physics june 2009 ms

8/6/2019 Physics June 2009 MS

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Mark Scheme Summer 2009

GCE

GCE Physics (8540/9540)

Edexcel Limited. Registered in England and Wales No. 4496750Registered Office: One90 High Holborn, London WC1V 7BH

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they need to help them deliver their education and training programmes to learners.For further information, please call our GCE line on 0844 576 0025, our GCSE team on0844 576 0027, or visit our website at www.edexcel.org.uk.

Summer 2009

Publications Code UA021601

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Mark scheme notes

Underlying principle

The mark scheme will clearly indicate the concept that is being rewarded, backed up byexamples. It is not a set of model answers.

For example:

(iii) Horizontal force of hinge on table top

66.3 (N) or 66 (N) and correct indication of direction [no ue][Some examples of direction: acting from right (to left) / to the left /West / opposite direction to horizontal. May show direction by arrow. Donot accept a minus sign in front of number as direction.]

1

This has a clear statement of the principle for awarding the mark, supported by some examplesillustrating acceptable boundaries.

1. Mark scheme format1.1 You will not see wtte (words to that effect). Alternative correct wording should be

credited in every answer unless the ms has specified specific words that must bepresent. Such words will be indicated by underlining e.g. resonance

1.2 Bold lower case will be used for emphasis.1.3 Round brackets ( ) indicate words that are not essential e.g. (hence) distance is

increased.1.4 Square brackets [ ] indicate advice to examiners or examples e.g. [Do not accept

gravity] [ecf].

2. Unit error penalties2.1 A separate mark is not usually given for a unit but a missing or incorrect unit will

normally cause the final calculation mark to be lost.

2.2 Incorrect use of case e.g. Watt or w will not be penalised.2.3 There will be no unit penalty applied in show that questions or in any other questionwhere the units to be used have been given.

2.4 The same missing or incorrect unit will not be penalised more than once within onequestion but may be penalised again in another question.

2.5 Occasionally, it may be decided not to penalise a missing or incorrect unit e.g. thecandidate may be calculating the gradient of a graph, resulting in a unit that is not onethat should be known and is complex.

2.6 The mark scheme will indicate if no unit error penalty is to be applied by means of [noue].

3. Significant figures

3.1 Use of an inappropriate number of significant figures in the theory papers will normallyonly be penalised in show that questions where use of too few significant figures hasresulted in the candidate not demonstrating the validity of the given answer.

3.2 Use of an inappropriate number of significant figures will normally be penalised in thepractical examinations or coursework.

3.3 Using g = 10 m s2 will not be penalised.



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4. Calculations

4.1 Bald (i.e. no working shown) correct answers score full marks unless in a show thatquestion.

4.2 If a show that question is worth 2 marks then both marks will be available for areverse working; if it is worth 3 marks then only 2 will be available.

4.3 use of the formula means that the candidate demonstrates substitution of physicallycorrect values, although there may be conversion errors e.g. power of 10 error.

4.4 recall of the correct formula will be awarded when the formula is seen or implied by

substitution.4.5 The mark scheme will show a correctly worked answer for illustration only.4.6 Example of mark scheme for a calculation:

Show that calculation of weight

Use of L W H

Substitution into density equation with a volume and density

Correct answer [49.4 (N)] to at least 3 sig fig. [No ue][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 conversionto kg is omitted and then answer fudged, do not give 3rd mark][Bald answer scores 0, reverse calculation 2/3]

Example of answer:

80 cm 50 cm 1.8 cm = 7200 cm3

7200 cm3 0.70 g cm-3 = 5040 g

5040 10-3 kg 9.81 N/kg

= 49.4 N

3

5. Quality of Written Communication5.1 Indicated by QoWC in mark scheme, placed as first mark.5.2 Usually it is part of a max mark.5.3 In SHAP marks for this are allocated in coursework only but this does not negate the

need for candidates to express themselves clearly, using appropriate physics terms.Likewise in the Edexcel A papers.

6. Graphs6.1 A mark given for axes requires both axes to be labelled with quantities and units, and

drawn the correct way round.6.2 Sometimes a separate mark will be given for units or for each axis if the units are

complex. This will be indicated on the mark scheme.6.3 A mark given for choosing a scale requires that the chosen scale allows all points to be

plotted, spreads plotted points over more than half of each axis and is not an awkwardscale e.g. multiples of 3, 7 etc.

6.4 Points should be plotted to within 1 mm. Check the two points furthest from the best line. If both OK award mark. If either is 2 mm out do not award mark. If both are 1 mm out do not award mark. If either is 1 mm out then check another two and award mark if both of these OK,

otherwise no mark.6.5 For a line mark there must be a thin continuous line which is the best-fit line for thecandidates results.



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6731 Unit Test PHY1

Question Number

Answer Mark

1 (a) How physical quantities change

Increase Decrease Stays the same

6 ticks right

5 ticks right

4 ticks right

3 ticks right

Total 8

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Question Number

Answer Mark

2. (a) Define principle of conservation of linear momentum

The total/net/sum momentum of a (closed) system does notchange or total/net/sum momentum before a collision equals thetotal/net/sum momentum after a collision(1) [Total/net/sum at least once. Accept but not all]

provided no resultant/net/unbalanced force acts [Accept noexternal force. Accept for total but not all.] (1)

(2)

(b)(i) As momentum

Equation for momentum seen or used(1)

Answer [ 144 kg m s-1. Apply ue. Allow -144 kg m s-1 and N s.](1)

Eg momentum = 48 kg x 3.0 m s-1 = 144 kg m s-1

(2)

(ii) State Bs momentum

Minus [Accept in opposite direction.](1)

144 kg m s-1 [Ecf answer from bi](1)

[Could get both marks for 144 kg m s-1/N s if answer in bi isnegative]

Explain with reference to principle of conservation of linearmomentum

(Since total) momentum before (separation) is zero the (total)momentum after (separation) must also be zero or pA+ pB = 0 .

(1) (For total momentum after separation to be zero) the momenta

(of A and B) must be equal (in magnitude) and opposite indirection or (A and B) have opposite momentum [Acceptmomentum of (A) is positive and (B) is negative] or (hence) pA = -pB (1)

(4)

Total 8



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Question Number

Answer Mark

3. (a) (i) The height of bounce

Attempt to measure area of a triangle between graph and time

axis.(1)

Answer [0.45 m, ue. Ignore negative sign](1)

Eg Area under graph =2

sm3xs3.0 -1

= 0.45 m

Or Use of correct equation(s) of motion (1) [If the equation involves g they must use 10 m s-2. Accept use of

s = gt2.] Answer [0.45 m. ue] Eg

s = ut + gt2 s =2

vu +t

= 3 m s-1 x 0.3 s + -10 m s-2 x 0.32 s2 =

2

0sm3 -1 +x 0.3

s = 0.45 m = 0.45 m

(2)



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(ii) The average speed

Attempt to measure the total area of both triangles formed

between graph and time axis or 2.25 m seen or use of correctmotion equations[allow g used in a(i)] (1) Answer [2.5 m s-1, ue. Ecf answer from a( i )]

(1)

Eg Total area = 0.45 m +2

sm6xs6.0 -1(= 2.25 m)

Average speed =s0.9

m25.2= 2.5 m s-1

(2)

(b) (i) Describe velocity change

Reduced magnitude and opposite direction(1)

[Accept (velocity changes by) (-)9 m s-1/ (+)6 m s-1 to -3 m s-1 /from 6 m s-1 in one direction to 3 m s-1 in the opposite direction /(velocity) decreases to zero then increases in the oppositedirection.]

(1)

(ii) Why velocity change as shown could not happen

The change in velocity is instantaneous / has taken zero

time[accept no time] or the ball would remain in contact withthe ground for a time (1)

This would mean infinite acceleration / force (which is impossible)or ball would (take time to) compress or the graph should not bevertical should have a (negative) gradient[accept description thatmeans this] where the change in velocity happens. (1)

[Accept reverse statements where appropriate for either mark]

(2)

Total 7



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Question Number

Answer Mark

4(a)(i) Resultant force

Use of F = ma(1) Answer [132 N, ue.]

(1)

Eg F = 110 kg x 1.2 m s-2 = 132 N

(2)

(ii) Resistive force

Answer [68 N, ue. Ecf answer from a(i).]

(1) Eg 200 N 132 N = 68 N

(1)

(b) Force B applies to A 100 N [If they do not write 100 N on the response line, but

mention it in their explanation give the mark](1)

Explain Either Box B is also accelerated 1.2 m s-2 (and same resistive force acts)

(1) (Since) box B is half the mass of A+B the force applied by A to B

must be half the value of the force applied to A (+ B)(1)

By Newtons 3rd law an equal but opposite force is applied by B toA. (1)

[Give this mark for a bald statement of N3 in terms of body A andbody B (even if the explanation is wrong or there is noexplanation)]

Or Correct substitution into F = ma / 66 N[unbalanced force on box B]

(1) Addition of 34 N[to get force of A on B] (1) [Allow this mark even if 68 N is used for friction] By Newtons 3rd law an equal but opposite force is applied by B to

A (1) [Apply this as above] [Note they can get these three marks for an answer of 132 N]

Eg F = 55 kg x 1.2 m s-2 = 66 N [unbalanced force on box B] Total force applied by A to B = 66 N + 34 N = 100N

(4)



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Hence by Newtons 3rd law force from B to A is the same

Or Correct substitution into F = ma/ 66 N [unbalanced force on

box A] (1)

Equates forces acting on box A (1) [Allow this mark even if 68 N is used for friction ] Equates both expressions to obtain force of Bon A (1) [ They could get these three marks for 66 N]

Eg F = 55 kg x 1.2 m s-1 = 66 N 200 N 34 N force of B on A = 66 N Force of B on A = 100 N

Total 7

Question Number

Answer Mark

5(a) Weight of pole

Use of w = mg(1)

Answer [ 2.9(4) N. Accept 3 N. Ue](1)

Eg w = 300 (x 10-3) kg x 9.81 N kg-1

= 2.94 N

(2)

(b) Distance of centre of gravity from end of handle Use of principle of moments [Allow one wrong distance]

(1) Distance from pivot [238 mm (allow 240 mm) or 233 mm (allow

230 mm)if 3 N used. Ecf candidates value for weight from b(i)](1)

Answer [273 mm or 268 mm. also 275 mm or 265 mm respectfully.Ecf, ue] (1)

Eg 20 N x 35 (x 10-3m) = 2.94 N x L

L = 238 x 10

-3

m Distance from end = 238 x 10-3 m + 35 x 10-3 m = 273 x 10-3 m

(3)

(c) Why force needs to be adjusted

QWOC(1)

The centre of gravity[allow centre of mass or weight] will movefurther from pivot / to the right (1)

[Allow c of g moves further from end of handle]

(therefore) the moment of the weight / the clockwise moment /the right hand side moment [accept the moment on this sidewhere it is clear what they mean] will increase or shown by a



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calculation / formula(1) (To maintain balance) the anticlockwise moment / the moment of

the force applied / the LHS moment[accept moment on the otherside] must increase [This must be stated not implied eg if acandidate shows cm has increased and states acm = cm this in

itself would not get this mark.] or justifies through calculation / formula why the force increases(1)

the force applied at the end of the stick must increase or shown ina calculation or formula (1)

(5)

Total 7

Question Number

Answer Mark

6(a) Velocity of stone

Use of appropriate equation(s) of motion(1)

Answer [11.8(m s-1) or 12 (m s-1) if g = 10 m s-2 is used.](1)

[Even though this is a show that accept 12 m s-1 when g = 10 m s-2 is used]

Eg v = u + at = 0 + 9.81 m s-2 x 1.2 s = 11.77(m s-1) (11.76 m s-1 when g = 9.8 m s-2)

(2)

(b)(i) Change in kinetic energy

Attempt to subtract two 2

1mv

2

values

(1)

[Allow this mark even if one is incorrect] Use of 0.6 kg for mass [in any equation] (1) Answer [(34.0 J 36 .0 J). Ue accept kg m 2 s-2. Ecf candidates

value for velocity from (a). Ignore negative in answer](1)

Eg EK= 0.5 x 0.6 kg x [(11.8 m s-1)2 (5 m s-1)2]

= 34.27 J

(3)

(ii)

Average resultant force Either EK set equal to work done

(1) Answer [(21.2 N - 22.5 N). Ue. Ecf candidates value for bi.]

(1)

Eg F x 1.6 m = 34.27 J F = 21.42 N

Or Use of v2 = u2 + 2as or two appropriate equations of motion and F

(2)



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= ma or F =t

mumv - (1)

Answer [(-)(21.3 N - 22.5 N). Ue. Ecf candidates value for bi.](1)

[allow both marks even if v and u are mixed up ie they will get a

positive value for acceleration]

Eg a =m1.6x2

sm0.5sm8.11 -22-222 += (-) 35.7 m s-2

F = 0.6 kg x (-) 35.7 m s-2 = (-)21.4 N

(c) Explain observation

Upward / resistive / viscous / drag friction force (due to water)equals weight of / downward force (due to stone) or Unbalanced/ resultant force is therefore zero. (1)

Hence acceleration is zero (1)

(2)

Total 9



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Question Number

Answer Mark

7 (a) What happens when an atom is ionised [Throughout this question; Do not accept particle of air; Do

accept recognisable symbols for alpha and beta] Alpha particle collides with (air) atom / molecule [or atom with

alpha] (1) As a result an electron is ejected / attracted out (of an

atom/molecule). [accept loses an electron even electrons](1)

leaving behind a positive ion (and a negative electron)(1)

[ Accept atom/molecule with a (net) positive charge]

Max(2)

(b) Calculate range of alpha particle

Either Determines number of ionisations throughout range / 1.9 x 105/shows correct calculation(1)

Divides the above value by 5 x 103(1)

Answer [38 mm. Ue.](1)

Eg No of ionisations =J10x7.4

J10x918-

-13

(= 1.9 x 105)

Range =3

5

10x510x1.9 mm

= 38(.3) mm Or Determines energy required for ionisations per mm ie 23.5 x 10-15 J

/ shows correct calculation(1)

Divides the above value into 9 x 10-13 J(1)

Answer [38 mm. Ue.](1)

Eg Energy per mm = 4.7 x 10-18 J x 5 x 103 ( = 23.5 x 10-15 J)

Range =J10x5.23

J10x915-

-13

= 38(.3) mm

(3)



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(c) Effect on range of alpha particle

The range is increased(1)

Number of atoms ionised per mm / per unit length would be lessor number of atoms per mm / unit length would be less or(average)distance between atoms would be greater[Accept atomswould be more spread out] or number of collisions per mm / perunit length would be less or alpha particles are less likely to hitatoms / molecules (1)

(2)

(d) Explain observation

(Beta particles) have a smaller (magnitude of) charge (and aretherefore less effective in removing electrons from atoms)(1)

(Beta particles) travel (much) faster (since they have a muchsmaller mass and yet similar kinetic energy) or spends less timeper mm / per unit length (1)

(As a result) they encounter / ionise a (much) smaller number ofatoms per mm / per unit length or beta particles are less likelyto hit atoms/molecules or beta particles lose less (kinetic) energyper mm / per unit length (1)

(Max 2)

Total 9



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Qu stioneNumber

Answer Mark

8. a)( Define half-lifeAverage time required (1)

r activity / intensity to halve or the number of

[Do accept mass / particles / atom / (radio)isotope / count /

Fo the count rate /(unstable) atoms/nuclei/ nuclides to halve. (1)not

sample / nuclide](2)

(b) (i) Determine half-lifef

2 hAnswer[ (1) (2)

Working that indicates that a 75% reduction indicates a time interval oalflives. (1)

24 days. Ue]

(b) (ii) Explain apparent contradiction[Do not accept particle for atom or nucleus]

is impossible to predict when/ which (single) nucleus/atoml dec

Idea that itl

e aat any instant

A =N apply `/ prediction iscept the number of

n half life/ e

wi ay (1)Idea that (with) a sample of radioactive material (where)there ar(very) largenumber of nuclei / atoms decayingmathematical laws (of probability)/possible / the rate of decay can be predicted [acatoms / nuclei decaying can be predicted] / they have a know

xponentially decay (1)

(2)

Total 6



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19/120

6732 Unit Test PHY2

Question Number

Answer Mark

1(a) Ammeter resistance

Current passes through ammeter or ammeter in series in circuit(1)

p.d. across ammeter needs to be zero, negligible or very small OR

Ammeter doesnt use any power [allow no power loss acrossammeter]

OR If it had resistance, it would reduce the current it is meant to be

measuring (1) [do not credit zero resistance so will not affect the current so max amount of current will flow through it]

2

1(b) Voltmeter resistance Voltmeter in parallel across component or it provides alternative

path for the current

(1)

Current in voltmeter needs to be zero, negligible or very small OR voltmeter doesnt use any power or doesnt reduce the p.d. it is

measuring. OR If its resistance was lower, it would reduce the p.d. it is meant to

be measuring (1)

[do not credit infinite so current keeps flowing in the circuit most of the current goes through the rest of the

circuit

2

Total for question 4



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Question Number

Answer Mark

2(a) Explanation of shape of graph

qowc(1)

Max Three from Initially the temperature is low OR wire not yet hot OR filament

heats up when turned on(1)

Current/electron flow causes heating(1)

Resistance of filament increases as temperature increases/wiregets hotter OR increased ion movement slows the electrons down(1)

(after 0.3 s, current constant) when temperature/resistance isconstant/maximum (1)

1

Max 3

2(b) It breaks because the current is large/ maximum current(1)

1

2(c) Use of sensor The changes occur very quickly OR in a short time OR you cant

take readings fast enough(1)

Sample rate of more 50 s-1 and less than 1000 s-1

(1) (every 0.001s to 0.02 s )

2




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Question Number

Answer Mark

3(a)(i) Current calculation Use of P=VI (1)

I = 52 A (1)

Example of answer I = 2500/48 I = 52.08 A

(use of 12 V gives I = 208 A scores 1/2)

2

3(ii) Charge calculation Use of Q = It (1) Q = 5.6 105 C (1) ecf their current

Example of answer Q = 52 3 3600 Q = 562680

2

3(iii) Energy calculation Use of E = Pt OR E = VQ OR E=VIt (1) Energy = 2.7 107 J (1) ecf their

charge

Example of answer E = 2500 33600 = 27000000 J

OR E = 48 5626800 = 27008640 J

2

3(b) Power needed is 30 kW (which is much greater than 4.0 kW ) OR calculation to show 40 kJ (which is much less than 300 kJ)

(1) Acceleration uses a lot of energy/power

(1) Range of car would be less

(1)

3




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Question Number

Answer Mark

4(a) Circuit diagram Labelled wire, battery and variable resistor in series OR correct

potentiometer circuit(1) Ammeter in series and voltmeter in parallel

(1)

2

4(b) Readings to be taken Current and potential difference

(1) Length of wire with a (metre) ruler

(1) Diameter of wire with micrometer screw gauge or

digital callipers / micrometer(1)

[do not accept just callipers or thickness or area]

3

4(c) Use of readings Use of R=V/I and =RA/l

(1) A= r2 OR A = (d/2)2

(1) Repetition of calculation using different lengths OR graphical

method (1)

Precautions Any two from: Reading of diameter at various places / different orientations Contact errors Zero errors on meters Wire straight when measuring length Wire not heating up/remains at constant temperature

Max 2 [credit for marking points for 4(b) might be found in answer to 4(c)

and vv.]

5




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Question Number

Answer Mark

5(a) Symbol definition n:number of charge carriers per unit volume OR number of charge

carriers m-3

OR charge carrier density.(1) Q: charge on the charge carrier

(1)

(ecf use of electron in n and Q can score mark for Q as charge onelectron)

2

5(b)

Z

W

n

n= 1

Reason: wires same material OR both wires made of copper(1)

Z

W

I

I= 1

Reason: wires in series(1)

Z

W

v

v= 2

Reason: Av = constant OR A inversely proportional to v OR v =I/nQA

OR halving A gives twice v(1)

3




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Question Number

Answer Mark

6(a) Meaning of absolute zero temperature

Temperature at which volume (or pressure) of a gas is zero OR Temperature at which kinetic energy/vibrations of

molecules/atoms of a gas is zero / minimum

(1)

(do not credit KE and PE zero) Measured in kelvin / K

(1)

2

6(b)(i) Pressure calculation Use of p/T = constant either Kelvin or Celsuis

(1) Conversion to Kelvin(1)

p = 1.37 105 Pa(1)

(accept 1.4 105 Pa)

3

6(b)(ii) Graph A Straight line with positive gradient

(1)

Joining (0, 1.00) to (100, 1.37) (ecf their value of p)(1)

(allow 1.36 to 1.40)

2

6(b)(iii) Graph B Straight line through (100. 2.73) (allow 2.70 to 2.80)

(1) Passing through (0, 2.00)

(1)

2




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Question Number

Answer Mark

7(a) Units of pV Unit of pressure = Nm -2 / kg m-1 s-2 and unit of volume = m 3

(1) Unit of energy = Nm / kg m 2 s-2

(1)

OR unit of n = mol

(1) unit of R = J K-1 mol -1 and unit of T = K

(1)

2

7(b) Gas pressure using kinetic theory Any four

Molecules have KE /are moving(1)

Molecules collide with the walls (of the container)(1)

Molecules exert a force(1)

Molecules undergo a change of direction/momentum(1)

Pressure = force/area(1)

There is a large number of molecules (hence constant pressure in

the gas). (1)

4




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Question Number

Answer Mark

8(a) Energy supplied to lead block Use of E = VIt (1) E = 315 (J) allow 310 or 320 (J) (1)

Example of answer E = 0.5 21 30 = 315 J

2

8(b) Specific heat capacity of lead Use of E = mc (1) c = 129 J kg -1 K -1 ( accept C -1) (1) (Full credit if 500 g used and units in J g-1 K-1) ( use of 300 J gives c = 122 J kg -1 K -1) Example of answer c = 315J / (0.5kg 4.9K) = 128.5 J kg -1 K -1

2

8(c) Application of U = W + Q W = 315/300 J

(1) Statement about work done on OR energy supplied to the

block/system/lead (1) ( do not credit work done on body)

Q = 0(1)

Block is thermally insulated from surroundings OR receives nothermal energy from surrounds nor gives any out(1)

must be equal to their W + Q ( U = 315/300 J)(1)

To balance the equation OR because Q = 0, all the energy suppliedbecomes internal energy.(1)

[ it is not enough to say because of conservation of energy]

6




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28/120

6733 Unit Test PHY3 (Topics)

Topic A Astrophysics

QuestionNumber

Answer Mark

1 (a) Star Properties

White dwarf: < 1.4M

(1)

Red Giant (1)

Neutron (star) [accept pulsar] (1)

Black hole: > 2.5 M

(1)(4)

(b)

(i)

Efficiency advantages

more sensitive (more photons detected) (1)

detect fainter stars (1)

detects more distant stars (1)

quicker image collection [accept reversed points] (1)Emulsion advantage

Resolution / pixel size may be smaller / pixilates if magnified(1)

(Any 2)

(1)

(ii) CCD linearity of response

Quality of written communication (1)

All detected wavelengths / frequecies responded to equally (1)

Image brightness corresponds to actual measured intensity (1) (3)

(c)

(i)

Nuclear fusion calculation E = m c2 and any m (1)

m substitution [(6.02136 5.98023) x 10-10 J OR 4.57 x 10-29

(kg)] (1)

4.11 x 10-12 (J) (1)

E = m c2= (4 x 1.6726 6.6447) x 10-27 kg x (3 x 108 m s-1 )2

= 4.11 x 10-12 J

(3)

(c)

(ii)

Solar luminosity calculation

Divide by four [must be shown] (1) 4.11 x 10-12 J x x 3.8 x 1038 = 3.9 x 1026 J [accept W or J s-1]

(1)

(2)



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(c)

(iii)

Intensity - Luminosity calculation

I = L / 4 D2 (1)

Correct substitution [e.c.f.] (1)

1.5 x 1011 m [e.c.f. to 3.0 x 1011 m] (1)

D = (L / 4 I)

= (3.9 x 1026 W / 4 (1370 W m-2)= 1.5 x 1011 m

(3)

(c)

(iv)

Wiens Law calculation

Use of Wiens law with T = 6100 K [or 6073 K] (1)

4.8 x 10-7 m [accept 477 nm but beware 500nm] (1)

max = 2.90 x 10-3 m K / 6100 K

= 4.8 x 10-7 m

(2)

(c)

(v)

Planck energy distribution curve

Correct shape curve [steeply rising, not at origin, less steepconcave fall] (1)

Peak at 480 nm [e.c.f.] (1)(2)

(d) Main Sequence definition

Fusing in core [accept burning] (1)

Hydrogen to helium [not atoms] (1) (2)

(e)

(i)

(ii)

(iii)

Hertzsprung-Russell Diagram

X clearly on 100 line and between 5000 K and 7000 K (1)

Diagonal falling line [or region] passing through X (1)

Steepens at both ends (1)

Upper right of X indicated and labelled (1)

(1)

(2)

(1)

(iv) White Dwarf radius

L = T4 4 r2 (1) L in range 3.9 x 1023 3.9 x 1025 (W) (1)

T in range 10 000 20 000 (K) (1)

Hence answer in range 2 70 x 106 m (1)

L = 4 T4 r 23.9 x 1026 W x 10-2 = 4 x 5.67 x 10-8 W m-2 K4 x (20 000 K)4 x r 2r = 6 x 106 m

(4)

Total 32



30/120


31/120

(d) Pre-stressed reinforced concrete beam


Steel / iron AND rod / cable / wire [at least one in diagram] (1)

Loaded / stressed AND force arrows shown in diagram (1)

Concrete cast / poured over (not cement) (1)

And allowed to set / solidify AND remove tension (1)

Steel in Tension AND Concrete in Compression (1)[cable returns to original length - max 4/5]

(1)

(Any 3)

(1)

(e)

(i)

Composite material definition

Made of more than one material (1)

To gain the (beneficial) properties of each material (1) (2)

(ii)Laminate and Particle Composite

Laminate = plywood AND Chipboard = particle composite (1)

Laminate diagram showing layers (1)

Laminate cross-grains / glued together [label or description]

(1)

Particle composite diagram with labelled particles (1)

(4)

(f)

(i)

(ii)

Hardening Processes

W.H. = beaten (repeatedly) [not just worked, accepthammered] (1)

Q.H. = heated and then cooled rapidly [or immerse in oil orwater] (1)

[max

if statements swapped or Q.H./W.H not mentioned]

Hardness and Brittleness only (1)

(2)

(1)

(g) Stress-strain graph

Strain = (2.5 x 108 / 2.0 x 1011) or correct gradient calculation(1)

Straight line to 1.25 x 10-3 and 2.5 x 108 Pa (1)

Plastic flow shown to strain of 3.0 x 10-3 (1) (3)

Total 32



32/120

Topic C - Nuclear and Particle Physics

QuestionNumber

Answer Mark

3 (a)

(i)

Particle classification

Electron, muon, tau-particle, any neutrino (1)

Proton, neutron, etc. (1)

Lepton (1)

Exchange particle [accept gauge boson or intermediate vectorboson] (1) (4)

(ii) Fundamental particles

Fundamental cannot be split into smaller particles AND Composite composed of smaller particles / can be sub-divided

(1)(1)

(iii) Charm, strange, top, bottom (quarks), muon, tau-particle

Any five correct (1)

Any six correct [symbols only scores maximum ] (1) (2)

(iv) Same mass /both are leptons (1)

Opposite charge / +1 and -1 charge (1)

Annihilate to produce gamma (ray(s) / photon(s)) (1)(3)

(b)

(i)

Sigma decay conservation laws and quark charges

B: (+1) = (+1) + (0) (permitted) (1)

Q: (+1) = (+1) + (0) (permitted) [zeros must be shown, accept(1)] (1)

(2)

(ii)

Q (uus) = +1 = (+

) + (+

) + qs AND qs = -

(1) (1)

(iii) Fundamental interactions

Gravitational, weak, strong, electromagnetic [this order appliesbelow] (1)

Mass, all particles, quarks, charged particles any three correct(1)

Mass, all particles, quarks, charged particles all four correct(1)

All four possible [ignore gravitational force] OR weak only (due

to flavour change) (1)(4)



33/120

(iv)

(v)

Naming Particles

Any of W+, W, Z0 (1)

X = proton [or +] (1)

(1)

(1)

(c)

(i)

Nuclear density

use of = mV (1)

Substitution: 4/3 x (5.84 x 10-15 m)3 x 2.29 x 1017 kg m-3 (1)

1.91 x 10-25 kg (1) (3)

(ii) Nucleon number

1.91 x 10-25 1.66 x 10-27 (e.c.f.) (1)

[OR multiply by 6.02 x 1023 OR r = r0 A1/3 route)

(1)

(d)

(i)

Beta minus decay equation_1

0 n p + + 1

1

0

1

n (1)1

0

p AND (1)1

1

0

1

(2)

(ii)

Beta energy spectrum

Correct shape [rise to peak, fall to concave maximum on x-axis,accept correction for nuclear attraction through origin] (1)

Number of beta (minus) / particles (1)

Kinetic energy (of beta particles) (1)(3)

(iii)

Antineutrino explanation


Constant energy change per decay expected / conservation ofenergy (1)

All beta should have same (kinetic) energy / beta have a rangeof energy (1)

Missing energy carried away by undetected particle (1) (4)

Total 32



34/120

Topic D - Medical Physics

QuestionNumber

Answer Mark

4(a) Uses of different sources

(X-ray) imaging / diagnosis / radiography (1)

Ultrasound [ignore kHz, MHz, do not accept keV, MeV] (1)

MeV X-rays [accept megavoltage X-rays, Cobalt-60, I-131, highenergy gamma rays] (1)

(Radiation) therapy / destruction of tissue (1)(4)

4(b)

(i)

Technetium decay and symbol

Mo m Tc + / e (1)9942

9943

01

(1)

(ii)

(iii)

Metastable (1)

Decays by gamma emission (to Tc ) (1)99

43

(0,) 1, 2, 3 days / (0,) 24, 48, 72 hours (1)

(2)

(1)

(iv)

(v)

Elution graph and process

Vertical line corresponds to an elution of the cell (1)

Removes radioactive material / Tc from the cell (hence lowersactivity) (1)


Flush [or similar description] with saline solution (1)

Tc dissolves and is removed (1)

Insoluble Mo remains in cell (1)

(2)

(4)

4(c)

(i)

(ii)

Anti-scatter grid

Lead (1)

Rays start from source and are straight lines to patient (1)

Any ray passing directly between grid, to film [AND noreflection from grid] (1)

Any ray being scattered within body and then absorbed by grid(1)

(1)

(3)



35/120

4(d)

(i)

Acoustic impedance

(specific) acoustic impedance (1)

of two materials / media (1)

(2)

(ii) [] = kg m-3 or [c] = m s-1 (1)

kg m-3 x m s-1 (1) (2)

(iii) Reflection coefficient calculation

Correct substitution: [(1.70 - 1.38) / (1.70 + 1.38)]2 (1)

1.1 x 10-2 / 1.1% [no unit, accept 0.0108, 0.01] (1) (2)

(iv)

(v)

Transmitted percentage

1 / 100% - % (1)

98.9% [accept 99%, 98.92%, e.c.f.] (1)

[range 1 2] x 106 kg m-2 s-1 (1)

(2)

(1)

(e)

(i)

(ii)

X-ray image energy and explanation

65 keV [or 1.04 x 10-14 J] (1)

White = bone, grey (darker) = tissue (1)

White (bone) = no X-rays reach film (to darken it) / Grey(tissue) = some X-rays reach film (1)

Absorption (strongly) dependent on proton number / Z (3) (1)

Hence greater attenuation for bone (20) than tissue (9) (1)

(1)

(4)

Total 32



36/120



37/120

6733/02 Practical Test PHY3

Group 1



38/120www.xtremepapers.netw w w .Xt r emePaper s.net


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Group 2



58/120



59/120



60/120



61/120

6734 Unit Test PHY4

QuestionNumber

Answer Mark

1.

(a)

(b)

(c)

(d)

Appropriate conditions

Velocity/speed (of wave) is constant / medium does not change(1)

[NOT velocity/speed of light/sound/any named wave OR medium must be avacuum]

(Meet) exactly/completely/directly out of phase / in antiphase / with aphase difference of [or any (2n+1)] or 180 / with a path differenceof [or any (n+)] / crest of one meets trough of the other

(1)[ignore reference to amplitude and any unlabelled diagrams]

Energy level separation/difference equals photon energy/hf(1)

E1 E2 = hfprovided E linked to energy levelEnergy difference between ground and an excited state equals photonenergy/hf

Electron/It has a greater momentum (than neutron)(1)

[NOT greater speedNOT greater speed or momentumBUT allow greater speed and momentum]

(4)

Total 4



62/120

QuestionNumber

Answer Mark

2. (a) (i) Formula and direction

F = mv

2

/r AND towards centre (of circular path)(1)[NOT just inwards]

(1)

(ii) Rewriting the formulaSubstitution of v = r into the correct equation to achieve required equation (1)[allow reverse show that back to correct equation using = v/r]OR stating F = ma with a = r2

(1)

(b) (i) Condition for forcemeter to read resultant force

No friction / no other horizontal force / board horizontal /forcemeter pulled horizontally/radially/parallel to trolley

(1)(1)

(ii) Calculation of force

Use of T = 28/5 OR f= 5/28 (1)

Use of = 2/T OR = 2f OR v =2r/T OR v =2rf(1)

Correct answer for F: 1.8 N(1)

e.g. T = 28/5 = 5.6 s = 2/(5.6 s) = 1.12 rad s-1F = 0.95 kg x 1.50 m x (1.12 rad s-1)2 = 1.79 N

[Note: 1.60 N means that not squared1.89 N means that mass has been omitted ... both score 1st 2 marks]

(3)



63/120

QuestionNumber

Answer Mark

2 (b) (iii) Experiment description

Any FIVE from:

1. QOWC (1)

2. Keep r (and m)constant (1)

3. (Measure) F and (n)T (NOT f) for various rotation speeds(1)

4. = 2/T (1)

5. Plot suitable graph e.g. F versus 2 / F versus 1/T2

(1)

OR suitable calculation e.g. F/2 / FT2

6. Hence graph is a straight line through origin(1)

OR hence calculation gives a constant value

7. Hence states that F is directly proportional to 2(1)

OR states correct expression for gradient/constante.g. F versus 2 gives mr

F versus 1/T2 gives 42mr

[Notes: points 5, 6 and 7 can be scored from a suitably labelled sketch graph ...origin assumed

points 6 and 7 can only be awarded if point 5 scored]

(5max)

Total 11



64/120

QuestionNumber

Answer Mark

3. (a) Points of maximum kinetic energy

X marked at both zero crossings (1)[Note: if Xs placed to one side then needs to be a clear indication that theyare referring to the zero-crossing points]

(1)

(b) Proof of formula

States or uses E = mv2

(1)

Substitute either vmax = 2fx0 AND f= 1/T OR vmax = 2x0/T to achieverequired equation(1)

[allow v = 2fx OR v = 2x/T]

(2)

(c) Determination of kinetic energy ratio

EITHERAttempt at substituting T = 2(M/k) in formula from part (b) for EOR uses T = 2(M/k) to show that TB = 2TA (1)

Hence shows that E = kxo2 OR states that E is independent of M

OR substitutes for TB into EB = 2TA2EA/TB

2(1)

Hence ratio = 1 (1)

ORInitial potential energy is the same, because same spring and amplitude

(1)

Hence maximum kinetic energy is the same since energy conserved(1)

Hence ratio = 1 (1)

(3)

Total 6



65/120

QuestionNumber

Answer Mark

4(a) Conditions for formula to be valid

Any TWO from:

Point source (1)

Source emits equally/uniformly/evenly in all directions(1)

No absorption [NOT just blocked] (1)

[Note: can score 2 marks for a single condition if two of the above pointsincluded]

(2max)

(b) Explanation of formula

Diagram showing sphere/circle with a labelled radius r around a source(1)

(Surface) area of sphere is 4r2

(1)

Intensity is power per unit area / Intensity = power/area / I = P/AOR Intensity is rate of flow of energy per unit area

(1)[NOT power per area]

[Note: no credit for simply explaining the effect of an inverse squarelaw]

(3)

(c) (i) Calculation of intensity

Use of Pythagoras (1)

Use of I = P/4r2 to get P [= 150 (W)] OR use of IQ/IP = (rP/rQ)2 (1)

Correct answer: 0.71 W m-2 (1)

e.g. Distance to Q = ((2.10 m)2 + (3.50 m)2)= 4.08 m

IQ = 2.7 W m-2 x (2.10 m / 4.08 m)2

= 0.715 W m-2

(3)

(ii) Why illumination will be more uniform

Light is reflected from walls/ceiling/another object in room(1)

[NOT from outside]

(1)

Total 9



66/120

QuestionNumber

Answer Mark

5(a) How diffraction leads to formation of pattern

Wave(front)s diffract from each slit and overlap(1)[scored either from text or from their diagram ... look for diffraction andoverlap only]

Superposition (or interference) occurs where waves overlap/meet(1)

[scored from text, including expanded labels on diagram]

Extra detail from text only: (1)slits act as coherent sourcesmaxima/bright fringes arise from constructive interference

minima/dark fringes arise from destructive interference

(3)

(b) Increasing the fringe width

EITHERi) Increase slits-to-screen distance (1)

ii) Fringes will be dimmer/less bright/less intense(1)

[ignore blurred/less clear/harder to observe]

ORi) Decrease slit separation (1)[NOT size]

ii) Slit separation will be harder to measure (precisely)(1)

ORi) an incorrect change e.g. reduce slit width

(0)

ii) correct disadvantage for their change e.g. fringes less bright (1)

(2)

(c) Problem of measuring fringe width

Hard to measure from/locate centre of fringe(1)

Measure across several fringes and divide by the number (1)[NOT measure several and find the average as this could mean severalindividual widths]

(2)



67/120

(d) Fringe pattern for narrower slits

Fringe spacing unchanged ... check peaks still 1 cm apart by eye(1)

Maximum intensity reduced ... check height of central maximum < 4 cm(1)

Intensity falls off more gradually away from centre ... check height oftheirouter maximum of their central maximum by eye

(1)

(3)

Total 10



68/120

QuestionNumber

Answer Mark

6(a) How the current arises

Radiation/light/photons ejects/releases (photo)electrons(1)[NOT just photoelectric emission]

Electrons cross/flow/travel/move to the anode(1)

[NOT accelerated to anode / attracted to anode / repelled by cathode]

(2)

(b) Calculation of photon energy

Use of E = hf (1)

Correct answer: 5.0 x 10-19

(J) [Minimum 2 sig fig] [no u.e.](1)

e.g.E = 6.63 x 10-34 J s x 7.5 x 1014 Hz

= 4.97 x 10-19 (J)

(2)

(c) Calculation of stopping potential

Use of eVs = Ephoton OR eVs = hf (1)

Correct answer 0.79 (V) [2 sig fig minimum] [no u.e.](1)

[Notes: 0.81 (V) if 5 x 10-19 J used for photon energyallow e.c.f. from part (b) for their photon energy value]

e.g.Vs = (4.97 x 10

-19 J 3.7 x 10-19 J)/(1.6 x 10-19 C)= 0.794 (V)

(2)

(d) Graph of current against reverse p.d.

Current decreases from point on I-axis(1)

[allow any shape of decrease BUT NOT major constant or increasing sections]

Current falls to zero at 0.79 V ... 1 small square(1)

[allow e.c.f. from part (c) for their stopping potential]

Current remains at zero after that point ... assume zero if nothing else shown (1)[accept currentgoing negative from that point if, and only if, they referto thenon-ideal behaviour of (some) real photocells]

(3)

Total 9



69/120

QuestionNumber

Answer Mark

7(a) Explanation of polarization and how it is used to represent the bit

i) (a transverse wave with) vibrations/oscillations/electric field/magneticfieldconfined to one plane

(1)[ignore unlabelled diagrams ... labels must include direction of vibration]

ii) vibrations in one plane/direction represent 0vibrations in another plane/direction represent 1

(1)[Accept any specific examplee.g. vertical vibrations represent 0, horizontal vibrations represent 1]

(2)

(b)(i) Type of radiation

Infrared / IR / ir (1) (1)

(b)(ii) Calculation of power

Use of E = hc/ OR Use of f= c/and E = hf(1)

Use of P = their E x 3 x 104 (1)

Correct answer: 3.8 x 10-15 W or J s1(1)

e.g. E = 6.63 x 10-34 J s x 3.00 x 108 m s-1 / (1.55 x 10-6 m)= 1.28 x 10-19 J

P = 1.28 x 10-19 J x 3 x 104 s-1= 3.85 x 10-15 W

(3)

Total 6



70/120

QuestionNumber

Answer Mark

8 (a) How recession velocities are measured

Compare light from galaxy with light from laboratory source/a source on Earth (1)

(Calculate v using) v/c = / where is identified as the change inwavelength/redshift (1)[allow in terms of frequency][Note: just stating equation insufficient for mark]

Extra detail: (1)Absorption spectrum from starSpectral lines comparedComparison between light from same element

(3)

(b) Arrows representing velocities

B: Arrow to right, length 2 cm [1 small square](1)

C: Arrow of correct length and direction (1)[Its components should be 1 cm to right, 1.5 cm up the page ... 1 smallsquare in each direction]

(2)

Total 5

Total for paper 60



71/120

6735/01 Unit Test PHY5

Question Answer Mark1 (a) Homogeneity [of this relationship; 0/2 for any other, e.g. g=GM/r2]

Correct units given (or reached) for all 4 of g, G,rand (1)

[Units of G may be quoted in base units here; g units N kg-1 or m s-2. Units donot need to be explicitly related to quantities; accept units without correlationand in any order.]

Convincing demonstration of homogeneity (consequent mark). (1)

[Most candidates will reduce RHS units to N kg-1 or m s-2 without furtherexplanation. Accept.]

(2)

(b) Value of

Substitution of g= 9.8(1) and r= 6.4 10n

(1)[accept g = 10, 5600]

Correct answer = 5500 (kg m-3) (No U.E.) (1)Example of answer:

3

62211

1

mkg5489

m104.6kgmN1067.64

kgN81.93

4

3

=

=

=Gr

g

(2)



72/120

(c) (i) g at sphere surface

Substitution of r= 0.15 and = 11300 (1)(or use of g = GM/r2 with r2 = 0.152 and M = 160kg)

Correct answer = 4.7(4) 10-7 (min 2 s.f.) (N kg -1) (1)[No U.E., bald correct answer, scores 2]

Example of answer:

17

3-2211

kgN1074.4

3

mkg11300m15.0kgmN1067.64

3

4

=

=

=Gr

g

(2)

(c) (ii) Force of attraction [Conclusion Student A automatically scores 0/2]

Force/(Weight) is 1560N -1600N (1)[Explicit statement of force value in this range]

Reason:

Newtons third law; referred to or applied [equality, if not opposite direction]ORDetailed reference to, or use of, F = GMm/r2, with M = ME and r = RE (1)

(2)

Total 8



73/120

QuestionNumber

Answer Mark

2 (a) (i)

(ii)

(iii)

Inverse square lawStatement ofprinciple; disallow if exponential used anywhere:

As separation [not radius] doubles/ is doubled/increases by factor n,force decreases to a quarter/ 1/n2 of its previous value, or similar reverseargument. OR F 1/r2 / F = k / r2 / F x r2 = k [Not F = 1/r2] (1)

Correct numericalwork, using values, taken from both A and B (ignore ^10-13):

(From A to B) rdoubles / rB:rA = 2:1 [factor 2 thus implicit], while force fallsfrom 3.6 0.9 (not from ~4 ~1). 0.9/3.6 = [factor (or 4 or 1/22)identified] [accept reverse argument B to A, with appropriate factors]

OR calculate value of k as 3.6 (x 10-26) at A and B. (1)

Coulombs law

2

21

4 r

QQF

o

= OR2

21

r

QkQF = quoted [symbols need not be defined.

Accept Qq, and d or x instead of r unexplained; any other letter must bedefined]

OR equivalent word equation (with ref. to k or 4o) [r not radius]

OR statement that force is directly proportional to product of charges andinversely proportional to the square of their separation [not radius] (1)

Force value

Use of F=2

0

21

4 r

QQ

, etc. (ecf a(ii) for first mark only, r= 4.7 10-14 m) (1)

[Allow r instead of r2.For the product Q1Q2, accept values arising from use ofe2, Z1Z2, A1A2, N1N2, Ze, Ae, etc.]

All values correctly substituted in correct equation (1)

Correct answer = 16 N (1)

Example of answer:

N3.16

m)10(4.7

C)101.6(2C)101.6(78CmN1099.8214-

19-19-2-29

2

21

=

=

=r

QkQF

N.B. Allow full credit for use of F 1/r2 principle leading to correct answer, i.e.taking reference force and separation values from graph, then using ratios:

e.g. 4.7 x 10-14/ rref = k (1)

(2)

(1)

(3)



74/120

F / Fref = 1 / k2 (1)

F = k2 x Fref = 16(.3) N (1)

(b) Repulsive force

Same sign charge/both positive/ same polarity/like charges repel (1)

[Not both charges are the same; not both atoms are positive.]

(1)

(c) Kinetic energy

Use of or reference to area of or under graph [even triangular]ORUse of any valid force distance using values attempted from graph[force value may be e.g. an average, extreme, median or difference] (1)

Value within broad range 1.6 - 2.3 10-13 (J) (1)[Do not penalise omission of unit for this 2nd mark]

Accurate answer in range 1.8 - 2.1 10-13 J (U.E.) (1)

[N.B.Use of FArA FBrB = 1.8 10-13 J scores zero]

[Some candidates will attempt an integration route. Credit this as analternative method for first mark, with same range criteria for subsequentmarks]

(3)

Total 10



75/120

QuestionNumber

Answer Mark

3 (a) (i)

(ii)

(iii)

ChargingTicks* for: To the left, moving right [3rd row] (1)

and To the right, moving left [6

th

row] (1)

[Lose one mark for each additional tick, to minimum zero. Inverse logic (just 2ticks in rows 1 and 4 only) loses first mark but gets second, ecf]

[*Accept unambiguous alternatives]

Minimum Emf

Use of 6 and 2.5 in V=Q/Cor its re-arrangement (1)[Beware confusion of Q with C]

2.4 (V) across capacitor (1)

2.4V (ecf their value, unless 0) + 0.2V = 2.6V (1)[2.4V as final answer scores 2]

Number of turnsSee 3/1.25 (10n) in the relevant re-arranged equation (1)[credit valid entry into equation in original form from data list]

240 [No unit. Apply U.E if one given] (1)

Example of answer:

240sWb101.25

V3.0

)()(

1-2-=

=

=

=

N

t

Nt

N

(2)

(3)

(2)

(b) Energy used

Use of2

21 CV=W or WV2 [e..g. see 5J] (1)

Correct answer / 0.75 / 75% / etc. [Accept 15/20] (1)[Answer of or equivalent scores 1 only]

Example of answer:

43

2015

22

12

21

22

12

21

Fraction

J15J5-J20W

J5)V0.2(F5.2

J20)V0.4(F5.2

==

==

===

===

ff

ii

CVW

CVW

(2)

Total 9



76/120

QuestionNumber

Answer Mark

4(a)(i) Rule

(Flemings) Left Hand OR (Flemings) Motor (Rule) [accept FLHR] (1) (1)

(ii) Polarity

X marked and / or Y marked + [accept anywhere on rail] (1)[accept alternative but not contradictory labelling of P and / or Q + ]

(1)

(b) Strength of field

Use of F= BIl. (Ignore 10x. Require l= 6 (cm) and F= 9 (mN)) (1)

See conversions; mN to N (unless answer in mT) and cm to m (1)

[Independent mark; l = 4cm 45mT or l = 10cm 18mT score this markonly to gain 1/3]

Correct answer 3.0 10-2 T / 30mT (1)[Accept alternative correct units, e.g. Wb m-2, N m-1 A-1, kg A-1 s-2]

(3)

(c) Graph

Straight line through origin to 9.0mN at 6.0cm (1)

Horizontal line at 9mN, starting at exactly 6.0cm, running to 10cm (canovershoot) (1)

[Max 1 mark if line from origin runs beyond 6.0cm, 9.0mN ]

[Line sections must be ruler-straight, accurate to +/- half square at originand given co-ordinates. If one of (6.0, 9.0) co-ordinates of knee is wrongbut behaviour otherwise correct, score 1 mark. If small curved region attransition between sections; score 1 mark]

(2)

Total 7



77/120

QuestionNumber

Answer Mark

5(a)(i) Single charge

Point to left of A, level with AB, labelled negative (1)[Allow negative charge placed at A; accept simply shownhorizontally to left of A]

(1)

(ii) Field shown as radial in form* with correct direction, minimum 4 lines,at least one arrow shown (ecf part (i); any position, either polarity) (1)(*ignore non-uniformity introduced to allow for presence of electron)

(1)

(b) (i) Path from C

Path shown fully to +Q, no spiral.

Path does not meet or cross given field lines (gauge by eye,ignore last few mm) (1)

Path remains equi-distant from field lines either side throughout itsjourney (gauge by eye, ignore last few mm where it can meet afield line, but not cross it) (1)

[If a path is shown starting from -Q, it must be marked with an arrow to theleft, and can score max 1.][If electron is shown rebounding from +Q, max 1 ] [both errors, score 0].

OR

Path shown as anticlockwise curve or spiral, or becoming orbit about +Q.

Path has slingshot or even slight anti-clockwise curve past +Q, spiral intowards +Q, or becomes orbit around +Q (1)

Path drifts towards and across outer field line (but not inner),having initially moved along a path equi-distant from both. (1)

[If a path is shown starting from -Q, it can score max 1.]

(2)

(b)(ii) Path from DSpeed increases [throughout] (1)

Acceleration decreases then increases (both needed) (1)[Do not accept slows down then speeds up. Decelerate is wrong]

(2)

Total 6

Total for paper 40



78/120


Group 1



95/120



102/120


111/120

Unit PHY6 6736/01

Question1

Answer Mark

(a)

equation 1: start with A = 209 (1

use of (1) on the leftn10

equation 2: end with Z= 84 (1)

use of / (1) on the righte0

1 0

1

(4)

(b)

graph goes through (0,100) and (135 5, 50 2 ) (1)

graph goes through (276,25) (1) smoothconcave curve not reaching the taxis (1)

(3)

c)(i) attempt to raise 0.5 / to the power of 365138 / 2.645 (1) 0.16 / 0.160 (no mark)

so A = 16 GBq (after one year) (1)

(2)

(c)(ii)

substituting = ln2/t at any stage (1)

ln = ln2 or ln(A0/A) =ln(A/A0) explicitly (1)

(2)

(d) mention of cloud or bubble chamber / GM tube / spark chamber

(1) Relevant absorbing medium named (1)

compare length of -tracks / compare thickness of absorbers tostop s / distance when detection stops (1)

(3)

(e)

use of t = ln2 with t in seconds (1)

= 5.81 108 s1 (no mark) use of A =N e.c.f. (1) A = 1.66 1014 Bq / s1 (no mark)

attempt to convert 5.41 MeV to J / correct use of 1.6 1019 J



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eV1 (1)

power = 144 / 143.9 (W) (1)

(4)

(f)

suitable: highly radioactive / high temperature / lightweight

source/ -source so all energy absorbed (1) produces high (initial) power per unit mass / volume

(1) limited: shortish half life (1)

power life limited (to a year at most) (1)

(4) (g)(i) four cell symbols in series with three groups of cells in parallel (1)

(1)

(g)(ii) (allow e.c.f. from incorrect arrangements) correct sum of Nin series as 0.30 N(1)

correct sum of Min parallel as 0.30 M(1)

(proper arrangement

0.4

)

(2)

(g)(iii)

lower resistance / larger current / still works if a tcouple fails (1)

(1)

(h)(i) qowc (1)

Thermocouples have a hot source and a cold sink (1) thermal / heat energy flows from hot to cold junction (1) mention of electrical work / electrical energy produced (1)

(4)

(h)(ii)

use of kelvin temperatures (here 353 K and 253 K) (1) efficiency = 0.28 / 28% (1)

(2)

Total

32



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Question2

Answer Mark

(a)

qowc (1) mention of ionising properties (of charged particles) (1) drops / bubbles (form on ions) (1) that can be illuminated / photographed (1) showing up the track of the charged particle (no mark) reference to saturated vapour or reduction in temperature /

superheated liquid or reduction in pressure (1)

(5)

(b)(i)

particle is moving from bottom to top of photograph / chamber (1) curvature of path changes (after passing through lead plate) (1)

(2)

(b)(ii)

using LHR / Fleming (with magnetic field down into plane of photo)

(1)

charge is positive, full e.c.f. from b(i) (1)

(2)

(b)(iii)

use of m2/r = Be (1)

momentum /m/p = Ber (accept if quoted) (1) substitution of 1.2 1020 N s and 1.5 T and 0.050 m (1) e= 1.6 1019 C (no mark)

(3)

(c) either T N A1 m1 (from F = BIL) / N C1 m1 s (from F = Bqv)

(1)

V J C1 (1)

J N m / A C s1 (1)

or use of E = N/t or VH = BLv (no mark)

E / VHhas unit V (1) A / L has unit m2 / m (1)

(3)



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t / vhas unit s / ms-1 (1)

Total 15



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Question3

Answer

Mark

(a)(i)

Recognition that at 16 N as x =40 cm / 0.40 m (1) use of Fxorusing spring constant F2/kor kx2 e.c.f. (1) 3.2 J (1)

(3)

(a)(ii)

values e.g. : x /m 0.10 0.20 0.30 ( F /N 4 8 12 ) E /J 0.2 0.8 1.8

at least one correct (non-zero) (not 16 N, 3.2 J) pair of values (1)

(1)

(a)(iii) axes of xand Elabelled (1)

with plotted values roughly correct (1) curve with increasing gradient / (1)

(3)

(b)

mention of potential difference / voltage (1)

applied across / to a capacitor (1) electrons / charges are moved / displaced (1)

the capacitor is charged (1)

max (3)

(c)

(each) spring extends by the same amount / (each) capacitor has

the same charge (implied by proof of total k = 3kand total C = C/3(1)

xis analogous to Q/Fis analogous to V (implied by F = kxanalogous to V = Q/C) (1) S diagram: springs share F/ C diagram: capacitors share V (may be labelling on diagrams showing newtons and / or volts) (1)

Both S and C store energy (not charge for C) (could be implied by equations for energy stored) (1)

(4)

(d)

discharge of a capacitor through a resistor / mention of R (e.g. RC)



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(1) analogous to radioactive decay / the decay of a radioactive source

(1)

(2)

Total 16



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Question4

Answer Mark

(a)(i)

an orbit that keeps a satellite above the same place / an orbit with atime period of 24 hours (1)

refer to the equator, e.g. on the equator / above the equator (1)

(2)

(a)(ii) use of m2/r = GmmE/r2(1)

2 = GmE/r (no mark)

(1)

(a)(iii) use of =2r/TOR v=r and = 2/T(1)

substitute for in2 = GmE/r(1)

r3 = GmET2 42 (accept any rearrangement.) (1)

(3)

(b)(i) use of c = f (no mark) =0.0146 m / 14.6 mm (1)

(1)

(b)(ii)

use of d = rwith in radians /d = rsinwith in degrees / d= rsinwithin degrees (1) d= 1000 km (all methods yield 1005 km) (1)

(2)

(c) reference to ripple tank / microwave apparatus / sound (1) use of a slit / double slit from 5 mm to 50 mm wide (1) diffraction is demonstrated by the spreading of waves (1) detection of waves in the shadow zone (1)

(all could be gained from a decent labelled diagram)

(4)

(d) reason 1: solar cells / panels are not 100 % efficient (1) not all the energy becomes electrical (1)

reason 2:solar panels do not always face / are not to Suns rays(1)

useful area less than 2.5m2 (1) reason 3: solar panels may receive no sunlight (1) when the satellite is in the Earths shadow (1) reason 4:less than 1.4kW m-2 is absorbed (1)

(4)



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some is reflected (1)

Total 17



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120/120

physics june 2009 ms

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