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
examination.
syllabuses.

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
examination.
syllabuses.

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.
examination.
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)kgm–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 = 12Nm 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 cos60° = 7.5ms–1
A1 [1]
(ii) vertical velocity = 15 sin60° = 13ms–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
= 4.6ms–1
A0 [1]
(c) (i) change in momentum = 60 × 10–3
[–4.6 – 7.5] C1
= (–)0.73Ns 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 = 15.3ms–1
A0 [2]
(ii) P = 0.5 m v2
/ t C1
m / t = 110 × 103
/ [0.25 × 0.5 × (15.3)2
] C1
= 3740kgs–1
A1 [3]
5 (a) ohm = volt / ampere B1 [1]
(b) ρ = RA / l or unit is Ωm C1
units: VA–1
m2
m–1
= NmC–1
A–1
m2
m–1
C1
= kgm2
s–2
A–1
s–1
A–1
m2
m–1
= kgm3
s–3
A–2
A1 [3]
(c) (i) ρ = [3.4 × 1.3 × 10–7
] / 0.9 C1
= 4.9 × 10–7
(Ωm) 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 = 80Nm–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 = 53.3Nm–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
examination.
syllabuses.

1 (a) average velocity = 540 / 30 C1
= 18ms–1
A1 [2]
(b) velocity zero at time t = 0 B1
positive value and horizontal line for time t = 5s to 35s B1
line / curve through v = 0 at t = 45s to negative velocity B1
negative horizontal line from 53 s with magnitude less than positive value and
horizontal line to time = 100s 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 = ½mv2
A1 [3]
(c) (i) work done = ½ × 1500 × [(30)2
– (15)2
] (=506250) C1
distance = WD / F = 506250 / 3800 = 133m A1 [2]
or F = ma a = 2.533 (ms–2
) C1
v2
= u2
+ 2as s = 133m 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
= 75000Vm–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
= 2500Hz 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
= 335ms–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 0.1c up to 0.99c 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
examination.
syllabuses.

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: 7.5cos40° / 7.5sin50° = 5.7(45) / 5.75 not 5.8N A1 [1]
(ii) vertically: 7.5sin40° / 7.5cos50° = 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 cos40° = T1 cos50° (B1)
T1 sin50° + T2 sin40° = 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 (6.32ms–2
)
and then s = (18 × 2.85) + ½ × 6.32 (2.85)2
= 25.7m}
(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 =7900N or F = ½ × 1250 × (18)2
/ 25.7 = 7900N 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 ms–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 = 2.8cm 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 = 3.96cm (4.0cm) 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
examination.
syllabuses.

1 (a) (ii) Value of k in range: 50cm ≤ k ≤ 100cm. [1]
(b) (iii) Values of d and D both with unit and d in range 4.0 ≤ d ≤ 6.0cm. [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 ≥ 40cm. [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 ± 0.05m–1
(0.0005cm–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 ± 5cm with consistent unit. [1]
[Total: 20]
2 (a) Measurement of all raw values of t to nearest 0.01mm or 0.001mm and t in range
0.10 ≤ t ≤ 0.50mm. [1]
(b) (i) Value of L in range 26.0cm ≤ L ≤ 30.0cm with consistent unit to nearest mm. [1]
(ii) Absolute uncertainty in L in range 1–2mm (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 0.7s ≤ T ≤ 1.5s with consistent unit. Supervisor help –1. [1]
Evidence of repeats. [1]
(f) Second value of L in range 5cm ≤ L ≤ 15cm. [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
maximum raw mark 40
examination.
syllabuses.

1 (a) (i) All raw values of d in the range 0.15mm ≤ d ≤ 0.40mm to 0.01mm with unit, or
supervisor's value ± 0.10mm [1]
(b) (iii) l and V with unit. Value of l in the range 40.0cm ≤ l ≤ 60.0cm. [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 ≥ 50cm. [1]
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).
All values of raw l must be given to 0.001m.
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]
square in both the x and y directions.
Quality: [1]
(0.002cm–1
) of 1/l of a straight line.
length.
candidate.

(d) (iii) Gradient: [1]
Intercept: [1]
Either:
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
Ωm
(0.2 × 10–6
– 2 × 10–6
Ωm). [1]
[Total: 20]
2 (a) Measurement of all raw w to nearest mm in range 2.0 ≤ w ≤ 3.5cm. [1]
(b) (iii) Value of l in range 25cm ≤ l ≤ 50cm 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 0.3s < T <1.5s. [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]

(g)
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
maximum raw mark 40
examination.
syllabuses.

1 (a) Measurement for V in range +0.10V to +0.90V, 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 0.33kΩ or less and 4.7kΩ or more.
There must be some distinguishing mark between the quantity and the unit.
All raw values of V must be given to the same precision and at least 2 d.p.
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.
Scale markings must be no more than 3 large squares apart.
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 ± 0.04V on the V axis from a straight line.
length.
candidate. There must be 5 points left after the anomalous point is disregarded.

(d) (iii) Gradient: [1]
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:
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 1.0V to 2.0V, with unit V. [1]
[Total: 20]
2 (a) Value of t in range 0.01 to 0.05mm, with unit. [1]
(b) (i) Value of w in range 5 to 15mm. 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 1mm [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
maximum raw mark 40
examination.
syllabuses.

1 (a) Raw value(s) of h to the nearest mm in range 5–15cm. [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]
There must be some distinguishing mark between the quantity and the unit,
e.g. m / kgm–1
but accept m (kg m–1
).
d d
All values of raw d must be given to the nearest mm.
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]
square in both x and y directions.
Quality: [1]
(0.005cm–1
) of 1/d of a straight line.
length.
candidate.

(iii) Gradient: [1]
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 0.08kg < m < 1.0kg with consistent unit. [1]
[Total: 20]
2 (a) (ii) Value of m in g or kg. 45g ≤ m ≤ 55g. [1]
(iii) Absolute uncertainty in m in range 1–5g 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]

(h)
‘Few readings’/ ‘take more
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.
0.1o
C, 0.5o
C
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
maximum raw mark 40
examination.
syllabuses.

1 (b) Measurement for H in range 0.200m to 0.900m. [1]
(c) (ii) First measurement of m, to nearest 0.001kg and in the range 0.045 to 0.055kg. [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 0.070kg or less, and 0.220kg or more.
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
.
All values of h must be given to the nearest mm.
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).
Plotting: [1]
All observations must be plotted.
square.
Quality: [1]
Scatter of points must be less than ± 50m─2
(± 0.005cm─2
) on the 1/y2
axis about a
straight line. All points must be plotted (at least 5) for this mark to be scored.
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
m─2
) and q (e.g. m─2
). [1]
[Total: 20]
2 (b) 0.250m ≤ a ≤ 0.350m and 0.450m ≤ b ≤ 0.550m, both with a correct and consistent [1]
unit.
Values of a and b given to nearest mm e.g. 0.350m or 35.0cm. [1]
(c) (ii) Value of R in range 0.05m to 0.50m (5cm to 50cm). [1]
Evidence of repeats (credit evidence here or in (f)). [1]
(d) Percentage uncertainty in R based on absolute uncertainty in range 0.002m to 0.01m [1]
(2mm to 10mm).
(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)
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
examination.
syllabuses.

Section A
1 (a) gravitational force provides the centripetal force B1
GMm/r 2
= mrω2
(must be in terms of ω) B1
r 3
ω2
= GM and GM is a constant B1 [3]
(b) (i) 1. for Phobos, ω = 2π/(7.65 × 3600) C1
= 2.28 × 10–4
rads–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
rads–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
3
kT 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
examination.
syllabuses.

Section A
1 (a) gravitational force provides the centripetal force B1
GMm/r 2
= mrω2
(must be in terms of ω) B1
r 3
ω2
= GM and GM is a constant B1 [3]
(b) (i) 1. for Phobos, ω = 2π/(7.65 × 3600) C1
= 2.28 × 10–4
rads–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
rads–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
3
kT 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

9702 PHYSICS
examination.
syllabuses.

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
ms–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
ms–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 2
x0
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 2
x0 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
(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 Nm–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
Js → 6.9 × 10–34
Js 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
Js → 6.9 × 10–34
Js (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, 5kΩ B1 [1]
(b) (i) ‘positive’ shown in correct position B1 [1]
(ii) V +
= (500/2200) × 4.5
≈ 1V 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
examination.
syllabuses.

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
(d) (i) [E5]
Uncertainty = best k –worst k

9702 PHYSICS
maximum raw mark 30
examination.
syllabuses.

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]
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]
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]
S Precaution linked to (large) heating of coil, e.g. switch off when not in use to avoid
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.
[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.
(c) (i) G1 Six points plotted correctly Must be within half a small square.
Do not allow ‘blobs’ (more than half a small
square).
U2 Error bars in lg T plotted correctly All error bars to be plotted. Must be
accurate to less than half a small square.
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.
Examiner judgement on worst acceptable
line. Lines must cross. Mark scored only if
error bars are plotted.
penalise POT.
U3 Uncertainty in gradient Method of determining absolute uncertainty
(c) (iv) C2 Negative y-intercept Must be negative. FOX does not score.
Check substitution into y = mx + c.
U4 Uncertainty in y-intercept Uses worst gradient and point on WAL.
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]

(c) (iii) Gradient [E3]
(iv) [E4]
Uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line
(d) (i) [E5]
Uncertainty = best k –worst k

9702 PHYSICS
maximum raw mark 30
examination.
syllabuses.

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]
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]
A Plot a graph of V against A. [1]
A Relationship valid if straight line through origin. [1]
S Precaution linked to (large) current in coil/heating, e.g. switch off when not in use to avoid
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.
[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 v2
.
Rows 1–4 to 3 s.f. or 4 s.f.
Rows 5–6 to 2 s.f. or 3 s.f.
(c) (i) G1 Six points plotted correctly Must be within half a small square. Do not
allow ‘blobs’ (more than half a small
square).
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.
of line should pass between (0.10, 1.16)
and (0.10, 1.24) and upper end of line
should pass between (0.45, 5.12) and
(0.45, 5.20). Allow ecf from points plotted
incorrectly – examiner judgement.
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.
penalise POT.
U3 Uncertainty in gradient Method of determining absolute uncertainty.
(d) C2 g = gradient/2h = gradient/1.2 Gradient must be used.
U4 Absolute uncertainty in g Uses worst gradient.
(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 = 0.25gradient ×
or v = 2
v 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]

(c) (iii) Gradient [U3]
(d) [U4]
Uncertainty = best g – worst g
Uncertainty = uncertainty in gradient/1.2
Uncertainty = g
m
m∆
(e) [U5]
Uncertainty = best v – worst v
Uncertainty = v
m
m∆
×
2
1
Uncertainty = v
g
g∆
×
2
1

9702 w11 ms_all

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