9702 s15 ms_all

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CAMBRIDGE INTERNATIONAL EXAMINATIONS

Cambridge International Advanced Subsidiary and Advanced Level

MARK SCHEME for the May/June 2015 series

9702 PHYSICS

9702/11 Paper 1 (Multiple Choice), maximum raw mark 40

Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2015 series for most Cambridge IGCSE

®, Cambridge International A and AS Level components and some

Cambridge O Level components.

Mark Scheme Syllabus Paper

Cambridge International AS/A Level – May/June 2015 9702 11

© Cambridge International Examinations 2015

Question Number

Key Question Number

Key

1 C 21 C

2 B 22 A

3 A 23 B

4 D 24 D

5 A 25 B

6 B 26 D

7 D 27 A

8 C 28 B

9 D 29 D

10 D 30 B

11 C 31 D

12 C 32 B

13 A 33 B

14 C 34 D

15 C 35 C

16 C 36 D

17 C 37 A

18 B 38 C

19 D 39 C

20 A 40 B





9702 PHYSICS








Question Number

Key Question Number

Key

1 C 21 A

2 D 22 B

3 C 23 C

4 C 24 C

5 D 25 B

6 C 26 C

7 D 27 B

8 D 28 B

9 C 29 B

10 C 30 B

11 D 31 C

12 C 32 A

13 A 33 A

14 B 34 D

15 C 35 A

16 D 36 B

17 D 37 A

18 D 38 A

19 A 39 C

20 B 40 C





9702 PHYSICS








Question Number

Key Question Number

Key

1 C 21 B

2 D 22 A

3 D 23 C

4 B 24 D

5 B 25 C

6 C 26 D

7 D 27 D

8 D 28 D

9 B 29 C

10 C 30 B

11 C 31 D

12 B 32 D

13 C 33 C

14 A 34 C

15 A 35 A

16 A 36 A

17 A 37 C

18 C 38 D

19 B 39 A

20 B 40 C





9702 PHYSICS

9702/21 Paper 2 (AS Structured Questions), maximum raw mark 60

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2015 series for most Cambridge IGCSE






1 (a) power = work / time or energy / time or (force × distance) / time B1

= kg m s–2

× m s–1

= kg m2

s–3

A1 [2]

(b) power = VI [or V2 / R and V = IR or I

2R and V = IR] B1

(units of V:) kg m2

s–3

A–1

B1 [2]

2 (a) speed = distance / time and velocity = displacement / time B1

speed is a scalar as distance has no direction and

velocity is a vector as displacement has direction B1 [2]

(b) (i) constant acceleration or linear/uniform increase in velocity until 1.1 s B1

rebounds or bounces or changes direction B1

decelerates to zero velocity at the same acceleration as initial value B1 [3]

(ii) a = (v – u) / t or use of gradient implied C1

= (8.8 + 8.8) / 1.8 or appropriate values from line or = (8.6 + 8.6) / 1.8 B1

= 9.8 (9.78) m s–2

or = 9.6 m s–2

A1 [3]

(iii) 1. distance = first area above graph + second area below graph C1

= (1.1 × 10.8) / 2 + (0.9 × 8.8) / 2 (= 5.94 + 3.96) C1

= 9.9 m A1 [3]

2. displacement = first area above graph – second area below graph C1

= (1.1 × 10.8) / 2 – (0.9 × 8.8) / 2

= 2.0 (1.98) m A1 [2]

(iv) correct shape with straight lines and all lines above the time axis or all below M1

correct times for zero speeds (0.0, 1.15 s, 2.1 s) and peak speeds

(10.8 m s–1

at 1.1 s and 8.8 m s–1

at 1.2 s and 3.0 s) A1 [2]

3 (a) 4.5 × 50 – 2.8 × M ( = ...) C1

(...) = –1.8 × 50 + 1.4 × M C1

(M = ) 75 g A1 [3]




(b) total initial kinetic energy/KE not equal to the total final kinetic energy/KE

or relative speed of approach is not equal to relative speed of separation

so not elastic or is inelastic B1 [1]

(c) force on X is equal and opposite to force on Y (Newton III) M1

force equals/is proportional to rate of change of momentum (Newton II) M1

time of collision same for both balls hence change in momentum is the same A1 [3]

4 (a) (i) two sets of co-ordinates taken to determine a constant value (F / x) M1

F / x constant hence obeys Hooke’s law A1 [2]

or gradient calculated and one point on line used (M1)

to show no intercept hence obeys Hooke’s law (A1)

(ii) gradient or one point on line used e.g. 4.5 / 1.8 × 10–2

C1

(k =) 250 N m–1

A1 [2]

(iii) work done or EP = area under graph or ½Fx or ½kx2

C1

= 0.5 × 4.5 × 1.8 × 10–2

or 0.5 × 250 × (1.8 × 10–2

)2 C1

= 0.041 (0.0405) J A1 [3]

(b) KE = ½mv2

½mv2 = 0.0405 or KE = 0.0405 (J) C1

(v = [2 × 0.0405 / 1.7]1/2

=) 0.22 (0.218) m s–1

A1 [2] 5 (a) very high/infinite resistance for negative voltages up to about 0.4 V B1

resistance decreases from 0.4 V B1 [2]

(b) initial straight line from (0,0) into curve with decreasing gradient but not to

horizontal M1

repeated in negative quadrant A1 [2]

(c) (i) R = 122

/ 36 = 4.0 Ω A1

or

I = P / V = 36 / 12 = 3.0 A and R = 12 / 3.0 = 4.0 Ω (A1) [1]




(ii) lost volts = 0.5 × 2.8 = 1.4 (V) or E = 12 = 2.8 × (R + r) C1

R = V / I = (12 – 1.4) / 2.8 or (R + r) = 4.29 Ω C1

= 3.8 (3.79) Ω or R = 3.8 Ω A1 [3]

(d) resistance of the lamp increases with increase of V or I B1 [1]

6 (a) diffraction is the spreading of a wave as it passes through a slit or past an edge B1

when two (or more) waves superpose/meet/overlap M1

resultant displacement is the sum of the displacement of each wave A1 [3]

(b) nλ = d sin θ and v = fλ C1

max order number for θ = 90°

hence n (= f / vN) = 7.06 × 1014

/ (3 × 108 × 650 × 10

3) M1

n = 3.6

hence number of orders = 3 A1 [3]

(c) greater wavelength so fewer orders seen A1 [1]

7 (a) a region/space/area where a (stationary) charge experiences an (electric) force B1 [1]

(b) (i) at least four parallel equally spaced straight lines perpendicular to plates B1

consistent direction of an arrow on line(s) from left to right B1 [2]

(ii) electric field strength E = V / d C1

E = (450 / 16 × 10–3

)

= 28 × 103 (28 125) V m

–1 A1 [2]

(iii) W = Eqd or Vq C1

q = 3.2 × 10–19

(C) C1

W = 28 125 × 3.2 × 10–19

× 16 × 10–3

or 450 × 3.2 × 10–19

= 1.4(4) × 10–16

J A1 [3]

(iv) 19

19

101.6450

103.2450ratio

−

−

×−×

××

= (evidence of working required)

= (–) 2 A1 [1]





9702 PHYSICS








1 (a) (work =) force × distance or force × displacement or (W =) F × d M1

units of work: kg m s–2

× m = kg m2

s–2

A1 [2]

(b) charge

forms) other to electrical (from ed)(transformenergy or (done) work ) (p.d. = B1 [1]

(c) R = V / I B1

units of V: kg m2

s–2

/ A s and units of I: A C1 or

R = P / I2 [or P = VI and V = IR] (B1)

units of P: kg m2

s–3

and units of I: A (C1)

or R = V2

/ P (B1) units of V: kg m

2 s

–2 / A s and units of P: kg m

2 s

–3 (C1)

units of R: (kg m

2 s

–2 / A

2 s =) kg m

2 s

–3 A

–2 A1 [3]

2 (a) speed decreases/stone decelerates to rest/zero at 1.25 s B1

speed then increases/stone accelerates (in opposite direction) B1 [2]

(b) (i) v = u + at (or s = ut + ½at2 and v2

= u2 + 2as) C1

= 0 + (3.00 – 1.25) × 9.81 C1

= 17.2 (17.17) m s–1

A1 [3]

(ii) s = ut + ½at2

s = ½ × 9.81 × (1.25)2 [= 7.66] C1

s = ½ × 9.81 × (1.75)2 [= 15.02] C1

(distance = 7.66 + 15.02)

[v = u + at = 0 + 9.81 × (2.50 – 1.25) = 12.26 m s–1

] or

s = ½ × 9.81 × (1.25)2 [= 7.66] (C1)

s = 12.26 × 0.50 + ½ × 9.81 × (3.00 – 2.50)2 [= 7.36] (C1)

(distance = 2 × 7.66 + 7.36)

Example alternative method:

s = (v2 – u2

) / 2a = (12.262 – 0) / 2 × 9.81 [= 7.66] (C1)

s = (v2 – u2

) / 2a = (17.172 – 12.26

2) / 2 × 9.81 [= 7.36] (C1)

(distance = 2 × 7.66 + 7.36)




22.7 (22.69 or 23) m A1 [3]

(iii) (s = 15.02 – 7.66 =) 7.4 (7.36) m (ignore sign in answer) A1

down A1 [2] (c) straight line from positive value of v to t axis M1

same straight line crosses t axis at t = 1.25 s A1

same straight line continues with same gradient to t = 3.0 s A1 [3]

3 (a) (i) (vertical component = 44 sin 30° =) 22 N A1 [1]

(ii) (horizontal component = 44 cos 30° =) 38(.1) N A1 [1]

(b) W × 0.64 = 22 × 1.60 C1

(W =) 55 N A1 [2]

(c) F has a horizontal component (not balanced by W)

or F has 38 N acting horizontally

or 38 N acts on wall or vertical component of F does not balance W

or F and W do not make a closed triangle of forces B1 [1]

(d) line from P in direction towards point on wire vertically above W and direction up B1 [1]

4 (a) (p =) mv C1

∆p (= – 6.64 × 10–27

× 1250 – 6.64 × 10–27

× 1250) = 1.66 × 10–23

N s A1 [2]

(b) (i) molecule collides with wall/container and there is a change in momentum B1

change in momentum / time is force or ∆p = Ft B1

many/all/sum of molecular collisions over surface/area of container produces

pressure B1 [3]

(ii) more collisions per unit time so greater pressure B1 [1]

5 (a) curved line showing decreasing gradient with temperature rise M1

smooth line not touching temperature axis, not horizontal or vertical anywhere A1 [2]

(b) (i) (no energy lost in battery because) no/negligible internal resistance B1 [1]




(ii) I = V / R

= 8 / 15 × 103 or 1.6 / 3.0 × 10

3 or 2.4 / 4.5 × 10

3 or 12 / 22.5 × 10

3 C1

= 0.53 × 10–3

A A1 [2]

(iii) p.d. across X = 12 – 8.0 – 3.0 × 103 × 0.53 × 10

–3 (= 2.4 V) C1

RX = 2.4 / (0.53 × 10–3

) C1

or

Rtot = 12 / 0.53 × 10–3

(= 22.5 × 103 Ω) (C1)

RX = (22.5 – 15.0 – 3.0) × 103

(C1)

4.5(2) × 103 Ω A1 [3]

(iv) resistance decreases hence current (in circuit) is greater M1

p.d. across X and Y is greater hence p.d across Z decreases A1

or explanation in terms of potential divider:

RZ decreases so RZ / (RX + RY + RZ) is less (M1) therefore p.d. across Z decreases (A1) [2]

6 (a) progressive waves transfer/propagate energy and stationary waves do not B1

amplitude constant for progressive wave and varies (from max/antinode to

min/zero/node) for stationary wave B1

adjacent particles in phase for stationary wave and out of phase for progressive

wave (B1) [2]

(b) (i) wave / microwave from source/S reflects at reflector/R B1

reflected and (further) incident waves overlap/meet/superpose B1

waves have same frequency/wavelength/period and speed (so stationary

waves formed) B1 [3]

(ii) detector/D is moved between reflector/R and source/S (or v.v.) B1

maximum, minimum/zero, (maximum… etc.) observed on

meter/deflections/readings/measurements/recordings B1 [2]

(iii) determine/measure the distance between adjacent minima/nodes or

maxima/antinodes or across specific number of nodes/antinodes B1

wavelength is twice distance between adjacent nodes/minima or maxima/

antinodes (or other correct method of calculation of wavelength from

measurement) B1 [2]




(c) v = fλ C1

f = 3.0 × 108

/ (2.8 × 10–2

) [= 1.07 × 1010

Hz] C1

11 (10.7) GHz A1 [3]

7 (a) 92 protons and 143 neutrons B1 [1]

(b)

value

a 1

b 0 (a and b both required) B1

c 141 B1

d 55 B1 [3]

(c) kinetic energy (of products) or gamma/γ (radiation or photon) B1 [1]

(d) (total) mass on left-hand side/reactants is greater than (total) mass on right-hand

side/products M1

difference in mass is (converted to) energy A1 [2]





9702 PHYSICS








1 (a) 150 or 1.5 × 102 Gm A1 [1]

(b) distance = 2 × (42.3 – 6.38) × 106 (= 7.184 × 10

7 m) C1

(time =) 7.184 × 107 / (3.0 × 10

8) = 0.24 (0.239) s A1 [2]

(c) units of pressure P: kg m s

–2 / m

2 = kg m

–1 s

–2 M1

units of density ρ: kg m–3

and speed v: m s–1 M1

simplification for units of C: C = v2 ρ / P units: (m

2 s

–2 kg m

–3) / kg m

–1 s

–2

and cancelling to give no units for C A1 [3]

(d) energy and power (both underlined and no others) A1 [1]

(e) (i) vector triangle of correct orientation M1

three arrows for the velocities in the correct directions A1 [2]

(ii) length measured from scale diagram 5.2 ± 0.2 cm or components of

boat speed determined parallel and perpendicular to river flow C1

velocity = 2.6 m s–1

(allow ± 0.1 m s–1

)

A1 [2]

2 (a) constant rate of increase in velocity/acceleration from t = 0 to t = 8 s B1

constant deceleration from t = 8 s to t = 16 s or constant rate of increase in

velocity in the opposite direction from t = 10 s to t = 16 s B1 [2] (b) (i) area under lines to 10 s C1

(displacement =) (5.0 × 8.0) / 2 + (5.0 × 2.0) / 2 = 25 m

or ½ (10.0 × 5.0) = 25 m A1 [2]

(ii) a = (v – u) / t or gradient of line C1

= (–15.0 –5.0) / 8.0

= (–) 2.5 m s–2

A1 [2]

(iii) KE = ½ m v2 C1

= 0.5 × 0.4 × (15.0)2 = 45 J A1 [2]

(c) (distance =) 25 (m) (= ut + ½ at

2) = 0 + ½ × 2.5 × t

2 C1

(t = 4.5 (4.47) s therefore) time to return = 14.5 s A1 [2]




3 (a) (power =) work done / time (taken) or rate of work done A1 [1]

(b) (i) F – R = ma C1

F = 1500 × 0.82 + 1200 C1 = 2400 (2430) N A1 [3]

(ii) P = Fv C1

= (2430 × 22) = 53 000 (53 500) W A1 [2]

(c) (there is maximum power from car and) resistive force = force produced by

car hence no acceleration

or suggestion in terms of power produced by car and power

wasted to overcome resistive force B1 [1]

4 (a) (i) diameter and extension: micrometer (screw gauge) or digital calipers B1

length: tape measure or metre rule B1 load: spring balance or Newton meter B1 [3]

(ii) to reduce the effect of random errors or to plot a graph to check for zero

error in measurement of extension or to see if limit of proportionality is

exceeded B1 [1]

(b) plot a graph of F against e and determine the gradient B1

E = (gradient × l) / [πd

2 / 4] B1 [2]

5 (a) R = ρl / A C1

= (5.1 × 10−7

× 0.50) / π(0.18 × 10−3

)2 = 2.5 (2.51) Ω M1 [2]

(b) (i) resistance of CD = 8 × resistance of AB = 20 (Ω) C1

circuit resistance = [1 / 5.0 + 1 / 20]−1

= 4.0 (Ω) C1

current = V / R = 6.0 / 4.0 C1 = 1.5 A A1 [4]

(ii) power in AB = I

2R or power = V

2 / R C1

= (1.2)2 × 2.5 = 3.6 W = (3.0)

2 / 2.5 = 3.6 W A1 [2]




(iii) potential drop A to M = 1.25 × 1.2 = 1.5 V M1

potential drop C to N = 3.0 V

p.d. MN = 1.5 V A1 [2]

6 (a) (i) coherent: constant phase difference B1

interference is the (overlapping of waves and the) sum of/addition of

displacement of two waves B1 [2]

(ii) wavelength = 3.2 m (allow ± 0.05 m) M1

f (= v / λ = 240 / 3.2) = 75 Hz A1 [2]

(iii) 90° (allow ± 2°) or π/2 rad A1 [1]

(iv) sketch has amplitude 3.0 ± 0.1 cm M1

correct displacement values at previous peaks to produce correct shape A1 [2]

(b) (i) λ = ax / D C1

x = (546 × 10–9

× 0.85) / 0.13 × 10–3

(= 3.57 × 10–3

m) C1

AB = 8.9 (8.93) × 10–3

m A1 [3] (ii) shorter wavelength for blue light so separation is less B1 [1] 7 (a) (i) (rate of decay) not affected by any external factors or changes in

temperature and pressure etc. B1 [1] (ii) two protons and two neutrons B1 [1]

(b) (i) (total) mass before decay/on left-hand side is greater than (total) mass M1

on right-hand side/after the decay

the difference in mass is released as kinetic energy of the products A1 [2]

(may also be some γ radiation) (to conserve mass-energy)

(ii) (6.2 × 106 × 1.6 × 10

−19 =) 9.9(2) × 10

−13 J A1 [1]





9702 PHYSICS

9702/31 Paper 1 (Advanced Practical Skills 1), maximum raw mark 40







1 (a) (ii) Value of w with unit, in range 45.0 cm to 55.0 cm. [1]

(c) (iii) Value of IB, with unit, to nearest 0.1 mA, in range 70.0 ≤ IB ≤ 100.0 mA. [1]

(d) Six sets of readings of w, IA and IB, different values, scores 5 marks, five sets scores 4 marks, etc. [5] Incorrect trend –1. Major help from Supervisor –2. Minor help from Supervisor –1. Range: [1] Range of w ≥ 60.0 cm. Column headings: [1]

Each column heading must contain a quantity and a unit. The presentation of quantity and unit must conform to accepted scientific

convention e.g. w / cm, w (cm), (IA+IB) / (IAIB) / A–1, (IA+IB) / (IAIB) / (1/A).

Do not allow (IA+IB) / (IAIB) / (A / A2). Consistency: [1] All values of w must be given to the nearest mm only. Significant figures: [1]

Every value of (IA+IB) / (IAIB) must be given to the same number of s.f. as (or one

more than) the least s.f. in the corresponding values of IA and IB. Calculated values: [1]

Values of (IA+IB) / (IAIB) calculated correctly to the number of significant figures given by the candidate.

(e) (i) Axes: [1]

Sensible scales must be used. Awkward scales (e.g. 3:10) are not allowed. Scales must be chosen so that the plotted points occupy at least half the graph grid in both x and y directions. Scales must be labelled with the quantity that is being plotted. Scale markings should be no more than three large squares apart.

Plotting of points: [1]

All observations in the table must be plotted. Diameter of points must be ≤ half a small square (no “blobs”). Plotted points must be accurate to within half a small square.

Quality: [1]

All points in the table must be plotted on the grid (at least 5) for this mark to be awarded. All points must be within ± 5 cm (± 0.05 m) on the w-axis from a straight line.

(ii) Line of best fit: [1]

Judge by balance of all points on the grid about the candidate’s line (at least 5 points). There must be an even distribution of points either side of the line along the full length. Allow one anomalous point only if clearly indicated by the candidate. Lines must not be kinked or thicker than half a square.




(iii) Gradient: [1]

The hypotenuse of the triangle must be greater than half the length of the drawn line. The method of calculation must be correct. Both read-offs must be accurate to half a small square in both the x and y directions.

y-intercept: [1]

Either: Correct read-offs from a point on the line and substituted into y = mx + c. Read-off must be accurate to half a small square in both x and y directions. Or: Check read-off of the intercept directly from the graph (accurate to half a small square).

(f) M = value of the candidate’s gradient and N = value of the candidate’s y-intercept. [1]

Do not allow substitution methods. Do not allow fractions. Unit for M correct (e.g. A–1

m–1 or A–1 cm–1 or A–1

mm–1 or mA–1 m–1 or mA–1

cm–1 or mA–1

mm–1) and unit for N correct (e.g. mA–1 or A–1). [1]

2 (a) (i) Value of L with unit, in range 55.0 cm ≤ L ≤ 65.0 cm. [1] (ii) Value of m to nearest gram or better, in range 10.0 g ≤ m ≤ 100.0 g. [1] (iv) Correct justification of significant figures in p linked to significant figures in L and m. [1] (b) (i) Value of M to the nearest gram or better, in range 90.0 g ≤ M ≤ 110.0 g. [1] (iii) Correct calculation of C. [1] (c) (ii) Value of x to the nearest mm, with unit, in range 5.0 cm ≤ x ≤ 20.0 cm. [1] (iii) Absolute uncertainty in x in range 2 – 5 mm. If repeated readings have been taken, then the uncertainty can be half the range (but not zero) if the working is clearly shown. Correct method of calculation to obtain percentage uncertainty. [1] (d) Second value of L. [1]

Second value of x. [1]

Correct trend for x with respect to L (x decreases as L decreases). [1] (e) (i) Two values of k calculated correctly. [1] (ii) Valid comment consistent with calculated values of k, testing against a criterion

specified by the candidate. [1]




(f) (i) Limitations (4 max.) (ii) Improvements (4 max.) Do not credit

A

Two readings not enough to draw a valid conclusion.

Take more readings (for different L) and plot a graph/ take more readings and compare k values.

“repeat readings”/ “too few readings”

B Difficult to measure x with reason, e.g. parallax/ruler not in line with wood/strip moves as touched while taking measurement/mass obscures end of rule/strip oscillates/balance achieved for a short time

Improved method to measure x e.g. attach mass to bottom of strip/mark scale on strip/mark strip at balance point/measure (L–x)/clamp ruler horizontally

Travelling microscope

Video

C Difficult to balance with reason, e.g. wind/air conditioning or pivot moves

Method to remove wind, e.g. turn off fans/close windows or method of fixing pivot to bench i.e. tape/heavier pivot

Sliding rule

Pivot size

D Problem with Blu-Tack, e.g. mass of Blu-Tack not taken into account

Method to overcome problem with Blu-Tack, e.g. measure mass of Blu-Tack and add to value of M or fix mass with named adhesive, e.g. tape/glue because this has less mass

E Difficult to know where centre of mass is with reason, e.g. slot in mass or Difficult to place centre of mass at end of strip

Detailed method of finding centre of mass

Method to attach mass on strip to ensure centre of mass is at the end of strip, e.g. hang mass from strip with thread

Mark centre of mass

Measure diameter

F Two strips have different density/p

Find mass or p of second strip Different thickness/cross-sectional area





9702 PHYSICS








1 (c) (ii) Value of h in the range 45.0 to 55.0 cm. [1] (iii) Value of x less than 50.0 cm. [1] (d) Six sets of readings of x and h scores 5 marks, five sets scores 4 marks etc. [5] Incorrect trend –1. Help from Supervisor –1.

Range: [1] xmax – xmin ≥ 60.0 cm. Column headings: [1] Each column heading must contain a quantity and a unit where appropriate. The presentation of quantity and unit must conform to accepted scientific convention. e.g. 1 / h / cm–1. x / h must have no unit. Consistency: [1] All values of h and all values of x must be given to the nearest mm. Significant figures: [1] Every value of x / h must be given to the same number of s.f. (or one more than) the least number of s.f. in the corresponding values of x and h as recorded in table. Calculation: [1] Values of x / h calculated correctly.

(e) (i) Axes: [1]

Sensible scales must be used. Awkward scales (e.g. 3:10) are not allowed. Scales must be chosen so that the plotted points occupy at least half the graph grid in both x and y directions. Scales must be labelled with the quantity that is being plotted. Scale markings should be no more than three large squares apart.

Plotting: [1] All observations must be plotted on the grid. Diameter of plotted points must be ≤ half a small square (no “blobs”). Plotted points must be accurate to within half a small square in both x and y directions. Quality: [1] All points in the table must be plotted (at least 5) for this mark to be awarded.

Scatter of points must be no more than ± 0.1 from a straight line in the x / h direction.

(ii) Line of best fit: [1]

Judge by balance of all points on the grid about the candidate’s line (at least 5 points). There must be an even distribution of points either side of the line along the full length. Allow one anomalous point only if clearly indicated (i.e. circled or labelled) by the candidate. Lines must not be kinked or thicker than half a square.




(iii) Gradient: [1] The hypotenuse of the triangle must be greater than half the length of the drawn line. The method of calculation must be correct. Both read-offs must be accurate to half a small square in both the x and y directions. y-intercept: [1] Either: Correct read-offs from a point on the line and substituted into y = mx + c or an equivalent expression. Both read-offs accurate to half a small square in both the x and y directions. Or: Intercept read directly from the graph, with read-off at x = 0 accurate to half a small square in y direction.

(f) Value of a = candidate’s gradient and value of b = candidate’s intercept. [1]

Units for a and b both correct and consistent with values. [1]

2 (a) (ii) All values of D to nearest 0.1 cm and in range 2.0 cm to 4.0 cm. [1]

Evidence of repeat readings of D. [1] (iii) Absolute uncertainty in D in range 0.2 to 0.5 cm and correct method of calculation

to obtain percentage uncertainty. If repeated readings have been taken, then the absolute uncertainty can be half the range (but not zero) if the working is clearly shown. [1]

(iv) Correct calculation of C with consistent unit. [1] (b) Justification for significant figures in C linked to significant figures in D only. [1] (d) (ii) r1 in range 5.0 cm to 25.0 cm, with unit, to nearest mm. [1] (v) r2 in range 5.0 cm to 25.0 cm. [1] (e) Second value of D. [1]

Second values of r1 and r2. [1] Second value of |r1 – r2| > first value of |r1 – r2|. [1]

(f) (i) Two values of k calculated correctly. [1] (ii) Sensible comment relating to the calculated values of k, testing against a

criterion specified by the candidate. [1]




(g) (i) Limitations (4 max.) (ii) Improvements (4 max.) Do not credit

A Two readings are not enough to draw a valid conclusion.

Take more readings and plot a graph/ obtain more k values and compare

“repeat readings”/ “few readings”/ only one reading/ take more readings and (calculate) average k

B Difficult to measure D (or there is uncertainty in D or C) because loop is not circular/not flat/deforms

Workable method of making a more circular loop, e.g. wrap loop around tube

Use micrometer Use vernier calipers Material weak Material flexible

C Parallax error with pointer/ pointer moves away from scale/ pointer (or spring) vibrates

Use shadow method

D Ruler not vertical Use set square to ensure ruler vertical/clamp ruler

E Difficult to judge reading when loop breaks away/ loop breaks away suddenly

Video with scale/ use maximum marker

Slow motion camera High speed camera Difficult to determine point (or moment) loop breaks away

F Difficult to lower beaker steadily Use adjustable-height stand

G Reading affected by contact between loop and beaker/ impurities in water

Use larger diameter container/ wider container Use distilled water

Larger beaker





9702 PHYSICS








1 (b) (iv) Value of y in the range 10.0 cm to 11.0 cm with unit. [1] (c) (ii) Value of y > value in (b)(iv). [1] (d) Six sets of readings of m and y scores 5 marks, five sets scores 4 marks etc. [5] Help from Supervisor –1.

Range: [1] Range of m to include m = 0 g and m = 50 g or 60 g.

Column headings: [1] Each column heading must contain a quantity and a unit where appropriate. The unit must conform to accepted scientific convention e.g. y(C + m) / cm g.

Consistency: [1] All values of raw y must be given to the nearest mm. Significant figures: [1] Every value of value of y(C + m) must be given to the same number of s.f. as (or one

more than) the least s.f. in the corresponding values of y, C and m as stated in the candidate’s table and (a)(ii).

Calculation: [1] Values of y(C + m) calculated correctly to the number of s.f. given by the candidate.

(e) (i) Axes: [1] Sensible scales must be used. Awkward scales (e.g. 3:10) are not allowed. Scales must be chosen so that the plotted points occupy at least half the graph grid in both x and y directions. Scales must be labelled with the quantity that is being plotted. Scale markings should be no more than three large squares apart.

Plotting: [1] All observations must be plotted. Diameter of plotted points must be ≤ half a small square (no “blobs”).

Plotted points must be accurate to within half a small square.

Quality: [1] All points in the table (at least 5) must be plotted on the grid for this mark to be awarded. All points must be within ± 40 g cm of a straight line in the y(C + m) direction.

(ii) Line of best fit: [1] Judge by balance of all points on the grid about the candidate’s line (at least 5 points). There must be an even distribution of points either side of the line along the full length. Allow one anomalous point only if clearly indicated by the candidate. Lines must not be kinked or thicker than half a square.




(iii) Gradient: [1] The hypotenuse of the triangle must be greater than half the length of the drawn line. The method of calculation must be correct. Both read-offs must be accurate to half a small square in both the x and y directions.

y-intercept: [1]

Either: Correct read-offs from a point on the line and substituted into y = mx + c. Read-offs must be accurate to half a small square in both x and y directions. Or: Check read-off of the intercept directly from the graph (accurate to half a small square).

(f) Value of A = candidate’s gradient and value of B = 2 × candidate’s intercept / A. [1] Do not allow fractions or final answer to 1 s.f. Unit for A (m, cm or mm) and B (g or kg) correct and consistent with value, with correct power of ten. [1] 2 (a) (i) Raw values of d and D to nearest 0.1 mm and with consistent SI unit, in ranges: 10.0 mm ≤=d ≤ 25.0 mm 20.0 mm ≤ D ≤ 40.0 mm. [1] (ii) Value of h with consistent unit in range 40.0 mm ≤ h ≤ 60.0 mm. [1]

(iii) Percentage uncertainty in d based on absolute uncertainty of 0.1 or 0.2 mm. If repeated readings have been taken, then the uncertainty can be half the range (but not zero) if the working is clearly shown. Correct method of calculation to obtain percentage uncertainty. [1] (b) (iii) Correct calculation of x. Answer must be correct when rounded to 2 s.f. [1] (c) Correct justification of s.f. in x linked to s.f. in D, d and h. [1] (d) (ii) Value of average t ≥ 0.5 s with unit. [1] Evidence of repeated readings (here or in (e)). [1]

(e) Second value of x. [1] Second value of t. [1] Second value of t < first value of t. [1] (f) (i) Two values of k calculated correctly. [1] (ii) Sensible comment relating to the calculated values of k, testing against a criterion specified by the candidate. [1]




(g) (i) Limitations (4 max.) (ii) Improvements (4 max.) Do not credit

A Two readings not enough to draw a valid conclusion

Take many readings and plot a graph/ take more readings and compare k values.

“repeat readings”/ “too few readings”

B Difficulty in release of cylinder from same position every time with reason, e.g. placing fingers in water, level of water surface changing, difficult to judge start point

Better method of holding and releasing cylinder e.g. stop gate/ use mark to ensure the water level is the same for each release

Clamps Force on release

C Cylinder doesn’t always fall vertically (i.e. path at an angle or cylinder tilted)/ hits sides on descent

Method of attaching string symmetrically/ method of symmetrical distribution of mass e.g. use glass beads or sand/ modelling clay distributed evenly/ glue marbles in symmetrically

Ignore string effects Sand on its own Narrow cylinder

D Times short/ large uncertainty in time

Use longer tube/ video with timer (or video and view frame by frame)

Reaction time is short “too fast/quick” High speed camera Light gate(s) Slow motion camera Terminal velocity

E Difficulty in identifying end point with reason e.g. refraction, glass curvature, tray in the way, bottom of cylinder not flat

Method to identify end point e.g. time to a mark on cylinder/listening to impact

Flat bottomed cylinder Sensors

F Limited number of marbles to fit in container/ different diameter marbles/ bubbles or air in container/ container deforms when measuring D

Use different shapes e.g. cubes/smaller spheres to occupy more space/ use sand/modelling clay to fill more space/ measure and account for different diameter of marbles in equation for x

Sand on its own without explanation





9702 PHYSICS








1 (b) (i) Value of r in the range 28.0 cm to 32.0 cm, with unit. [1] (c) (ii) Value of T in range 2.0 s to 4.0 s. If out of range, allow Supervisor’s value ±20%. [1]

Evidence of repeat measurements for T. [1] (d) Six sets of readings of r and T scores 4 marks, five sets scores 3 marks etc. [4] Incorrect trend –1. Help from Supervisor –1.

Range: [1] rmax – rmin ≥ 30 cm. Column headings: [1] Each column heading must contain a quantity and a unit. The presentation of quantity and unit must conform to accepted scientific convention e.g. r2

/ m2. Consistency: [1] All values of r must be given to the nearest mm. Significant figures: [1] The number of significant figures for every value of T

3 must be the same as, or one more than, the number of significant figures in the corresponding time. Calculation: [1] Values of T

3 calculated correctly to the number of significant figures given by the candidate.

(e) (i) Axes: [1]

Sensible scales must be used. Awkward scales (e.g. 3:10) are not allowed. Scales must be chosen so that the plotted points occupy at least half the graph grid in both x and y directions. Scales must be labelled with the quantity that is being plotted. Scale markings should be no more than three large squares apart. Plotting: [1] All observations in the table must be plotted on the grid. Diameter of plotted points must be ≤ half a small square (no “blobs”). Plotted points must be accurate to within half a small square. Quality: [1] All points in the table must be plotted (at least 5) for this mark to be awarded.

All points must be within ± 2 s3 of a straight line in the T

3 direction. (ii) Line of best fit: [1]

Judge by balance of all points on the grid about the candidate’s line (at least 5 points). There must be an even distribution of points either side of the line along the full length. Allow one anomalous point only if clearly indicated (i.e. circled or labelled) by the candidate. Lines must not be kinked or thicker than half a square.




(iii) Gradient: [1] The hypotenuse of the triangle must be greater than half the length of the drawn line. The method of calculation must be correct. Both read-offs must be accurate to half a small square in both the x and y directions.

y-intercept: [1] Either: Correct read-offs from a point on the line substituted into y = mx + c or an equivalent expression. Read-offs must be accurate to half a small square in both x and y directions. Or: Intercept read directly from the graph, with read-off accurate to half a small square.

(f) Value of a = candidate’s gradient and value of b = candidate’s intercept. [1]

Units for a and b are correct (e.g. s3 m–2 for a and s3 for b). [1]

2 (a) (ii) Value for t in range 0.10 cm to 0.90 cm and given to nearest 0.01 cm. [1] Value for D in range 3.0 cm to 6.0 cm. [1] Value for h less than t. [1]

(b) Correct calculation of R. [1]

Value of R given to 2 or 3 significant figures. [1] (c) (ii) Value for f in range 13.0 cm to 17.0 cm or 28.0 to 32.0 cm. [1] (iii) Absolute uncertainty in f in range 0.2 cm to 0.5 cm and correct method of

calculation to obtain percentage uncertainty. If repeated readings have been taken, then the absolute uncertainty can be half the range (but not zero) if the working is clearly shown. [1]

(d) Second values for t, D and h. [1]

Second value for f. [1] (e) (i) Two values of k calculated correctly. [1]

Quality: Both k values in range 0.50 to 1.50. [1]

(ii) Sensible comment relating to the calculated values of k, testing against a

criterion specified by the candidate. [1]




(f) (i) Limitations (4 max.) (ii) Improvements (4 max.) Do not credit

A Two readings are not enough to draw a valid conclusion

Take more readings and plot a graph / obtain more k values and compare

“repeat readings”/ “few readings”/ only one reading/ take more readings and (calculate) average k

B Reason for difficulty in measuring t, h or D e.g. jaws of calipers slip off ends of lens/jaws too short and cannot reach centre of lens

Use a travelling microscope References to parallax

C h is small/large uncertainty in h Use micrometer/travelling microscope

D Difficult to obtain sharp image/hard to focus/blurred image

Use a dark(ened) room/ turn off lights/ use point/more compact source of light

E Difficult to measure f/take measurement with ruler/measure distance, with reason e.g. difficult to keep lens steady/screen not vertical/lens not vertical/ruler not perpendicular to lens or screen

Mount lens in holder/clamp/ fix lens to bench with e.g. Blu-Tack/ use optical bench

Flexible/bendy screens





9702 PHYSICS

9702/41 Paper 4 (A2 Structured Questions), maximum raw mark 100





Cambridge International AS / A Level – May / June 2015 9702 41


Section A 1 (a) (gravitational) force proportional to product of masses and inversely proportional

to square of separation M1 reference to either point masses or particles or ‘size’ much less than separation A1 [2]

(b) gravitational force provides/is the centripetal force B1

GMNm / r

2 = mrω2 (or mv

2 / r) M1

2π / T (or v = 2πr / T) leading to GMN = 4π2r

3 / T

2 A1 [3]

(c) MN / MU = (3.55 / 5.83)3 × (13.5 / 5.9)2 x3 factor correct C1 T2 factor correct C1 ratio = 1.18 (allow 1.2) A1

alternative method: mass of Neptune = 1.019 × 1026 kg (C1)

mass of Uranus = 8.621 × 1025 kg (C1)

ratio = 1.18 (A1) [3]

2 (a) (sum of) potential energy and kinetic energy of molecules/atoms/particles M1 mention of random motion/distribution A1 [2] (b) (i) pV = nRT

either at A, 1.2 × 105 × 4.0 × 10−3 = n × 8.31 × 290

or at B, 3.6 × 105 × 4.0 × 10−3 = n × 8.31 × 870 C1 n = 0.20 mol A1 [2]

(ii) 1.2 × 105 × 7.75 × 10–3 = 0.20 × 8.31 × T or T = (7.75 / 4.0) × 290 C1 T = 560 K A1 [2]

(Allow tolerance from graph: 7.7–7.8 × 10–3 m3)

(c) temperature changes/decreases so internal energy changes/decreases B1 volume changes (at constant pressure) so work is done B1 [2] 3 (a) (numerically equal to) quantity of (thermal) energy/heat to change state/phase of

unit mass M1 at constant temperature A1 [2]

(allow 1/2 for definition restricted to fusion or vaporisation)

(b) (i) at 70 W, mass s–1 = 0.26 g s–1 A1 at 110 W, mass s–1 = 0.38 g s–1 A1 [2]




(ii) 1. P + h = mL or substitution of one set of values C1 (110 – 70) = (0.38 – 0.26)L C1 L = 330 J g–1 A1 [3]

2. either 70 + h = 0.26 × 330

or 110 + h = 0.38 × 330 C1 h = 17 / 16 / 15 W A1 [2] 4 (a) (i) frequency at which object is made to vibrate/oscillate B1 [1] (ii) frequency at which object vibrates when free to do so B1 [1] (iii) maximum amplitude of vibration of oscillating body B1 when forced frequency equals natural frequency (of vibration) B1 [2] (b) e.g. vibration of quartz/piezoelectric crystal (what is vibrating) M1 either for accurate timing or maximise amplitude of ultrasound waves (why it is useful) A1 [2] (c) e.g. vibrating metal panels (what is vibrating) M1 either place strengthening struts across the panel or change shape/area of panel (how it is reduced) A1 [2] 5 (a) (magnitude of electric field strength is the potential gradient B1 use of gradient at x = 4.0 cm M1

gradient = 4.5 × 104 N C–1 (allow ± 0.3 × 10

4) A1 or

x

QV

04πε

= and 2

04 x

QE

πε

= leading to x

VE = (B1)

E = 1.8 × 103 / 0.04 (M1)

= 4.5 × 104 N C–1 (A1) [3]

(b) (i) 3.6 × 103 V A1 [1]

(ii) capacitance = Q / V C1

= (8.0 × 10–9) / (3.6 × 103)

= 2.2 × 10–12 F A1 [2] 6 (a) (i) gravitational B1 [1] (ii) gravitational and electric B1 [1] (iii) magnetic and one other field given B1 magnetic, graviational and electric B1 [2]




(b) (i) out of (plane of) paper/page (not “upwards”) B1 [1] (ii) B = mv / qr C1

= (3.32 × 10–26 × 7.6 × 104) / (1.6 × 10–19

× 6.1 × 10–2) C1 = 0.26 T A1 [3] (c) sketch: semicircle with diameter < 12.2 cm B1 [1] 7 (a) can change (output) voltage efficiently or to suit different consumers/appliances B1 by using transformers B1 [2] (b) for same power, current is smaller B1 less heating in cables/wires or thinner cables possible or less voltage loss in cables B1 [2]

8 (a) (i) p = h / λ

= (6.63 × 10–34) / (6.50 × 10–12) C1

= 1.02 × 10–22 N s A1 [2]

(ii) E = hc / λ or E = pc

= (6.63 × 10–34 × 3.00 × 108) / (6.50 × 10–12) C1

= 3.06 × 10–14 J A1 [2]

(b) (i) 0.34 × 10–12 = (6.63 × 10–34) / (9.11 × 10–31 × 3.0 × 108) × (1 – cos θ) C1

θ = 30.7° A1 [2] (ii) deflected electron has energy M1 this energy is derived from the incident photon A1

deflected photon has less energy, longer wavelength (so ∆λ always positive) B1 [3] 9 (a) nucleus/nuclei emits M1 spontaneously/randomly A1

α-particles, β-particles, γ-ray photons A1 [3] (b) (i) N – ∆N A1 [1] (ii) ∆N / ∆t A1 [1] (iii) ∆N / N A1 [1] (iv) ∆N / N∆t A1 [1] (c) graph: smooth curve in correct direction starting at (0,0) M1

n at 2t½ is 1.5 times that at t½ (± 2 mm) A1 [2]




Section B

10 (a) (i) (potential =) 1.2 / (1.2 + 4.2) × 4.5 = +1.0 V A1 [1] (ii) (for VIN > 1.0 V) V+ > V– B1 output (of op-amp) is +5 V or positive M1 diode conducts giving +5 V across R or Vout is +5 V A1

(for VIN < 1.0 V) output of op-amp –5 V / negative so diode does not conduct, giving Vout = 0 or 0 V across R A1 [4]

(b) (i) square wave with maximum value +5 V and minimum value 0 M1 vertical sides in correct positions and correct phase A1 [2] (ii) re-shaping (digital) signals/regenerator (amplifier) B1 [1] 11 (a) change/increase/decrease anode/tube voltage B1 electrons striking anode have changed (kinetic) energy/speed B1 X-ray/photons/beam have different wavelength/frequency B1 [3]

(b) (i) I = I0 e–µx B1 [1] (ii) contrast is difference in degree of blackening (of regions of the image) B1

µ (very) similar so similar absorption of radiation (for same thickness) so little contrast A1 [2]

12 (a) (i) loudspeaker/doorbell/telephone etc. B1 [1] (ii) television set/audio amplifier etc. B1 [1] (iii) satellite/satellite dish/mobile phone etc. B1 [1] (b) e.g. lower attenuation/fewer repeaters more secure less prone to noise/interference physically smaller/less weight lower cost greater bandwidth (any two sensible suggestions, 1 each) B2 [2]

(c) (i) ratio = 25 + (62 × 0.21) C1 = 38 dB A1 [2] (ii) ratio / dB = 10 lg (P2 / P1) C1

38 = 10 lg (P / 9.2 × 10–6)

P = 58 mW or 5.8 × 10–2 W A1 [2]

(allow 1/2 for missing 10 in equation)




13 (a) (i) to align nuclei/protons B1 to cause Larmor/precessional frequency to be in r.f. region B1 [2] (ii) Larmor/precessional frequency depends on (applied magnetic) field strength B1 knowing field strength enables (region of precessing) nuclei to be located M1 by knowing the frequency A1 [3]

(b) E = 2.82 × 10–26 × B

6.63 × 10–34 × 42 × 106 = 2.82 × 10–26 × B C1 B = 0.99 T A1 [2]





9702 PHYSICS








1 (a) (i) 1. F = Gm1m2 / x2

= (6.67 × 10–11 × 2.50 × 5.98 × 1024) / (6.37 × 106)2 M1 = 24.6 N (accept 2 s.f. or more) A1 [2]

2. F = mxω2 or F = mv

2 / x and v = ωx (accept x or r for distance) C1

= 2.50 × 6.37 × 106 × (2π / 24 × 3600)2

= 0.0842 N (accept 2 s.f. or more) A1 [2] (ii) reading = 24.575 – 0.0842 B1 = 24.5 N (accept only 3 s.f.) A1 [2] (b) gravitational force provides the centripetal force M1 gravitational force is ‘equal’ to the centripetal force

(accept Gm1m2 / x2 = mxω2 or FC = FG) M1 ‘weight’/sensation of weight/contact force/reaction force is difference between FG and FC which is zero A1 [3]

2 (a) mean speed = 1.44 × 103

m s–1 A1 [1] (b) evidence of summing of individual squared speeds C1 mean square speed = 2.09 × 106

m2 s–2 A1 [2]

(c) root-mean-square speed = 1.45 × 103

m s–1 A1 [1] (allow ECF from (b) but only if arithmetic error) 3 (a) (numerically equal to) quantity of heat/(thermal) energy to change state/phase of


(allow 1/2 for definition restricted to fusion or vaporisation) (b) (i) constant gradient/straight line (allow linear/constant slope) B1 [1] (ii) Pt = mL or power = gradient × L C1 use of gradient of graph (or two points separated by at least 3.5 minutes) M1 110 × 60 = L × (372 – 325) × 10–3

/ 7.0 L = 9.80 × 105

J kg–1 (accept 2 s.f.) (allow 9.8 to 9.9 rounded to 2 s.f.) A1 [3] (iii) some energy/heat is lost to the surroundings or vapour condenses on sides M1 so value is an overestimate A1 [2] 4 (a) displacement (directly) proportional to acceleration/force M1 either displacement and acceleration in opposite directions or acceleration (always) towards a (fixed) point A1 [2]




(b) (i) ⅓π rad or 1.05 rad (allow 60° if unit clear) A1 [1]

(ii) a0 = –ω2 x0

= (–) (2π / 1.2)2 × 0.030 C1 = (–) 0.82 m s–2 A1 [2] (special case: using oscillator P gives x0 = 1.7 cm and a0 = 0.47 m s–1 for 1/2)

(iii) max. energy ∝ x0

2 ratio = 3.02

/ 1.72 C1 = 3.1 (at least 2 s.f.) A1 [2]

(if has inverse ratio but has stated max. energy ∝ x0

2 then allow 1/2) (c) graph: straight line through (0,0) with negative gradient M1 correct end-points (–3.0, +0.82) and (+3.0, –0.82) A1 [2] 5 (a) work done bringing/moving per unit positive charge M1 from infinity (to the point) A1 [2] (b) (i) slope/gradient (of the line/graph/tangent) B1 [1]

(allow dV / dx, but not ∆V / ∆x or V / x) (allow potential gradient) (negative sign not required) (ii) maximum at surface of sphere A or at x = 0 (cm) B1 zero at x = 6 (cm) B1 then increases but in opposite direction B1 [3] (any mention of attraction max. 2/3) (c) (i) M shown between x = 5.5 cm and x = 6.5 cm B1 [1] (ii) 1. ∆V = (570 – 230) = 340 V (allow 330 V to 340 V) A1 [1]

2. q(∆)V = ½mv

2 or change/loss in PE = change/gain in KE or ∆EK = ∆EP B1 4.8 × 107 × 340 = ½v

2 C1 v2 = 3.26 × 1010 v = 1.8 × 105

m s–1 (not 1 s.f.) A1 [3] 6 (a) packet/quantum/discrete amount of energy M1 of electromagnetic energy/radiation/waves A1 [2] (b) (i) arrow below axis and pointing to right B1 [1]




(ii) 1. E = hc / λ = (6.63 × 10–34 × 3.0 × 108) / (6.80 × 10–12) C1 = 2.93 × 10–14

J (accept 2 s.f.) A1 [2] 2. energy of electron = (3.06 – 2.93) × 10–14 = 1.3 × 10–15

J C1

speed = )/(2 mE C1

= 5.4 × 107 m s–1 A1 [3]

(c) momentum is a vector quantity B1 either must consider momentum in two directions or direction changes so cannot just consider magnitude B1 [2] 7 (a) moving magnet gives rise to/causes/induces e.m.f./current in solenoid/coil B1 (induced current) creates field/flux in solenoid that opposes (motion of) magnet B1 work is done/energy is needed to move magnet (into solenoid) B1

(induced) current gives heating effect (in resistor) which comes from the work done B1 [4] (b) current in primary coil give rise to (magnetic) flux/field B1 (magnetic) flux / field (in core) is in phase with current (in primary coil) B1

(magnetic) flux threads/links/cuts secondary coil inducing e.m.f. in secondary coil B1 (there must be a mention of secondary coil) e.m.f. induced proportional to rate of change/cutting of flux/field so not in phase B1 [4] 8 (a) (i) energy = 5.75 × 1.6 × 10–13 = 9.2 × 10–13

J A1 [1] (ii) number = 1900 / (9.2 × 10–13 × 0.24) C1 = 8.6 × 1015

s–1 A1 [2] (b) (i) decay constant = 0.693 / (2.8 × 365 × 24 × 3600) C1 = 7.85 × 10–9

s–1 (allow 7.8 or 7.9 to 2 s.f.) A1 [2]

(ii) A = λN 8.6 × 1015 = 7.85 × 10–9 × N C1 N = 1.096 × 1024 C1 mass = (1.096 × 1024

× 236) / (6.02 × 1023) M1 = 430 g A1 [4] (c) 0.84 = 1.9 exp(–7.85 × 10–9

t) C1 t = 1.04 × 108

s = 3.3 years A1 [2]




Section B 9 (a) VB = 1000 mV C1 when strained, VA = 2000 × 121.5 / (121.5 +120.0) = 1006.2 mV M1 change = 6.2 mV (allow 6 mV) A1 [3] (b) (i) 1. resistor between VIN and V– and V+ connected to earth B1 resistor between V– and VOUT B1 [2] 2. P / + sign shown on earth side of voltmeter B1 [1] (ii) ratio of RF / RIN = 40 M1

RIN between 100 Ω and 10 kΩ A1 [2] (any values must link to the correct resistors on the diagram) 10 (a) product of density (of medium) and speed (of ultrasound) M1 in the medium A1 [2] (b) (i) 7.0 × 106 = 1.7 × 103 × speed C1 speed = 4.12 × 103

m s–1 wavelength = (4.12 × 103) / (9.0 × 105) m C1 = 4.6 mm (2 s.f. minimum) A1 [3]

(ii) for air/tissue boundary, IR / I ≈ 1 M1 for air/tissue boundary, (almost) complete reflection/no transmission A1

for gel/tissue boundary, IR / I = 0.12 / 3.12

= 1.04 × 10–3 (accept 1 s.f.) M1 gel enables (almost) complete transmission (into the tissue) A1 [4] 11 (a) (i) metal (allow specific example of a metal) B1 [1] (ii) e.g. provides ‘return’ for the signal shields inner core from interference/reduces cross-talk/reduces noise increased security (any two sensible suggestions, 1 each) B2 [2] (b) (i) (gradual) loss of power/intensity/amplitude B1 [1] (ii) dB is a log scale B1 either large (range of) numbers are easier to handle (on a log scale) or compounding attenuations/amplifications is easier B1 [2] (c) attenuation = 190 × 11 × 10–3 = 2.09 dB C1 –2.09 = 10 lg(POUT / PIN) C1 ratio = 0.62 A1 [3]




12 handset transmits (identification) signal to number of base stations B1 base stations transfers (signal) to cellular exchange B1 (idea of stations needed at least once in first two marking points) computer at cellular exchange selects base station with strongest signal B1 computer at cellular exchange selects a carrier frequency for mobile phone B1 [4] (idea of computer needed at least once in these two marking points)





9702 PHYSICS








Section A 1 (a) (gravitational) force proportional to product of masses and inversely proportional

to square of separation M1 reference to either point masses or particles or ‘size’ much less than separation A1 [2]

(b) gravitational force provides/is the centripetal force B1

GMNm / r

2 = mrω2 (or mv

2 / r) M1

2π / T (or v = 2πr / T) leading to GMN = 4π2r

3 / T

2 A1 [3]

(c) MN / MU = (3.55 / 5.83)3 × (13.5 / 5.9)2 x3 factor correct C1 T2 factor correct C1 ratio = 1.18 (allow 1.2) A1

alternative method: mass of Neptune = 1.019 × 1026 kg (C1)

mass of Uranus = 8.621 × 1025 kg (C1)

ratio = 1.18 (A1) [3]

2 (a) (sum of) potential energy and kinetic energy of molecules/atoms/particles M1 mention of random motion/distribution A1 [2] (b) (i) pV = nRT

either at A, 1.2 × 105 × 4.0 × 10−3 = n × 8.31 × 290

or at B, 3.6 × 105 × 4.0 × 10−3 = n × 8.31 × 870 C1 n = 0.20 mol A1 [2]

(ii) 1.2 × 105 × 7.75 × 10–3 = 0.20 × 8.31 × T or T = (7.75 / 4.0) × 290 C1 T = 560 K A1 [2]

(Allow tolerance from graph: 7.7–7.8 × 10–3 m3)

(c) temperature changes/decreases so internal energy changes/decreases B1 volume changes (at constant pressure) so work is done B1 [2] 3 (a) (numerically equal to) quantity of (thermal) energy/heat to change state/phase of


(allow 1/2 for definition restricted to fusion or vaporisation)

(b) (i) at 70 W, mass s–1 = 0.26 g s–1 A1 at 110 W, mass s–1 = 0.38 g s–1 A1 [2]




(ii) 1. P + h = mL or substitution of one set of values C1 (110 – 70) = (0.38 – 0.26)L C1 L = 330 J g–1 A1 [3]

2. either 70 + h = 0.26 × 330

or 110 + h = 0.38 × 330 C1 h = 17 / 16 / 15 W A1 [2] 4 (a) (i) frequency at which object is made to vibrate/oscillate B1 [1] (ii) frequency at which object vibrates when free to do so B1 [1] (iii) maximum amplitude of vibration of oscillating body B1 when forced frequency equals natural frequency (of vibration) B1 [2] (b) e.g. vibration of quartz/piezoelectric crystal (what is vibrating) M1 either for accurate timing or maximise amplitude of ultrasound waves (why it is useful) A1 [2] (c) e.g. vibrating metal panels (what is vibrating) M1 either place strengthening struts across the panel or change shape/area of panel (how it is reduced) A1 [2] 5 (a) (magnitude of electric field strength is the potential gradient B1 use of gradient at x = 4.0 cm M1

gradient = 4.5 × 104 N C–1 (allow ± 0.3 × 10

4) A1 or

x

QV

04πε

= and 2

04 x

QE

πε

= leading to x

VE = (B1)

E = 1.8 × 103 / 0.04 (M1)

= 4.5 × 104 N C–1 (A1) [3]

(b) (i) 3.6 × 103 V A1 [1]

(ii) capacitance = Q / V C1

= (8.0 × 10–9) / (3.6 × 103)

= 2.2 × 10–12 F A1 [2] 6 (a) (i) gravitational B1 [1] (ii) gravitational and electric B1 [1] (iii) magnetic and one other field given B1 magnetic, graviational and electric B1 [2]




(b) (i) out of (plane of) paper/page (not “upwards”) B1 [1] (ii) B = mv / qr C1

= (3.32 × 10–26 × 7.6 × 104) / (1.6 × 10–19

× 6.1 × 10–2) C1 = 0.26 T A1 [3] (c) sketch: semicircle with diameter < 12.2 cm B1 [1] 7 (a) can change (output) voltage efficiently or to suit different consumers/appliances B1 by using transformers B1 [2] (b) for same power, current is smaller B1 less heating in cables/wires or thinner cables possible or less voltage loss in cables B1 [2]

8 (a) (i) p = h / λ

= (6.63 × 10–34) / (6.50 × 10–12) C1

= 1.02 × 10–22 N s A1 [2]

(ii) E = hc / λ or E = pc

= (6.63 × 10–34 × 3.00 × 108) / (6.50 × 10–12) C1

= 3.06 × 10–14 J A1 [2]

(b) (i) 0.34 × 10–12 = (6.63 × 10–34) / (9.11 × 10–31 × 3.0 × 108) × (1 – cos θ) C1

θ = 30.7° A1 [2] (ii) deflected electron has energy M1 this energy is derived from the incident photon A1

deflected photon has less energy, longer wavelength (so ∆λ always positive) B1 [3] 9 (a) nucleus/nuclei emits M1 spontaneously/randomly A1

α-particles, β-particles, γ-ray photons A1 [3] (b) (i) N – ∆N A1 [1] (ii) ∆N / ∆t A1 [1] (iii) ∆N / N A1 [1] (iv) ∆N / N∆t A1 [1] (c) graph: smooth curve in correct direction starting at (0,0) M1

n at 2t½ is 1.5 times that at t½ (± 2 mm) A1 [2]




Section B

10 (a) (i) (potential =) 1.2 / (1.2 + 4.2) × 4.5 = +1.0 V A1 [1] (ii) (for VIN > 1.0 V) V+ > V– B1 output (of op-amp) is +5 V or positive M1 diode conducts giving +5 V across R or Vout is +5 V A1

(for VIN < 1.0 V) output of op-amp –5 V / negative so diode does not conduct, giving Vout = 0 or 0 V across R A1 [4]

(b) (i) square wave with maximum value +5 V and minimum value 0 M1 vertical sides in correct positions and correct phase A1 [2] (ii) re-shaping (digital) signals/regenerator (amplifier) B1 [1] 11 (a) change/increase/decrease anode/tube voltage B1 electrons striking anode have changed (kinetic) energy/speed B1 X-ray/photons/beam have different wavelength/frequency B1 [3]

(b) (i) I = I0 e–µx B1 [1] (ii) contrast is difference in degree of blackening (of regions of the image) B1

µ (very) similar so similar absorption of radiation (for same thickness) so little contrast A1 [2]

12 (a) (i) loudspeaker/doorbell/telephone etc. B1 [1] (ii) television set/audio amplifier etc. B1 [1] (iii) satellite/satellite dish/mobile phone etc. B1 [1] (b) e.g. lower attenuation/fewer repeaters more secure less prone to noise/interference physically smaller/less weight lower cost greater bandwidth (any two sensible suggestions, 1 each) B2 [2]

(c) (i) ratio = 25 + (62 × 0.21) C1 = 38 dB A1 [2] (ii) ratio / dB = 10 lg (P2 / P1) C1

38 = 10 lg (P / 9.2 × 10–6)

P = 58 mW or 5.8 × 10–2 W A1 [2]

(allow 1/2 for missing 10 in equation)




13 (a) (i) to align nuclei/protons B1 to cause Larmor/precessional frequency to be in r.f. region B1 [2] (ii) Larmor/precessional frequency depends on (applied magnetic) field strength B1 knowing field strength enables (region of precessing) nuclei to be located M1 by knowing the frequency A1 [3]

(b) E = 2.82 × 10–26 × B

6.63 × 10–34 × 42 × 106 = 2.82 × 10–26 × B C1 B = 0.99 T A1 [2]





9702 PHYSICS

9702/51 Paper 5 (Planning, Analysis and Evaluation), maximum raw mark 30







1 Planning (15 marks) Defining the problem (3 marks)

P V is the independent variable, or vary V and f is the dependent variable, or

measure f. Or f is the independent variable, or vary f and V is the dependent variable, or

measure V. [1]

P Change f (allow V) until the mass leaves/gap between plate. [1]

P Keep the position of the mass constant. (Do not allow keep mass constant.) [1]

Methods of data collection (5 marks)

M Labelled diagram showing signal generator/a.c. supply connected to vibrator with

two wires with mass on plate. At least two labels needed. [1]

M Voltmeter/c.r.o. connected in parallel with vibrator in a workable circuit. [1]

M Measure f or T from signal generator/c.r.o. (Allow detailed use of motion

sensor/stroboscope.) [1]

M Detail regarding mass leaving the plate: listen to noise, look for gap. [1]

M Repeat each experiment for the same value of V (allow f if consistent with above)

and average. [1]

Method of analysis (2 marks)

Plot a graph of:

A

f

2

against

1 / V

1 / V

against

f

2

f against

1 / V

1 / V

against

f

lg V

against

lg f

lg f against

lg V

or or or or

V

against

1 / f

2

1 / f

2

against

V

V

against

1 / f

1 / f against

V

[1]

A k =

2gradient π× gradient

2π

=kk =

22gradient π×

2

2

gradient

π=k

k = c102×π k = c22 10×π [1]

Safety considerations (1 mark)

S Precaution linked to mass leaving vibrating plate, e.g. use safety

screen/goggles/sand tray. [1]




Additional detail (4 marks)

D Relevant points might include [4]

1 Wait for vibrator to oscillate evenly

2 Method to determine period of oscillation from c.r.o., i.e. one time period × time-base

3 Method to determine f from c.r.o. having determined T, i.e. f = 1 / T

4 Method to determine V from c.r.o, i.e. amplitude (height) × y-gain

5 Relationship is valid if the graph is a straight line passing through the origin

[For lg – lg graph the gradient must be correct (–2 or –0.5)]

6 Determine f (allow V if consistent with above) by increasing and decreasing V or f 7 Clean surfaces of metal plate/small mass

8 Spirit level to keep plate horizontal/eye level to look for gap

Do not allow vague computer methods.




2 Analysis, conclusions and evaluation (15 marks)

Mark Expected Answer Additional Guidance

(a) A1 gradient = m

y-intercept = lg k

(b) T1

T2

1.70 or 1.699 1.312 or 1.3118

1.79 or 1.785 1.204 or 1.2041

1.85 or 1.851 1.114 or 1.1139

1.90 or 1.903 1.041 or 1.0414

1.95 or 1.954 0.98 or 0.978

2.00 or 1.996 0.90 or 0.903

Allow a mixture of significant figures.

T1 (first column) and T2 (second column)

must be values in table.

U1 From ±0.01 to ±0.03 Allow more than one significant figure.

(c) (i) G1 Six points plotted correctly Must be within half a small square.

Do not allow “blobs”.

Ecf allowed from table.

U2 Error bars in lg P plotted correctly All error bars to be plotted. Must be

accurate to less than half a small square.

(ii) G2 Line of best fit Upper end of line must pass between

(1.75, 1.24) and (1.75, 1.255) and lower

end of line must pass between (2.00, 0.900)

and (2.00, 0.915).

G3 Worst acceptable straight line.

Steepest or shallowest possible

line that passes through all the

error bars.

Line should be clearly labelled or dashed.

Examiner judgement on worst acceptable

line. Lines must cross. Mark scored only if

error bars are plotted.

(iii) C1 Gradient of line of best fit Must be negative. The triangle used should

be at least half the length of the drawn line.

Check the read-offs. Work to half a small

square. Do not penalise POT. (Should be

about –1.35.)

U3 Uncertainty in gradient Method of determining absolute uncertainty:

difference in worst gradient and gradient.

(iv) C2 y-intercept Check substitution into y = mx + c.

Allow ecf from (c)(iii). (Should be about 4.)

Do not allow read-off of false origin.




U4 Uncertainty in y-intercept Uses worst gradient and point on worst

acceptable line.

Do not check calculation. Do not allow if

false origin used.

(d) (i) C3 k = 10y-intercept

C4 m = gradient and given to 2 or 3 s.f.

and in the range –1.30 to –1.44

Must be negative.

Allow –1.3 or –1.4 (2 s.f.)

(ii) U5 Percentage uncertainty in k

Uncertainties in Question 2

(c) (iii) Gradient [U3]

uncertainty = gradient of line of best fit – gradient of worst acceptable line

uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)

(iv) [U4]

uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line

uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)

(d) (ii) [U5]

max k = 10max y-intercept

and min k = 10min y-intercept

percentage uncertainty = 100max

×

−

k

kk = 100

min×

−

k

k k =

( )100

minmax2

1

×

−

k

k k





9702 PHYSICS








1 Planning (15 marks)

Defining the problem (3 marks)

P t is the independent variable and I (or amplitude of reflected signal) is the dependent

variable, or vary t and measure I (or amplitude of reflected signal). [1]

P Keep distance from the wall/foam to the speaker/microphone constant. [1]

P Keep the amplitude or intensity I0 of the sound before reflection constant. [1]


M Labelled diagram of workable experiment including speaker, microphone/sound detector,

foam and wall. [1]

M Signal generator/a.c. power supply connected to speaker. [1]

M Microphone connected to oscilloscope or sound (intensity) meter. [1]

M Measure the thickness with a rule/micrometer/vernier calipers. [1]

M Method to determine the density; ρ = m / V. [1]


A Plot a graph of ln I against t.

(Allow log I against t and lg I against t graphs.) [1]

A α = – gradient / ρ (must be consistent with graph plotted) [1]


S Precaution linked to loud sounds, e.g. use ear plugs/muffs/defenders.

Allow switch off sound source to prevent damage to ears. [1]



1 Keep the frequency constant

2 Carry out experiment in a quiet room/no other sources of sound

3 Method to keep angles constant/positions of speaker and microphone constant.

4 Method and explanation to detect reflected sound from foam only, e.g. barrier, tube or

method to avoid reflections

5 Method to determine mass, e.g. use scales/balance and method to determine volume

6 Relationship is valid if the graph is a straight line (ignore reference to y-intercept)

7 Method to check that emitted sound I0 is constant or method to check y-intercept is ln I0.

8 Intensity is proportional to the amplitude2.






Expected Answer Additional Guidance

(a)

A1 gradient = d

Efε

(b) T1 X / 10–2

m2

T2

4.80 or 4.800

5.40 or 5.400

6.30 or 6.300

7.20 or 7.200

8.10 or 8.100

9.00 or 9.000


Must be table values.





U2 Error bars in X plotted

correctly

All error bars to be plotted. Must be accurate to

less than half a small square.

(ii) G2 Line of best fit Lower end of line must pass between (5.1, 5.0)

and (5.3, 5.0) and upper end of line must pass

between (8.5, 8.5) and (8.8, 8.5).


Steepest or shallowest

possible line that passes

through all the error bars.


Examiner judgement on worst acceptable line.

Lines must cross. Mark scored only if error bars

are plotted.

(iii) C1 Gradient of best fit line The triangle used should be at least half the

length of the drawn line. Check the read-offs.

Work to half a small square. Do not penalise

POT. (Should be about 1 × 10–4

.)



(d) (i) C2 ε = 6.25 × 10–7

× gradient Do not penalise POT.

(Should be about 6 or 7 × 10–11

.)

C3 F m–1

or C V–1

m–1

Allow A m–1

V–1

Hz–1

or A s m–1

V–1

or A2

s4

kg–1

m–3

.

Power of 10 must be correct.

(ii) U4 Percentage uncertainty in ε 10.83% + percentage uncertainty in gradient




(e) C4 f in the range 73.0 to 84.4 and

given to 2 or 3 s.f.

Allow 73 to 84 for 2 s.f.

f =

ε

9100.5

−

×

U5 Absolute uncertainty in f Clear working needed.

Allow ecf from (d)(ii).





(d) (ii) [U4]

max ε = fE

d

min min

max gradientmax

×

×

min ε = fE

d

max max

min gradient min

×

×

% uncertainty = 100gradient

gradient×

∆+

∆+

∆+

∆

E

E

f

f

d

d

= 10012.0

0.2

400

10

0.0030

0.0002

gradient

gradient×

+++

∆

(e) [U5]

max f = E X

d

minminmin

maxmax

××

×

ε

I

min f = E X

d

maxmaxmax

minmin

××

×

ε

I

∆f =

∆+

∆+

∆+

∆+

∆

ε

ε

E

E

d

d

l

l2

I

If =

∆++++

ε

ε

12.0

0.2

0.500

0.0012

0.0030

0.0002

5.0

0.1f =

∆+

ε

ε

0.107 f

∆f = f

+

100

10.7 (d)(ii)

+= correct is if

100

gradientiny uncertaint %21.5(d)(ii)f





9702 PHYSICS








1 Planning (15 marks) Defining the problem (3 marks)

P V is the independent variable, or vary V and f is the dependent variable, or

measure f. Or f is the independent variable, or vary f and V is the dependent variable, or

measure V. [1]

P Change f (allow V) until the mass leaves/gap between plate. [1]

P Keep the position of the mass constant. (Do not allow keep mass constant.) [1]


M Labelled diagram showing signal generator/a.c. supply connected to vibrator with

two wires with mass on plate. At least two labels needed. [1]

M Voltmeter/c.r.o. connected in parallel with vibrator in a workable circuit. [1]

M Measure f or T from signal generator/c.r.o. (Allow detailed use of motion

sensor/stroboscope.) [1]

M Detail regarding mass leaving the plate: listen to noise, look for gap. [1]

M Repeat each experiment for the same value of V (allow f if consistent with above)

and average. [1]


Plot a graph of:

A

f

2

against

1 / V

1 / V

against

f

2

f against

1 / V

1 / V

against

f

lg V

against

lg f

lg f against

lg V

or or or or

V

against

1 / f

2

1 / f

2

against

V

V

against

1 / f

1 / f against

V

[1]

A k =

2gradient π× gradient

2π

=kk =

22gradient π×

2

2

gradient

π=k

k = c102×π k = c22 10×π [1]


S Precaution linked to mass leaving vibrating plate, e.g. use safety

screen/goggles/sand tray. [1]






1 Wait for vibrator to oscillate evenly

2 Method to determine period of oscillation from c.r.o., i.e. one time period × time-base

3 Method to determine f from c.r.o. having determined T, i.e. f = 1 / T

4 Method to determine V from c.r.o, i.e. amplitude (height) × y-gain

5 Relationship is valid if the graph is a straight line passing through the origin

[For lg – lg graph the gradient must be correct (–2 or –0.5)]

6 Determine f (allow V if consistent with above) by increasing and decreasing V or f 7 Clean surfaces of metal plate/small mass

8 Spirit level to keep plate horizontal/eye level to look for gap






Mark Expected Answer Additional Guidance

(a) A1 gradient = m

y-intercept = lg k

(b) T1

T2

1.70 or 1.699 1.312 or 1.3118

1.79 or 1.785 1.204 or 1.2041

1.85 or 1.851 1.114 or 1.1139

1.90 or 1.903 1.041 or 1.0414

1.95 or 1.954 0.98 or 0.978

2.00 or 1.996 0.90 or 0.903


T1 (first column) and T2 (second column)

must be values in table.





U2 Error bars in lg P plotted correctly All error bars to be plotted. Must be

accurate to less than half a small square.

(ii) G2 Line of best fit Upper end of line must pass between

(1.75, 1.24) and (1.75, 1.255) and lower

end of line must pass between (2.00, 0.900)

and (2.00, 0.915).


Steepest or shallowest possible

line that passes through all the

error bars.


Examiner judgement on worst acceptable

line. Lines must cross. Mark scored only if

error bars are plotted.

(iii) C1 Gradient of line of best fit Must be negative. The triangle used should

be at least half the length of the drawn line.

Check the read-offs. Work to half a small

square. Do not penalise POT. (Should be

about –1.35.)



(iv) C2 y-intercept Check substitution into y = mx + c.

Allow ecf from (c)(iii). (Should be about 4.)

Do not allow read-off of false origin.




U4 Uncertainty in y-intercept Uses worst gradient and point on worst

acceptable line.

Do not check calculation. Do not allow if

false origin used.

(d) (i) C3 k = 10y-intercept

C4 m = gradient and given to 2 or 3 s.f.

and in the range –1.30 to –1.44

Must be negative.

Allow –1.3 or –1.4 (2 s.f.)

(ii) U5 Percentage uncertainty in k





(iv) [U4]

uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line

uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)

(d) (ii) [U5]

max k = 10max y-intercept

and min k = 10min y-intercept

percentage uncertainty = 100max

×

−

k

kk = 100

min×

−

k

k k =

( )100

minmax2

1

×

−

k

k k

9702 s15 ms_all

Education