difference is (n + ½)λ / phase diff π /180o / waves in antiphase / out of phase

• “crest” from one wave cancels “trough” from other 3

Reason for changes


• movement changes path of reflected waves

• so changes path difference

• A movement of 75 cm is about ¼ wavelength

• waves reflected so path difference changed to ½ wavelength

• enough to change from antiphase to in phase / change in phase difference

• causes constructive interference/superposition 3[8]

658. Warm river

How radioactive nuclei heat,

e.g. by decay/ionising/nuclear radiation 1

α , β and γ radiation

α helium nucleus [or equivalent] (1)

β (fast) electron (1)

γ electromagnetic wave (1)

[Accept an answer that fully differentiates between the types of radiationby describing their properties] 3

Most hazardous nuclei

α emitting (1)

When ingested, α particles damage body cells

[e.c.f. from previous β or γ linked to penetration & damage] (1) 2

Source of radioactivity

e.g. rocks, Sun, cosmic radiation 1[7]

659. Device

Potential divider or potentiometer 1

Voltmeter reading

A 9.0 V (1)

B 0 V (1) 2

Diagram

Label X two thirds of the way down from A [Allow e.c.f.] 1

Explanation

Any 3 points from the following:

• lamp in parallel with lowest 1/3 of AB

• when resistors in parallel, resistance decreases

• p.d. across lamp reduced to below 3 V

• current divides

• no longer enough current to light lamp 3[7]

660. Properties of gold

Malleable (1)

Ductile (1) 2

E.g. elastic, stiff 1

Definitions

Hard: material not readily scratched/indented (1)

Plastic behaviour: material remains in stretched/deformed shape when force removed (1) 2

[5]

661. Speed of ultrasound

Use of υ = s/t (1)

= 150 × 10–3 (m) ÷132 × 10–6 (s)

= 1140 m s–1 (1) 2

Change of trace

Extra pulse(s)

OR

Reflected pulse moves closer 1

Principle of Doppler probe

3 points from:

• Arrange probe so that soup is approaching

• Soup reflects ultrasound

• with changed frequency/wavelength

• change in frequency/wavelength depends on speed

• Probe detects frequency of reflected ultrasound

• Use of diagrams showing waves 3

Determination of speed

1 point from:

• Frequency/wavelength change

• Angle between ultrasound direction and direction of flow of soup 1

Comment

Lumps give larger reflections

Lumps travel slower 1[8]

662. Wavelength range

465 – 720 nm (± ½ square) 1

Sketch graph

Scale (No more than 90 – 100%)

AND all graph between 96% and 99% (1)

Inversion – in shape with 2 peaks (at 510 and 680 nm) (1) 2

Wavelength

(µ = υ 1 / υ 2 = fλ 1 /fλ 2) λ 1 = 360 nm × 1.38 (1)

(= 497 nm) 1

Explanation

Thickness = λ /4 OR path difference = 180 nm (1)

Path difference = λ /2 (1)

Minimum reflection needs destructive interference betweenreflected rays from front and back of coating (1) 3

Difference between unpolarised and plane polarised light

Unpolarised light consists of waves vibrating in all planes(perpendicular to direction of propagation) (1)

Polarised light consists of waves vibrating in one plane only (1)

OR

Diagrams showing:

Waves / rays in 1 plane (1)

Waves / rays in many planes (1) Max 2[9]

663. Energy of photon of light

E = hf = 6.63 × 10–34 J s × 6.0 × 1014 Hz = 3.98 × 10–19 (J) 1

Graph

Points correct (± ½ square) (2)

Single straight line of best fit (NOT giving intercept below 4.5 × 1014) (1)

Line drawn as far as f axis (1) 4

Value for h

Large triangle [at least 7 cm on K.E. axis] (1)

Gradient = e.g. (6.05 – 4.55) × 10 14 / 1.0 × 10–19 = 1.5 × 1033 (1)

h = 1/gradient = 6.67 × 10–34 J s (1) 3

Value of φ

Reading co-ordinates of a fixed point on graph (e.g. 0, 4.55 × 1014) (1)

φ from equation, e.g.

so φ = frequency intercept × h

= e.g. 4.55 × 1014 × 6.67 × 10–34

= 3.03 × 10–19 J (1) 2

Explanation

Not enough energy [OR frequency too low]

For 2nd mark, numerical/added detail required,

e.g calculation: E = 6.63 × 10–34 × 4.5 × 1014 Hz = 2.98 × 10–19 < φ

OR threshold frequency read from graph 2[12]

664. Quarks: What is meant by “charge = + 2/3”

sign: +/positive/sign same as proton/sign opposite to electron (1)

size: 2/3 charge on a proton / electron (1) 2

Mass of strange quark in kilograms

m = 0.2 GeV/c2

= 0.2 × 109 × 1.6 × 10–19 (1)

/ 9 × 1016

= 3.6 × 10–28 kg (1) 2

Charge and mass of anti – particle to the charmed quark

Charge: –3

2 (1)

Mass: 1.3 GeV/c2 [No unit penalty for omitting GeV/c2] (1) 2

Prediction of top quark

Symmetry of the model / 3rd generation partner / other valid statement 1

Reason for length of time to find experimental evidence for top quark

High energy needed (to create it) / needs a big accelerator/other valid reason 1

Use of conservation law to explain prediction

Momentum (in context)

Total momentum = 0 OR mtυ t = mbυ b OR in words

mt >> mb → υ t « υ b / greater mass (→ lower velocity) 3[11]

665. How movement of magnet produces voltage shown on c.r.o screen

Any 4 from:

• Boxes correct

• Mention of Faraday’s law/equation/word description

• Flux max when magnet vertical / box 1 / box 3

• Flux zero when magnet horizontal / box 2 / box 4

• When flux max, not changing, V = 0

• When flux changing fastest, V max

• Appropriate comment about sense of voltage, e.g.,when poles reversed, V reversed 4

Differences between figures (i) and (ii)

Qualitative points: (max 2)

(Faster turning, giving) td

dφ↑ (1)

= V ↑and f ↑ OR T ↓ (1)

OR

Quantitative points: (max 3)

(f × 2 =) td

dφ (max) × 2 (1)

= V × 2 (1)

f × 2 (OR T ÷ 2) (1) 3

Flux at each end of magnet

Area 1 big square = 100 µ (Vs) or 100 × 10–6 (Vs)

OR area of 1 little square = 4 µ (Vs) or 4 × 10–6 (Vs)

OR area = 32 little squares (29 – 35)

OR area = 4/3 big squares (1.2 – 1.4) (1)

Area = 130 × 10–6 (Vs) (120 – 140) (1)

Φ = Area / 2 × 240

= 2.7 × 10–7 Wb (2.5 – 2.9) (1) 3

Magnetic flux density at end of bar magnet

B = Φ /A OR Φ = BA OR A = 0.01 × 0.005 OR A = 5 × 10–5 m2 (1)

= 3.0 × 10–7 / 5.0 × 10–5

= 6.0 × 10–3 T (accept Wb m–2) (1) 2[12]

666. Magnitude of charges

Value of v or ω (v = 1023 ms–1, ω = 2.7 × 10 –6 s–1) (1)

Value of a (a = 2.7 × 10–3 ms–2) (1)

F = r

m 2υ (1)

Value of F (F = 2 × 1020 N) (1)

F = 2

1 2

r

QkQ (1)

Charge = 5.7 × 1013 C (1) 6

[Use of 2r

GMm to calculate F:

Allow Me in range 1024 – 1025 kg without penalty, otherwise max 4 for question.]

667. Magnitude of gravitational force on Cassini

F = GMm/r2 1

Expression

g = F/m (1)

so g = GM/r2 (1) 2

Maximum acceleration

Appreciation that acceleration = g-field (1)

Addition of orbital height to radius of Venus (1)

g = G × 4.87 × 1024 kg / (6384 × 103)2

= 7.97 m s–2 3

Effect of acceleration on velocity of Cassini

Any 2 from:

• Acceleration is at right angles to direction of motion

• Speed unchanged

• (Velocity changed since) direction changed Max 2

Percentage change

Use of ∆ f / f =υ /c (1)

= 12/3 × 108

so percentage change = 1200/3 × 108 = 4 × 10–6 (1) 2[10]

668. Simple harmonic

Acceleration (OR force) ∝ displacement (1)

and in opposite direction (1)

[OR F = – kx (1) OR x = A cos ω t (1) symbols defined (1)] 2

Graph

(i) Ek inverse of potential energy curve

(ii) T horizontal line at 2 × 10–6 J 2

Stiffness

Use of E = ½ kx2 (1)

2 × 10–6 = ½ k 0.102

k = 4 × 10–4 J m–2 (1)

[accept N m–1] 2

Variation in potential energy

Any 3 points from the following:

• I150 / I100 = 1002 /1502 = 0.44

• So Emax = 0.44 × 2 × 10–6 = 0.88 × 10–6 (J) [no u.e]

• Curve through (0,0), max 0.8 – 0.9

• Curve the same

• I ∝(amplitude)2 Max 3[9]

669. Equation

Recall of pV = nRT 1

Moles of air

Estimate of temperature ≈ 20 oC [Range 0 – 39] (1)

Use of equation, including conversion of temperature to K AND sensible volume (1)

Evaluation: n = pV/RT = 1.0 × 105 Pa × 20 × 10–6 m3 /8.31 × 293

= 8 × 10–4 (1) 3

Volume of bubble

(V ∝ T since p and n constant) (1)

so volume smaller (1) 1[5]

670. Nuclear equations

X = 2 (1)

Z = 2 (1) 2

Physics of nuclear fission and fusion


• either/both transform mass into energy

• products formed have greater binding energy/nucleon

• either/both reaction(s) have a mass defect

• fission – splitting nucleus; fusion joining nuclei together

• fission used in nuclear reactors; fusion reactors not yet available[ or only in bombs/stars]

• fusion needs high pressure and temperature/high energy particles

• fission forms radioactive products AND fusion forms stable products

• explanation of either involving strong nuclear force Max 5[7]


(Variable) resistor symbol (1)

Voltmeter in parallel with cell/resistor (1)

Ammeter in series [even if R missing] (1) 3

A V

Maximum power available

Use of P = IV (1)

Any pair of values which round to 1.4 W (1) 2

Analysis of data

1000 W m–2 P = 1.4 W

100 W m–2 P = 0.11 W OR at least one further value of P (1)

Inspection of ratio

[e.g. 100 ÷ 1000, 0.11 ÷ 1.4, 1000 ÷ 1.4] (1)

Comment based on candidate’s result

[e.g. Yes as ratio is similar] (1) 3

Graph

E/V + scale : 2 large squares = 0.05 V (1)

Points (1)

Straight line good fit (1)

t/°C + scale: 1 large square = 10° (1)

[No penalty if t vs E] 4

Determination of mathematical relationship

Intercept = 0.640 →0.655 (1)

A gradient evaluated (1)

= 2.1 → 2.3 × 10–3 (1)

E = –2.2 × 10–3t + 0.65 [e.c.f.] (1) 4

Other axes:

Intercept 290 → 310 (1)

Gradient (1)

440 → 460 (1)

t = –450E + 300 (1)

Determination of light power from the sun

Attempted evaluation of an area (1)

= 0.13 → 0.17 [–1 if 10–6] (2)

[0.10 → 0.20 (1)]

Their answer × 4.0 = ……….(W) [no u.e.] (1) 4[20]

672. Explanation of “coherent”

In / constant phase (difference) (1)

symbol 51 \f "Monotype Sorts" \s 123 (1) 1

Power delivered by laser

P = 1510400

40−× (1)

= 1 × 1014 W (1) 2

Energy level change

υ = fλ / f = 9

8

101050

103−×

× [–1 if omit 10–9] (1)

Use of E = hf / 6.6 × 10–34 × 9

8

101050

103−×

× (1)

[If f = 1/T used, give this mark]

= 1.9 × 10–19 J (1) 3[6]

673. Diagram:Shown and labelledTicker timer at top or Strobe light (1)Tape from trolley through timer or camera [consequent] (1)ORMotion sensor pointing at trolley or video (1)Connection to datalogger/computer or rule [both consequent] (1)ORThree or more light gates (1)Connection to datalogger/computer [consequent] (1)[Two light gates connected to ‘timer’ – max 1][Rule and stop clock - max 1]

Values for υ and a : 2

0.95 m s–1 [2 s.f.] (1)Use of gradient or formula (1)

0.79 m s–2 [no e.c.f. if u = 0] (1)Distance AB: 3AB = ‘area’ under graph, or quote appropriate equation of motion (1)Physically correct substitutions (1)0.86 m [allow 0.9 m] [e.c.f. wrong u or a] (1)Graph: 3Smooth curve rising from origin, getting steeper (1)Initial gradient non-zero [consequent] (1)(0.70, 0.86) matched (e.c.f. on distance) (1) 3

[11]

674. Compositions of nuclei:Different number of neutrons (1)Same number of protons / proton no. (1)

Physical property:Boiling point/melting point/density/ [ not mass; heavier; RAM] (1) 3

Nuclear Equation:

( )31 H (1)

( )32 He (1)

( )01− β (1) 3

Experiment:GM tube [allow ionisation chamber, cloud chamber] (1)

How to check no alpha:Source close/next to/near/up to 5 cm to GM or ionisation /cloud chamber,insert paper, no change in ‘count rate’

OR

Source close to GM, move away, no sudden drop in count rate (1)

How to check no gamma:Insert a few mm aluminium, count rate reduced to zero

OR

Apply E or B field, GM tube fixed, count rate to zero (1)

Correcting/Allowing for background (i.e. measure it, and look forcount reducing to background in “no gamma” test) (1) 4

[10]

675. Half-life:Use of t1/2 λ = 0.69 (1)13 (1)

Initial number of nuclei:Use of A = λ N (ignore wrong time units) (1)

1.0 × 1015 (1) 4

Graph:Horizontal line from same initial point (1)[max drop 1 small square]

Initial activity marked as 6.4 × 108 Bq or equivalent scale (1)

Their half-life marked where A = 3.2 × 108 Bq, or equivalent scale (1) 3[7]

676.

Base unit Derivedunit

Basequantity

Derivedquantity

Mass

Charge

Joule

Ampere

Volt

(1)

(1)

(1)

(1)

(1)

5[5]

677. Explanation:As the temperature rises, the resistance decreases (1)As the resistance decreases, so the ammeter reading/current increases (1)[No mention of resistance 0/2][Current controls temperature → controls R is wrong physics – 0/2][If T changes so R changes OR vice versa so I changes 1 mark only][Correct static relationship (extremes) 1 mark only]

Reading on milliammeter:At 20 °C R = 1.4 (kΩ ) (1)Substitute correctly in V = IR i.e. 6 V = I × 1400 Ω (1)

[Allow their incorrect R; ignore 10x] (1)Milliammeter reading = 0.0043 A OR 4.3 mA [no e.c.f.] (1)[Accept 4 mA/4.2 mA] 5

[5]

678. Current:

Conversion, i.e. 0.94 × 10–3 m s–1 (1)

Use of 1.6 × 10–19 C (1)Answer 3.0 A

1.0 × 1029 m–3 × 0.20 × 10–6 m2 × 1.6 × 10–19 C × 0.94 × 10–3 mm s–1 (1)

Current = 3.0 A [Accept 2.8 A if 0.9 × 10–3 used.] 3

Resistance:

Recall R = A

lρ (1)

Substitution:

R = 26-

8

m10 0.20

m4.0mΩ107.1

××× −

(1)

Resistance = 0.34 Ω (1) 3

Potential difference:Potential difference = 3.0 A × 0.34 Ω (1)= 1.0 V (1.02 V)[Mark for correct substitution of their values or for the answer of 1.0 V] 1

Explanation:(Increasing resistivity) increases resistance (1)Leads to a smaller current (1) 2

Comparison:Drift velocity decreases (in second wire) (1) 1[Allow V1/V2 = I1/I2][Allow e.c.f. answer consistent with their current answer][Resistivity up, current down

ρ up, I down / 2 (2nd mark)][10]

679. E.m.f.Use of intercept mentioned/indicated on graph/when I = 0 (1)e.m.f. = 1.5 V (1) 2

Use of graph:Internal resistance: mention use of gradient/use of numbers/triangle on graph (1)Internal resistance = 0.5 Ω (1) 2

[Finds r and/or V by substitution, can score answer mark, but NOT method mark]

[Gradient = Ω5.00.1

0.15.1=

−

They might write gradient = 5.10.1

5.1 = Ω OR gradient = 2.1

5.1 - ignore signs]

Graph:Negative gradient of a straight line starting anywhere (1)from (0.0, 3.0) [No e.c.f.] (1)heading for (1.0, 2.0[1.9 → 2.1])/gradient of −1 [Consequent mark] 1 3

Filament lamp: any two ofif the variable resistor is set to zero [NOT, as RVR down] (1)the lamp prevents I from becoming too large (1)and overloading/damaging the ammeter (1)bulb acting like a fuse OR prevents short circuit (1)bulb means there is still resistance in circuit (1) Max 2

[9]

680. Variables:Temperature (of gas) (1)Amount of gas/mass of gas/number of molecules or moles (1) 2

Diagram to include any three of the following:

• trapped gas/fixed mass of gas (1)

• scale [or see dashed lines] (1)

• method of varying pressure [accept unlabelled syringe] (1)

• measurement of pressure [must label pump; accept P.G.] (1) Max 3

[Balloons drawn – no marksAny unworkable apparatus – 1 max i.e. e.o.p.Accept standard apparatus/syringes with pressure gauge/masses on moveable pistons.Ignore water baths.Heating experiment scores zero.]

Results:Reference to finding volume from their measurements [Accept volume scale labelled on diagram] (1)Label axes (1) 2e.g. P → 1/V or V → 1/P: [Accept p ≈ 1/L where L has been identified.Ignore unit errors on graph]

[7]

681. Peak wavelengths:

β Ori 2.9 × (10–3) m K

1.1 × (104) K (1)

= 2.6 × 10–7 m OR 260 nm (263 nm) (1)

α Cet 8.1 × 10–7 m OR 810 nm (805 nm, 800 nm) (1)

[Penalise “Not nanometres” once only] 3

Power per square metre:

Attempt to use σ T4 (5.67 and 1.14 substituted) (1)

= 8.3 × 108 (W m–2) (1) 2

Labelled graph shows:Ori peak at ~ 260 nm [e.c.f. their value] [Obviously to left of 400 nm] (1)Cet peak at ~ 800 nm [e.c.f.] [Obviously to right of 700 nm] (1)Area Ori » area Cet [e.c.f. their power] (1)[Accept > 4 × height][irrespective of λ ] 3

Explanation:[NB Refer to candidate’s graph]Ori at blue end of spectrum; Ceti at red end (1)BOTH outside visible region [e.g. in ultraviolet, in infra-red, “near”] (1)[Apply e.c.f. if wrong λ s above] 2

[10]

682. (a) Parallax diagram:

θ

S t a rL i g h t f r o m d i s t a n t s t a r

EE S

α

*

Labels on any FOUR of the following:

• distant stars labelled or implied e.g. by parallel lines [Accept without arrow heads]

• light from (nearby) star

• Earth orbiting Sun shown OR labelled June-January

• Angles α , θ labelled

• Angle P = distance Sun-Earth divided by distance to star OR trigonometry involving appropriate angle, sine and tangent, 1 all and D (4) 4

Suitability:The difference in angles is too small to measure for distant stars/distant stars do not move OR equivalent/we canonly measure angles to 1/100 arc sec (1) 1

(b) Supernova:Quality of language (1)Star → or begins as red giant (1)H burning or fusion ceases/fuel used up (1)Star collapses/contracts/shrinks/implodes (1)Shock wave (1) Blows off outer layers (1) Max 4

Remnant:Becomes neutron star/black hole (1) 1

[10]

683. Completion of table:

Material Young

modulus/1010 Pa

Ultimate tensile

stress/108 N m–2Nature

A 1.2 or 1.25 3.1 or 3.15(< 3.2)

Tough (2)

B 3.0 3.6 Brittle

2

Line drawn on graph: straight and stops suddenly (1)

at stress 3.6 × 108 N m–2 ) if not brittle, then peaks at this value) (1)(and strain 0.012) Correct gradient for straight regione.g. through 0.01, 3.0. (1) 3

Hooke’s law marked up to stress 2.7 to 2.9 × 1010 Pa [must be labelled] (1) 1

Energy stored:

[Accept stress in range 2.4 – 2.5 × 1010 Pa]Factor ½ × (1)Extension = 0.020 × 2.5 m [0.05 m] (1)

F = 2.4 × (108) N m–2 × 8.8 × (10–7) m2 (1)

[210 N; 220 N if stress = 2.5 × 1010 Pa]

= 5.25 J [5.5 J if using stress = 2.5 × 1010 Pa] [ue] (1) 4

[For middle 2 marks candidates may use stress× strain× volume, credit 1

mark for calculating stress× strain 2.4 × (108) N m–2 × 0.020 and 1 mark

for volume 8.8 × (10–7) m2 × 2.5 m][10]

684. Strain energy:is the energy stored /used/created/added/neededwhen a material is under stress/strained/loaded/stretchedOR in stretched bonds OR is work done when stressed (1) 1

Why stress should not be beyond elastic limit:Must return to original length when unstressedOR must give up stored energy when unstressed OR must not deform permanently/plastically (1) 1

Car calculation:

Substitution in mgh 1200 kg × 10 m s–2 [9.81]× 0.03 m (1)= 360 [350] J (1)3 kg steel store 390 J OR 360 J needs 2.8 kg steel (1) 3[Can also be argued in terms of 390 J → 3.3 cm]

Tendons:Will store 0.4 m × 2500 N /(1000) J (1)∆ h = 1.3 m [1.4 m ] [e.c.f. their energy ÷ (75 × 10[9.81]) (1) 2

[u.e.][7]

685. Atom is neutral (1)Quark composition is duu (1)

Antiproton is (−2/3) + (−2/3) + (+1/3) (= −1) (1) 3

Explanation:As soon as it touches the container/matter (1)(Matter and antimatter) annihilate (1)[Not “cancel”; not “react”] 2

Completion of table:

Quarks Charge

up charm TOP +2/3

down strange BOTTOM −1/3

1

[OR TRUTH & BEAUTY][Both needed for 1 mark]

(i) Neutral strange meson: ds OR d s (1)

(ii) Positive charmed meson: c d OR c s (1)

(iii) Neutral strange baryon: uss/css/uds/cds OR any of their antiparticles,e.g. u s s (1) 3

[9]

686. Conservation laws:

(i) Charge: (−1) + (+1) = (0) + (−1) + (+1) + (0) (1)Baryon number: (0) + (+1) = (+1) + (0) + (0) + (0) (1)[So possible, no mark]

(ii) Charge: (+1) + (+1) = (+1) + (+1) + (+1) + (−1) (1)Baryon number: (+1) + (+1) = (+1) + (+1) + (+1) + (−1) (1) 4[So possible, no mark]

[4]

687. (a) (i) l to nearest mm and between 850 mm → 860 mmor centre value ± 3 mm (1) 1d ± 0.02 mm of supervisor’s value and from ≥ 2 readings, to 0.01 mm or better (2) 2[± 0.02 mm from 1 reading 1 mark (1) ± 0.05 mm from ≥ 2 readings 2 mark][Unit error −1 once only]Variation in diameter 1

(ii) Correct calculation and unit and ≥ 2 significant figures (1)[Allow e.c.f.]Correct calculation and unit and 2 significant figures (1)[Allow e.c.f.]

Value 8.5 – 9.5 g cm–3 [No e.c.f.] (1) 3

(b) (i) m giving 2.0 ≤ t ≤ 6.0 s with measurements to better than 1 s [Do not award these 2 marks if systematic error, e.g. 0.0305 s] (2)4 values of t averaged (2)[2/3 values 1 mark]Correct calculation of a [Allow e.c.f.] (1)2/3 significant figure + unit from correct calculation (1) 6

(ii) If trolley not at rest at start position, then 0/3 (1)Start and stop positions marked in some way (1)Eye level at start and end/touch marker at start and end (1)Stopwatch started as trolley released and stopped after 0.500 m/use same point on trolley at start and stop (1) 3

(c) (i) Zero tabulated or plotted (1)4 other results (or results up to N = 0) tabulated and reasonable (1)

Graph:Scale and plots and correct orientation (1)Good curve (2)[For candidates who re-start at 48 every time: scale (probably in range 20 to 30), plots and correct orientation (1);horizontal straight line through mean position (2)] 5

(ii) Not representationThrows discrete – time continuous (1)48 is too small to represent a large number of atoms (1)Statistical error is very large with such a small number (1)

Reasonable representation:Probability of decay per throw = ½ /curve decreases by equal amountsin equal times/one throw is equivalent to one half life (1)Both decays are random (1) Max 3

[24]

Sample results:

(a) (i) l = 0.855 mNo zero error on micrometerd = 0.38, 0.38 mmAverage d = 0.38 mmReduces random error due to variation in diameter of the wire

(ii) V = 5.8510

0.38π

4

12

×

= 0.097 cm3

Mass = 0.84 g

p = 097.0

84.0 = 8.7 g cm–3

(b) (i) m = 50 gt = 2.84, 3.16, 3.06, 3.16, 3.02Av. t = 3.05 s

a = 2)05.3(

5.02 × = 0.108 m s–2

(c) (i) x N

0 481 282 13 94 3

Graph:

5 0

4 0

3 0

2 0

1 0

0

N

0 1 2 3 4x

688. Table:

Description Type of wave

A wave capable of causing photo-electric emission of electrons

Ultraviolet (1)

A wave whose vibrations are parallel to the direction of propagation of the wave

Sound (1)

A transverse wave of wavelength

5 × 10–6 m

Infrared (1)

The wave of highest frequency Ultraviolet (1)

4[4]

689. Explanation:

• waves diffracted from each slit/each slit acts as a source

• these superpose/interfere (1)

• maxima/reinforcement – waves in phase/pd = nλ (1)[or on a diagram][crest & crest] (1)

• minima/cancellation – waves in antiphase/pd = (n+1/2)λ (1)[or on a diagram][crest and trough] [not just ‘out of phase’] (1)

• phase or path difference changes as move around AB (1) Max 4

Determination of wavelength:Use of wavelength = p.d. [incorrect use of xs/D 1/3 max] (1)3 × (path difference. e.g.78 − 66 mm) (1)= 36 mm[Range 30 – 42 mm] (1) 3

Explanation:Less/No diffraction/spreading (1)∴ waves will not superimpose/overlap as much (1) 2

Explanation:Fixed phase relationship/constant phase difference (1)Both waves derived from single source [transmitter ⇒] (1) 2

[11]

690. Graphs:

T T

0 0θ m (2) 2

Description:Time for a number of cycles ÷ by no. of cycles (1)[accept swings]Count from centre of swing/repeat timing and average/keep amplitude small (1)Repeat for different lengths AND plot Graph of T v √l (1)[allow for ratio method]should be straight line through origin [consequent] (1)[allow for ratio method] 4

Calculation (based on graph):Attempt to find gradient (1)

Rate of change = 0.103 –0.106 s2 m–1 (1)

Rate of change of l plus comment on answer:

9.6 m s–2 [1/their value above] [no ue] [ecf] (1)close to/roughly/approx. acceleration of free fall/g (1)

[only if range 8.8 to 10.8 m s–2] 4[10]

691. Calculation:

E = hc/λ [seen or implied] (1)physically correct substitutions (1)

÷ 1.6 × 10–19 eV J–1 (1)5.78 eV (1) 4

Maximum kinetic energy:3.52 eV [ecf but not if –ve.] (1)

Stopping potential:3.52 V [Allow e.c.f., but not signs] (1) 2

Annotated graph:Position of S (1)Cuts V axis between origin and existing graph (1)Similar shape [I levels off up/below existing line] (1) 3

[9]

692. Discussion:No equilibrium or there is a resultant force (1)Direction changing or otherwise would move in a straight line (or off at an tangent) (1)acceleration or velocity changing (1)Force towards centre or centripetal (1)The tension provides this force [consequent] (1)[OR for last 2 marks: weight of ball acts downwards (1)vertical component of tension balances it (1)] 5

Free-body diagram:

W/weight/mg/gravitational ‘attraction’ [not ‘gravity’] (1) 1[6]

693. Explanation:Doppler shift:change in frequency/wavelength (1)due to motion of source/galaxy/observer (1)

Galaxies:The shift of a spectral line or use formula to find v. (1)‘Red shift’ ⇒ receding or ‘Blue shift’ ⇒ approaching (1)Quality of written communication (1) 5

Graph:Shape rough parabola; must hit time axis. (1) 1

Experimental difficulties:υ = Hd[No mark]d difficult to measure for distant galaxies (1)Hence H is inaccurate/uncertain. [consequent] (1)υ fairly accurately measured or H is squared so error bigger (1) 3

[9]

694. Identification of vector and scalar:

Vector = force, lift, horizontal distance, height, weight (1)

Scalar = mass, time, distance (1) 2

Calculation of weight:W = mg OR = 0.1 kg × 9.81 N kg–1 (1)

[Max 1 mark for g = 10 N kg–1]

= 0.98 N (1) 2

Calculation of vertical acceleration:

s = ut + ½ at2

1 m = 0 + ½ × a × (2.5 s)2

a = 2 m ÷ 6.25 s2 = 0.32 ms–2

State u = 0 (1)

Substitute correct distance (1)

a = 0.32 m s–2 [no u.e.] (1) 3

Calculation of resultant vertical force:

F = ma (stated or implied) (1)

= 0.1 kg × 0.32 m s–2 [Allow e.c.f.]

= 0.032 N (1) 2

Calculation of lift force:

[F = ma mark may be awarded here]

Weight – lift = resultant (1)

Lift = 0.98 N – 0.032 N = 0.948 N [Allow e.c.f.]

= 0.95 N [Allow –0.95 N] (1) 2[11]

695. Completion of circuit:

Ammeter and voltmeter used [correct symbols required] (1)

Ammeter in series, voltmeter in parallel (1)

[Do not penalise variable resistor in series] 2

Explanation of difference voltages:

Any two from:

• Internal resistance of cell/battery

• When current flows, energy transferred to / lost by internal resistance/heating in cell

• Hence voltage across internal resistance/ “lost volts”

• Reduced terminal p.d. / V = E – Ir / E = V + Ir 2

Show that internal resistance is about 0.6 Ω :

R = V/I

= (1.5 – 1.25) (1)

¸ 0.4 A

= 0.63 Ω [No u.e.] (1) 2

Calculation of resistance of bulb:

R = V/I (1)

= 1.25 V ÷ 0.4 A

= 3.1 Ω (1) 2

Explanation of lower resistance with ohmmeter:

Identify lower temperature with ohmmeter (1)

Lattice ions’/atoms’ vibrations impede electrons (1)

[Allow converse argument] 2[10]

696. Forces on diagram:

Tension/T in cable on both sides (1)

Weight / W / mg / 18 000 N / 18 000 [not “gravity”] (1)

[Penalise each wrong force in addition to the 3 but ignore upthrust] 2

Calculation of tension:

Net vertical force = zero

W = 2T (1)

× sin 2.5°[allow cos 87.5°] [θ wrong = eop] (1)

T = 206 000 N (1) 3

Show that total k.e. is about 500 000 J:

k.e. = ½ mυ 2 (1)

= 0.5 × 54 × 1250 kg × (4 m s–1)2 (1)

= 540 000 J [No u.e.] (1) 3

Calculation of time to reach maximum speed:

P = W/t

t = W/P = 540 000 J ÷ 500 000 W [Allow e.c.f.] (1)

= 1.1 s [Allow 1.0 s from 500 000 J] (1) 2

Suggest a reason why time longer:

Friction / air resistance reduces acceleration / resultant forceOR Friction / air resistance reduces useful power 1

[11]

697. Show that the energy supplied heating milk is about 60 000 J:

E = Pt OR = 140 W × 420 s (1)

= 59 000 J [No u.e.] (1) 2

Calculation of thermal energy transferred to milk:

E = mc∆ T

= 0.2 kg × 3900 J kg–1 K–1 × (38.5 – 13) (*) K (1)

[(*)Allow a different recognisable temp. change for this mark only] = 19 900 J (1) 2

Calculation of efficiency:

Efficiency = (useful energy output ¸ total energy input) × 100%

= (19 900 J ÷ 58 800 J) × 100% [Allow e.c.f.] (1)

= 34% [Accept fraction e.g. 0.34] (1) 2

Explanation of increase in efficiency:

Surrounding water must also be heated / only heating milk

[Not just heat lost to the surroundings] 1

Explanation of preference for method:

e.g. slower / more gradual temperature increase / easier to control

/ hygiene – keeps milk sterile / more uniform heating /

bottle at correct temperature also

[Do not accept prevents overheating / milk would be too hot.] 1[8]

698.

Desirable Not desirable

Reason

Elastic (1) Rod returns to original shape/position on unloading (1)

Brittle (1) Rod should not snap/shatter(when lifting a heavy fish) (1)

Hard (1) Rod will not scratch/dent with a large force (1)

Tough (1) Rod can withstand a sudden impact or dynamic load (1)

[Two marks for each line][8]

699. Crosses on graph:

P at end of straight line section (1)

Y – accept between P and maximum force value (1) 2

Young modulus calculation:

Calculation of stress (with force up to 1000 N)

= 25–

3

m103

N101

××

=A

F= 3.3 × 107 (N m–2) (1)

Calculation of strain (with corresponding extension from straight line graph)

= 05.0m0.4

m2.0 ==L

x (1)

E = =×

=0.05

mN103.3

strain

stress –27

6.7 × 108 Pa (1) 3

Lines on graph:

(i) L graph: twice extension for given force [Ignore end] (1)

(ii) A graph: 3 × force for given extension [Ignore end] (1)

Reasoning based on rearranged equation e.g. F = Eax/L and E constant implied in:

2L gives 2x for same F (1)

3A gives 3F for same x (1) 4

Energy stored in the rope:

Energy = ½ Fx = ½ × 1000 N × 0.2 m (1)

= 100 J (1) 2

Why a longer rope is less likely to break:

Any one point from:

• greater extension for same force

• larger area under graph

• more energy stored 1[12]

700. Energy of photon of green light:

f = m105.58

sm1037–

–18

××

=λc

= 5.38 × 1014 Hz (1)

E = hf = 6.63 × 10–34 J s × 5.38 × 1014 Hz (1)

= 3.56 × 10–19 J 2

Diagram:

Larger gap identified (1)

Downwards arrow between levels of same element (1)

2[4]

701. Diffraction:

The spreading out of waves when they pass through a narrow slit or around an object (1)

Superposition:

Two or more waves adding (1)

to give a resultant wave [credit annotated diagrams] (1)

Quantum:

A discrete/indivisible quantity (1) 4

Particles:

Photon/electron (1) 1

What the passage tells us:

Any 2 points from:

• large objects can show wave-particle duality

• quantum explanations now used in “classical” solutions

• quantum used to deal with sub-atomic particles andclassical with things we can see Max 2

[7]

702. Isotope of lead:

Pb20682

1

Other particles:

(82) electrons 1

How appropriate number of quarks can combine:

3 quarks involved (1)

2 × + 2/3 + 1 × –1/3 = + 1 (1) 2

Explanation:

High energy is needed/high temperature/high speed (1)

Mention of E → m OR E = mc2 (1) 2

Description:

Relates to electron (1)

e.g. charge +1/antiparticle/annihilates with (1) 2[8]

703. Calculation of total momentum:

In 1 s, p = 1400 J /3 × 108 m s–1

= 4.7 × 10–6 N s (kg m s–1)

[No u.e.] 2

Force exerted on whole sail:

Momentum change/time [OR symbols] (1)

= 4.7 × 10–6 × 1.5 × 106

= 7.0 (7.1) N (1) 2

Explanation of why force is doubled:

Photons bounce back (1)

So their change of momentum is doubled (1) 2

Calculation of maximum increase in speed:

a = F/m = 7/1200 m s–2 [allow e.c.f.] (1)

(= 0.006 m s–2 )

∆ υ = a ∆ t = 0.006 m s–2 × 604 800 s (1)

= 3500 m s–1 (1) 3[9]

704. Calculation of energy:

E = ½ CV2 (1)

= ½ × 100 × 10–6 F× (4 V)2

= 8 × 10-4 J (1) 2

Parts of circuit which will transfer energy to surroundings:

Ss / the wires between it and C 1

Discussion:

No – negligible energy 1

Completion of graph:

Convex curve up from 0 (1)

through (200, 4) (1)

Then drops to zero (1) 3

6

4

2

0 1 0 0 2 0 0 3 0 0 t / s

p . d . / V

Approximate value for R :

Time constant = RC (1)

⇒R = 200 s/100µ F

= 2× 106 Ω [Allow 1.6 – 2.2 MΩ ] (1) 2

Discussion of effect of increasing resistance, R :

R goes up ⇒ T goes up (1)

T goes up means longer toasting (1) 2[11]

705. (a) The origin of the induced e.m.f:

Faraday’s law (1)

As conductor cuts field lines (1)

Electrons experience force along wire (1)

⇒ move to one end ⇒ e.m.f. (1) 3

(b) Reduction in orbit height due to flow of current:

Current + field ⇒force OR Fleming L H rule (1)

Lenz’s law: (1)

Force opposes motion (1)

Orbiting craft lose energy/speed (1) 3[Max 5]

706. Calculation of minimum height x :

a = r

2υ (1)

⇒ r

2υ ≥ g at top (1)

At top: ½ mυ 2 = mgx [OR mgh 1 mark] (2)

⇒ υ 2 = 2gx (Note: derived from υ 2 = u2 + 2as = 0)

[For υ 2 = 2g(x + 1) (1)(speed at bottom)]

⇒ 2gx/r ≥ g at top

⇒ x ≥ r/2

⇒ x ≥ 0.25 m

[≥ or =] (1)[5]

707. Calculation of resultant force:

[a = (υ – u)/t = 16 m s–1[(4 × 60) s]

= 0.0666 m s–2

F = ma = 84 000 kg × 0.0666 m s–2 = 5600 N]

OR

Use of t

u)( −υuse of mυ (1)

Use of F = ma use of t

mυ(1)

5600 N 5600 N (1) 3

Free-body force diagram:

Diagram [truck can be just a blob] showing:

8 4 0 0 0 0 N

8 4 0 0 0 0 N

1 1 2 0 0 N1 6 8 0 0 N

823 200 – 840 000 N down (1)

same as down up (1)

11 200 N either way (1)

correct resultant to left

[e.c.f.]

4

[Ignore friction. Each extra force –1]

Calculation of average power:

Power = KE gained/time = ½mv2/t OR KE = 3.84 × 108 J (1)

= 3.84 × 108 J/(4 × 60) s (1)

= 1.60 × 106 W [OR J s–1] (1)

3

Other credit-worthy responses:

½ mυ 2 Fv

t

Fd (1)

24016103

21

26 ××× 3 × 106 × 0.666 × 8240

1920666.01036 ×××

[e.c.f. 0.666 and 1920possible]

(1)

1.6 × 106 W 1.6 × 106 W 1.6 × 106 W (1)

3[10]

708. Principle of conservation of linear momentum:

[Not just equation – symbols must be defined]

Sum of momenta/total momentum remains constant (1)

[Equation can indicate]

[Not “conserved”]

If no (resultant) external force acts (1)

[Not “closed/isolated system”]

2

Laws of Motion:

2nd and 3rd laws (1)

1

Description:

Measure velocities/speeds before and after collision (1)

Suitable technique for measuring velocity

e.g. ticker tape/ticker timer

light gate(s)

motion sensor (1)

[Not stop clock or just datalogger]

How velocity is found from their technique

[Need distance ÷ time + identify distance and time. Could get with (1)stopclock method.] (1)

υ after = ½ υ before/ calculate mυ before and after/check e.g. m1υ 1 = (m1 + m2) υ 2

4

Reason for discrepancy:

Friction/air resistance (1)

[Ignore any reference to energy] (1)

Explanation:

The Earth (plus car) recoils (1)

With same momentum as the car had (1)

2[10]

709. Explanation of why it is a good approximation:

Resistance of connecting lead is (very) small (1)

So I × R(very) small = (very) small p.d./e–1s do little work so p.d. small/r small (1)

compared with rest of the circuit so p.d. small

2

Circumstances where approximation might break down:

If current is large OR resistance of rest of circuit is small (1)

[Not high voltage/long lead/thin lead/high resistivity lead/hot lead]

1

Calculation:

Use of R = A

lρ with A attempted × sectional area (1)

Correct use of 16 (1)

Use of V = IR (1)

0.036 V (1)

4[10]

710. Relationship between current and charge:

Current is the rate of flow of charge/rate of change of charge OR current is charge per second

OR I = Q/t (with or without d or ∆ ) but with symbols defined (1)

1

Explanation:

Since I is constant, Q o on capacitor (= It) increases at a steady rate OR charge flows at a constant rate (1)

Since V ∝ Q, V also increases at a steady rate (1)

OR

V = Q/C = It/C (1)

and V = (I/C) × t compared with y = (m) × x

2

Determination of current, using graph:

Use of Q = CV Attempt to get grad (1)

Use of I = Q/t Use of I = C × grad (1)

= 1.1 mA 1.1 mA (1)

3

Explanation:

Decrease [If increase, 0/3] (1)

As capacitor charges, VR decreases (1)

R must decrease because I = VR/R OR R must decrease to prevent I

falling (1)

3

Second graph:

Line added to graph showing:

Any curve getting less steep with time [from origin; no maximum] (1)

And with same initial gradient as original straight line (1)

2[11]

711. Calculation of energy released for each fission:

∆ m = 0.2035u (1)

Convert their ∆ m to kg [× (1.66 × 10–27) kg] OR × 931 (1)

Convert their kg to J [× (3 × 108 m s–1)2] OR × 1.6 × 10–13 (1)

= 3.04 × 10–11 (J) (1)

4

Calculation of power output:

Energy per mole = 3.04 × 10–11 J × 6 × 1023 (1)

Full e.c.f. their energy = 1.8 × 1013 J

Power = s5

JenergyAny [No e.c.f.] (1)

= 3.6 × 1012 W [Accept J s–1] (1)

3[7]

712. Nuclear equation:

–01–

01–

–3216

3215 eeββSP +→ [Ignore + γ + υ ] (1)

1

Description:

Take background count (1)

Take count close to source, then insert paper/card and count (1)

Little/no change (1)

[OR absorption in air: Take close reading and move counter back; no sudden reduction (1)(1)]

Insert sheet aluminium and count (1)

Down to background, or zero (1)

Max 4

Diagram: Any region above dots [show (1) or (X)] (1)

1

1 2 0

1 0 0

8 0

6 0

4 0

2 0

00 2 0 4 0 6 0 8 0

N u m b e r o fn e u t r o n s N

N u m b e r o f p r o t o n s Z

X XX X X

Explanation:

1 β – decay involves a neutron → a proton Any two from: (1)

Any two from:

2. on the diagram this means ↓–1 ( +1 / diagonal movement

3. so nuclide moves towards dotted line

4. decay means greater stability (1)(1)

[β – in wrong region, (1) and (4) only available.

Decay towards drawn N = Z line 1 and 2 only available][9]

713. Range of extensions where Hooke’s law is obeyed:

(From 0 to) 9 (mm) or 9.5 (mm) (1)

1

Addition to diagram:

Horizontal ruler fixed to bench with marker anywhere on wire

OR

Vertical ruler with pointer on load/hanger OR closely aligned with ruler (1)

Length to be measured, as shown on diagram:

Length from double blocks to marker on wire

OR

L e n g t h f r o m d o u b l e b l o c k s t o j u s t a b o v e t h e p o i n t w h e r e m a s s h a n g e r i s h u n go n p u l l e y

(1)

2

Young modulus:

Use of E = Ae

Fl OR

le

AF ÷ (1)

F, e valid pair on straight line region consistent with their answer to point 1 (1)

[Do not allow 10 mm 44 N. Ignore 10n error]

= 1.2 × 1011 N m–2/Pa/kg m–1 s–2 (1)

[1.1 – 1.3]

3

Energy stored in wire:

Use of ½ Fx/area up to 7 mm OR count squares ≈ 50 (1)

0.1 J [Accept Nm] (1)

2

One energy transformation:

GPE → elastic potential energy (1)

1

Tensile strength of brass:

Attempt to calculate stress i.e F/A (1)

46/47 N Fmax off graph (1)

= 3.5 × 108 (N m–2) [No u.e.] (1)

[3.5 – 3.62]

3[12]

714. Wavelength of the microwaves:

λ = 442 mm – 420 mm (1)

= 22 mm [2.2 cm, 0.22 m] (1)

Frequency of microwaves:

Use of c = fλ with λ from above substituted OR if no attempt, then (1)

C = 3. × 108 substituted

1.4 × 1010 Hz [e.c.f. λ above] (1)

4

Maximum Q and minimum D marked on diagram:

Either Q (1)

Any D (1)

2

I n c o m i n g

m i c r o w a v e s

M e t a l s h e e tw i t h t w o s l i t s

4 2 0 m m

4 4 2 m m

P

R

D

D

D

D

Q

Q

Q

X

X

Why a maximum would not be detected at P:

Wavelength of sound wave = 0.3 m (1)

Path difference at P is not whole wavelength (1)

2

[OR valid reference to phase difference OR λ sound greater so no

diffraction with this slit width OR valid reference to λ = xs/D][8]

715. Ionisation energy of atomic hydrogen:

13.6 eV OR 2.18 × 10–18 J [– sign, X] (1)

1

Why energy levels are labelled with negative numbers:

Work/energy is needed to raise the electrons/atoms to an energy of 0 eV, somust start negative (1)(1)

OR

Work/energy is given out when the electrons/atoms move to the groundstate, so energy now less than 0, i.e. negative (1)(1)

OR

the ground state is the most stable/lowest energy level of theelectrons/atoms and must be less than 0, i.e. negative (1)(1)

2

[1st mark essential: e– highest/maximum/surface/ionised/free has energy = 0eV

2nd mark: raising levels means energy in OR falling levels means energy out ∴ negative levels]

Wavelength of photon:

∆ E = 1.89 (eV) (1)

Convert ∆ E to joules, i.e. ×(1.6 × 10–19)

OR

λ = )106.1(89.1

1031063.619

834

−

−

×××××

[Their E] (1)

= 6.6 × 10–7 (m) [6.5 – 6.7] (1)

3

Production of line spectrum of atomic hydrogen in a laboratory:

Source – hydrogen discharge tube/hydrogen lamp/low p hydrogen with high V across (1)

(view through) diffraction grating/prism/spectrometer/spectroscope (1)

2

Sketch:

A few vertical lines on a blank background OR sharp bands

Dark on light/light on dark NOT equally spaced (1)

1

Absorption spectrum:

White light through gas in container (1)Diffraction grating/prism/spectrometer (1)Must be dark lines on bright background (1)

[9]

716. Sketch graph showing:

p decreasing as V increases [Accept straight line] (1)

Smooth curve, asymptotic to both axes [Not touching or going to touch] (1)

2

Explanation of shape of graph:

As V increases:

packing density of the molecules decreases

OR molecules travel further between collisions (1)

[Look for change in molecular spacing]

Collision (rate) with walls decreases OR change in number of collisionswith walls [Ignore reference to intermolecular collisions] (1)

2

How to calculate pressure of air in the syringe:

[NB Not gauges/manometers/pV method]

Weight (of mass) ÷ area of piston [no need for Xle] (1)

plus atmospheric pressure (1)

[Penalise if wrong area]

2

Suggested possible source of error:

Any one from:

• temperature not constant

• leakage of air OR mass of gas not constant

• weight of piston not included

• friction (1)

1

[Not non-uniformity of tube/dead space][7]

717. Calculation of potential difference:

Use of E = V/d [d in m or cm] (1)

V = 90 kV (1)

Calculation of maximum kinetic energy:

Use of × 1.6 × 10–19 [in E = qV e.c.f. value of V] 1.4 × 10–14 (J) (1)

[e.c.f. their V × 1.6 × 10–19] (1)

4

Maximum speed of one of these electrons:

Use of k.e. = ½ mυ 2 with m = 9.1 × 10 –31 kg (1)

[Full e.c.f. their k.e. possible; make sure v is speed term]

= 1.8 × 108 m s–1 [u.e. but only once] (1)

Diagram:

2

At least 3 radial lines touching object (1)

Direction towards electron (1)

2

Expression for electric potential V:

V = r

19106.1

επ

−

0

××

41

OR re

04π ε OR r

91044.1 −×

[not k unless defined]

definedunless

4 0

Qr

QNot

επ

[With or without “–” sign] (1)

1[9]

718. Word equation:

Force proportional to product of masses and inversely proportional to (distance / separation) squared (1)(1)

2

[No force 0/2]

OR

F = 2)distance(21 massmass ××G

(1)

[or (separation)2 instead of bottom line]

Calculation of force:

From Newton’s law OR idea that force = weight = mgplanet (1)

F = 226

232211

m)1040.3(

kg1kg1042.6kgmN1067.6

××××× −−

[Substitution in correct equation only]

OR

gmars = 226

23

m)104.3(

kg1042.6G

×××

(1)

= 3.7 N (1)

Smaller (1)

3

Explanation of reasoning:

g is less, but ρ is similar/same [so R is less] (1)

[2nd mark is consequential on first mark]

2[7]

719. Flux through closed window:

Flux = 20 × 10–6 T × (1.3 × 0.7) m2

[if two equations, must use (1.3 × 0.7) each time] (1)

BH chosen OR area correct (1)

= 1.8 × 10–5 Wb/T m2

2

Average e.m.f. induced:

E = s8.0

Wb108.1 5−× [e.c.f.] (1)

= 2.3 × 10–5 V [5.6 × 10–5 V if Bv used] (1)

2

Effect on induced e.m.f. of converting window: (1)

Zero induced e.m.f. [Not “very small”]

No change in flux linkage OR no flux cut OR e.m.fs. in opposite sides cancel out 2/2 [Consequential] (1)

2[6]

720. Momentum of driver:

Correct use of p = mυ [OR with numbers] (1)

= 1500 N s OR 1500 kg m s–1 (1) 2

Average resultant force:

Correct choice of F × t = ∆ p OR F = ma (1)

F × 0.07 (s) = 1500 (N s) F = 50 × 429/50 × 30/0.07 (1)

= 21 kN = 21 kN (1) 3

[Ignore sign of answer]

Why resultant force is not the same as force exerted on driver by seatbelt:

Air bags /floor/friction/seat/steering wheel (1)[Named force other than weight/reaction] 1

[6]

721. Completion of table:

Force Description of force Body which exerts force

Body the force acts on

A Gravitational Earth Child

B (Normal) reaction OR contact OR E/M (1)

Earth/ground Child(1) for both

C Gravitational [Not gravitational weight] (1)

Child Earth(1) for both

4

Why A and B are equal in magnitude:

Child is at rest/equilibrium OR otherwise child would move/accelerate (1)[NB use of N3 would contradict this]

Why must forces B and D be equal in magnitude:

Newton’s third law OR action + reaction equal and opposite (1)[NB use of N1 or N2 here would contradict this] [Not Newton pair] 2

What child must do to jump and why he moves upwards:

Push down, increasing D (1)

∴ B increases [must be clearly B or description of B] (1)

and is > A OR there is a resultant upward force [clearly on child] (1)[Not “movement”] 3

[9]

722. Force:

200 N (1) 1

Free body force diagram:

[Unlabelled arrows = 0; ignore point of application of force][Double arrows = 0] [Arrows required for marks]

W / weight / 400 N /mg/ pull of Earth / gravitational force (1)[Not gravity]

Two tensions OR Two × 300 N OR two × 311 N (1)[Accept T for tension OR any label that is not clearly wrong,e.g. R/W/N 200 N] 2

Applied force:

Attempt to resolve vertically (1)

2T sin 40 = 400 (1)

[400 ×cos 40 → 306 N(no marks)400 × sin 40 → 257 N (no marks)

200/cos 40° = 261 N → gets 1 out of 3 (attempted to resolve)]

T = 310 (N) OR 311 (N) [No unit penalty] (1) 3

Two reasons why first method is easier:

Force applied is smaller/feels lighter/tension smaller [Not weighs less] (1)

They are not pulled sideways/forces only upwards/pulling against each other (1)[Answer must be in terms of forces] 2

Why solution is not sensible:

Because the tension (or description of tension) would be greater (1)OR bigger sideways force 1

[Do not accept bigger force][9]

723. Completion of diagram:

U s e f u l w o r kd o n e b y m o t o r

( I n c r e a s e )

m g h

i n g p eO R w . d . a g a i n s tg r a v i t y / [ N o tw . d . o n c a r ]

(1) 1

(i) Useful work done by motor:

Correct substitution in mgh, i.e. 3400 (kg) 9.81 (m s–2) × 30 (m) (1)

= 1.00 MJ OR M Nm [1.02 MJ] (1)

(ii) Power output of motor:

Power = above (J) / 15 (s) (1)

= 67 kW [e.c.f.] (1) 4

Overall energy conversion occurring as vehicle travels from B to C:

( G r a v i t a t i o n a l )p o t e n t i a l e n e r g y

w . d . b y g r a v i t yO R

K i n e t i c e n e r g y( a n d g . p . e )

(1) 1

Speed of vehicle at point C:

∆ h = 18/(30 – 12) (1)

Use of ½ mv2 = g.p.e. lost (1)

[If get height wrong, can only get second mark]

υ = 19 m s–1 [18.8 m s–1] 3

How speed at C would be expected to differ from previous answer:

Same speed/no effect [If this is wrong, no marks] (1)

GPE and KE both symbol 181 \f “ 12µ m ORg same for all masses OR ms cancel (1) 2

[Not g is constant][11]

724. Sources of background radiation:

Radioactive rocks/radon gas/cosmic rays or solar wind (1)

Fall out/leaks from nuclear installations

named materials, e.g. uranium/granite/14C (1) 2

Nuclear equation:

01

2210

2211 ++→ NeNa e [Accept β ]

22 and 0 (1)10 and +1 OR 10 and 1 (1)

20 – 26 Na [NOT 22Na] [Must have correct proton number, if given] (1) 3

Decay constant of sodium–22 in s–1:

λ =0.69/2.6 [Ignore conversion to seconds] [Not 0.69/1.3] (1)

λ = 8.4 × 10–9 [No unit, no e.c.f.] (1)

Number of nuclei:

2.5 = 8.4 × 10–9 N (1)

N = 3.0 × 108 (1) 4

Whether salt is heavily contaminated:

(No.) This is a small number (compared to no. of atoms in a spoonful of salt)

OR

Rate < background (1) 1[10]

725. Uranium correctly marked at (92, 142) (1)

Beta decay: SE at 45° [One box] into the uranium (1)

Alpha decay: Proton number down 2 (1)

Neutron number down 2 (1)

[NB No arrows needed, but lines must be labelled appropriately; lines not essential if clear][4]

726. Number of carriers or electrons per unit volume / per m3 /carrier density/electron density (1)

[Not charge density / concentration]

Drift velocity OR drift speed OR average/mean/net/overall velocity (1) 2

[Not just velocity; not speed unless drift]

m–3 (1)

m2 As m s–1 (1)

Multiply and reduce to A (1) 3

[Base units not needed][Mixed units and symbols could get the third mark]

[mA = m–1 loses 1 mark]

Metal:

M : l a r g e s o t h e r e i s a c u r r e n tn

I n s u l a t o r

I : z e r o ( n e g l i g i b l e ) / v e r y s m a l l s o l e s sc u r r e n t ( o r z e r o c u r r e n t )

n

n : i n m e t a l l a r g e rn m u c h

C u r r e n t i n m e t a l i s l a r g e r

( 1 )

( 1 )

2

[Ignore anything about v. Allow e.g. electron density for n][7]

727. Use R = ρ l/A OR correct rearrangement OR plot R → l gradient = ρ /A (1)[Symbols or words]

With A = tw (1) 2

l = RA/ρ [Rearrangement mark symbols or numbers] (1)

Use of A = tw (1)

[Correct physical quantities substituted but ignoring unit errors, powers of 10]

= 110 m

[111 m] (1) 3

Reduce width/w of strip OR use thinner/t foil [Not reduce A; not increase T, V, I] (1)

Smaller w/t/A will be less accurate OR have larger error OR larger R will be more accurate (1) 2

[Increase w or t, could give e.c.f. to increased accuracy][7]

728. I 2 R / (ε I – I 2 r) / R

Ir 2)( −ε (1)

I 2 r / (ε I – I 2 r) R

Ir 2)( −ε(1)

ε I OR I 2 R + I 2 r / ε 2 / (R + r) (1)

ε I = I 2 R + I 2 r OR (It = I 2 RT + I 2 rt / their (iii) = their (i) + their (ii) (1)

Cancel I (OR I and t) and arrange [only if energy equation is correct] (1) 5

Maximum current occurs when R = 0 (1)

Imax = ε /r (1) 2

OR larger r means smaller I (1 mark)

1 MΩ [Could be underlined OR circled] (1)

It gives the smallest current (1)

[If 100 kΩ this reason: 1 only] 2[9]

729. No, because V is not proportional to I OR not straight line through origin / (1)only conducts above 0.5 V / resistance changes 1

Use of R = 0.74 / current from graph (1)

= 9.25 Ω [9.0 – 9.5 Ω ] [Minimum 2 significant figures] (1) 2

Calculation of p.d. across R [8.26]

Calculation of total resistance[109 – 115]

Ratio R: ratio V E = Σ IR (1)

÷ I – diode resistance [9] Correct substitutions

Correct substitutions (1)

103 Ω [100 – 106] (1)

3

[If not vertical line, 0/2]

0 . 7 0 . 7≠ 0 . 7

A n y t h i n g ( g a p , c u r v e , b e l o w a x i s )

( 1 )( 1 )

( 1 )( 0 )

( 1 )( 0 ) (1)(1) 2

[Otherwise 0 0 ][8]

730. ρ = Nm/V (1) 1

p = 2/3 (N/V) ½ m <c2> / <c2> ∝ k.e. / m <c2> ∝ k.e. (1)[Full backwards argument can get 1 / 2; full qualitative argument scores 1 / 2]

k.e. (or ½ m <c2> or <c2>) ∝ T, ∴ p ∝ T (1) 2

V constant (1)

N constant / for fixed mass / fixed number of moles [Not fixed amount] (1)

[Near ideal conditions, specified, can replace one of the above][Fixed density, 1 mark] 2

See (273, 308) or 404 (1)

Use P1/T1 = P2/T2 (kelvin temp) (1)

= 456 kPa (1)

[355 kPa gets 3 marks]

[303 kPa → 342 kPa, 101 kPa → 114 kPa gets 2 marks] 3[8]

731. Power = ORtime

energyOR

time

work rate of doing work OR rate of

transfer of energy (1) 1

[Symbols, if used, must be defined]

Unit = Watt OR J s–1 (1) 1

Base units:

kg m2 s–3 (1)(1) 2

[If incorrect, possible 1 mark for energy or work = kg m2 s–2 or for J = Nm][6]

732. H–R diagram:

Circle S on main sequence at L = 10° (1)

Circle M on main sequence at top left (1) 2

Numbers on temperature axis showing increase ← (1)

Coolest 3000 ± 1000; hottest 20 000 – 50 000[Both for the mark] (1) 2

Large mass stars:

They are brighter OR have greater luminosity OR are hotter (than the Sun) (1)

and burn up fuel/hydrogen quickly [Not energy] (1) 2

Calculation:

See E = mc2 (1)

See 218

26

)sm103(

watt109.3−×

× s (1)

= 4.3 × 109 kg s–1 (1) 3[9]

733. Red giants:

They are cool high volume stars OR cool large surface area OR cool high luminosity/bright/cool big stars (1)

Parallax displacement:

Compares the angle between a star and a distant star OR the position of a star relative to a distant star/relative to fixed background (1)

Viewed six months later (1) 3

[These latter two marks could be obtained from a suitably labelled diagram, but probably only 1 out of 2 for just a diagram.]

Period:

5 (days) ± half day (1) 1

• Luminosity of B greater than luminosity of A (1) 1

B must be further away (1)

Convincing reasoning, for example: [consequent on correct distance]

I same, L greater, hence D greater OR reference to formula

I = L/4π D2 OR reference to inverse square law (1)[e.c.f. wrong conclusion at *] 2

Two forces:

Gravitational/gravitation/gravity (1)

(Force due to) radiation/photon/electromagnetic wave pressure/forces (1) 2[9]

734. Hooke’s law:Extension proportional to (∝) force/load OR F = k∆ x with F, x defined (1)

below the elastic limit OR below limit of proportionality (1) 2

Ultimate tensile stress = 2.3 (× 108 Pa)

Young modulus = stress/strain [No mark]

= any pair off linear region between 0.8, 1 and 1.6, 2.1 (1)

= 1.3 × 1011 (Pa/N m–2) [1.2 – 1.4] (1) 3

Attempt to calculate 26 m107.1

N250−×

OR P correctly plotted (1)

Elastic because on straight line/equivalent (1) 2

Point P on line at stress = 1.5 × 108 Pa [e.c.f their value of stress] (1) 1

Extension of wire:

Determine strain = 1.1 × (10–3 ) [OR 1.2] (1)

[Either by calculation or by reading off graph]

Extension = 3 × strain [e.c.f.]= 3.3 (3.6) × 10–3 m 2[10]

735. Graph:

F

x

Axes and shape (1)

Arrow heads or labels [if axes inverted, arrows must be reversed] (1) 2

Warmer because:

Area represents energy or work done [may be labelled on graph] (1)[Must refer to graph]

Converted to heat (in rubber band) (1) 2[4]

736. N–Z grid

Sr at 38, 52 (1)

Y at 39, 51 [e.c.f. Sr incorrect → 1 diagonal move] (1) 2

Rb at 37, 45 (1)

Decays by β + emission/positron/α (1) 2[4]

737. Charge on strange quark = – 1/3 (1) 1

Conservation law:

Charge – (–1) + (+1) → (0) + X/by charge conservation (1)

X is neutral (1) 2

Particle X is a meson (1)

Baryon number conservation (0) + (+1) → (+1) + (0) (1) 2

OR discussion in terms of total number of q + q = 5 OR Σ q – q = 3

Composition of X is s d [0/3 if not q q ](1)

Justify S quark:

This is not a weak interaction/only a weak interaction can change quark type/this isa strong interaction/strangeness is conserved/ quark flavour cannot change (1)

Justify d quark:

X neutral; s – 1/3; d + 1/3. [e.c.f. if s = – 1/3 in first line.]

For the third mark accept any q q pair that creates a mesonof the charge deduced for X above. (1) 3

[The justification for both q and q can be done also by tracking individual quarks][8]

738. Potential difference = charge

ywork/energ OR

current

power

OR in words: work done in moving 1 coulomb of charge between two points. (1) 1

Unit: volt OR J C–1 OR V (1) 1

Base units: kg m2 A–1 s–3 (1)(1) 2

[2/2 possible even if final answers wrong for recognising that As = C J = Nm][4]

739. (a) (i) Table with columns headed with quantities and units (1)

Reading every ½ minute or better OR timed for every 1°C fall (1)

Start ≥ 80 °C and finish ≤ 70 °C (1)

Concave curve established (1) 4

(ii) Graph:

Scale [more than ½ grid in both directions, avoiding scales of 3 etc; (1)allow 60 s per square]

Axes [labelled, with units] (1)

Plots [accurate to ½ division] (1)

Line [smooth curve] (1) 4

(iii) Good tangent at 75°C [must be a curve for this mark] (1)

Large ∆ [∆ x∆ y ≥ 64 cm2] (1)

Correct calculation + unit and > 2 significant figures [ignore sign] (1)

Correction + unit [W or J s–1] (1) 4

(b) (i) Temperature sensor (1)

Data logger (1)

Computer (1)[–1 if no arrows]

≤ 30 s (1) 4

(ii) (9.3 – 9.4) V + unit (1)

Potential divider [not series rheostat] OR labelled [or with arrow]variable power supply (1)

Heater (1)

Voltmeter (1)If voltmeter in series with heater can only get first mark] 4

(iii) Any two from:

Leave datalogger operating overnight OR set time for 12 – 24 h (1)

Sampling rate ≤ 30 min (1)

Download data to computer next morning (1)

Any two reasons:· (1)(1)

• variation of room temperature

• evaporation of water

• fluctuations in power supply 4

[NB “heat lost from beaker” is not a reason][24]

Sample results:

(a) (i)

t/min θ °C

0.00.51.01.52.02.53.03.54.04.55.0

85.083.081.079.578.077.075.574.573.572.571.5

(ii) Graph:

8 5

8 0

7 5

7 00 1 2 3 4 5 t / m i n

θ / º C

(iii)0.05.5

0.700.82

Δt

Δ

−−=θ

K min–1

= – 2.2 K min–1

= – 0.036 K s–1

P = 900 × 0.036 J s–1

= 32 W

(b) (i)

T e m p e r a t u r es e n s o r

D a t al o g g e r

C o m p u t e r

Sampling rate of every 10 s would give 30 readings in 5 minutes

(ii) p.d. = 9.4 V

Diagram:

Set p.d. to 9.4 V as read off from curve

(iii) Set data logger to record for 24 h at sampling rate of every 30 min

Reasons:

• variation of room temperature

• evaporation of water

• fluctuations in power supply

740. (a) Average of t found from ≥ 3 runs and average found with units (1)(1)

[2 runs →(1)]

2.0 s – 5.0 s (1)

[Systematic error in t, do not award third mark]

[All measurements to nearest second –1]

Correct calculation and unit to ≥ 2 significant figures (1) 4

(b) Diagram clearly showing distance travelled[could be from base of mass to ground] (1)

Using same point on trolley or falling mass (1)

Good method for determining starting and stopping positions of trolley,e.g. eye level, block on bench, etc. (1) 3

(c) Correct re–arrangement (1)

Correct substitution (SI units) (1)

Correct calculation with unit and ≥ 2 significant figures (1) 3

(d) Sensible ∆ m > 30 g OR max value of m(100 g) (1)

Average t found from ≥ 3 runs (1)(1)

[From 2 →(1)]

Correct calculation (of a) and F, with units for F and ≥ 2 significant figures (1) 4

(e) Correct percentage difference with mean or either value as denominator (1)

Sensible conclusion based on where less or more than 10% (1) 2

[Allow modulus if one F positive and the other negative]

(f) (i) Keep M constant OR keep x constant [or measure x] (1)

Use range of value of m (1)

Find corresponding t (1)

Calculate acceleration (1)

(ii) Axes labelled (1)

Correct line (1)

–F shown on graph OR dotted extrapolation OR F with negative signstated in part (iii) (1)

(iii) Gradient = g (1)

Intercept = –F [Ignore sign] (1)

[Accept numerical values from (c) or (d) without units] Max 8

Sample results:

(a) m = 40 g [i.e. 30 g added to hanger]

t/s: 3.9, 4.3, 4.2, 3.9, 4.1 average = 4.08 s

a = 208.4

500.02 ×

= 0.060 m s–2

(b) Diagram:

S t a r t m a r ko n b e n c h

F i n i s h m a r ko n b e n c h

0 . 5 0 0 m

Pulley may be used as end point

(c) (M = 2.42 kg)

F = mg – (M + m) a

= (0.04 × 9.81) – (2.42 + 0.04) × 0.060

= 0.392 – 0.147 N

= 0.24 N

(d) m = 80 g

t/s = 2.0, 2.3, 2.1, 2.4, 2.3 Average t = 2.22 s

a = 222.2

500.02 ×

= 0.20 m s–2

F = 0.08 × 9.81 – (2.42 + 0.08) × 0.20

= 0.785 – 0.500

= 0.28 N

(e) Percentage difference = %10026.0

24.028.0 ×−

= 15% (> 10%)

So F cannot be considered as constant

(f) (i) Keep M constant OR keep x constant [or measure x]

Use range of value of m

Find corresponding t

Calculate acceleration

(ii) (M+m)a

0m

– F

(iii) Gradient = g

Intercept = –F[24]

741. Unit for power:

Watt OR W OR J/s OR any correct S.I. equivalent (1) 1

Average speed:

500 m /120 s

= 4.2 OR 4.17 m s–1 [allow 4.1 or 4.16 or 4 but not 4.0] (1) 1

Average resistive force:[1st mark is for the formula in any arrangement] (1)F = P/υ [accept P = work/t or P = F × d/t]= 230 W / 4.2 m s–1 (1)= 55 N 2

Initial acceleration:

Tangent at t = 0 drawn (1)

Acceleration = gradient (1)

⇒ value in range 0.50 – 0.85 m s–2 (1) 3

To travel 500 m:

Distance travelled = area under graph (1)

Valid attempt to evaluate appropriate area (1)

Answer in range 105 s – 119 s (1)[allow use of s = υ × t for full marks] 3

Why speed becomes constant:

a = 0 OR F net/total = 0 OR equilibrium/balanced forces (1)

for a = 0, resistive F = forward F [accept friction or drag for resistive] (1)[do not accept “forces are equal” or “has reached terminal velocity”]

frictional force increases as velocity increases (1) Max 2[12]

742. Current in motor:

V

PI = = 300 000 W / 420 V

= 714A [allow710][no u.e.] (1) 1

Problem:

Overheating in wires OR circuit/motor becomes hot

OR Need thick/large/heavy cables

OR other sensible comment (1) 1

Why e.m.f. of battery must be more than 420 V:

Mention of internal resistance (1)

Detail e.g. loss of p.d. inside battery when current delivered/ lost volts (1)

OR equations used correctly 2 marks 2

Overall efficiency of motor:

K.E. gained = ½ mυ 2

= ½ × 1160 ×107 2 J

= 6.64 M J (1)

Energy input = P × t

= 300 000 × 100 J

= 30 M J (1)

⇒ efficiency = out/in × 100

= 6.64/30 × 100 =22% (1) 3

Reasons for energy losses: (1)

Work (done) against air resistance (1)

Work (done) against friction (1)

Heating in wires of circuit (1)

Heating, in battery (1)

Heating in motor coils (1) Max 2

OR other sensible comments e.g. sound

[Friction or heat loss scores zero unless detailed e.g. heat due to friction/air

resistance √ but heat to surroundings ×]

[9]

743. Speed:

υ 2 = u2 + 2 as (OR ½ mυ 2 = mgh)

⇒υ 2 = 2 × 9.81 × 55 (1)

⇒υ = 33 (32.8) m s –1 (1) 2

Assumption:

No air resistance

OR no friction

OR energy conserved/ no energy loss

OR constant acceleration [not “gravity constant”] (1) 1

Value for g:

s = ut + ½ at2(1)

⇒ g = 2

2

t

s OR 21.2

302 × (1)

=13.6 s –2 2

Explanation:

Imprecise timing– [not “human error” on its own] (1)

t too small ⇒ g too large (1)

OR

Clock started late/ stopped early OR shuttle already moving (1)[ not “air resistance”]t too small ⇒ g too large (1) 2

Improvements:

Any two from

Repeat timing, then average (1)

Time using video (1)

Electronic method e.g. light gates, sensors in context (1)

OR other sensible suggestion e.g. time to bottom (1) Max 2

Maximum resultant force:

F = ma (1)

= 5000 × 4.5 × 9.81 N

= 2.2 × 105 N (1) 2[11]

744. Explanation of formula:

(For fundamental) λ = 2 l (1)

⇒ υ = λ × f [stated or used]H3 2 × B3 × D3 (1) 2

How value is calculated:

Volume = π r2 × l

3

23–

m12

105.2π ×

××= (1)(1)

OR 23–

2

10mmindiameterπ

×

OR P1 * (0.001 * C5/2) Λ 2

OR similar valid route

[ for 2

)diam( 2

× π , for factor 10–3] 2

Value in G4:

Mass/metre = ρ × volume/metre

OR

= 1150 × 0.000 000 79 kg (1)

= 0.00091 kg m –1 [no u.e.] (1) 2

Formula in cell I3:

υ = µ/T

⇒ T = µ υ 2 (1)

⇒ I3 = H3 * H3 * G3

OR H3 Λ 2 * G3 (1) 2

Comment:

No + reason (e.g. 133 >> 47) (1)

OR

Yes + reason (e.g. 47, 64, 133 all same order of magnitude) (1)

More detail, e.g, f changes by factor 32, OR l by factor of 15. T only by factor 2.5⇒ similar Ts. (1)(1)

OR other sensible points. 3[11]

745. Speed of raindrop:

υ = u + at = 0 + 9.81 m s–2 × 0.2 s = 1.96 m s–1 ≈ 2 m s–1 (1) 1

Explanation:

Air resistance (1)Drag force increases with (speed) (1)

So resulting accelerating force/acceleration drops (1)

Terminal velocity when weight = resistance (+ upthrust) (1) Max 2

Mass of raindrop:

Mass = volume × density

substitute 1.0 × 10 3 kg –3 × 4π × (0.25 × 10–3 m)3 /3 (1)

6.5 × 10–8 (kg) (1) 2

Terminal velocity:

Viscous drag = weight (1)

VT = (6.54 × 10–8 kg × 9.81 m s–2) / (6π × 1.8 × 10–5 kg m–1 s–1 × 2.5 ×10 –4 m) (1)[Allow e.c.f. for m and r]

So terminal velocity = 7.56 m s–1 (1) 3

Graph:

Line drawn which begins straight from (0,0) (1)Then curves correctly (1)to horizontal (1)Scale on velocity axis (1)[More than 2 sensible values and unit] Max 3

Explanation:

VT increases (because of greater mass) (1) 1[12]

746. Why resistance changes:

Wire lengthens OR cross–sectional area OR diameter reduces (1)

Use of R = ρ l/A to explain [R and l, R ∝ 1/A (1) 2

Advantage:

A long length of wire OR small area OR multiple stretching (1) 1

Diagram:

Circuit with ammeter in series (1)voltmeter in parallel (with strain gauge) (1)

OR multimeter across strain gauge (1)(1)[Multimeter with power supply – 1 only] Max 2

Resistance:

R = ρ l /A

= 4.9 × 10–7 Ω m × 0.2 m/π × (2 × 10 –4 m /2)2 (1)[i.e. area = (1)] (1)

= 3.1Ω (1) 3[8]

747. Diagrams:

Diagram showing 2 waves π radians out of phase (1)

Adding to give (almost) zero amplitude (1)

Reference to destructive interference (1) Max 2

Wavelength of red light:

For example, red wavelength is 1.5 times blue wavelength (1)[OR red wavelength is 50% more than blue wavelength]

= 1.5 × 460 nm = 690 nm (1) 2

Dark bands :

Spacing = 4.0 mm (1) 1

Explanation of pattern:

Sunlight has a range of frequencies/colours (1)

Gaps between part of feather (act as slits) (1)

Different colours [OR gap width] in the sunlight diffracted by different amounts (1)

Red light bends more [OR blue less] hence coloured edges (1)[No colours linked to refraction] Max 3

[8]

[8]

748. Threshold wave:

Electron requires certain amount of energy to escape from surface (1)

This energy comes from one photon (1)

Use of E = hf (1)

(So photon needs) minimum frequency (1)

Hence maximum wavelength

OR use of E = hc/λ (1) Max 4

Work function:

f = c/λ = 3.0 × 108 / 700 × 10–9 m (1)

= 4.28 × 10 14 Hz (1)

E = hf = 6.63 ×10–34 J s × 4.28 × 10 14 Hz = 2.84 × 10–19 (J) [Allow e.c.f.] (1) 3

Circuit :

Circuit showing resistors only in series (1)

Potentials labelled (1)[Use of potential divider – allowed]Resistor values 1: 1: 1 OR 1:2 (1) Max 2

Suggestion:

Cosmic rays travel more slowly than light (1) 1[10]

749. How stiffness calculated:

Stiffness = force/deformation OR k = F/x (1) 1

Difference :

Stiffness is for one particular sample (1)

Young modulus is for any sample of a materialusing E = Fx/A∆ x (1) 2

Stress :

F/A = 30 kg × 9.8 m s–1 (1)/π (2 × 10 –2 m)2 [A = 1.26 × 10–3 m 2] (1)= 2.3 × 105 N m–2 (1) 3

Suitability of material:

The polymer is not so stiff/will undergo too much compression and (1)will unbalance the body (1)

Young modulus smaller [OR different] (1) Max 1

Hence not suitable (1) 1[8]

750. Light from sky:

Light is polarised (1) 1

Change in intensity:

Filter allows through polarised light in one direction only (1)

When polarised light from the sky is aligned with filter, light is let through (1)

When polarised light is at right angles with polarising filter, less light passes (1)

Turn filter so that polarised light from blue sky isnot allowed through, so sky is darker (1) Max 2

Clouds:

Light from clouds must be unpolarised (1) 1

Radio waves:

Radio waves can be polarised OR transverse (1) 1

Why radio waves should behave in same way as light:

Both are electromagnetic waves/transverse (1)[Transverse only, credited for 1 answer] 1

[6]

751. Momentum: mass × velocity [accept defined symbols] 1

Physical quantity:

(Net) force (1)

on lorry (1) 2

[“Rate of change of momentum” scores one only]

Magnitude:

s40

61Ns = 1.5 (3) × 104 N

Gradient measurements (1)

Correct calculation (1) 2

[Accept 1.4 – 1.7 to 2 s.f.]

Explanation of shape:

Force decreases as speed increases (1)

[Allow “rate of change of momentum”]

Any one of:

• Air resistance increases

• Transmission friction increases

• Engine force reduces 2[7]

752. Velocity of protons:

p=Bqr => υ = m

Bqr(1)

υ = 27–

19–

1067.1

5.11060.12.0

××××

= 2.9 × 107 ms–1 (1)

[must have 2.9]

≈10

103 8× (ms–1) (1) 3

Time for last semi-circle of orbit:

t = 71087.2

5.1

××= π

υd

(1)

1.6(4) × 10–7 s (1) 2

Frequency of accelerating p.d.

f =t

1 = 3.0 MHz [allow ecf] (1) 1

[6]

753. Exponential shape (1)

Value at RC > 1.5 V [only if shape correct] (1)

Levels off at 3 V (1) 3

Why movement of diaphragm causes p.d:

No movement, no change in C, no signal (1)

OR moving diaphragm changes C

As C changes so V changes (1)

Vc + IR is constant (1)

Hence IR changes – signal (1) 4

OR for last 3 marks

As C changes Q changes

Q flows through R

hence V = IR for resistor as signal 7

754. Wavelength of photon:

2E (1)

= 135 × 106 × 1.6 × 10–19 (1)

⇒ E = 1.08 × 10–11 J (1)

E = hf =λch

(1)

⇒ λ = E

hc

λ = 1.84 × 10–14 m (1) 5[5]

755. Credit to be given for all good, relevant Physics

Examples of mark scoring points [each relevant formula is also worth 1 mark]:

Between plates field is uniform

Acceleration is constant

Energy gained = 2000e

All ions have same F or same energy

From hole to detector is zero field/force

Ion travels at constant speed

g negligible

time proportional to 1 /velocity

time proportional to 1 /mass

in a vacuum there are no collisions or friction forces[Max 7]

756. Frequency:

Natural frequency/fundamental frequency (1) 1

[Not resonant]

Explanation:

Resonance occurs when driving frequency = natural frequency (1)

causing maximum energy transfer(1)

increased/maximum amplitude (1) Max 2

Graph:

Undamped - marked A

Acceptable shape - narrow peak (1)

Resonant frequency marked under graph max. (1)

Damped - marked B

B - entire graph below A (1)

[Accept touching graphs] (1) 4

Peak covers greater frequency range than A

Prevention of resonance:

Damps oscillations (1)

Fewer forced oscillations (1)

Explanation of damping [e.g. in terms of energy transfers] (1) Max 2[9]

757. Displacement-time graph:

Cosine curve (1)

Constant period and amplitude (1) 2

D i s p l a c e m e n t

T i m e0

Velocity-time and acceleration-time graphs:

Velocity: sine curve; 90° out of phase with displacement-time (1)

V e l o c i t y

T i m e0

Acceleration: cosine curve; 180° out of phase with displacement-time (1) 2

A c c e l e r a t i o n

T i m e0

Two requirements:

Force towards centre (1)

a (or F) ∝ x (1)

opposite direction [acceleration and displacement acceptable] (1) Max 2

Displacement-time and acceleration-time:

Starts positive, curved, always > 0 (1)

Period same as velocity (1) 2

D i s p l a c e m e n t

T i m e0

Acceleration constant not equal to 0 (1)

Sharp peak when ball in contact with floor (1) 2

A c c e l e r a t i o n

T i m e0

Explanation:

Not simple harmonic motion (1)

Reason e.g. acceleration constant when ball in free fall / period not constant / acceleration not ∝ displacement. (1) 2

[12]

758. Sketch graph:

Acceptably shaped exponential decay curve drawn (1)

Activity halving every 8 days (1) 2

A c t i v i t y / M B q

T i m e / d a y s

0 5 1 0 1 5 2 0 2 5

8 0

6 0

4 0

2 0

0

Description and differences

I contains 53 protons (1)

and 78 neutrons (1)

eg. β are fast electrons/γ electromagnetic waves (1)

β charged; γ uncharged (1) 4

Explanation:

Cat emits β , γ which are hazardous to employees (1) 1

Calculation:

λ = ln2 / half life

= ln2 / (8 × 24 × 60 × 60) (1)

= 1.0 × 10–6 s–1 (1)

Use of activity = (initial activity) e–λ t

So t = ln (80/50) ÷ 1.0 × 10–6s

=4.7 × 105s (1) 4

(= 5.4 days)

Assumption:

Cat does not excrete any 131I (1) 1

or daughter product not radioactive[12]

759. Show that:

F = GMm/r2 (1)

= 6.9 × 1024 (N) (1) 2

Calculation:

a= F/M= 3.1 × 10–6 m s –2 (1) 1

Explanation:

Planet exerts gravitational force on star (1)

Planet revolves around star, so direction of force changes with time (1)

Diagram showing force (or effect of force on star due to planet) (1) 3

Speed of star:

Using υ = 2π r/ T and a = υ / r (1)

r = υ T/2π so a = 2π υ / T

so υ = aT / 2π (1)

= 3.1 × 10–6 × 9.2 × 107 ÷ 2π [allow ecf for a]

= 45.4 m s–1 3

Calculation:

∆ λ = λ υ /c = 656 ×10–9 × 45 / 3.0 × 108(1)

= 9.8 × 10 –14m

[Accept 2 × ∆ λ for maximum marks] (1) 2[11]

760. Discussion:

Credit to be given for all good, relevant Physics

Examples of mark scoring points:

Universe may continue to expand (1)

or may collapse back on itself (1)

Fate depends on mass of Universe (1)

Since mass determines force/deceleration on moving stars (1)

So far, not enough mass has been found to stop expansion (1)

There may be matter present which is currently undetectable (1)

Nature of this dark matter

e.g. neutrinos, very hard to detect (1)

black holes, no light escapes (1)

WIMPS explained (1)

neither expand nor collapse (1)

explained in terms of energy (1)[Max 6]

761. (a) Either extra column of logs of Ton and V

OR I/V2 (1)

Correct values of above (1)

Axes : log Ton / log V or Ton/ I/V2 (1)

Scales [Are points spread across more than ½ page?] (1)

Points placed correctly (1)

Axes labelled and in second case with units (1)

Straight line (1) 7

(b) Graph in a straight line through origin (1)

Ton is (directly) proportional to 2

1

V/ Ton ∝ I/V2 (1)

OR

log T v. log V straight line gradient –2 (1)

Ton ∝ I/V – 2 (1) 2

(c) ν intercept OR large triangle on graph (1)

inv. log = OR gradient = 20.1 (×10–3) (1)

k = 7.7 × 109 (V2 N–1) [–1 if milli forgotten; if gradient = 50, –1] (2)

[–1 for outside 7.6 → 7.8 OR losing 109] 4

[Evidence of substitution (1) to correct answer (1), max 2]

(d) Same liquid crystal / substance and voltage/V (1)

Same temperature / viscosity/T/ η (1)

Log Ton and log d OR Ton and d 2 (1) 3[16]

762. Demonstration of how statement leads to equation:

Momentum = mass × velocity (1)

Therefore force ∝ mass × rate of change of velocity (1)

Therefore force ∝ mass × acceleration (1)

Definition of newton or choice of units makes the proportionality constant equal 1 (1) 4

[Standard symbols, undefined, OK; “=“ throughout only loses mark 4. No marks for just manipulating units. If no ∆ υ (e.g. mυ /t),can only get marks 1 and 4]

Effect on time:

Time increases (1) 1

Explanation:

Acceleration smaller/momentum decreases more slowly/F = t

p

∆∆

(1)

[Need not say ∆ p = constant]

So force is smaller (1) 2

[Independent mark, but must be consistent with previous argument][If no previous argument, this becomes fully independent mark] 7]

763. Calculation of work done:

Work = area under graph/average force × distance (1)= 2

1 × 0.040 m × 22 N (1)= 0.44 J (1) 3

[Allow any correct unit, e.g. N m. Penalise unit once only]

[Fd → + 0. 88 J gets 1/3]

Calculation of energy:

GPE = 0.024 kg × 9.81 (or 10) m s–2 × 0.60 m (1)

= 0.14 J (1) 2

Comparison:

Some energy transferred to some other form (1)

Reason [a mechanism or an alternative destination for the energy], e.g. (1)

FrictionAir resistanceHeat transfer to named place [air, frog, surroundings etc]Internal energyVibrational energy of springSound

OR quantitative comparison (0.3 J converted)

[No e.c.f. if gpe > work] 2[7]

764. Definition of symbols:

n = number of electrons/carriers per unit volume (per m3)ORelectron (or carrier) density (1)

υ = average (OR drift) velocity (OR speed) (1) 2

Ratio Value Explanation

x

y

n

n 1 Same material (1) (1)

x

y

l

l 1 Connected in series/Kirchoff’s 1st law/conservation of charge/current is the same (1) (1)

x

y

v

v 2 A is halved so ν double[Accept qualitative, e.g. A ↓ so v ↑, or goodanalogy] (1) (1)

6

[Accept e.g. ny = nx.....]

[No e.c.f ]

[NB Mark value first, without looking at explanation. If value correct, mark explanation. If value wrong, don’t mark explanation except: if υ y/υ x = ½ or 1:2, see if explanation is correct physics, and if so give (1). No e.c.f.]

[8]

765. Explanation:

Changing direction/with no force goes straight on (along tangent) (1)

Acceleration/velocity change/momentum change (1) 2

Identification of bodies:

A: Earth [Not Earth’s gravitational field] (1)

B: scales [Not Earth/ground] (1) 2

Calculation of angular speed:

Angular speed = correct angle ÷ correct time [any correct units] (1)

= 4.4 × 10–3rad min–1 / 0.26 rad h–1/ 2π rad day–1 etc (1) 2

Calculation of resultant force:

Force = mrω 2 (1)

= 55 kg × 6400 ×103 m × (7.3 × 10–5 rad s–1)2 (1)

= 1.9 N (1) 3

[No e.c.f here unless ω in rad s–1]

Calculation of value of force B:

Force B = 539N – 1.9N (1)

= 537 N (1) 2

[e.c.f. except where R.F = 0]

Force:

Scales read 537 N (same as B) [allow e.c.f.]

Newton’s 3rd law/force student exerts on scales (1) 1[12]

766. Estimation of charge delivered:

Charge = area under graph (1)

= a number of squares × correct calculation for charge of one square i.e. correct attempt at area e.g. single triangle (1)

= (3.5 to 4.8) × 10–3 C (A s, µ A s) (1)

[Limit = triangle from 41 µ A → 300 s]

OR

Charge = average current × time (1)

= (something between 10 and 20 µ A) × 300 s (1)

= (3.5 to 4.8) × 10–3 C (1) 3

[But Q = It → 0/3, e.g. 41 µ A × 300 s]

Estimation of capacitance

C = calculated charge/9.0V time constant ≈ 100 s (1)

= 390 to 533 µ F C = 100 s/220 k.Ω = 450 µ F (1) 2[5]

767. Demonstration that resistance is 0.085 Ω :

R = ρ l/A (1)

= 1.7 ×10–8 Ω m ×20 m / (4.0 ×10–6 m2) (1) 2

Calculation of voltage drop:

V = 37 A × 0.085 Ω (1)

= 3.1 V × 2 = 6.3 V [Not if Vshower then found] (1) 2

[Only one conductor, leading to 3.1 V, gets 1st mark][Nothing if wires in parallel]

Explanation:

Lower resistance/R = 0.057 Ω /less voltage drop/new V=3

2old V (1)

Power dissipated in cable/energy wasted/wire not so hotOR more p.d/current/power to showerOR system more efficient (1) 2

[6]

768. Labels of elements:

B i n d i n g e n e r g yp e r n u c l e o n ×

×

×

D

F e

u

0 1 0 0 2 0 0N u c l e o n n u m b e r

D close to O: AND U > 200 (1)

Fe at peak (1) 2

Meaning of binding energy:

Energy needed to split/separate a nucleus (1)

into protons and neutrons/nucleons (1)

OR

Energy released when nucleus formed (1)

from protons and neutrons/nucleons (1)

OR

Energy released due to mass change/defects (1)

Sum of masses of protons and neutrons > mass of nucleus (1)

[In each of the cases above, the second mark is consequent upon the first]

Explanation:

Uranium (1)

Binding energy per nucleon of products is higherORProducts/atoms/element/nuclei nearer peak (1)

Therefore more stable (1) 5[7]

769. Calculation of kinetic energy:

f = λ

1–8103 sm×(E = hf = 1.63 ×10–17 J) (1)

φ converted to J: 6.20 × 1.6 × 10–19 OR Photon energy converted toeV: 1.63 ÷ 1.6 × 10

(Subtract to obtain kinetic energy)

Kinetic energy = (1.5 – 1.56) × 10–17 J[OR 95.7/97.4 eV]

[Beware 1.6398 0/3; > 101 eV 0/3]

Demonstration of speed of electrons:

1.53 × 10–17 J = ½ × 9.11 × 10–31 kg × υ 2 (1)[e.c.f their kinetic energy in joules]

υ = 5.8 × 106 m s–1 (1)[If υ is not between 5 and 7 must comment to get mark]

[5]

770. Why example is resonance:

These are forced vibrations, i.e. 2 systems [driver and driven) (1)

The vibrations have max amplitude at one particular frequency or decrease both sides (1)

When forcing frequency = natural frequency of steering wheel car/ “system” (1) 3

Calculation of maximum acceleration:

Acceleration = (2π × 2.4)2 s–2 × 6 mm [accept any attempted conversion

to m, e.g. 6 × 10–2] (1)

= 1400 (1360) mm s–2 [no e.c.f.] (1)

(1.4 m s–2) 2[5]

771. Calculation of air pressure at 100 °C:

Pressure = 1.00 × 105 Pa × 373 K / 273 K (1)[If T in °C → 0/2]

= 1.37 × 105 Pa / N m–2 (1) 2

Graphs to show how air pressure varies with temperature (line A) and how different pressure then varies over same temperature range (line B):

4

3

2

1

00 2 0 4 0 6 0 8 0 1 0 0

P r e s s u r e / 1 0 P a5

2 . 7 3

1 . 3 7

A

B

( 2 . 6 – 2 . 8 )

( 1 . 3 – 1 . 4 )

T e m p e r a t u r e / ° CL i n e A :

Line A:

Any rising straight line (1)

through correct points [e.c.f end point] (1) 2

Line B:

Rising straight line above line A for all its length (1)through correct points [e.c.f both points] (1) 2

[6]

772. Estimate of time constant, using graph:

V / 3 O R V / e O R t

3 V 3 . 3 V 1 6 s

2 6 s 2 3 . 2 4 s 2 3 s

M e t h o d

V a l u e 2 3 2 6 s

1 / 2

( 1 )

( 1 ) 2

Calculation of resistance and hence capacitance:

R = i

V OR 3–1019.0

9

× (1)

Resistance = 47 kΩ [ue] (1)

Substitute in t = RC [e.c.f their t, their R] OR answer 300 µ F

Capacitance = 500 µ F (1) 3

Addition to graph of line showing how potential difference varies with time:

A curve of shape shown below, i.e. getting less steep (1)

Any convex curve ending at ≈ 7.5 V, crossing at ≈ 15 s (1) 2[7]

773. Calculation of e.m.f. induced across falling rod:

Correct use of E = Blυ (1)

υ = 25 m s–1 (1)

e.m.f. = 7.3 – 7.4 × 10–4 V (1) 3

Explanation of why magnitude of vertical component is not required:

Earth’s field is parallel to direction of fall/body falls vertically (1)

Therefore no flux cut (1) 2[5]

774. No mark scheme available



777. Addition of forces to produce a free-body diagram for trolley:

[Unlabelled arrows = 0; ignore point of application of force] (1)R/N/Reaction/push of slope/ normal contact force[NOT normal]should be approximately perpendicular to slope]

F/friction/drag/air resistance (1) 3

W/weight/mg/pull of Earth/gravitational force[Not gravity] [Should be vertical by eye]

Evidence that trolley is moving with constant velocity: 1

Trolley travelled equal distances (in same time)

Acceleration of trolley down slope:

0 / no acceleration 1

What value of acceleration indicates about forces acting on trolley:

Forces balance OR Σ F = 0 OR in equilibrium OR zero resultant 1

[Forces are equal 0/1][6]

778. Calculation of how long wheel takes to complete one revolution:

Time = 2π × 60 m/0.20 m s–1 (1)

= 1900 s/1884 s/31.4 min (1) 2

Change in passenger’s velocity:

Direction changes OR up (N) → down (S) OR + → – (1)

OR 180° (1) 2

0.40 m s–1

[0.40 m s–1 without direction = 2/2]

Calculation of mass:

(G)pe = mgh

m = 80 × 103 J/9.81 m s–2 × 120 m) (1)[This mark is for rearranging the formula; accept 10 instead (1)of 9.81 and 60 instead of 120 but do not e.c.f. to next mark]m = 68 kg (1) 3

Sketch graph:

G P E / k J G P E / k J

8 0 8 0

9 5 0 9 5 0t / s t / s

O R

Labelled axes and line showing PE increasing with timeSinusoidal shape (1)(950 s, 80 kJ) (1)[Accept half the time they calculated at start of question (1)instead of 950 s as e.c.f.][PE v h 0/3] 3

Whether it is necessary for motor to supply the gpe:

No, because passenger on other side is losing gpe (1)If wheel equally loaded OR balanced with people (1)

ORYes, because no other passengers (1)so unequally loaded (1) 2

[12]

779. Definition of linear momentum:

Mass × velocity [Words or defined symbols; NOT ft] (1) 1

Newton’s second law:

Line 3 only (1) 1

Newton’s third law:

Line 2 OR 1& 2 (1) 1

Assumption:

No (net) external forces/no friction/drag (1)

In line 3 (he assumes the force exerted by the other trolley is

the resultant force) [Only if 1st mark earned] (1) 2

Description of how it could be checked experimentally that momentum is conserved in a collision between two vehicles:

Suitable collision described and specific equipment tomeasure velocities [e.g. light gates] (1)

Measure velocities before and after collision (1)

How velocities calculated [e.g. how light gates used] (1)

Measure masses / use known masses/equal masses (1)

Calculate initial and final moment a and compare ORfor equal trolleys in inelastic collision, then ν 1 = ½ ν 2 (1) Max 4

[9]

780. How to determine background radiation level in laboratory:

Source not present/source well away from GM tube [> 1 m] (1)

Determine - count over a specific period of time> 1 min OR repeats (1) 2

How student could confirm that sample was a pure beta emitter:

To demonstrate no γ :

A1 between tube and source: reading → 0 or background (1)

No γ / γ not stopped by Al (1)

To demonstrate no α :

GM moved from very close (or ≈ 1 cm) to sourceto ≈ 10 cm: count rate does not drop (or no sudden drop) (1)

No α / α stopped by a few cm air (1)

Clarity: Only available if at least 2 of above 4 marksawarded. Use of bullet points acceptable. (1) 5

[7]

781. Demonstration that speed is about 7 m s–1

υ = d/t = 130 m / 18 s(= 7.2 m s–1) (1) 1

Calculation of average deceleration:

υ 2 = u2+ 2as (1) 1

→ 0 = 7.22 [OR 72] + 2 × a × 110 → a

= 0.24 [OR 0.22] m s–2 (1) 1

[Ignore signs]

Calculation of average decelerating force if combined mass is 99 kg:

F = ma

= 99 × 0.24 (0.22) N (1) 1

= 24 (22) N (1) 1

Discussion:

Need balanced forces OR forward force = resistive force(s) (1) 1

F needed > 24 (22) N (1) 1

Max 2

since that value was average force (1)

[Any 2 points from three]

Energy transferred: 1

3 J (1)[7]

782. Explanation: .

I= E/r + R (1) 1

Appropriate formula for cell E9:

C9 * D9 OR RI OR 1 Ω × 4 A (1) 1

Appropriate formula for cell F 11

D11 *E11 OR VI OR 3A × 6V OR C11 * D11 *D11OR RI2 OR 2 Ω × (3 A)2 (1) 1

Short circuit current:

6 A (1) 1

Explanation:

r and R in series → potential division (1) 1

as R ↑, r constant → R has greater share of 12 V (1) 1

OR other valid argument

Sketch graph of power against resistance:

P / W

0 2 1 0 R / Ω

1 8

18 (1)2 (1)Shape including asymptote (1) 3

Comment:

Maximum when R = r (1)

in accordance with maximum power theorem (1)

OR P → 0 as R → ∞ (1) Max 2[11]

783. Explanation of words:

Coherent

Same frequency and constant phase relationship (1) 1

Standing wave


Superposition/interference

Two (or more) wavetrains passing through each other

Having equal A, f, λ

+ system of nodes and antinodes (1) (1) 2

Position of one antinode marked on diagram

Correctly marked A (in centre of rings – hot zone) (1) 1

Wavelength demonstration:

λ = c/f = 3 × 108 /2.45 × 109 m

= 12.2 cm (1) 1

Path difference:

(22.1 + 14) – (20 + 10) cm

= 6.1 cm (1) 1

Explanation:

6.1 cm = ½ × λ (1) 1

Waves at X in antiphase/ destructive interference (1) 1

→ node (1) 1

Explanation of how two separate microwave frequencies overcomesuneven heating problem:

Different wavelengths (1) 1

So a path difference which gives destructive interference atone wavelength may not do so at another (1) 1

[11]

784. Demonstration that initial vertical component is about 7 m s–1:

υ = υ sin θ = 12.9 × sin 34.5° m s–1 (1)

= 7.31 m s–1 (1) 1

Calculation of time:

t υ /g = 7.31/9.81 s (1)

= 0.745 s (1) 2

Comment:

More than twice (1)

Because falls further than rises (1)

OR

Same time to fall to same level (1) 2

OR

Similar reasoning

Calculation of horizontal distance:

d =υ × t = 12.9 m s–1 × cos 34.5° × 1.71 s (1)

= 18.2 m (1) 2

Demonstration:

K.E. = ½ mυ 2 = ½ × 5 × (12.9)2J

(= 416 J) (1) 1

Two reasons why figure is low:

Gravitational p.e. gained by shot (1)

Kinetic energy and/or gravitational p.e. gained by athlete's body (1) (1)

[At least ONE of the above needed to gain two marks]

Work against air resistance during acceleration (1)

Other sensible point (1) Max 2

Suggestion for measuring υ .

Use video + frame analysis OR computer analysis package (1)

Further detail,e.g. υ = distance ÷ time interval OR gradient of svt graph (1) 2

[12]

785. Explanation of line spectra:

Specific frequencies or wavelengths (1)

Detail, e.g. absorption/emission (1)

OR within narrow band of wavelengths 2

Explanation how line spectra provide evidence for existence or energy levels in atoms:

Photons (1)

Associated with particular energies (1)

Electron transitions (1)

Discrete levels (to provide line spectra) (1) 3[5]

786. Why warm surface water floats:

Cold water is denser than warm water (1) 1

Explanation of why ultrasound waves reflect thermocline:

This is surface separating layers of different density (1) 1

Explanation of why submarine is difficult to detect:

Ultrasound from ship partially reflects upwards fromthermocline so little is transmitted (1)

Any reflected sonar from submarine partially reflectsdownwards from thermocline (1) 2

Explanation of why sonar cannot be used from a satellite:

Lack of medium to transmit sound waves from satellite 1

Calculation of time between emission and detection of radar pulse:

2s /c (1)

= 2 × 6.0 × 107 m ÷ 3.0 × 108 ms–1 = 0.4 s (1) 2

Calculation of minimum change in height of ocean:

Minimum observable distance

= ct = 3.0 × 108 m s–1 × 1.0 × 10–9 s = 0.30 m (1)

so change in ocean height = 0.15 m (1) 2

Possible problem:

Sensible answer eg (1)

atmospheric pressure could change ocean height

bulge not large enough compared with waves

tidal effects

whales 1[10]

787. Use of definitions:

Stress × strain = length

extension

area

force × (1)

= beltofvolume

beltstretchtoenergy (1) (1) 3

Use of graph to show energy stored per unit volume of seat belt material:

Energy stored = area under graph (1)

≈ 15 ½ (± 1) large squares (2 cm × 2 cm) (1)

≈ 15.5 × 8 × 106 J M–3 = 1.2 × 108 J m–3 (1) 3

(i) Kinetic energy = ½mυ 2 = ½ × 60 kg × (20 ms–1)2 =12 000 J 1

(ii) Volume = energy ÷ energy per unit volume (1)

= 12 000 J ÷ 1.2 × 108 J m–3 = 1.0 × 10–4 m3 (1) 2

(iii) Area = volume ÷ 2 m = 1.0 × 10–4 m3 ÷ 2 m

= 5 × 10–5 m2 (1)

so dimensions could be 50 mm × 1 mm (1) 2

[Width limits: 24 – 100 mm, thickness limits: 0.4 – 2 mm][11]

788. Meaning of whorl:

An eddy/circular flow/whirlpool OEP (1) 1

Diagram and description of flow patterns:

Laminar

At least 3 reasonably parallel and straight lines (1)

No abrupt change in direction/no whorls/eddies (1) 2

Turbulent flow

No order shown in the flow/small broken circular shapes or similar (1)

Mixing between layers of liquid/whorls/whirlpools/eddies occur along the flow (1) 2

Explanation of statement in terms of energy transfers:

Kinetic energy, of motion of eddies becomes kinetic energy of

molecules in liquid; overall kinetic energy reduces and flow

slows/ordered kinetic energy → disordered kinetic energy (1) (1) 2[7]

789. Explanation:

Light hits glass–juice boundary at less than the critical angle (1)

And is refracted into the juice (1) 2

Marking angles on diagram:

the critical angle C – between ray and normal on prism/liquid face (1)

an incident angle i – between incident ray and normal atair/ glass or glass air interface (1)

a refracted angle r – between refracted ray and normalat air/glass or glass air interface (1) 3

Explanation of term critical angle:

The angle in the more (1)

dense medium where the refracted angle in the less dense medium is 90 (1) 2

Plot of results on grid:

[NB Axes are labelled on the grid]

Scales: y–axis (1)

x–axis (1)

Points correctly plotted (1) 4

Best fit line (curve expected) (1)

Refractive index found from graph:

Value = 1.400 ± 0.002 (1) 1[12]

790. Completion of nuclear equation:

One mark for top line all correct (1)

One mark for bottom line all correct

73 Li + 1

1 p → 74 Be + 1

0 n 2

Calculation of energy transfer

P = V × I = 2.8 × 106 V × 2.0 × 10–3 A= 5.6 × 103 W OR 5.6 kW

One mark for value (1)

One mark for power of ten and unit (1) 2

Demonstration that energy is absorbed at rate of 17 GW per cubic metre:

23–6–

3

m101.2m10280

W105.6

Volume

Power

××××= = 1.66 × 1010 W m–3

= 17 GW m–3

Substitution (1)Calculation (1) 2

Suggested problem:

Very hot/target overheats/vaporises/difficult to cool OR other good

relevant physics (1) 1[7]

791. Forces acting on molecule, shown on diagram A:

Forces not collinear and sense correct (1) 1

Explanation of why molecules align with field:

Forces not in same line (1)

Hence turning effect [OR torque] 2

Field lines shown on diagram B:

At least three lines drawn equidistant(1)

Direction correct (1) 2

Calculations of electric field strength:

m100.1

V5.15–×

==d

VE (1)

= 1.5 × 105 V m–1 (1) 2[7]

792. Demonstration that water must be thrown backwards at about 13 m s–1:

Force = t

waterofmomentum

∆∆ )(

(1)

8 × 105 N = 6 × 104 × ∆ V (1)

∆ V= 4

5

106

108

××

= 13 m s–1 (1) 3

Calculation of power expended:

P = F × υ = 8.0 × 105 N × 20 m s–1 (1)

1.6 × 107 W OR 15 MW (1) 2

Calculation of rate at which water gains kinetic energy:

½ × m/t × υ 2 = ½ 6 × 104 kg/s × (13 M s–1)2 (1)

= 5.07 × 106 W OR 5.1 MW (1)

[Allow 5.3 MW if 13.33 m s–l used] 2

Overall efficiency:

Power in = 1.6 s

l × 3.4 × 107

l

J = 5.44 × 107 W (1)

[Intermediate value not explicitly needed]

Power out = 16.0 × 106 + 5.4 × 106 = 21.4 × 106 (1)

Efficiency = 5.54

4.21 = 0.39 (39%) (1) 3

[10]

793. Capacitors and storage of energy:

E = ½ cV2 (1)

Ew = ½ × 68 × 10–3 F × (16 V)2 = 8.7 J (1)

E2 = ½ × 1 × 10–3 F × (400 V)2 = 80 J (1) 3

[Allow calculations in proportion using 2

V

V

F

C

µ

Range of actual value:

900 µ F < C < 1500 µ F (1) (1) 2

(i) Calculation of charge:

Q=CV = 68 × 10–3 F × 16 V= 1.1 C (1) 1

(ii) Demonstration that maximum leakage is about 3000 µ A:

I = 0.003 × 10–6 A/µ F V × 68 000 µ F × 16 V (1)

= 3.26 × 10–3 A [3.3 mA] (1) 2

(iii) Estimate of time for capacitor to discharge with reasoning:

s334A1026.3

C09.13–

0

=×

=I

Q

[This is time constant] (1)

Numerical example such as: for less than 0.7% remaining t = 5τ =

1670 s OR well-reasoned estimate showing t >> 300 s (1) 2[10]

794. Explanation of how it can be deduced that magnetic field acts out of the plane:

Current flow in opposite direction to e– movement/same as e+ movement (1)

(Force acts into spiral) hence Fleming’s left-hand rule (gives field out of

paper) (1) 2

Explanation of which e– moves faster:

(the “atomic” electron) since path is straighter so r larger and

BQ

mr

υ= (1) 1

Calculation of momentum:

p = BQr = 5.4 × 10–3 T × 1.6 × 10–19 C × 0.048 m (1)

= 4.1 × 10–23 N s (1) 2

Explanation of why path of the positron is a spiral:

Positron continually losing speed/energy (by ionising) 1

Discussion of conservation of two properties:

Charge:

e+ and e– (1)

recoiling electron and stationary positive ion (1)

Energy:

e+ and e– creation (1)

since E = mc2 (1)

EK of recoiling electron (1)

EK of e+ and e– pair (1)

Ionisation energy (1)

Momentum:

Incoming photon momentum goes to recoil electron (mostly) (1)

After collision:

Momentum up = momentum down (1)

2 go up (one slightly) and only one goes down so down one is faster (1) Max 5[11]

795. Why large voltage is generated in secondary circuit:

Faraday’s Law in words including ‘flux linkage’

Current flow in primary (1)

causes magnetic flux in core (1)

Flux links secondary (1)

Opening switch S causes flux to reduce (1)

Changing flux in. secondary induces e.m.f (1)

Many turns on secondary means large flux linkage (1)

Hence rate of change of flux linkage is large

reduction time is short (1)

Hence induced e.m.f. is large (1) Max 6[6]

796. What happens when switch moved from A to B:

Ball released/drops

Clock starts 2

Time for ball to fall 1.00 m:

x = gt2/2 [sufficient use of equation of motion]

t2 = 2 × 1.00 m/9.81 m s–2 [correct substitution in all equations]

t = 0.45 s 3

Time for ball to reach ground, with reason:

0.45 s/same as before (e.c.f)

Horizontal and vertical motion independent/same (vertical) height

Same vertical distance/same (or zero) initial velocity

[NOT same force or acceleration] 2

Speed at which ball was fired:

2.00 m/0.45 s (e.c.f)

4.4 m s–1 2[9]

797. (a) Mark the method before marking the circuit

Suitable circuit A

V

Short circuit option

V

A

What is measured

Set of readings of V and I

V and I Two sets of V and I

V and I

What is then done

Plot V against I Record V for open circuit

Substitute inV = E – Ir

Record V for open circuit

Finding E and r

E = interceptr = – gradient

E = open circuit voltager from V = E–Ir

Solve simultaneous equations

E=open circuit voltage r from r=E–Ir

-

Suitable circuit

V RA

R

Potentiometer

T oP o t

What is measured

V for known R I for known R Two sets of I and R

l for known R

What is then done



Substitute in E = I(R + r)

l’ for open circuit

Finding E and r E = open circuit voltage r fromE/V = (R + r)/R

E = open circuitvoltager fromE = I(R + r)

Solve simultaneous equations

E from l’ (calibrated)

R

r)(R

l

l +=

′

Mark other procedures in a similar way 4

[Mark text, then tick for circuit if it does the job described.

If diagram alone, ask if it can do the job and give mark if yes]

(b) (i) p.d. across battery:

V = E − Ir

= 12.0 V − 3.0 A × 3.0 Ω (substitution)

= 3.0 V 2

(ii) Straight line from (0,12) to (3,3) (e.c.f.) 1

Current: 2.05 to 2.10 A 1[8]

[Allow correct intersection of their line (ignore shape), ± 0.05 A, of the characteristic with their graph, even if theirs is wrong. A line MUST be drawn for the last mark.]

798. Explanation of variation shown on the graph:

More electrons set free. Any one from: as temperature increases; thermal energy/vibration increases/resistance decreases/current increases 2

Resistance of thermistor:

V (across thermistor) = 1.20 VResistance ratio = voltage ratioR = 495 Ω

or

I = 0.80 V/330 Ω (substitution)= 0.002424 AV across thermistor = 1.20 VR = 1.20 V/0.002424 A= 495 Ω

or

I = 0.80 V/330 Ω= 0.002424 AR(total) = 2.0 V/0.002424 A

= 825 ΩR = 825 Ω − 330 Ω= 495 Ω 3

Explanation:

Thermistor resistance lowWhy: thermistor hotter/more current, power, charge carriersWhy v. small: thermistor takes smaller fraction of p.d. or ratio of p.d. 3

[8]

799. In this experiment alpha particles werescattered by thin films of metals such as gold.

The experiment led to the conclusion that the atom had a

positively charged nucleus of diameter approximately 10–15 m and containing

most of the mass of the atom[5]

800. Half-life of radionuclide:One value for half-life: 33 → 36 sRepeat and average/evidence of two values (u.e.) 2

Decay constant:ln2 ÷ their value for t1/2 calculated correctly

= (0.02) s–1 (u.e.) 1

Rate of decay:

Tangent drawn at N = 3.0 × 1020 Attempt to find gradient, ignore “−“ sign

= 5.5 → 6 × 1018

[or Use of N = Noe−λ t , calculate λ , or other graphical means]

[NB 6.25 × 1018 = 0/3 as use of coordinates] 3

Decay constant:

Substitute in dN/dt = − λ N

e.g. 6 × 1018 = (−) λ × 3 × 10 20 [their above]= (0.02) [their λ correctly calculated] 2

Methods: 1

Either value chosen with a valid reason

e.g. 1st because can take several and average

1st because difficult to draw tangent[9]

801.

4 0

2 0

00 5 1 0 1 5

e / m m

F / N A

B

1

2

It breaks/fractures at greater force/stressBrittle material is A/straight line/linearJust elastic/no plastic deformation 4

Wire B because area greaterConvincing argument comparing area 1 with area 2 e.g. could show vertical at 11 mm

[Last mark consequent upon previous mark] 2[6]

802. A body oscillates with simple harmonic motion when the resultant force F acting on it and its displacement x are related by the expression

F = − kx or F ∝ − x.

The acceleration of such a body is always directed towards the centre of the oscillation or in the opposite direction to displacement/x =

0 /equilibrium/similar

The acceleration of the body is a maximum when its displacement is

maximum and its velocity is maximum when its displacement is zero. 4

Force constant:= 14

N m–1 or kg s–2 2[6]

803. Circumstances under which two progressive waves produce a stationary wave:

Both transverse/longitudinal/same typeWaves have same frequency/wavelengthand travel/act in opposite directions/reflected back. Max 2 marks

Experiment using microwaves to produce stationary waves:

T r a n s m i t t e r M e t a l o rp l a t e b a c k w a r d s t r a n s m i t t e r

Adjust distance of transmitter/plateHow it could be shown that a stationary wave had been produced:Note readings on probe/detector/receiver form a series of maximum or minimum readings or zero 3

[5]

804.

M M M

RRC

C

One of compression C and one rarefaction R marked as above.

Wavelength of wave = 11 - 11.6 cm (u.e.)

One of maximum displacement M marked as above [M, 5th, 6th, 7th ].Amplitude of wave = 8 (± 1 mm) [consequent mark]

[4]

805. Explanation:Photons/quanta Photon releases / used electronEnergy/frequency of red < energy/frequency of ultra violetRed insufficient energy to release electrons so foil stays 4

Ultraviolet of greater intensity: foil/leaf collapses quicker/fasterRed light of greater intensity: no change/nothing 2

Observations if zinc plate and electroscope were positively charged:Foil rises or Foil stays same/nothing

as electrons released it becomes more Released electrons attracted back by

positive positive plate/more difficult to

release electrons 2[8]

806. Energy level diagram:

0 e V

– 1 . 5 1 e V

– 3 . 3 9 e V

– 1 3 . 6 V

− 13.6 → 0

− 1.51 → 0 AND − 3.39 → 0 ONLY 2

Why level labelled − 13.6 eV is called ground state:Correct reference to stability/lowest energy state/level of the electron/ atom/hydrogen 1

Transition which would result in emission of light of wavelength 660 nm:

Correct use of c =fλ or E = hc/λ or f = m10660

ms1039–

–18

××

Correct use of eV/J i.e. ÷ 1.6 × 10–19

∆ E = 1.88

Transition = 1.5 → 3.39

[May be a downward arrow on diagram] 4[7]

807. Maximum acceleration of mass:

a = (−) ω 2x with x = 6.0 mm used or a = (−)(2π f)2 x

ω = 2.3

2π or f =

2.3

2π

= 23 mm s–2 [u.e.] 3

Graph:

0 . 0 2

– 0 . 0 2

– 6 6

x

x

a / m s – 2

Straight lineNegative gradient4 quadrants: line through 0,0Line stops at 6, 0.023 [e.c.f. x, a] 4

Reason why mass may not oscillated with simple harmonic motion:F not proportional to x or a not proportional to xSpring past elastic limit: K not constant: spring may swing as well as bounce.Other possibilities, but not air resistance, energy losses 2

[9]

808. (i) Reference to (individual) nuclei/atoms/particlesEach has a chance of decay/cannot predict which/when will decay 2

(ii) Use of λ t ½ = ln 2

→ λ = ln 2 ÷ 600 s = 1.16/1.2 × 10–3 s–1

∴ A = (1.16 × 10–3 s–1) (2.5 × 105) [Ignore minus sign]

= 288/290 Bq/s–1 [c.a.o.] [Not Hz] [17 300 min–1]

(iii) )(// 136

01

01

137 eXCeN νβ ++→ [N/O/C/X] [e.c.f. β –]

[β + on left, max 1/2] 5[7]

809. Description of force C which forms a Newton’s third law pair with A

Man pulling Earth upwards

with a gravitational force 2

Similarities and differences

Similarities [any 3]:

Magnitudes or equalKind (or type) of force or gravitational forcesLine of action [but not same plane, or point, or parallel]Time interval or durationConstant [not true in general but true in this instance] Max 3

Differences:

On different bodies [must say “bodies” or equivalent]

Direction [again, it answers this particular question] or opposite 2

Two forces which show whether or not man is in equilibrium:

A and B 1[8]

810. Completion of a correct circuit diagram:

Ammeter in series with lamp and supply [Ignore voltmeter position]

Voltmeter across lamp and ammeter [and maybe with ammeter 2

A

V+

–

0 - 1 2 V

Measurements:

Record voltmeter reading

Record corresponding ammeter reading [“corresponding” may beimplied]

Repeat for range of supply voltage settings [or currents] 3Labelled sketch:

I2 A

1 2 V V

Label axes I and V [with or without units]Graph line with correct curvature [overlook any tendency of the currentvalue to saturate]

Show 12 V, 2 A correctly [Allow 12 and 2 if units are labelled on axes] 3

[The second mark is lost if axes are not labelled, unless 2 A and 12 V arepresent, with the units, to make sense of the axes.]

[8]

811. Calculation of total amount of energy released during flight:

1.71 × 105 litres × (38 MJ litre–1)

= 6.5 ×106 MJ 2

Calculation of input power to engines:

6.5 × 1012J ÷ (47 × 3600 s) [Allow e.c.f for energy released]

= 38 MW 2

Calculation of aircraft’s average speed:

(41 000 km) ÷ (47 h) or (41 ×106 m)/(47 × 3600 s)

= 870 km h–1 or 240 m s–1

Multiply maximum thrust by average speed and comment on answer:

One engine: (700 kN × 870 km h–1) or (700 kN × 240 m s–1)

Two engines: (2 × 700 kN × 870 km h–1)

or (2 × 700 kN × 240 m s–1) = 340 MW

[Allow any correct unit with corresponding arithmetic, eg kN km h–1)

Statement recognising that the product is a power.

Either a comparison of the two powers or a comment on the engine thrusts. 6[10]

812. Diagram of torch circuit:

The lamp will light

Correct circuit 2

[Circuit showing one cell only is allowed one mark only unless the cell is labelled 4.5 V. If a resistor is included, allow first mark only unless it is clearly labelled in some way as an internal resistance.]

3 . 5 V / 3

0 . 3 A

3 . 5 V

Voltage across each circuit component and current in lamp:Either 3.5 V/3 shown across the terminals of one cell or 3.5 V acrossall three cells3.5 V shown to be across the lamp0.3 A flowing in the lamp [i.e. an isolated 0.3 A near the lamp does notscore] 3

Calculation of internal resistance of one of the cells:

Lost volts = 4.5 V - 3.5 V or 1.5 V – 3

V5.3

or total resistance = (4.5 V)/0.3 A) = 15 KΩ

Internal resistance of one cell = [(1.0 V)/(0.3 A)] ÷ 3

or [(0.33 V) (0.3 A)] or lamp resistance = (3.5 V) / (0.3 A)11.7 Ω= 1.1 Ω or = (3.3Ω )/3 = 1.1 Ω 3

[Some of these latter marks can be read from the diagram if it is so

labelled][8]

813. Calculation of time bullet takes to reach top of its flight and statement of any assumption made:

– 9.8 m S–2 = (0 m s–1 – 450 m s–1)/t

t = 46 s

Assumption: air resistance is negligible, acceleration constant or equivalent 3

Sketch of velocity-time curve for bullet’s flight:

Label axes

Show the graph as a straight line inclined to axis

+ 450 m s–1 and 46 s shown correctly

– 450 m s–1 or 92 s for a correctly drawn line

4 5 0 m s

– 4 5 0 m s

4 6 s9 2 s t

– 1

– 1υ

Explanation of shape of graph:

Why the line is straight - acceleration constant or equivalent or why the velocity changes sign or why the gradient is negative

Calculate the distance travelled by bullet, using graph:

Identification of distance with area between graph and time axis or implied in calculation

20 700 m for g = 9.8 ms–2 or alternative answers from different but

acceptable “g” values. 7

[Allow e.c.f with wrong time value.][10]

814. Type of radioactive decayα -decay

Nuclear equation for decay

HE4/α4Y143147226062

+×

[1 mark for letters, 1 mark for numbers]

Addition of arrow to diagram

8 8

8 6

8 4

8 2

8 05 8 6 0 6 2 6 4

N

Z

X

Y

W

P

Point P on diagram

[5]

815. Design of experiment to find what types of radiation are emitted:

Soil in container with opening facing detector

Take background count /or shield apparatus

With detector close to soil, insert paper

or take close reading then at, ≈ + 5 cm; count rate reduced so α present

Insert aluminium foil: further reduction ∴ β present

Insert lead sheet: count rate still above background or count rate reduced to zero,∴ ϒ present.or, if no count after aluminium foil, no ϒor, if count rate above background with thick aluminium, then ϒ present

[5]

816. Region on graph where copper wire obeys Hooke’s law:

Hooke’s law region up to (9,15)

Additional information needed:

Length and cross-sectional area

Estimate of energy stored in wire:

Sensible attempt at area up to 20 mm

Answer in range 250 → 270

0.26 J[5]

817. Use of graph to estimate work function of the metal:

φ = (6.63 × 10–34 J s) (6.0 × 1014 Hz) – (some value)

Value in brackets: (1.6 × 10–19 × 0.5 J)

3.2 × 10–19 J or 2 eV 3

Addition to axes of graph A obtained when intensity of light increased:

A starts at –0.5

A → larger than /max

Addition to axes of graph B obtained when frequency of light increased:

B starts at less than – 0.5

B → same of lower than /max 4[7]

818. Observations on voltmeter:

(a) Movement which implies brief or pulsed then V reads zero

(b) Negative reading with respect to direction above

(c) Alternating reading positive to negative and continuous[5]

819. Sketch of two graphs:

a

0 t

SinusoidalNegative start

a

dO

Linear through 0,0Negative gradient 4

Amplitude of tide = 3.1 m

Next mid-tide at 12.00 (noon)

Next low tide at 15.00 (3 pm) 3

Calculation of time at which falling water levels reaches ring R:

x = x0 sin

12

2 tπ [Allow cosine]

1.9 m = 3.1 m sin

h12

2 tπ

[Error carried forward for their amplitude above; not 1.2 ml

t = 1.26 h or t = 4.25 h if cosine used

Time at R = 12.00 h + 1.26 h = 13.26 h (1.16 pm) 4[11]

820. Description:

EitherTwo connected dippers just touching/above the water

OrDipping beam or single source (1)reaches two slits (1)

Vibrated electrically (1)Level tank/shallow water/sloping sides (1)

-

EitherIlluminate project on to screen

OrUse stroboscope (1)to freeze the pattern (1)

Max 5

Diagram:(i) Correct line A - centre line (1)(ii) Correct line B (above or below A) (1)(iii) Correct line C (between A and B) (1)

both B and C correct (1) 4[Total 9 marks]

821. Ionisation energy:

2810 eV (4.5 ×10–16 J) (1)

Calculation of maximum wavelength:Energy in eV chosen above converted to joules (1)Use of λ = c/f (1)Maximum wavelength = 4.4 × 10–10 m (1)

Part of electromagnetic spectrum:γ -ray / X-ray (1) 5

Calculation of the de Broglie wavelength:λ = h/p p identified as momentum (1)Either m or υ correctly substituted (1)Wavelength = 1.1 × 10–13 m (1) 3

[Total 8 marks]

822. Displacement-time graph:

Sine or cosine (1)

Max/min at +1.8 and –1.8 (1)

T = 0.05 s and at least one cycle (1) 3

Calculation of maximum speed:

Correct use of υ = ω x0 or 2π fx0

= 2π × 10 s–1 × 1.8 ×10–2 m (1)

Maximum speed = 2.3 m s–1 (226 cm s–1) (1) 2

Any two places correctly marked M (1) 1[Total 6 marks]

823. Total magnetic flux through the loop when 30 mm from end of magnet:Flux = B × A= 1 × 10-3 T × 16 × 10-4 m2 (1)

[Substitution of 1, 16. Ignore × 10 here]= 1.6 × 10-6 Wb (1)

Total magnetic flux through the loop when 10 mm from end of magnet:Flux = 30 ×10–3 T × 16 × 10–4 m2

= 4.8 × 10–5 Wb [Unit penalty once only] (1) 3

Average speed of movement of the loop:E = ∆ φ /∆ t

∆ t = V1015

Wb104.466

6

−

−

××

(1)

= 3.1 s

Use of speed = distance ÷ time = 20 mm ÷ 3.1 s (1)= 6.5 mm s–1 (1) 3

Slow down nearer to the magnet (1) 1[Total 7 marks]

824. You are to investigate the behaviour of a simple pendulum when its swing is interrupted.

(a) A simple pendulum of length 1 = 0.500 m is provided. Take measurements to determine its period of oscillation, T0.

10T0/s: 14.15, 14.09, av. 14.12

0 = 1.41 s

T0 = 1.42 ± 0.03 s found from >10 oscillations, with unit and > 2 decimal places (1)

(1 mark)

(b) Now place the rod at an initial distance h = 0.250 m as shown in the diagram so that it will interrupt the swing of the pendulum; when the pendulum is at rest the rod should just touch the string.

S i d e v i e w

h

l

F r o n t v i e w

Determine the period T for oscillations of small amplitude.10T/s: 11.94, 11.88, av 11.91T = 1.19 s

Keeping l constant at 0.500 m, determine T for four more values of h. Tabulate all your results below, together with values of T2.

Table, with units, including correct values for T2 to > 2 decimal places (1)Repeat readings (1)Σ T > 20 (1)5 good values (3)(± 0.02 s2 from line of best fit)[4 gets (2), 3 gets (1)]

Range of h > 250 mm (1) 7

-h/m 10T/s T/s T2/s2

0.050

0.150

0.250

0.300

13.77, 13.78

12.83, 12.87

11.94, 11.88

11.38, 11.41

1.38

1.29

1.19

1.14

1.90

1.65

1.42

1.30

0.350 10.85, 10.91 1.09 1.18

[Penalties: wrong h – 2, wrong T (e.g. 10T, 1/T) – 2]

(7 marks)

(c) Plot a graph of T2 against h on graph paper.Graph:Scale (1)Axes labelled, with units (1)Plots (1)Line (2) 4

2 . 0

1 . 9

1 . 8

1 . 7

1 . 6

1 . 5

1 . 4

1 . 3

1 . 2

0 . 0 0 . 1 0 . 2 0 . 3 h / m

T / s2 2

(4 marks)

(d) Theory suggests that

g

h

g

lT

222 24 ππ −=

where g = gravitational field strength.

Discuss the extent to which your results support this theory.Straight line of negative slope (1)

Intercept (1)

Related to T0 in (a) (1)

[T02 plotted or compared to 4π 2l/g]

[First (2) marks can be obtained from curve (1), so does not support theory (1) or comparison with y =mx + c]

The graph is a straight line of negative slope and intercept of T02 = 2.01 s2.

This gives a value of T0 = 1.42 s, which agrees very closely with the 1.41 s found in (a).

(3 marks)

(e) Determine the gradient of your graph.Large ∆ (1)(∆ x∆ y > 80 cm2)Correct calculation with sign > 2 significant figures (1)[Ignore units]

Gradient = –00.035.018.101.2

−−

= – 2.37 (s2 m-1)

(2 marks)

(f) Use your answer to (e) to find a value for the gravitational field strength g.Correct formula for g (1)Correct calculation with appropriate unit (1)Value 6.0 → 10.5 m s-2 and 1 or 2 decimal places (1)(deduced correctly from gradient)[Ignore minus signs]

Gradient = –gπ2 2

→ g = –gradient

π2 2

= –37.2

π2 2

−

= 8.3 m s-2 (N kg-1)

(3 marks)[Total 20 marks]

825. The following statements apply to a body orbiting a planet at constant speed and at constant height. Indicate whether each statement is true (√) or false (x).

Statement True/False

The body is travelling at constant velocity. x

The body is in equilibrium because the centripetal force is equal and opposite to the weight.

x

The only force acting on the body is its weight. √

The body’s acceleration towards the planet equals the gravitational field strength at the position of the body.

√

(Total 4 marks)

826. State the difference between scalar and vector quantities.Scalar quantities are non-directional (1)Vector quantities are direction (1)

A lamp is suspended from two wires as shown in the diagram. The tension in each wire is 4.5N.

4 0 º4 0 º4 . 5 N4 . 5 N

Calculate the magnitude of the resultant force exerted on the lamp by the wires.4.5 N cos 40° (1)× 2 (1)Resultant force = 6.9 N (1)

What is the weight of the lamp? Explain your answer.6.9 N (1)Weight = Supporting force or it is in equilibrium (1)

[Total 7 marks]

827. Draw diagrams to represent

(i) the gravitational field near the surface of the Earth,

Direction

Lines: at least 3 parallel perpendicular equally spaced

(ii) the electric field in the region of an isolated negative. point charge.

Direction

Lines: at least 3 radial equally spaced

(4 marks)

How does the electric field strength E vary with distance r from the point charge?

E ∝ r²

1 (1)

(1 mark)

Give an example of a region in which you would expect to find a uniform electric field.Between charged parallel plates (1).

(1 mark)[Total 6 marks]

828. State Lenz's law of electromagnetic inductionDirection of induced emf is such as to oppose the charge producing it (2)

(2 marks)

An exhibit at a science centre consists of three apparently identical vertical tubes, T1, T2 and T3, each about 2 m long. With the tubes are three apparently identical small cylinders, one to each tube.

When the cylinders are dropped down the tubes those in ~T, and ~T2 reach the bottom in less than I second, while that in ~T3 takes a few seconds.

Explain why the cylinder in T3 takes longer to reach the bottom of the tube than the cylinder in T1

In T3 magnetic flux cuts copper tube (1)induction occurs (1)current in copper tube (1)creates magnetic field (1)opposite to magnet’s which repels slows magnet

T1 is plastic so no induction/no current forms (1)

(5 marks)

Explain why the cylinder in T2 takes the same time to reach the bottom as the cylinder in T1

In T2 falling cylinder unmagnetised so no flux cut or no induction (1)

Both T1 and T2 have only force of gravity acting on them (1)


829. Each of the following graphs can be used to describe the motion of a body falling from rest. (Air resistance may be neglected.)

A B C D E

Which graph shows how the kinetic energy of the body (y-axis) varies with the distance fallen (x-axis)?

Graph C (1)

Explain your answer.Since kinetic energy gained = potential energy lost, (1)

Kinetic energy gained ∝ distance fallen (1)

(3 marks)

Which graph shows how the distance fallen (y-axis) varies with the time (x-axis)?

Graph E (1)

Explain your answer.Speed increases with time (1)

So gradient increases with distance (1)

(3 marks)

Which graph shows the relationship between acceleration (y-axis) and distance (x-axis)?

Graph A (1)

Explain your answer.Acceleration is constant (1)

throughout the motion (1)


830. Experiments on the photoelectric effect show that

• the kinetic energy of photoelectrons released depends upon the frequency of the incident light and not on its intensity,

• light below a certain threshold frequency cannot release photoelectrons.

How do these conclusions support a particle theory but not a wave theory of light?Particle theory: E = hf implied packets/photons (1)

One photon releases one electron giving it k.e. (1)

Increase f ⇒ greater k.e. electrons (1)

Lower f; finally ke = O ie no electrons released Waves (1)

Energy depends on intensity / (amplitude)2 (1)

More intense light should give greater k.e–NOT SEEN (1)

More intense light gives more electrons but no change in maximum kinetic energy (1)

Waves continuous ∴ when enough are absorbed electrons should be released–NOT SEEN (1)

(6 marks)

Calculate the threshold wavelength for a metal surface which has a work function of 6.2 eV.6.2eV × 1.6 × 10–19 C (1)

Use of Ehc=λ (1)

Threshold wavelength = 2.0 × 10–7 m (1)

To which part of the electromagnetic spectrum does this wavelength belong?UV ecf their λ (1)


831. A light aluminium washer rests on the end of a solenoid as shown in the diagram.

I

I

A l u m i n i u mw a s h e r

S o l e n o i d

A large direct current is switched on in the solenoid. Explain why the washer jumps and immediately falls back.

B field produced by solenoid (1)

Flux lines CUT washer (1)

Induced current/e.m.f. in washer (1)

B field of solenoid opposite to B field washer (1)

Repulsive force lifts washer (1)

Steady current so no changing of flux/no induction (1)

OR explain by force on current carrying conductor in B field (LH rule)

[Total 5 marks]

832. An athlete of mass 55 kg runs up a flight of stairs of vertical height 3.6 m in 1.8 s. Calculate the power that this athlete develops in raising his mass.

Power = s)8.1(

m)(3.6N/kg)(9.81kg)55( ××

Numerator correct (1)

Denominator correct (1)

Power = 1080 W (1)

(3 marks)

One way of comparing athletes of different sizes is to compare their power-to-weight ratios. Find a unit for the power-to-weight ratio in terms of SI base units.

Units correctly attached to a correct equation (1)

e.g. N

s m N

weight

power 1−

= (1)

= m s–1 (1)

(0) if m s–1 is derived wrongly

(2 marks)

Calculate the athlete’s power-to-weight ratio.

Power to weight ratio = )s m (9.81kg) 55(

W)1080(2−×

(1)

Power-to-weight ratio = 2 [m s–1] (1)

(Unit error not penalised in final part)


833. The diagram shows a velocity-time graph for a ball bouncing vertically on a hard surface.

t / sv / m s – 1

+ 5 . 0

0

– 5

1 . 0 2 . 0

At what instant does the graph show the ball to be in contact with the ground for the third time?2.05 s ≤ t ≤ 2.10 s (2)

OR

2.00 ≤ t ≤ 2.20 s (1)

(2 marks)

The downwards-sloping lines on the graph are straight. Why are they straight?Acceleration of the ball or force on the ball or gravitational field strength

is constant or uniform (2 or 0)

(2 marks)

Calculate the height from which the ball is dropped.Relevant equation or correct area (1)

Substitution correct (1)

Answer (1.2 m ≤ height ≤ 1.3 m) (1)

(3 marks)

Sketch a displacement-time curve on the axes below for the first second of the motion.

Dis

plac

emen

t/m

– 1 . 2 5

0 . 50

1 . 0 t /s

Displacement scale agreeing with above (1)

First half of curve correct (1)

Second half correct with reduced height (1)

(3 marks)

What is the displacement of the ball when it finally comes to rest?–1.25 m (correct magnitude and direction)

(Look at candidate’s displacement origin)


834. A satellite orbits the Earth once every 120 minutes. Calculate the satellite’s angular speed.Correct substitution into angle/time (1)

Answer with correct unit (1)

r.p.m. etc. not allowed

Angular speed = e.g. 0.052 rad min-1 180°h-1

(2 marks)

Draw a free-body force diagram for the satellite.

(1 mark)(If the Earth is shown, then the direction must be correct)

The satellite is in a state of free fall. What is meant by the term free fall? How can the height of the satellite stay constant if the satellite is in free fall?

Free fall – when gravitational force is the only force acting on an object (1)

Height – (1) for each clear and relevant physics statement (1) + (1)


835. A student was studying the motion of a simple pendulum the time period of which was given by T = 2π (l/g)½.

He measured T for values of l given by

l/m = 0.10, 0.40, 0.70, 0.70, 1.00

and plotted a graph of T against √l in order to deduce a value for g, the free-fall acceleration. Explain why these values for l are poorly chosen.

An inadequacy PLUS a reason why (many possibilities) e.g. some values too short to produce accurate T values; when the values are square rooted; spacing is unsatisfactory. (1)

(1 mark)

How would the student obtain a value of g from the gradient of the graph?

Gradient = g

π2 (1)

Hence g = 2

2

)(

4

gradient

π (1)

(2 marks)

The graph below shows three cycles of oscillation for an undamped pendulum of length 1.00 m.

D i s p l a c e m e n t T i m e / s

2 4 6

Add magnitudes to the time axis and on the same axes show three cycles for the same pendulum when its motion is lightly air damped.

T = 2 s (1)

Tdamped = Toriginal (1)

Amplitude reduced (1)

Continuous reduction in amplitude (1)


836. A body oscillates with simple harmonic motion. On the axes below sketch a graph to show how the acceleration of the body varies with its displacement.

a

d

Straight line through origin (1)

Negative gradient (1)

(2 marks)

How could the graph be used to determine T, the period of oscillation of the body?Reference to gradient of line (1)

Gradient = (-) ω 2 or = (-)(2π f)²

OR T = gradient)(

2

−π

(1)

(2 marks)

A displacement-time graph from simple harmonic motion is drawn below.

D i s p l a c e m e n t T i m e

H

L

2 . 0 0 8 . 0 0 p . m .

( i ) H a n d L

( i i )

( i i i ) 2 . 0 0 a n d 8 . 0 0 p . m .

The movement of tides can be regarded as simple harmonic, with a period of approximately 12 hours.

On a uniformly sloping beach, the distance along the sand between the high water mark and the low water mark is 50 m. A family builds a sand castle 10 m below the high water mark while the tide is on its way out. Low tide is at 2.00 p.m.

On the graph

(i) label points L and H, showing the displacements at low tide and next high tide,

(ii) draw a line parallel to the time axis showing the location of the sand castle,

(iii) add the times of low and high tide.(3 marks)

Calculate the time at which the rising tide reaches the sand castle.Use of x = x sin ω t

15 = 25 sin ω t

ω = π6

or ω t = 37° / t = 1.23 hours

Time = 6.14 p.m. ANY THREE LINES (3)

Full error carried forward from wrong diagram

Alternative using graph:

Identify coordinates (1)

Convert to time (1)

Add to reference time (1)


837. The diagram below shows a loudspeaker which sends a note of constant frequency towards a vertical metal sheet. As the microphone is moved between the loudspeaker and the metal sheet the amplitude of the vertical trace on the oscilloscope continually changes several times between maximum and minimum values. This shows that a stationary wave has been set up in the space between the loudspeaker and the metal sheet.

M e t a ls h e e tL o u d s p e a k e r

M i c r o p h o n e

T o o s c i l l o s c o p e( t i m e b a s e o f f )

S i g n a lg e n e r a t o r

How has the stationary wave been produced?by superposition/interference (1)

with a reflected wave/wave of same speed and wavelength in opposite direction (1)

(2 marks)

State how the stationary wave pattern changes when the frequency of the signal generator is doubled. Explain your answer.

Maxima/nodes/equivalent are closer together (1)since wavelength is halved (1)

(2 marks)

What measurements would you take, and how would you use them, to calculate the speed of sound in air?

Measure distance between minima/equivalent (1)

Repeat/take average (1)

Method of finding frequency (1)

λ = 2 × (node – node)/equivalent (1)

V = f × λ (1)

(Four marks maximum)

Other methods eligible for full marks.

(4 marks)

Suggest why the minima detected near the sheet are much smaller than those detected near the loudspeaker.

Near the sheet there is almost complete cancellation (1)

since incident and reflected waves are of almost equal amplitude (1)


838. The graph shows how the height above the ground of the top of a soft bouncing ball varies with time.

H e i g h t o ft o p o fb o u n c i n gb a l l

A

BC

D

A B C Dt t t t

B a l l

Describe briefly the principal energy changes which occur between the times

tA and tB

The ball loses gravitational potential energy and gains kinetic energy

(2 marks)

tB and tCThe kinetic energy is transformed into elastic potential energy when the ball deforms on the ground.

(3 marks)

tC and tDThe elastic potential energy is converted back into kinetic energy


839. Define capacitanceCapacitance = Charge / Potential difference.

(2 marks)

An uncharged capacitor of 200 µ F is connected in series with a 470 kΩ resistor, a 1.50 V cell and a switch. Draw a circuit diagram of this arrangement.

4 7 0 k Ω

2 0 0 F µ1 . 5 0 V

(1 mark)

Calculate the maximum current that flows.Current = 1.5 V/470 kΩ

Current = 3.2 µ A

(2 marks)

Sketch a graph of voltage against charge for your capacitor as it charges. Indicate on the graph the energy stored when the capacitor is fully charged.

S h a d e d a r e ae q u a l se n e r g y s t o r e d

V

Q(4 marks)

Calculate the energy stored in the fully-charged capacitor.½CV2 = ½ (200 µ F) (1.5 V)2

Energy = 2.25 µ J


840. The graph below shows the behaviour of a material A subjected to a tensile stress.

( N O T E L O W E R S T R A I N )S t r a i n

S t r e s s / P a

M a t e r i a l A

B

C

B r i t t l e ( 1 )P l a s t i c ( 1 )

B o t h Y o u n g m o d u l i ( 1 )

How would you obtain the Young modulus of material A from the graph?Find gradient of the linear region (1)Read values off graph and divide stress by strain/equivalent (1)

(2 marks)

What is the unit of the Young modulus?Pa/N m–2/kg m–1 s–2

(1 mark)

On the same graph, draw a second line to show the behaviour of a material B which has a greater Young modulus and is brittle.

Draw a third line to show the behaviour of a material C which has a lower value of Young modulus and whose behaviour becomes plastic at a lower strain.

[Total 6 marks]

47429190 unit 5 ans excellent loads of questions physics edexcel

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