9702 w11 ms_all

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS

GCE Advanced Subsidiary Level and GCE Advanced Level

MARK SCHEME for the October/November 2011 question paper

for the guidance of teachers

9702 PHYSICS

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

Mark schemes must be read in conjunction with the question papers and the report on the examination.

• Cambridge will not enter into discussions or correspondence in connection with these mark schemes.

Cambridge is publishing the mark schemes for the October/November 2011 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Mark Scheme: Teachers’ version Syllabus Paper

GCE AS/A LEVEL – October/November 2011 9702 11

© University of Cambridge International Examinations 2011

Question Number

Key Question Number

Key

1 D 21 C

2 B 22 C

3 C 23 B

4 B 24 B

5 B 25 C

6 A 26 B

7 D 27 A

8 B 28 D

9 B 29 A

10 C 30 C

11 A 31 A

12 A 32 C

13 A 33 A

14 C 34 D

15 C 35 A

16 C 36 B

17 A 37 D

18 B 38 B

19 C 39 B

20 A 40 D





9702 PHYSICS








Question Number

Key Question Number

Key

1 C 21 C

2 D 22 A

3 D 23 B

4 D 24 D

5 B 25 B

6 C 26 B

7 B 27 B

8 D 28 A

9 D 29 D

10 B 30 A

11 A 31 A

12 D 32 C

13 D 33 D

14 B 34 A

15 B 35 A

16 B 36 C

17 A 37 B

18 D 38 A

19 A 39 C

20 A 40 C





9702 PHYSICS








Question Number

Key Question Number

Key

1 C 21 B

2 D 22 B

3 B 23 C

4 B 24 C

5 B 25 B

6 D 26 C

7 A 27 D

8 B 28 A

9 B 29 C

10 A 30 A

11 C 31 C

12 A 32 A

13 C 33 D

14 A 34 A

15 C 35 A

16 A 36 D

17 C 37 B

18 C 38 B

19 A 39 D

20 B 40 B





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.


• Cambridge will not enter into discussions or correspondence in connection with these mark schemes. Cambridge is publishing the mark schemes for the October/November 2011 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.




1 (a) density = mass / volume B1 [1] (b) density of liquids and solids same order as spacing similar / to about 2× B1

density of gases much less as spacing much more or density of gases much lower hence spacing much more B1 [2]

(c) (i) density = 68 / [50 × 600 × 900 × 10–9] C1

= 2520 (allow 2500) kg m–3 A1 [2] (ii) P = F / A C1

= 68 × 9.81 / [50 × 600 × 10–6] C1 = 2.2 × 104 Pa A1 [3]

2 (a) torque is the product of one of the forces and the distance between forces M1

the perpendicular distance between the forces A1 [2] (b) (i) torque = 8 × 1.5 = 12 N m A1 [1] (ii) there is a resultant torque / sum of the moments is not zero M1

(the rod rotates) and is not in equilibrium A1 [2] (c) (i) B × 1.2 = 2.4 × 0.45 C1

B = 0.9(0) N A1 [2]

(ii) A = 2.4 – 0.9 = 1.5 N / moments calculation A1 [1]

3 (a) (i) horizontal velocity = 15 cos 60° = 7.5 m s–1 A1 [1] (ii) vertical velocity = 15 sin 60° = 13 m s–1 A1 [1] (b) (i) v2 = u2 + 2as

s = (13)2 / (2 × 9.81) = 8.6(1) m A1 [1] using g = 10 then max. 1

(ii) t = 13 / 9.81 = 1.326 s or t = 9.95 / 7.5 = 1.327 s A1 [1]

(iii) velocity = 6.15 / 1.33 M1

= 4.6 m s–1 A0 [1] (c) (i) change in momentum = 60 × 10–3 [–4.6 – 7.5] C1

= (–)0.73 N s A1 [2] (ii) final velocity / kinetic energy is less after the collision or

relative speed of separation < relative speed of approach M1 hence inelastic A0 [1]




4 (a) electrical potential energy (stored) when charge moved and gravitational potential energy (stored) when mass moved B1 due to work done in electric field and work done in gravitational field B1 [2]

(b) work done = force × distance moved (in direction of force)

and force = mg M1

mg × h or mg × ∆h A1 [2] (c) (i) 0.1 × mgh = ½ mv2 B1

0.1 × m × 9.81 × 120 = 0.5 × m × v2 B1 v = 15.3 m s–1 A0 [2]

(ii) P = 0.5 m v2 / t C1

m / t = 110 × 103 / [0.25 × 0.5 × (15.3)2] C1 = 3740 kg s–1 A1 [3]

5 (a) ohm = volt / ampere B1 [1]

(b) ρ = RA / l or unit is Ω m C1 units: V A–1

m2 m–1 = N m C–1

A–1 m2

m–1 C1 = kg m2

s–2 A–1

s–1 A–1

m2 m–1

= kg m3 s–3

A–2 A1 [3]

(c) (i) ρ = [3.4 × 1.3 × 10–7] / 0.9 C1

= 4.9 × 10–7 (Ω m) A1 [2] (ii) max = 2.(0) V A1

min = 2 × (3.4 /1503.4) = 4.5 × 10–3 V A1 [2]

(iii) P = V2 / R or P = VI and V = IR C1 = (2)2 / 3.4 = 1.18 (allow 1.2) W A1 [2]

(d) (i) power in Q is zero when R = 0 B1 [1] (ii) power in Q = 0 / tends to zero as R = infinity B1 [1]




6 (a) extension is proportional to force (for small extensions) B1 [1] (b) (i) point beyond which (the spring) does not return to its original length when the

load is removed B1 [1] (ii) gradient of graph = 80 N m–1 A1 [1] (iii) work done is area under graph / ½ Fx / ½ kx2 C1

= 0.5 × 6.4 × 0.08 = 0.256 (allow 0.26) J A1 [2] (c) (i) extension = 0.08 + 0.04 = 0.12 m A1 [1] (ii) spring constant = 6.4 / 0.12 = 53.3 N m–1 A1 [1] 7 (a) nuclei with the same number of protons B1 and a different number of neutrons B1 [2] (b) (i) (mass + energy) (taken together) is conserved (B1)

momentum is conserved (B1) one point required max. 1 B1 [1]

(ii) a = 1 and b = 0 B1

x = 56 B1 y = 92 B1 [3]

(c) proton number = 90 B1

nucleon number = 235 B1 [2]





9702 PHYSICS








1 (a) average velocity = 540 / 30 C1 = 18 m s–1 A1 [2] (b) velocity zero at time t = 0 B1 positive value and horizontal line for time t = 5 s to 35 s B1 line / curve through v = 0 at t = 45 s to negative velocity B1 negative horizontal line from 53 s with magnitude less than positive value and

horizontal line to time = 100 s B1 [4] 2 (a) (i) force is rate of change of momentum B1 [1] (ii) work done is the product of the force and the distance moved in the direction

of the force B1 [1] (b) (i) W = Fs or W = mas or W = m(v2 – u2) / 2 or W = force × distance s A1 [1] (ii) as = (v2 – u2) / 2 any subject M1 W = mas hence W = m(v2 – u2) / 2 M1 RHS represents terms of energy or with u = 0 KE = ½mv

2 A1 [3] (c) (i) work done = ½ × 1500 × [(30)2 – (15)2] (=506250) C1 distance = WD / F = 506250 / 3800 = 133 m A1 [2] or F = ma a = 2.533 (m s–2) C1 v2 = u2 + 2as s = 133 m A1 (ii) the change in kinetic energy is greater or the work done by the force has to

be greater, hence distance is greater (for same force) A1 [1] allow: same acceleration, same time, so greater average speed and greater

distance 3 (a) (i) stress = force / (cross-sectional) area B1 [1] (ii) strain = extension / original length or change in length / original length B1 [1] (b) point beyond which material does not return to the original length / shape / size

when the load / force is removed B1 [1]




(c) UTS is the maximum force / original cross-sectional area M1 wire is able to support / before it breaks A1 [2] allow one: maximum stress the wire is able to support / before it breaks (d) (i) straight line from (0,0) M1 correct shape in plastic region A1 [2] (ii) only a straight line from (0,0) B1 [1] (e) (i) ductile: initially force proportional to extension then a large extension for small change in force B1 brittle: force proportional to extension until it breaks B1 [2] (ii) 1. does not return to its original length / permanent extension (as entered

plastic region) B1 2. returns to original length / no extension (as no plastic region / still in

elastic region) B1 [2] 4 (a) electric field strength = force / positive charge B1 [1] (b) (i) at least three equally spaced parallel vertical lines B1 direction down B1 [2] (ii) E = 1500 / 20 × 10–3 = 75000 V m–1 A1 [1] (iii) F = qE C1 (W = mg and) qE = mg C1 q = mg / E = 5 × 10–15 × 9.81 / 75000 = 6.5 × 10–19

C A1 negative charge A1 [4] (iv) F > mg or F now greater B1 drop will move upwards B1 [2]

5 (a) (i) І1 + І3 = І2 A1 [1]

(ii) E1 = І2R2 + І1R2 + І1R1+ І1r1 A1 [1] 2 2 (iii) E1 – E2 B1

= –І3r2 + І1 (R1 + r1 + R2 / 2) B1 [2] (b) p.d. across BJ of wire changes / resistance of BJ changes B1 there is a difference in p.d across wire and p.d. across cell E2 B1 [2] 6 (a) waves overlap B1 (resultant) displacement is the sum of the displacements of each of the waves B1 [2]




(b) waves travelling in opposite directions overlap / incident and reflected waves

overlap (allow superpose or interfere for overlap here) B1 waves have the same speed and frequency B1 [2] (c) (i) time period = 4 × 0.1 (ms) C1 f = 1 / T = 1 / 4 × 10–4 = 2500 Hz A1 [2] (ii) 1. the microphone is at an antinode and goes to a node and then an

antinode / maximum amplitude at antinode and minimum amplitude at node B1 [1]

2. λ / 2 = 6.7 (cm) C1

v = fλ C1 v = 2500 × 13.4 × 10–2 = 335 m s–1 A1 [3]

incorrect λ then can only score second mark 7 (a) (i) the half life / count rate / rate of decay / activity is the same no matter what

external factors / environmental factors or two named factors such as temperature and pressure changes are applied B1 [1]

(ii) the observations of the count rate / count rate / rate of decay / activity /

radioactivity during decay shows variations / fluctuations B1 [1] (b)

property α-particle β-particle γ-radiation

charge (+)2e –e 0

mass 4u 9.11 × 10–31 kg 0

speed 0.01 to 0.1 c up to 0.99 c c

one mark for each correct line B3 [3] (c) collision with molecules B1 causes ionisation (of the molecule) / electron is removed B1 [2]





9702 PHYSICS








1 (a) scalar has magnitude/size, vector has magnitude/size and direction B1 [1] (b) acceleration, momentum, weight B2 [2]

(–1 for each addition or omission but stop at zero) (c) (i) horizontally: 7.5 cos 40° / 7.5 sin 50° = 5.7(45) / 5.75 not 5.8 N A1 [1] (ii) vertically: 7.5 sin 40° / 7.5 cos 50° = 4.8(2) N A1 [1] (d) either correct shaped triangle M1

correct labelling of two forces, three arrows and two angles A1 or correct resolving: T2 cos 40° = T1 cos 50° (B1) T1 sin 50° + T2 sin 40° = 7.5 (B1) T1 = 5.7(45) (N) A1 T2 = 4.8 (N) A1 [4] (allow ± 0.2 N for scale diagram)

2 (a) 1. constant velocity / speed B1 [1]

2. either constant / uniform decrease (in velocity/speed) or constant rate of decrease (in velocity/speed) B1 [1]

(b) (i) distance is area under graph for both stages C1

stage 1: distance (18 × 0.65) = 11.7 (m) stage 2: distance = (9 × [3.5 – 0.65]) = 25.7 (m) total distance = 37.(4) m A1 [2] (–1 for misreading graph) for stage 2, allow calculation of acceleration (6.32 m s–2) and then s = (18 × 2.85) + ½ × 6.32 (2.85)2 = 25.7 m

(ii) either F = ma or EK = ½mv2 C1

a = (18 – 0)/(3.5 – 0.65) EK = ½ × 1250 × (18)2 C1 F = 1250 × 6.3 =7900 N or F = ½ × 1250 × (18)2 / 25.7 = 7900 N A1 [3] or initial momentum = 1250 × 18 (C1) F = change in momentum / time taken (C1) F = (1250 × 18) / 2.85 = 7900 (A1)

(c) (i) stage 1: either half / less distance as speed is half / less

or half distance as the time is the same or sensible discussion of reaction time B1 [1]

(ii) stage 2: either same acceleration and s = v2 / 2a or v2 is ¼ B1

¼ of the distance B1 [2]




3 (a) (i) power = work done per unit time / energy transferred per unit time / rate of work done B1 [1]

(ii) Young modulus = stress / strain B1 [1]

(b) (i) 1. E = T / (A × strain) (allow strain = ε) C1 T = E × A × strain = 2.4 × 1011 × 1.3 × 10–4 × 0.001 M1 = 3.12 × 104

N A0 [2] 2. T – W = ma C1 [3.12 × 104 – 1800 × 9.81] = 1800a C1 a = 7.52 m s–2 A1 [3]

(ii) 1. T = 1800 × 9.81 = 1.8 × 104

N A1 [1] 2. potential energy gain = mgh C1 = 1800 × 9.81 × 15 = 2.7 × 105

J A1 [2] (iii) P = Fv C1

= 1800 × 9.81 × 0.55 C1 input power = 9712 × (100/30) = 32.4 × 103

W A1 [3] 4 (a) p.d. = energy transformed from electrical to other forms

unit charge B1 e.m.f. = energy transformed from other forms to electrical unit charge B1 [2]

(b) (i) sum of e.m.f.s (in a closed circuit) = sum of potential differences B1 [1]

(ii) 4.4 – 2.1 = I × (1.8 + 5.5 + 2.3) M1

I = 0.24 A A1 [2]

(iii) arrow (labelled) I shown anticlockwise A1 [1]

(iv) 1. V = I × R = 0.24 × 5.5 = 1.3(2) V A1 [1]

2. VA = 4.4 – (I × 2.3) = 3.8(5) V A1 [1]

3. either VB = 2.1 + (I × 1.8) or VB = 3.8 – 1.3 C1 = 2.5(3) V A1 [2]




5 (a) transverse waves have vibrations that are perpendicular / normal to the direction of energy travel B1 longitudinal waves have vibrations that are parallel to the direction of energy travel B1 [2]

(b) vibrations are in a single direction M1

either applies to transverse waves or normal to direction of wave energy travel or normal to direction of wave propagation A1 [2]

(c) (i) 1. amplitude = 2.8 cm B1 [1]

2. phase difference = 135° or 0.75π rad or ¾π rad or 2.36 radians (three sf needed) numerical value M1 unit A1 [2]

(ii) amplitude = 3.96 cm (4.0 cm) A1 [1] 6 (a) (i) greater deflection M0

greater electric field / force on α-particle A1 [1] (ii) greater deflection M0

greater electric field / force on α-particle A1 [1] (b) (i) either deflections in opposite directions M1

because oppositely charged A1

or β less deflection (M1)

β has smaller charge (A1) [2]

(ii) α smaller deflection M1 because larger mass A1 [2]

(iii) β less deflection because higher speed B1 [1] (c) either F = ma and F = Eq or a = Eq / m C1

ratio = either (2 × 1.6 × 10–19) × (9.11 × 10–31) (1.6 × 10–19) × 4 × (1.67 × 10–27) or [2e × 1 / 2000 u] / [e × 4u] C1 ratio = 1 /4000 or 2.5 × 10–4 or 2.7 × 10–4 A1 [3]





9702 PHYSICS

9702/31 Paper 3 (Advanced Practical Skills 1), maximum raw mark 40







1 (a) (ii) Value of k in range: 50 cm ≤ k ≤ 100 cm. [1]

(b) (iii) Values of d and D both with unit and d in range 4.0 ≤ d ≤ 6.0 cm. [1] (c) Six sets of readings of d and D scores 5 marks, five sets scores 4 marks etc. Incorrect trend then –1. Supervisor’s help –1. [5] Range of d: ∆d ≥ 40 cm. [1] Column headings: [1] Each column heading must contain a quantity and a unit where appropriate. There must be some distinguishing mark between the quantity and the unit, for

example d / m, d (m), 1/D (m–1). Consistency of presentation of raw readings: [1] All values of raw d and D must be given to the nearest mm. Significant figures: [1] Significant figures for 1/D must be the same as, or one more than, the number used

in D. Calculation: (D – d)/D calculated correctly. [1] (d) (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 which is being plotted. Scale markings must be no more than three large squares apart.

Plotting of points: [1]

All observations in the table must be plotted. Check that the points are correctly plotted. Work to an accuracy of half a small square in both x and y directions. Do not accept ‘blobs’ (points with diameter greater than half a small square).

Quality: [1] All points in the table must be plotted (at least 5) for this mark to be scored. Scatter

of points must be less than ± 0.05 m–1 (0.0005 cm–1) of 1/D of a straight line. (ii) Line of best fit: [1]

Judge by balance of all the points on the grid (at least 5) about the candidate's line. 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.




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

Intercept: [1]

Either: Check correct read-off from a point on the line and substitution into y = mx + c. Read-off must be accurate to half a small square in both x and y directions. Allow ecf of gradient value. Or: Check the read-off of the intercept directly from the graph.

(e) A = value of gradient, B = – (value of y-intercept). [1]

A/B = k ± 5 cm with consistent unit. [1]

[Total: 20] 2 (a) Measurement of all raw values of t to nearest 0.01 mm or 0.001 mm and t in range

0.10 ≤ t ≤ 0.50 mm. [1] (b) (i) Value of L in range 26.0 cm ≤ L ≤ 30.0 cm with consistent unit to nearest mm. [1] (ii) Absolute uncertainty in L in range 1–2 mm (but if repeated readings have been

taken then the absolute uncertainty could be half the range unless zero). Correct method shown to find the percentage uncertainty. [1] (c) (ii) Correct calculation of V with consistent unit. Allow ecf. [1] (e) Value of T in range 0.7 s ≤ T ≤ 1.5 s with consistent unit. Supervisor help –1. [1]

Evidence of repeats. [1]

(f) Second value of L in range 5 cm ≤ L ≤ 15 cm. [1] (g) Second value of T. [1] Quality: second value of T < first value of T. [1] (h) (i) Two values of k calculated correctly. [1] (ii) Justification of s.f. in k linked to raw data in L and T/t. [1] (iii) Sensible comment relating to the calculated values of k, testing against a criterion

specified by the candidate. [1]




(i)

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

A Two readings are not enough (to draw a conclusion)

Take more readings and plot a graph/calculate more k values (and compare)

‘Few readings’/‘take more readings and calculate average k’/‘only one reading’

B Card does not swing freely/ friction between pivot and card

Make hole bigger/bush or bearing idea

C Difficult to judge end/start/a complete swing

Use of fiducial marker/pointer Reaction time error/human reaction/difficult to know when to start/stop timer

D Irregular/uneven/unusual swings/not in same vertical plane/centre of bottom rule not fixed

Method of keeping shape aligned vertically/turn off fans

E Oscillations die out quickly/ heavy damping

Use increased thickness of card

F T short/large uncertainty in T Improved method of timing e.g. video and timer/frame-by-frame. Increase l/length of card

Use of computer/light gates/ camera/high speed camera/ too fast/time too fast/time more swings/time large no. of swings

[Total: 20]





9702 PHYSICS








1 (a) (i) All raw values of d in the range 0.15 mm ≤ d ≤ 0.40 mm to 0.01 mm with unit, or supervisor's value ± 0.10 mm [1] (b) (iii) l and V with unit. Value of l in the range 40.0 cm ≤ l ≤ 60.0 cm. [1] (c) Six sets of readings of l and V scores 5 marks, five sets scores 4 marks etc. Incorrect trend –1. Major help from supervisor –2. Minor help from supervisor –1. [5]

Range of l: ∆l ≥ 50 cm. [1] Column headings: [1] Each column heading must contain a quantity and a unit where appropriate. There must be some distinguishing mark between the quantity and the unit, e.g. 1/V / V–1. Accept 1/V (V–1) but do not allow 1/V (V)–1 or 1/l (m). Consistency of presentation of raw readings: [1] All values of raw l must be given to 0.001 m. Significant figures: [1] Significant figures for 1/ V must be to the same as, or one more than, the least number

of significant figures used in raw V. Calculation: 1/ V calculated correctly. [1] (d) (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 which is being plotted. Scale markings must be no more than three large squares apart. Plotting of points: [1] All observations in the table must be plotted.

Check that the points are correctly plotted. Work to an accuracy of half a small square in both the x and y directions.

Do not accept ‘blobs’ (points with diameter greater than half a small square). Quality: [1] All points in the table must be plotted (at least 5) for this mark to be scored. Scatter

of points must be less than ± 0.2 m–1 (0.002 cm–1) of 1/l of a straight line. (ii) Line of best fit: [1] Judge by balance of all the points on the grid (at least 5) about the candidate's line.

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.




(d) (iii) Gradient: [1] The hypotenuse of the triangle used must be at least half the length of the drawn

line. Both read-offs must be accurate to half a small square in both x and y directions. The method of calculation must be correct.

Intercept: [1] Either: Check correct read-off from a point on the line and substitution into y = mx + c. Read-off must be accurate to half a small square in both x and y directions. Allow ecf of gradient value. Or: Check the read-off of the intercept directly from the graph. (e) (i) M = value of gradient, N = value of y-intercept. [1]

(ii) Substitution into equation and answer for ρ in range 2 – 20 × 10–7 Ω m

(0.2 × 10–6 – 2 × 10–6 Ω m). [1]

[Total: 20] 2 (a) Measurement of all raw w to nearest mm in range 2.0 ≤ w ≤ 3.5 cm. [1] (b) (iii) Value of l in range 25 cm ≤ l ≤ 50 cm with unit. [1] (c) (ii) Correct calculation of d. Write in correct value if incorrect. Unit not needed. [1] (iii) Absolute uncertainty in d in the range 3–15 mm (but if repeated readings have

been taken then the absolute uncertainty could be half the range, unless zero). Correct method shown used to find the percentage uncertainty. [1] (d) Value of T in range 0.3 s < T <1.5 s. [1] Evidence of repeats. [1] (e) Second value of l. [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) Justification of s.f. in k linked to time AND d or l. [1] (iii) Sensible comment relating to the calculated values of k, testing against a criterion





(g)



Take more readings and plot a graph/ calculate more k values (and compare)


B Loops not the same size/ ruler(s) not horizontal/

Method to ensure loops same size, e.g. use glue or tape to secure ends together/tie around fixed points/template/mark string

Difficult to tie knots Use of spirit level/use of plumbline Rubber bands or other material Use of Blu-Tack String stretching

C Difficult to judge end/start/ centre of swing/difficult to judge complete swing

Use of fiducial marker/pointer Reaction time error Human reaction Difficult to know when to start/ stop timer

D Irregular/uneven/unusual swings/not in same horizontal plane/centre of bottom rule not fixed

Method of ensuring correct release, e.g. (two) stop(s) at either end

Fans/switch off fans Amplitude changes

E Loops slide Method, e.g. glue, tape to fix loop to rule/ drill hole to attach string/ make grooves to hold string

‘Glue’ or ‘tape’ on its own

F T or time short/large uncertainty in T

Improved method of timing, e.g. video and timer/frame-by-frame/increase d or l, decrease distance between loops, correct position of motion sensor linked to data logger

Use of computer Light gates Camera High speed camera Too fast Time too fast Time more swings Time large no. of swings.

G Reason for calculation of d inaccurate, e.g. different w/ thickness of rule not taken into account

Measure d directly/measure w for both rules/measure and allow for thickness.

Parallax error Use of set squares Vernier callipers Just ‘different w’

[Total: 20]





9702 PHYSICS








1 (a) Measurement for V in range +0.10 V to +0.90 V, with unit. [1] (c) (ii) Six sets of values for R and V scores 6 marks, five sets scores 5 marks etc. [6]

Incorrect trend –1. Major help from supervisor –2, minor help –1. Range: [1]

R values must include 0.33 kΩ or less and 4.7 kΩ or more. Column headings: [1] Each column heading must contain a quantity and a unit where appropriate. There must be some distinguishing mark between the quantity and the unit. Consistency of presentation of raw readings: [1] All raw values of V must be given to the same precision and at least 2 d.p. Significant figures: [1] R/(R + 1) must be given to the same as or one more than the s.f. for R. Calculation: [1] R/(R + 1) calculated correctly.

(d) (i) Axes: [1] Sensible scales must be used, no awkward scales (e.g. 3:10).

Scales must be chosen so that the plotted points must occupy at least half the graph grid in both x and y directions. Scales must be labelled with the quantity which is being plotted. Scale markings must be no more than 3 large squares apart. Plotting of points: [1] All observations in the table must be plotted. Check that the points are correctly plotted. Work to an accuracy of half a small square. Do not accept ‘blobs’ (points with diameter greater than half a small square).

Quality: [1] All points in the table must be plotted (at least 5) for this mark to be scored. Scatter of points must be less than ± 0.04 V on the V axis from a straight line.

(ii) Line of best fit: [1] Judge by balance of all the points on the grid (at least 5) about the candidate's line. 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. There must be 5 points left after the anomalous point is disregarded.




(d) (iii) Gradient: [1] The hypotenuse of the triangle used must be at least half the length of the drawn line. Both read-offs must be accurate to half a small square or better in both x and y directions. The method of calculation must be correct. Intercept: [1] Either: Check correct read-off from a point on the line and substitution into y = mx + c. Read-off must be accurate to half a small square in both x and y directions. Allow ecf of gradient value. Or: Check the read-off of the intercept directly from the graph

(e) a = value of gradient, b = –(value of intercept). Do not allow a fraction. [1]

Value of b is in range 1.0 V to 2.0 V, with unit V. [1]

[Total: 20] 2 (a) Value of t in range 0.01 to 0.05 mm, with unit. [1] (b) (i) Value of w in range 5 to 15 mm. Raw reading(s) must be to nearest mm. [1]

Evidence of repeated readings of w. [1] (ii) Percentage uncertainty in w based on absolute uncertainty of 1 mm [1]

(but if repeated readings have been taken then the absolute uncertainty could be half the range, unless this is zero). Correct method used to find the % uncertainty.

(c) Correct calculation of A using candidate’s values from (a) and (b). [1] (d) (iii) At least three measurements of F used. [1]

Average calculated correctly, with unit. [1] (e) Second value of w. [1] Second value of F. [1] F increases as w increases. [1] (f) (i) Two values of k calculated correctly. [1] (ii) Sensible comment relating to the calculated values of k, testing against a [1] criterion specified by the candidate..




(g)





B Difficult to see maximum/breaking F/break happens suddenly

Video (plus ‘slow motion’ or ‘to view force’ or ‘to view newton-meter’) use maximum-hold newton-meter/ use weights (e.g. sand) to measure F

Just ‘use video camera’

C Difficult to see ends of cuts/difficult to measure w because strip is transparent/ same colour as background

Use contrasting background/mark ends of cuts

‘Difficult to measure w’/use coloured polythene

D w measurement has low precision

Improved method of measuring w e.g. use vernier calliper or use travelling microscope/use larger w

E t not constant Measure t between cuts Micrometer squashes polythene

F Large (%) uncertainty/error in t

Improved method of measuring t e.g. measure several layers or use digital micrometer for better precision

G Sellotape detaches from bench

Improved method of fixing to bench e.g. use clamp or use wider tape or use glue or use stickier tape

‘use stronger tape’

H t (or w) changes as strip stretches/as F increases

Measure just before or after strip breaks

Do not allow ‘repeated readings’ Do not allow ‘use a computer to improve the experiment’ [Total: 20]





9702 PHYSICS








1 (a) Raw value(s) of h to the nearest mm in range 5–15 cm. [1] (b) (ii) Value of d with unit: d < h. [1] (d) Six sets of readings of m and d scores 5 marks, five sets scores 4 marks etc. Incorrect trend –1. Supervisor’s help –1. [5]

Range of m: ∆m ≥ 60 g. [1] Column headings: [1] Each column heading must contain a quantity and a unit where appropriate. There must be some distinguishing mark between the quantity and the unit,

e.g. m / kg m–1 but accept m (kg m–1). d d Consistency of presentation of raw readings: [1] All values of raw d must be given to the nearest mm. Significant figures: [1] Significant figures for 1 must be to the same as, or one more than, the number of d significant figures in d. Calculation: m/d calculated correctly. [1] (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 which is being plotted. Scale markings must be no more than three large squares apart. Plotting of points: [1] All observations in the table must be plotted. Check that the points are correctly plotted. Work to an accuracy of half a small

square in both x and y directions. Do not accept ‘blobs’ (points with diameter greater than half a small square). Quality: [1]

All points in the table must be plotted (at least 5) for this mark to be scored. Scatter of points must be less than ± 0.5 m–1 (0.005 cm–1) of 1/d of a straight line.

(ii) Line of best fit: [1] Judge by balance of all the points on the grid (at least 5) about the candidate's line.

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.




(iii) Gradient: [1] The hypotenuse of the triangle used must be at least half the length of the drawn

line. Both read-offs must be accurate to half a small square in both x and y directions. The method of calculation must be correct.

Intercept: [1] Either: Check correct read-off from a point on the line and substitution into

y = mx + c. Read-off must be accurate to half a small square in both x and y directions. Allow ecf of gradient value.

Or: Check the read-off of the intercept directly from the graph. (f) Values of A = –gradient and B = intercept. [1]

Substitution of d = h shown and 0.08 kg < m < 1.0 kg with consistent unit. [1] [Total: 20] 2 (a) (ii) Value of m in g or kg. 45 g ≤ m ≤ 55 g. [1] (iii) Absolute uncertainty in m in range 1–5 g with unit. Correct method shown to find the percentage uncertainty. [1] (b) (iii) Value of V to at least 1 d.p. with unit. Supervisor help –1. [1]

(c) Raw value(s) of θ1 to nearest °C. [1]

(d) (ii) Value of θ2 > θ1 with unit. [1]

(iii) Calculation of (θ2 – θ1). [1]

(e) Second value of V > first value of V. [1]

(f) Second values of θ2 and θ1. [1]

Second value of (θ2 – θ1) > first value of (θ2 – θ1). [1] (g) (i) Two values of k calculated correctly. [1]

(ii) Justification of s.f. in k linked to raw data in V and (θ2 – θ1). [1] (iii) Sensible comment relating to the calculated values of k, testing against a criterion





(h)




‘Few readings’/ ‘take more readings and calculate average k’/ ‘only one reading’

B Heat loss (to surroundings or beaker)

Method to reduce heat loss, e.g. lagging, lid

Switch off fans to reduce convection

C Small value of (θ2 – θ1)/

% uncertainty in (θ2 – θ1) is large

Method to increase (θ2 – θ1) e.g. higher voltage, lower resistance, increased time, less water

D Low precision of thermometer Either: thermometer with specified better precision, e.g. 0.1

oC, 0.5

oC Or: named device such as thermocouple or resistance thermometer.

Not accuracy

E Resistor/bulb of thermometer is not completely immersed

Use narrower beaker

F Water is left behind in measuring cylinder

Method to measure mass of water, e.g. subtract mass of empty beaker from mass of beaker with water

Just “weigh water”

G Resistor continues to give out heat when switched off/ temperature continues to rise after switching off

Wait until temperature reaches a maximum before reading

Do not credit: precision of measuring cylinder; different starting temperatures of water; uneven temperature distribution in beaker; parallax errors in reading volume or temperature; reaction time error in timing.

[Total: 20]





9702 PHYSICS








1 (b) Measurement for H in range 0.200 m to 0.900 m. [1] (c) (ii) First measurement of m, to nearest 0.001 kg and in the range 0.045 to 0.055 kg. [1] (d) Six sets of values for h and m scores 5 marks, five sets scores 4 marks etc. [5] Incorrect trend then –1. Help from supervisor –1. Range: [1] m values must include 0.070 kg or less, and 0.220 kg or more. Column headings: [1]

Each column heading must contain a quantity and a unit where appropriate. There must be some distinguishing mark between the quantity and the unit.

e.g. y–2/m–2, 1/m2(1/kg2) but not 22 kg/

1

m

.

Consistency of presentation of raw readings: [1] All values of h must be given to the nearest mm. Significant figures: [1] Every value of 1/y2 must be given to the same s.f. as (or one more than) the s.f. in y. Calculation: [1] 1/y2 calculated correctly.

(e) (i) Axes: [1] Sensible scales must be used, no awkward scales (e.g. 3:10). 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 which is being plotted.

Plotting: [1]

All observations must be plotted. Check that the points are correctly plotted. Work to an accuracy of half a small square. Do not accept ‘blobs’ (points with diameter greater than half a small square).

Quality: [1] Scatter of points must be less than ± 50 m2 (± 0.005 cm2) on the 1/y2 axis about a straight line. All points must be plotted (at least 5) for this mark to be scored.

(ii) Line of best fit: [1]

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




(iii) Gradient: [1] The hypotenuse must be at least half the length of the drawn line. Both read-offs must be accurate to half a small square or better in both x and y directions.

The method of calculation must be correct. Do not allow ∆x/∆y.

Intercept: [1] Either: Check correct read-off from a point on the line, and substitution into

y = mx + c. Read-off must be accurate to half a small square or better in both x and y directions. Allow ecf of gradient value.

Or: Check the read-off of the intercept directly from the graph.

(f) p = value of gradient and q = (value of intercept). [1] Both values must be from (e)(iii). Do not allow fractions. Correct consistent units for p (e.g. kg2

m2) and q (e.g. m2). [1] [Total: 20] 2 (b) 0.250 m ≤ a ≤ 0.350 m and 0.450 m ≤ b ≤ 0.550 m, both with a correct and consistent [1] unit. Values of a and b given to nearest mm e.g. 0.350 m or 35.0 cm. [1] (c) (ii) Value of R in range 0.05 m to 0.50 m (5 cm to 50 cm). [1]

Evidence of repeats (credit evidence here or in (f)). [1] (d) Percentage uncertainty in R based on absolute uncertainty in range 0.002 m to 0.01 m [1] (2 mm to 10 mm). (If repeated readings have been done then the absolute uncertainty could be half the

range, unless this is zero.) Correct method to get % uncertainty. (e) Correct calculation of v with consistent unit. [1] (f) (ii) Second values of a and b. [1]

Second value of R. [1] Second R less than first R. [1] Correct calculation of second v. [1]

(g) (i) Two values of k calculated correctly. [1] (ii) Valid conclusion based on the variation in k being within (or outside) a stated [1] criterion.




(h)



Take more readings and plot a graph/calculate more k values (and compare).

Few readings/only one reading/take more readings and calculate average k /‘repeat readings’

B Difficult to locate start position/measure R owing to parallax

Method to locate start point e.g. plumb line/clamped vertical rule using set square to bench

‘Parallax error’/parallax error linked to a or b

C Difficult to locate end point /measure R owing to ball bouncing/skipping/sinking/rule displaced from ball

Method to locate end point of R e.g. vertical clamped pointer/tray without lip (so rule can be placed on sand)/sand on bench/carbon paper /painted ball/video with playback plus scale in shot/ detailed hot spot

Vague video methods/ball moves/smooth sand/change depth of sand

D Difficult to release ball from rest/without exerting a force

Method of improving release e.g. use an electromagnet

Use a release mechanism

E (Vertical) distance fallen is less than a

Method of measuring a to surface of sand/correcting the value of a by measuring depth of sand

F Difficult to make tube horizontal (as not flexible enough)/judge horizontal/ clamp blocks horizontally

Method to ensure tube is horizontal e.g. use reference line (window sill)/spirit level /measure several heights from bench.

G Ball sticks in tube/slows down due to e.g. sand in tube/bend in tube/kink in tube/too much friction

Method to overcome sticking e.g. use new ball each time /clean ball with cloth before putting back in tube/use wider tube/smaller ball/open track

Lubricate/clean tube

Do not allow ‘rule is not perpendicular to bench’. Do not allow unspecified computer methods. [Total: 20]





9702 PHYSICS

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







Section A 1 (a) gravitational force provides the centripetal force B1

GMm/r 2 = mrω2 (must be in terms of ω) B1 r 3ω

2 = GM and GM is a constant B1 [3] (b) (i) 1. for Phobos, ω = 2π/(7.65 × 3600) C1

= 2.28 × 10–4 rad s–1 (9.39 × 106)3 × (2.28 × 10–4)2 = 6.67 × 10–11 × M C1 M = 6.46 × 1023 kg A1 [3]

2. (9.39 × 106)3 × (2.28 × 10–4)2 = (1.99 × 107)3 × ω2 C1

ω = 7.30 × 10–5 rad s–1 C1 T = 2π/ω = 2π/(7.30 × 10–5) = 8.6 × 104 s = 23.6 hours A1 [3]

(ii) either almost ‘geostationary’

or satellite would take a long time to cross the sky B1 [1] 2 (a) e.g. moving in random (rapid) motion of molecules/atoms/particles

no intermolecular forces of attraction/repulsion volume of molecules/atoms/particles negligible compared to volume of

container time of collision negligible to time between collisions (1 each, max 2) B2 [2]

(b) (i) 1. number of (gas) molecules B1 [1] 2. mean square speed/velocity (of gas molecules) B1 [1] (ii) either pV = NkT or pV = nRT and links n and k

and <EK> = ½m<c2> M1

clear algebra leading to <EK> = 2

3kT A1 [2]

(c) (i) sum of potential energy and kinetic energy of molecules/atoms/particles M1

reference to random (distribution) A1 [2] (ii) no intermolecular forces so no potential energy B1

(change in) internal energy is (change in) kinetic energy and this is proportional to (change in ) T B1 [2]




3 (a) (i) amplitude remains constant B1 [1] (ii) amplitude decreases gradually M1

light damping A1 [2] (iii) period = 0.80 s C1

frequency = 1.25 Hz (period not 0.8 s, then 0/2) A1 [2] (b) (i) (induced) e.m.f. is proportional to M1

rate of change/cutting of (magnetic) flux (linkage) A1 [2] (ii) a current is induced in the coil M1

as magnet moves in coil A1 current in resistor gives rise to a heating effect M1 thermal energy is derived from energy of oscillation of the magnet A1 [4]

4 (a) (i) zero field (strength) inside spheres B1 [1] (ii) either field strength is zero

or the fields are in opposite directions M1 at a point between the spheres A1 [2]

(b) (i) field strength is (–) potential gradient (not V/x) B1 [1] (ii) 1. field strength has maximum value B1

at x = 11.4 cm B1 [2] 2. field strength is zero B1

either at x = 7.9 cm (allow ±0.3 cm) or at 0 to 1.4 cm or 11.4 cm to 12 cm B1 [2]

5 (a) (i) Bqv(sinθ) or Bqv(cosθ) B1 [1] (ii) qE B1 [1] (b) FB must be opposite in direction to FE B1

so magnetic field into plane of paper B1 [2]




6 (a) (i) period = 1/50 C1 t1 = 0.03 s A1 [2]

(ii) peak voltage = 17.0 V A1 [1] (iii) r.m.s. voltage = 17.0/√2

= 12.0 V A1 [1] (iv) mean voltage = 0 A1 [1] (b) power = V 2/R C1

= 122/2.4 = 60 W A1 [2]

7 (a) each line represents photon of specific energy M1

photon emitted as a result of energy change of electron M1 specific energy changes so discrete levels A1 [3]

(b) (i) arrow from –0.85 eV level to –1.5 eV level B1 [1] (ii) ∆E = hc /λ C1

= (1.5 – 0.85) × 1.6 × 10–19 C1 = 1.04 × 10–19 J λ = (6.63 × 10–34 × 3.0 × 108)/(1.04 × 10–19) = 1.9 × 10–6 m A1 [3]

(c) spectrum appears as continuous spectrum crossed by dark lines B1

two dark lines B1 electrons in gas absorb photons with energies equal to the excitation energies M1 light photons re-emitted in all directions A1 [4]

8 (a) (i) time for initial number of nuclei/activity M1

to reduce to one half of its initial value A1 [2] (ii) λ = ln 2/(24.8 × 24 × 3600) M1

= 3.23 × 10–7 s–1 A0 [1] (b) (i) A = λN C1

3.76 × 106 = 3.23 × 10–7 × N N = 1.15 × 1013 A1 [2]

(ii) N = N0 e

–λt = 1.15 × 1013 × exp(–ln 2 × 30/24.8) C1 = 4.97 × 1012 A1 [2]

(c) ratio = (4.97 × 1012)/(1.15 × 1013 – 4.97 × 1012) C1

= 0.76 A1 [2]




Section B 9 (a) e.g. reduced gain

increased stability greater bandwidth or less distortion (allow any two sensible suggestions, 1 each, max 2) B2 [2]

(b) (i) V – connected to midpoint between resistors B1

VOUT clear and input to V+ clear B1 [2] (ii) gain = 1 + RF/R

15 = 1 + 12000/R C1 R = 860 Ω A1 [2]

(c) graph: straight line from (0,0) to (0.6,9.0) B1

straight line from (0.6,9.0) to (1.0,9.0) B1 [2] (d) either relay can be used to switch a large current/voltage M1

output current of op-amp is a few mA/very small A1 [2] or relay can be used as a remote switch (M1) for inhospitable region/avoids using long heavy cables (A1)

10 (a) e.g. large bandwidth/carries more information

low attenuation of signal low cost smaller diameter, easier handling, easier storage, less weight high security/no crosstalk low noise/no EM interference (allow any four sensible suggestions, 1 each, max 4) B4 [4]

(b) (i) infra-red B1 [1] (ii) lower attenuation than for visible light B1 [1] (c) (i) gain/dB = 10 lg(P2/P1) C1

26 = 10 lg(P2/9.3 × 10–6) P2 = 3.7 × 10–3 W A1 [2]

(ii) power loss along fibre = 30 × 0.2 = 6.0 dB C1

either 6 = 10 lg(P/3.7 × 10–3) or 6 dB = 4 × 3.7 × 10–3 or 32 = 10 lg(P/9.3 × 10–6) input power = 1.5 × 10–2 W A1 [2]




11 (a) (i) switch M1 so that one aerial can be used for transmission and reception A1 [2]

(ii) tuning circuit M1

to select (one) carrier frequency (and reject others) A1 [2] (iii) analogue-to-digital converter/ADC M1

converts microphone output to a digital signal A1 [2] (iv) (a.f.) amplifier (not r.f. amplifier) M1

to increase (power of) signal to drive the loudspeaker A1 [2] (b) e.g. short aerial so easy to handle

short range so less interference between base stations larger waveband so more carrier frequencies (any two sensible suggestions, 1 each, max 2) B2 [2]





9702 PHYSICS








Section A 1 (a) gravitational force provides the centripetal force B1

GMm/r 2 = mrω2 (must be in terms of ω) B1 r 3ω

2 = GM and GM is a constant B1 [3] (b) (i) 1. for Phobos, ω = 2π/(7.65 × 3600) C1

= 2.28 × 10–4 rad s–1 (9.39 × 106)3 × (2.28 × 10–4)2 = 6.67 × 10–11 × M C1 M = 6.46 × 1023 kg A1 [3]

2. (9.39 × 106)3 × (2.28 × 10–4)2 = (1.99 × 107)3 × ω2 C1

ω = 7.30 × 10–5 rad s–1 C1 T = 2π/ω = 2π/(7.30 × 10–5) = 8.6 × 104 s = 23.6 hours A1 [3]

(ii) either almost ‘geostationary’

or satellite would take a long time to cross the sky B1 [1] 2 (a) e.g. moving in random (rapid) motion of molecules/atoms/particles

no intermolecular forces of attraction/repulsion volume of molecules/atoms/particles negligible compared to volume of

container time of collision negligible to time between collisions (1 each, max 2) B2 [2]

(b) (i) 1. number of (gas) molecules B1 [1] 2. mean square speed/velocity (of gas molecules) B1 [1] (ii) either pV = NkT or pV = nRT and links n and k

and <EK> = ½m<c2> M1

clear algebra leading to <EK> = 2

3kT A1 [2]

(c) (i) sum of potential energy and kinetic energy of molecules/atoms/particles M1

reference to random (distribution) A1 [2] (ii) no intermolecular forces so no potential energy B1

(change in) internal energy is (change in) kinetic energy and this is proportional to (change in ) T B1 [2]




3 (a) (i) amplitude remains constant B1 [1] (ii) amplitude decreases gradually M1

light damping A1 [2] (iii) period = 0.80 s C1

frequency = 1.25 Hz (period not 0.8 s, then 0/2) A1 [2] (b) (i) (induced) e.m.f. is proportional to M1

rate of change/cutting of (magnetic) flux (linkage) A1 [2] (ii) a current is induced in the coil M1

as magnet moves in coil A1 current in resistor gives rise to a heating effect M1 thermal energy is derived from energy of oscillation of the magnet A1 [4]

4 (a) (i) zero field (strength) inside spheres B1 [1] (ii) either field strength is zero

or the fields are in opposite directions M1 at a point between the spheres A1 [2]

(b) (i) field strength is (–) potential gradient (not V/x) B1 [1] (ii) 1. field strength has maximum value B1

at x = 11.4 cm B1 [2] 2. field strength is zero B1

either at x = 7.9 cm (allow ±0.3 cm) or at 0 to 1.4 cm or 11.4 cm to 12 cm B1 [2]

5 (a) (i) Bqv(sinθ) or Bqv(cosθ) B1 [1] (ii) qE B1 [1] (b) FB must be opposite in direction to FE B1

so magnetic field into plane of paper B1 [2]




6 (a) (i) period = 1/50 C1 t1 = 0.03 s A1 [2]

(ii) peak voltage = 17.0 V A1 [1] (iii) r.m.s. voltage = 17.0/√2

= 12.0 V A1 [1] (iv) mean voltage = 0 A1 [1] (b) power = V 2/R C1

= 122/2.4 = 60 W A1 [2]

7 (a) each line represents photon of specific energy M1

photon emitted as a result of energy change of electron M1 specific energy changes so discrete levels A1 [3]

(b) (i) arrow from –0.85 eV level to –1.5 eV level B1 [1] (ii) ∆E = hc /λ C1

= (1.5 – 0.85) × 1.6 × 10–19 C1 = 1.04 × 10–19 J λ = (6.63 × 10–34 × 3.0 × 108)/(1.04 × 10–19) = 1.9 × 10–6 m A1 [3]

(c) spectrum appears as continuous spectrum crossed by dark lines B1

two dark lines B1 electrons in gas absorb photons with energies equal to the excitation energies M1 light photons re-emitted in all directions A1 [4]

8 (a) (i) time for initial number of nuclei/activity M1

to reduce to one half of its initial value A1 [2] (ii) λ = ln 2/(24.8 × 24 × 3600) M1

= 3.23 × 10–7 s–1 A0 [1] (b) (i) A = λN C1

3.76 × 106 = 3.23 × 10–7 × N N = 1.15 × 1013 A1 [2]

(ii) N = N0 e

–λt = 1.15 × 1013 × exp(–ln 2 × 30/24.8) C1 = 4.97 × 1012 A1 [2]

(c) ratio = (4.97 × 1012)/(1.15 × 1013 – 4.97 × 1012) C1

= 0.76 A1 [2]




Section B 9 (a) e.g. reduced gain

increased stability greater bandwidth or less distortion (allow any two sensible suggestions, 1 each, max 2) B2 [2]

(b) (i) V – connected to midpoint between resistors B1

VOUT clear and input to V+ clear B1 [2] (ii) gain = 1 + RF/R

15 = 1 + 12000/R C1 R = 860 Ω A1 [2]

(c) graph: straight line from (0,0) to (0.6,9.0) B1

straight line from (0.6,9.0) to (1.0,9.0) B1 [2] (d) either relay can be used to switch a large current/voltage M1

output current of op-amp is a few mA/very small A1 [2] or relay can be used as a remote switch (M1) for inhospitable region/avoids using long heavy cables (A1)

10 (a) e.g. large bandwidth/carries more information

low attenuation of signal low cost smaller diameter, easier handling, easier storage, less weight high security/no crosstalk low noise/no EM interference (allow any four sensible suggestions, 1 each, max 4) B4 [4]

(b) (i) infra-red B1 [1] (ii) lower attenuation than for visible light B1 [1] (c) (i) gain/dB = 10 lg(P2/P1) C1

26 = 10 lg(P2/9.3 × 10–6) P2 = 3.7 × 10–3 W A1 [2]

(ii) power loss along fibre = 30 × 0.2 = 6.0 dB C1

either 6 = 10 lg(P/3.7 × 10–3) or 6 dB = 4 × 3.7 × 10–3 or 32 = 10 lg(P/9.3 × 10–6) input power = 1.5 × 10–2 W A1 [2]




11 (a) (i) switch M1 so that one aerial can be used for transmission and reception A1 [2]

(ii) tuning circuit M1

to select (one) carrier frequency (and reject others) A1 [2] (iii) analogue-to-digital converter/ADC M1

converts microphone output to a digital signal A1 [2] (iv) (a.f.) amplifier (not r.f. amplifier) M1

to increase (power of) signal to drive the loudspeaker A1 [2] (b) e.g. short aerial so easy to handle

short range so less interference between base stations larger waveband so more carrier frequencies (any two sensible suggestions, 1 each, max 2) B2 [2]





9702 PHYSICS








Section A 1 (a) (i) weight = GMm/r 2 C1

= (6.67 × 10–11 × 6.42 × 1023 × 1.40)/(½ × 6.79 × 106)2 C1 = 5.20 N A1 [3]

(ii) potential energy = –GMm/r C1

= –(6.67 × 10–11 × 6.42 × 1023 × 1.40)/(½ × 6.79 × 106) M1 = –1.77 × 107 J A0 [2]

(b) either ½mv 2 = 1.77 × 107 C1

v 2 = (1.77 × 107 × 2)/1.40 C1 v = 5.03 × 103 m s–1 A1 or ½mv 2 = GMm/r (C1) v 2 = (2 × 6.67 x 10–11 × 6.42 × 1023)/(6.79 × 106/2) (C1) v = 5.02 × 103 m s–1 (A1) [3]

(c) (i) ½ × 2 × 1.66 × 10–27 × (5.03 × 103)2 = 2

3 × 1.38 × 10–23 × T C1

T = 2030 K A1 [2] (ii) either because there is a range of speeds M1

some molecules have a higher speed A1 or some escape from point above planet surface (M1) so initial potential energy is higher (A1) [2]

2 (a) temperature scale calibrated assuming linear change of property with

temperature B1 neither property varies linearly with temperature B1 [2]

(b) (i) does not depend on the property of a substance B1 [1] (ii) temperature at which atoms have minimum/zero energy B1 [1] (c) (i) 323.15 K A1 [1] (ii) 30.00 K A1 [1]




3 (a) acceleration proportional to displacement/distance from fixed point M1 and in opposite directions/directed towards fixed point A1 [2]

(b) energy = ½mω 2x0

2 and ω = 2πf C1

= ½ × 5.8 × 10–3 × (2π × 4.5)2 × (3.0 × 10–3)2 C1 = 2.1 × 10–5 J A1 [3]

(c) (i) at maximum displacement M1

above rest position A1 [2]

(ii) acceleration = (–)ω 2x0 and acceleration = 9.81 or g C1

9.81 = (2π × 4.5)2 × x0 x0 = 1.2 × 10–2 m A1 [2]

4 (a) e.g. storing energy

separating charge blocking d.c. producing electrical oscillations tuning circuits smoothing preventing sparks timing circuits (any two sensible suggestions, 1 each, max 2) B2 [2]

(b) (i) –Q (induced) on opposite plate of C1 B1

by charge conservation, charges are –Q, +Q, –Q, +Q, –Q B1 [2] (ii) total p.d. V = V1 + V2 + V3 B1

Q/C = Q/C1 + Q/C2 + Q/C3 B1 1/C = 1/C1 + 1/C2 + 1/C3 A0 [2]

(c) (i) energy = ½CV 2 or energy = ½ QV and C = Q/V C1

= ½ × 12 × 10–6 × 9.02 = 4.9 × 10–4 J A1 [2]

(ii) energy dissipated in (resistance of) wire/as a spark B1 [1]




5 (a) supply connected correctly (to left & right) B1 load connected correctly (to top & bottom) B1 [2]

(b) e.g. power supplied on every half-cycle

greater average/mean power (any sensible suggestion, 1 mark) B1 [1]

(c) (i) reduction in the variation of the output voltage/current B1 [1] (ii) larger capacitance produces more smoothing M1

either product RC larger or for the same load A1 [2]

6 (a) unit of magnetic flux density B1

field normal to (straight) conductor carrying current of 1 A M1 force per unit length is 1 N m–1 A1 [3]

(b) (i) force on particle always normal to direction of motion M1

(and speed of particle is constant) magnetic force provides the centripetal force A1 [2]

(ii) mv 2/r = Bqv M1

r = mv/Bq A0 [1] (c) (i) the momentum/speed is becoming less M1

so the radius is becoming smaller A1 [2] (ii) 1. spirals are in opposite directions M1

so oppositely charged A1 [2] 2. equal initial radii M1

so equal (initial) speeds A1 [2]




7 (a) (i) packet/quantum of energy M1 of electromagnetic radiation A1 [2]

(ii) minimum energy to cause emission of an electron (from surface) B1 [1]

(b) (i) hc/λ = Φ + Emax M1 c and h explained A1 [2]

(ii) 1. either when 1/λ = 0, Φ = –Emax or evidence of use of x-axis intercept from graph

or chooses point close to the line and substitutes values of 1/λ and

Emax into hc/λ = Φ + Emax C1 Φ = 4.0 × 10–19 J (allow ±0.2 × 10–19 J) A1 [2]

2. either gradient of graph is 1/hc C1

gradient = 4.80 × 1024 → 5.06 × 1024 M1 h = 1/(gradient × 3.0 × 108)

= 6.6 × 10–34 J s → 6.9 × 10–34 J s A1

or chooses point close to the line and substitutes values of 1/λ and

Emax into hc/λ = Φ + Emax (C1)

values of 1/λ and Emax are correct within half a square (M1)

h = 6.6 × 10–34 J s → 6.9 × 10–34 J s (A1) [3] (Allow full credit for the correct use of any appropriate method) (Do not allow ‘circular’ calculations in part 2 that lead to the same value of Planck constant that was substituted in part 1)

8 (a) (i) probability of decay (of a nucleus) M1

per unit time A1 [2]

(ii) λt½ = ln 2

λ = ln 2/(3.82 × 24 × 3600) M1 = 2.1 × 10–6 s–1 A0 [1]

(b) A = λN C1 200 = 2.1 × 10–6 × N C1 N = 9.5 × 107 ratio = (2.5 × 1025)/(9.5 × 107) = 2.6 × 1017 A1 [3]




Section B 9 (a) any value greater than, or equal to, 5 kΩ B1 [1] (b) (i) ‘positive’ shown in correct position B1 [1] (ii) V + = (500/2200) × 4.5

≈ 1 V B1 V – > V + so output is negative M1 green LED on, (red LED off) A1 [3] (allow full ecf of incorrect value of V +)

(iii) either V + increases or V + > V – M1

green LED off, red LED on A1 [2] 10 quartz/piezo-electric crystal B1

p.d. across crystal causes either centres of (+) and (–) charge to move or crystal to change shape B1 alternating p.d. (in ultrasound frequency range) causes crystal to vibrate B1 crystal cut to produce resonance B1 when crystal made to vibrate by ultrasound wave M1 alternating p.d. produced across the crystal A1 [6]

11 (a) sharpness: ease with which edges of structures can be seen B1

contrast: difference in degree of blackening between structures B1 [2]

(b) (i) I = I0 e–µx C1

I/I0 = exp(–0.20 × 8) = 0.20 A1 [2]

(ii) I/I0 = exp(–µ1 × x1) × exp(–µ 2 × x2) (could be three terms) C1

I/I0 = exp(–0.20 × 4) × exp(–12 × 4) C1

I/I0 = 6.4 × 10–22 or I/I0 ≈ 0 A1 [3] (c) (i) sharpness unknown/no B1 [1] (ii) contrast good/yes (ecf from (b)) B1 [1]




12 (a) e.g. carrier frequencies can be re-used (without interference) (M1) so increased number of handsets can be used (A1) e.g. lower power transmitters (M1) so less interference (A1) e.g. UHF used (M1) so must be line-of-sight/short handset aerial (A1) (any two sensible suggestions with explanation, max 4) B4 [4]

(b) computer at cellular exchange B1

monitors the signal power B1 relayed from several base stations B1 switches call to base station with strongest signal B1 [4]





9702 PHYSICS

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







1 Planning (15 marks) Defining the problem (3 marks) P r is the independent variable, B is the dependent variable or vary r and measure B. [1] P Keep the number of turns on the coil(s) constant. [1]

Do not accept same coil. P Keep the current in the coil constant. [1] Methods of data collection (5 marks) M1 Diagram showing coil and labelled Hall probe positioned in the centre of a coil.

Solenoids will not be credited. [1] M2 Circuit diagram for coil connected to a (d.c.) power supply. [1] M3 Connect Hall probe to voltmeter/c.r.o. Allow galvanometer but do not allow ammeter. [1] M4 Measure diameter or radius with a ruler/vernier callipers. [1] M5 Method to locate centre of coil. e.g. determine max VH; cross rules; projection [1] Method of analysis (2 marks) A Plot a graph of B against 1/r [allow lg B against lg r or other valid graph] [1] A Relationship is valid if the graph is a straight line passing through the origin [if lg-lg then straight line with gradient = –1 (ignore reference to y-intercept)] [1] Safety considerations (1 mark) S Precaution linked to (large) heating of coil, e.g. switch off when not in use to avoid

overheating coil; do not touch coil because it is hot. [1] Additional detail (4 marks) D Relevant points might include [4] 1 Use large current/large number of turns to create a large magnetic field. 2 Use of rheostat to keep current constant in coil. 3 Monitor constant current with ammeter to check current is constant. 4 Hall probe at right angles to direction of magnetic field/plane of coil. 5 Reasoned method to keep Hall probe in constant orientation (e.g. use of set square, fix to

rule, optical bench or equivalent). 6 B is proportional to voltage across Hall probe/calibrate Hall probe in a known magnetic field. 7 Repeat experiment with Hall probe reversed and average. 8 Repeat measurement for r or d and average. Do not allow vague computer methods. [Total: 15]




2 Analysis, conclusions and evaluation (15 marks)

Part Mark Expected Answer Additional Guidance

(a) A1 1.5 or 3/2 Ignore y-intercept (incorrect y-intercept will be penalised in (d)(i)).

(b) T1 T2

8.111 or 8.1106 4.38

8.258 or 8 2577 4.62

8.625 or 8.6253 5.188 (5.19)

8.827 or 8.8267 5.483 (5.48)

9.029 or 9.0294 5.771 (5.77)

9.274 or 9.2742 6.152 (6.15)

Allow a mixture of decimal places. T1 must be table values. T2 must be a minimum of 2 d.p. Ignore rounding errors.

U1 From ± 0.07 or ± 0.08, to ± 0.005 Allow more than one significant figure.

(c) (i) G1 Six points plotted correctly Must be within half a small square. Do not allow ‘blobs’ (more than half a small square). Ecf allowed from table.

U2 Error bars in lg T plotted correctly All error bars to be plotted. Must be accurate to less than half a small square.

(c) (ii) G2 Line of best fit If points are plotted correctly then lower end of line should pass between (8.0, 4.20) and (8.0, 4.28) and upper end of line should pass between (9.4, 6.32) and (9.4, 6.38). Allow ecf from points plotted incorrectly – examiner judgement.

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.

(c) (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.

U3 Uncertainty in gradient Method of determining absolute uncertainty Difference in worst gradient and gradient.

(c) (iv) C2 Negative y-intercept Must be negative. FOX does not score. Check substitution into y = mx + c. Allow ecf from (c)(iii).

U4 Uncertainty in y-intercept Uses worst gradient and point on WAL. Do not check calculation. FOX does not score.

(d) (i) C3 Method to determine k k= 102 × y-intercept [k is about 10–16, if FOX 108]

U5 Uncertainty in k Best k – worst k using y-intercept. Allow ecf for method from (c)(iv).

(d) (ii) C4 M between 2.36 × 1026 and 2.36 × 1028 given to 2 or 3 s.f.

Must be in range. Allow between 2.4 × 1026 and 2.4 × 1028 for 2 s.f.

[Total: 15]




Uncertainties in Question 2 (c) (iii) Gradient [E3] Uncertainty = gradient of line of best fit – gradient of worst acceptable line Uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient) (iv) [E4] Uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line Uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient) (d) (i) [E5] Uncertainty = best k –worst k





9702 PHYSICS








1 Planning (15 marks) Defining the problem (3 marks) P r is the independent variable, B is the dependent variable or vary r and measure B. [1] P Keep the number of turns on the coil(s) constant. [1]

Do not accept same coil. P Keep the current in the coil constant. [1] Methods of data collection (5 marks) M1 Diagram showing coil and labelled Hall probe positioned in the centre of a coil.

Solenoids will not be credited. [1] M2 Circuit diagram for coil connected to a (d.c.) power supply. [1] M3 Connect Hall probe to voltmeter/c.r.o. Allow galvanometer but do not allow ammeter. [1] M4 Measure diameter or radius with a ruler/vernier callipers. [1] M5 Method to locate centre of coil. e.g. determine max VH; cross rules; projection [1] Method of analysis (2 marks) A Plot a graph of B against 1/r [allow lg B against lg r or other valid graph] [1] A Relationship is valid if the graph is a straight line passing through the origin [if lg-lg then straight line with gradient = –1 (ignore reference to y-intercept)] [1] Safety considerations (1 mark) S Precaution linked to (large) heating of coil, e.g. switch off when not in use to avoid

overheating coil; do not touch coil because it is hot. [1] Additional detail (4 marks) D Relevant points might include [4] 1 Use large current/large number of turns to create a large magnetic field. 2 Use of rheostat to keep current constant in coil. 3 Monitor constant current with ammeter to check current is constant. 4 Hall probe at right angles to direction of magnetic field/plane of coil. 5 Reasoned method to keep Hall probe in constant orientation (e.g. use of set square, fix to

rule, optical bench or equivalent). 6 B is proportional to voltage across Hall probe/calibrate Hall probe in a known magnetic field. 7 Repeat experiment with Hall probe reversed and average. 8 Repeat measurement for r or d and average. Do not allow vague computer methods. [Total: 15]






(a) A1 1.5 or 3/2 Ignore y-intercept (incorrect y-intercept will be penalised in (d)(i)).

(b) T1 T2

8.111 or 8.1106 4.38

8.258 or 8 2577 4.62

8.625 or 8.6253 5.188 (5.19)

8.827 or 8.8267 5.483 (5.48)

9.029 or 9.0294 5.771 (5.77)

9.274 or 9.2742 6.152 (6.15)

Allow a mixture of decimal places. T1 must be table values. T2 must be a minimum of 2 d.p. Ignore rounding errors.



U2 Error bars in lg T plotted correctly All error bars to be plotted. Must be accurate to less than half a small square.



Line should be clearly labelled or dashed. Examiner judgement on worst acceptable line. Lines must cross. Mark scored only if error bars are plotted.


U3 Uncertainty in gradient Method of determining absolute uncertainty Difference in worst gradient and gradient.

(c) (iv) C2 Negative y-intercept Must be negative. FOX does not score. Check substitution into y = mx + c. Allow ecf from (c)(iii).

U4 Uncertainty in y-intercept Uses worst gradient and point on WAL. Do not check calculation. FOX does not score.

(d) (i) C3 Method to determine k k= 102 × y-intercept [k is about 10–16, if FOX 108]

U5 Uncertainty in k Best k – worst k using y-intercept. Allow ecf for method from (c)(iv).

(d) (ii) C4 M between 2.36 × 1026 and 2.36 × 1028 given to 2 or 3 s.f.

Must be in range. Allow between 2.4 × 1026 and 2.4 × 1028 for 2 s.f.

[Total: 15]




Uncertainties in Question 2 (c) (iii) Gradient [E3] Uncertainty = gradient of line of best fit – gradient of worst acceptable line Uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient) (iv) [E4] Uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line Uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient) (d) (i) [E5] Uncertainty = best k –worst k





9702 PHYSICS








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

P A is the independent variable and V is the dependent variable or vary A and measure V. [1] P Keep the number of turns on coil Y or coil X constant. [1] P Keep the current in coil X constant. [1] Methods of data collection (5 marks) M1 Two independent coils labelled X and Y; coil Y wound over coil X. [1] M2 Alternating power supply/signal generator connected to coil X. [1] M3 Coil Y connected to voltmeter/c.r.o. in a workable circuit. [1] M4 Measure diameter/radius/lengths with a ruler/vernier callipers. [1] M5 Method to determine area. [1] Method of analysis (2 marks) A Plot a graph of V against A. [1] A Relationship valid if straight line through origin. [1] Safety considerations (1 mark) S Precaution linked to (large) current in coil/heating, e.g. switch off when not in use to avoid

overheating coil; do not touch coil because it is hot. [1] Additional detail (4 marks) D Relevant points might include [4] 1 Use large current in coil X/large number of turns/high frequency a.c. to produce measurable

e.m.f. 2 Detail on measuring e.m.f., e.g. height × y-gain on CRO. 3 Keep frequency of power supply constant. 4 Use of rheostat to keep current constant in coil X. 5 Monitor with a.c. ammeter. 6 Avoid other alternating magnetic fields. 7 Repeat measurement for r or d or lengths and average. Do not allow vague computer methods. [Total: 15]






(a) A1 2gh Allow 2hg.

(b) T1 T2

0.111 or 0.1111 1.320 – 1.321

0.200 or 0.2000 2.295 – 2.310

0.273 or 0.2727 3.189 – 3.204

0.333 or 0.3333 3.842 – 3.846

0.385 or 0.3846 4.41 – 4.45

0.429 or 0.4286 4.84 – 4.94

T1 for ratio values: Ignore sf in 2nd row. T2 for v

2. Rows 1–4 to 3 s.f. or 4 s.f. Rows 5–6 to 2 s.f. or 3 s.f.



U2 Error bars in v2 plotted correctly All error bars to be plotted. Check third and fourth plot. Must be accurate to less than half a small square.



Line should be clearly labelled or dashed. Should pass from top of top error bar to bottom of bottom error bar or bottom of top error bar to top of bottom error bar. Mark scored only if error bars are plotted.


U3 Uncertainty in gradient Method of determining absolute uncertainty. Difference in worst gradient and gradient.

(d) C2 g = gradient / 2h = gradient / 1.2 Gradient must be used. Allow ecf from (c)(iii).

U4 Absolute uncertainty in g Uses worst gradient. Do not check calculation.

(e) C3 Ratio = 0.6/(0.6 + 1.8) = 0.25 Expect to see 1.00 added and largest m.

C4 Between 1.66 and 1.70 given to 2 or 3 s.f.

v = 6.025.02 ××× g = g×3.0

or v = 0.25 gradient ×

or v = 2v read from graph for ratio 0.25.

Must be in range. Allow 1.7.

U5 Determines absolute uncertainty Allow ecf. Expect to see difference between best and worst values.

[Total: 15]




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) (d) [U4] Uncertainty = best g – worst g Uncertainty = uncertainty in gradient / 1.2

Uncertainty = gm

m∆

(e) [U5] Uncertainty = best v – worst v

Uncertainty = v

m

m∆×

2

1

Uncertainty = vg

g∆×

2

1

9702 w11 ms_all

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