9702 s12 ms_all

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
GCE Advanced Subsidiary Level and GCE Advanced Level
MARK SCHEME for the May/June 2012 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 May/June 2012 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 – May/June 2012 9702 11
© University of Cambridge International Examinations 2012
Question
Number
Key
Question
Number
Key
1 A 21 D
2 D 22 B
3 C 23 C
4 D 24 B
5 C 25 B
6 C 26 C
7 D 27 B
8 B 28 B
9 D 29 C
10 D 30 C
11 B 31 D
12 A 32 C
13 C 33 C
14 D 34 A
15 D 35 C
16 B 36 B
17 C 37 A
18 B 38 C
19 D 39 A
20 D 40 B

9702 PHYSICS
examination.
syllabuses.

Question
Number
Key
Question
Number
Key
1 C 21 C
2 D 22 C
3 D 23 C
4 B 24 B
5 B 25 C
6 A 26 D
7 A 27 C
8 D 28 D
9 A 29 C
10 A 30 B
11 B 31 C
12 D 32 C
13 A 33 D
14 B 34 B
15 D 35 D
16 D 36 D
17 B 37 D
18 D 38 C
19 B 39 B
20 D 40 B

9702 PHYSICS
examination.
syllabuses.

Question
Number
Key
Question
Number
Key
1 C 21 B
2 A 22 C
3 D 23 B
4 C 24 D
5 D 25 D
6 D 26 B
7 C 27 C
8 D 28 C
9 B 29 B
10 A 30 C
11 D 31 D
12 B 32 A
13 D 33 C
14 C 34 C
15 D 35 B
16 C 36 C
17 B 37 C
18 D 38 A
19 B 39 B
20 B 40 A

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) (i) V units: m3
(allow metres cubed or cubic metres) A1 [1]
(ii) Pressure units: kgms–2
/ m2
(allow use of P = ρgh) M1
Units: kgm–1
s–2
A0 [1]
(b) V / t units: m3
s–1
B1
Clear substitution of units for P, r4
and l M1
msm
msmkg
8 13
421
1 −
−−
−
==
ltV
rP
C
4
π
Units: kgm–1
s–1
A1 [3]
(8 or π in final answer –1. Use of dimensions max 2/3)
2 (a) (i) v = u + at C1
= 4.23 + 9.81 × 1.51 M1
= 19.0(4) ms–1
(Allow 2 s.f.) A0 [2]
(Use of –g max 1/2. Use of g = 10 max 1/2. Allow use of 9.8. Allow 19 ms–1
)
(ii) either s = ut + ½ at2
(or v2
= u2
+ 2as etc.)
= 4.23 × 1.51 + 0.5 × 9.81 × (1.51)2
C1
= 17.6m (or 17.5m) A1 [2]
(Use of –g here wrong physics (0/2))
(b) (i) F = ∆P / ∆t need idea of change in momentum C1
= [0.0465 × (18.6 + 19)] / 12.5 × 10–3
C1
= 140N A1
(Use of – sign max 2/4. Ignore –ve sign in answer)
Direction: upwards B1 [4]
(ii) h = ½ × (18.6)2
/ 9.81 C1
= 17.6 m (2 s.f. –1) A1 [2]
(Use of 19 ms–1
, 0/2 wrong physics)
(c) either kinetic energy of the ball is not conserved on impact
or speed before impact is not equal to speed after hence inelastic B1 [1]
3 (a) Resultant force (and resultant torque) is zero B1
Weight (down) = force from/due to spring (up) B1 [2]
(b) (i) 0.2, 0.6, 1.0s (one of these) A1 [1]
(ii) 0, 0.8s (one of these) A1 [1]
(iii) 0.2, 0.6, 1.0s (one of these) A1 [1]

(c) (i) Hooke’s law: extension is proportional to the force (not mass) B1
Linear/straight line graph hence obeys Hooke’s law B1 [2]
(ii) Use of the gradient (not just F = kx) C1
K = (0.4 × 9.8) / 15 × 10–2
M1
= 26(.1) Nm–1
A0 [2]
(iii) either energy = area to left of line or energy = ½ ke2
C1
= ½ × [(0.4 × 9.8) / 15 × 10–2
] × (15 × 10–2
)2
C1
= 0.294J (allow 2 s.f.) A1 [3]
4 (a) (i) R = V2
/ P or P = IV and V = IR C1
= (220)2
/ 2500
= 19.4Ω (allow 2 s.f.) A1 [2]
(ii) R = ρl / A C1
l = [19.4 × 2.0 × 10–7
] / 1.1 × 10–6
C1
= 3.53m (allow 2 s.f.) A1 [3]
(b) (i) P = 625, 620 or 630W A1 [1]
(ii) R needs to be reduced C1
Either length ¼ of original length
or area 4× greater
or diameter 2× greater A1 [2]
5 (a) (i) sum of e.m.f.’s = sum of p.d.’s around a loop/circuit B1 [1]
(ii) energy B1 [1]
(b) (i) 2.0 = I × (4.0 + 2.5 + 0.5) C1
I = 0.286A (allow 2 s.f.) A1 [2]
(If total resistance is not 7Ω, 0/2 marks)
(ii) R = [0.90 / 1.0] × 4 (= 3.6) C1
V = I R = 0.286 × 3.6 = 1.03V A1 [2]
(If factor of 0.9 not used, then 0/2 marks)
(iii) E = 1.03V A1 [1]
(iv) either no current through cell B
or p.d. across r is zero B1 [1]
6 (a) (i) coherence: constant phase difference M1
between (two) waves A1 [2]
(ii) path difference is either λ or nλ
or phase difference is 360° or n × 360° or n2π rad B1 [1]

(iii) path difference is either λ/2 or (n + ½) λ
or phase difference is odd multiple of either 180° or π rad B1 [1]
(iv) w = λD / a C1
= [630 × 10–9
× 1.5] / 0.45 × 10–3
C1
= 2.1 × 10–3
m A1 [3]
(b) no change to dark fringes B1
no change to separation/fringe width B1
bright fringes are brighter/lighter/more intense B1 [3]
7 (a) (i) 2 protons and 2 neutrons B1 [1]
(ii) e.g. positively charged 2e
mass 4u
constant energy
absorbed by thin paper or few cm of air (3cm → 8cm)
(not low penetration)
highly ionizing
deflected in electric/magnetic fields
(One mark for each property, max 2) B2 [2]
(b) mass-energy is conserved B1
difference in mass ‘changed’ into a form of energy B1
energy in the form of kinetic energy of the products / γ-radiation
photons / e.m. radiation B1 [3]

9702 PHYSICS
examination.
syllabuses.

1 (a)
l8
4
C
rP
t
V π
=
C = [π × 2.5 × 103
× (0.75 × 10–3
)4
] / (8 × 1.2 × 10–6
× 0.25) C1
= 1.04 × 10–3
N s m–2
A1 [2]
(b) 4 × %r C1
%C = %P + 4 × %r + %V/t + %l
= 2% + 5.3% + 0.83% + 0.4% (= 8.6%) A1
∆C = ± 0.089 × 10–3
N s m–2
A1 [3]
(c) C = (1.04 ± 0.09) × 10–3
N s m–2
A1 [1]
2 (a) (i) v2
= u2
+ 2as
= (8.4)2
+ 2 × 9.81 × 5 C1
= 12.99 m s–1
(allow 13 to 2 s.f. but not 12.9) A1 [2]
(ii) t = (v – u) / a or s = ut + ½at2
= (12.99 – 8.4) / 9.81 or 5 = 8.4t + ½ × 9.81t2
M1
t = 0.468s A0 [1]
(b) reasonable shape M1
suitable scale A1
correctly plotted 1st
and last points at (0,8.4) and (0.88 – 0.96,0)
with non-vertical line at 0.47s A1 [3]
(c) (i) 1. kinetic energy at end is zero so ∆KE = ½ mv2
or ∆KE = ½ mu2
– ½ mv2
C1
= ½ × 0.05 × (8.4)2
= (–) 1.8J A1 [2]
2. final maximum height = (4.2)2
/ (2 × 9.8) = (0.9 (m))
change in PE = mgh2 – mgh1 C1
= 0.05 × 9.8 × (0.9 – 5) C1
= (–) 2.0J A1 [3]
(ii) change is – 3.8 (J) B1
energy lost to ground (on impact) / energy of deformation of the ball /
thermal energy in ball B1 [2]
3 (a) A body continues at rest or constant velocity unless acted on by a resultant
(external) force B1 [1]
(b) (i) constant velocity/zero acceleration and therefore no resultant force M1
no resultant force (and no resultant torque) hence in equilibrium A1 [2]
(ii) component of weight = 450 × 9.81 × sin 12° (= 917.8) C1
tension = 650 + 450g sin12° = (650 + 917.8) C1
= 1600 (1570)N A1 [3]

(iii) work done against frictional force or friction between log and slope M1
output power greater than the gain in PE / s A1 [2]
4 (a) total resistance = 20 (kΩ) C1
current = 12 / 20 (mA) or potential divider formula C1
p.d. = [12 / 20] × 12 = 7.2V A1 [3]
(b) parallel resistance = 3 (kΩ) C1
total resistance 8 + 3 = 11 (kΩ) C1
current = 12 / 11 × 103
= 1.09 × 10–3
or 1.1 × 10–3
A A1 [3]
(c) (i) LDR resistance decreases M1
total resistance (of circuit) is less hence current increases A1 [2]
(ii) resistance across XY is less M1
less proportion of 12V across XY hence p.d. is less A1 [2]
5 (a) E = stress / strain B1 [1]
(b) (i) 1. diameter / cross sectional area / radius
2. original length B1 [1]
(ii) measure original length with a metre ruler / tape B1
measure the diameter with micrometer (screw gauge) B1 [2]
allow digital vernier calipers
(iii) energy = ½ Fe or area under graph or ½ kx2
C1
= ½ × 0.25 × 10–3
× 3 = 3.8 × 10–4
J A1 [2]
(c) straight line through origin below original line M1
line through (0.25, 1.5) A1 [2]
6 (a) two waves travelling (along the same line) in opposite directions overlap/meet M1
same frequency / wavelength A1
resultant displacement is the sum of displacements of each wave /
produces nodes and antinodes B1 [3]
(b) apparatus: source of sound + detector + reflection system B1
adjustment to apparatus to set up standing waves – how recognised B1
measurements made to obtain wavelength B1 [3]
(c) (i) at least two nodes and two antinodes A1 [1]
(ii) node to node = λ / 2 = 34cm (allow 33 to 35cm) C1
c = fλ C1
f = 340 / 0.68 = 500 (490 to 520)Hz A1 [3]

7 (a) W = 1 and X = 0 A1 [1]
Y = 2 A1 [1]
Z = 55 A1 [1]
(b) explanation in terms of mass – energy conservation B1
energy released as gamma or photons or kinetic energy of products or
em radiation B1 [2]

9702 PHYSICS
examination.
syllabuses.

1 (a) displacement is a vector, distance is a scalar B1
displacement is straight line between two points / distance is sum of lengths
moved / example showing difference B1 [2]
(either one of the definitions for the second mark)
(b) a body continues at rest or at constant velocity unless acted on by a resultant
(external) force B1 [1]
(c) (i) sum of T1 and T2 equals frictional force B1
these two forces are in opposite directions B1 [2]
(allow for 1/2 for travelling in straight line hence no rotation / no resultant
torque)
(ii) 1. scale vector triangle with correct orientation / vector triangle with correct
orientation both with arrows B1
scale given or mathematical analysis for tensions B1 [2]
2. T1 = 10.1 × 103
(± 0.5 × 103
)N A1
T2 = 16.4 × 103
(± 0.5 × 103
)N A1 [2]
2 (a) weight = 452 × 9.81
component down the slope = 452 × 9.81 × sin14° M1
= 1072.7 = 1070N A0 [1]
(b) (i) F = ma C1
T – (1070 + 525) = 452 × 0.13 C1
T = 1650 (1653.76)N any forces missing 1/3 A1 [3]
(ii) 1. s = ut + ½at2
hence 10 = 0 + ½ × 0.13t2
C1
t = [(2 × 10) / 0.13]1/2
= 12.4 or 12s A1 [2]
2. v = (0 + 2 × 0.13 × 10)1/2
= 1.61 or 1.6ms–1
A1 [1]
(c) straight line from the origin B1
line down to zero velocity in short time compared to stage 1 B1
line less steep negative gradient B1
final velocity larger than final velocity in the first part – at least 2× B1 [4]
3 (a) V = h × A
m = V × ρ B1
W = h × A × ρ × g B1
P = F / A B1
P = hρg
P is proportional to h if ρ is constant (and g) B1 [4]
(b) density changes with height B1
hence density is not constant with link to formula B1 [2]

4 (a) electric field strength is the force per unit positive charge (acting on a stationary
charge) B1 [1]
(b) (i) E = V / d C1
= 1200 / 14 × 10–3
= 8.57 × 104
Vm–1
A1 [2]
(ii) W = QV or W = F × d and therefore W = E × Q × d C1
= 3.2 × 10–19
× 1200
= 3.84 × 10–16
J A1 [2]
(iii) ∆U = mgh C1
= 6.6 × 10–27
× 9.8 × 14 × 10–3
= 9.06 × 10–28
J A1 [2]
(iv) ∆K = 3.84 × 10–16
– ∆U
= 3.84 × 10–16
J A1 [1]
(v) K = ½mv2
C1
v = [(2 × 3.8 × 10–16
) / 6.6 × 10–27
]1/2
= 3.4 × 105
ms–1
A1 [2]
5 (a) (i) sum of currents into a junction = sum of currents out of junction B1 [1]
(ii) charge B1 [1]
(b) (i) ΣE = ΣIR
20 – 12 = 2.0(0.6 + R) (not used 3 resistors 0/2) C1
R = 3.4Ω A1 [2]
(ii) P = EI C1
= 20 × 2
= 40W A1 [2]
(iii) P = I2
R C1
P = (2)2
× (0.1 + 0.5 + 3.4)
= 16W A1 [2]
(iv) efficiency = useful power / output power C1
24 / 40 = 0.6 or 12 × 2 / 20 × 2 or 60% A1 [2]

6 (a) (i) diffraction bending/spreading of light at edge/slit B1
this occurs at each slit B1 [2]
(ii) constant phase difference between each of the waves B1 [1]
(iii) (when the waves meet) the resultant displacement is the sum of the
displacements of each wave B1 [1]
(b) dsinθ = nλ
n = d / λ = 1 / 450 × 103 × 630 × 10–9
C1
n = 3.52 M1
hence number of orders = 3 A1 [3]
(c) λ blue is less than λ red M1
more orders seen A1
each order is at a smaller angle than for the equivalent red A1 [3]
7 (a) thin paper reduces count rate hence α B1
addition of 1cm of aluminium causes little more count rate reduction hence only
other radiation is γ B1 [2]
(b) magnetic field perpendicular to direction of radiation B1
look for a count rate in expected direction / area if there were negatively
charged radiation present. If no count rate recorded then β not present. B1 [2]

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

1 (a) (iii) Value of x in the range 0.50 – 0.60 m. [1]
(b) (ii) Value of T with unit: 0.9s < T <1.3s. [1]
Evidence of repeats. [1]
(c) Six sets of readings of x and T scores 4 marks, five sets scores 3 marks etc.
Incorrect trend –1. Minor help from Supervisor –1; major help –2. [4]
Range of x at least 25cm. [1]
Column headings:
Each column heading must contain a quantity and a unit where appropriate. [1]
The unit must conform to accepted scientific convention e.g. x/m or x(m) or x in m.
Consistency of presentation of raw readings: [1]
All values of x must be given to the nearest mm.
Significant figures: [1]
Significant figures for √x should be the same as, or one more than, s.f. for x.
Calculation: √x 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 on the grid occupy at least half the
graph grid in both x and y directions.
Scales must be labelled with the quantity that is being plotted.
Scale markings should not be greater than three large squares apart.
Plotting of points: [1]
All the observations in the table must be plotted.
Check the points are plotted correctly.
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. Judge by
the scatter of all the points about a straight line. All points must be within 0.04 m½
(0.4 cm½
) on the √x axis from a straight line.
(ii) Line of best fit: [1]
Judge by the 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 if clearly indicated (e.g. circled or labelled) by the candidate.
Line must not be kinked or thicker than half a small square.

(iii) Gradient: [1]
The hypotenuse of the triangle must be at least half the length of the drawn line.
Both read-offs must be accurate to half a small square in both the x and y directions.
Do not allow ∆x/∆y.
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 in both the x and y directions. Allow ecf of
gradient value.
Or:
Check the read-off of the intercept directly from the graph.
(e) Value of P = candidate’s gradient and Q = value of candidate’s intercept. Do not allow
fractions. [1]
Unit for P (sm–½
or scm–½
or smm–½
) consistent with value, and Q (s). [1]
[Total: 20]
2 (a) (iii) Value of F0 with unit. [1]
(iv) Absolute uncertainty in F0 in range 0.4 – 1N.
If repeated readings have been taken, then the uncertainty can be half the range.
Correct method of calculation of percentage uncertainty. [1]
(v) Value of µ given to 2 or 3 s.f. [1]
(b) (ii) Value of θ with unit to the nearest degree. [1]
(iii) Correct calculation of (sin θ + µ cos θ). [1]
(c) (ii) Value of F. [1]
(d) Second value of θ. [1]
Second value of θ < first value of θ. [1]
Second value of F < first value of F. [1]
Allow F2 > F1 if θ2 > θ1.
(e) (i) Correct calculation of two values of k. [1]
(ii) Sensible comment relating to the calculated values of k, testing against a specified
criterion. [1]

(f)
(i) Limitations 4 max. (ii) Improvements 4 max. No credit/not enough
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 some parts of board rougher
than others/
surface of board is uneven/
board not flat
method to ensure same
section of board used in each
experiment (e.g. mark one
section)
board is rough/
there is friction between the
block and the board/
use a smoother surface/
references to oil/lubricants
C large (percentage) uncertainty
in F
use larger/heavier masses values of F very similar
D difficulty in arranging newton-
meter parallel to board/pulling
in line with board
use (long) piece of string to
connect the newton-meter to
the block
newton-meter touching board
when attached
E block moves suddenly/without
warning (so difficult to read
newton-meter at the instant
the block starts to move)
value of F changes when
block moves
use system of pulley and
weights/ sand to measure F/
use a newton-meter with a
max hold facility/
use video and playback/
use force sensor and
datalogger/computer
F board tends to slip/
board not stable/
supporting block can topple
method described to secure
board/block/support e.g.
clamp the board, fix the
supporting block to the bench
with tape/blu-tack
G cannot zero newton-meter
when used horizontally
use system of pulley and
weights/ sand to measure F/
use force sensor and
datalogger/computer
zero error in newton-meter
Ignore ‘parallax problems’, ‘use assistant’ or references to draughts, fans, a.c.
[Total: 20]

9702 PHYSICS
maximum raw mark 40
examination.
syllabuses.

1 (a) Value of L in range 0.80m > L > 0.60m. Consistent with unit. [1]
(b) (iii) Value of h0, less than 50cm, to the nearest mm. [1]
(c) Six sets of readings of d and h scores 5 marks, five sets scores 4 marks etc.
Help from Supervisor –1. [5]
Range of d: [1]
To include 25.0cm (0.250m) or more and 10.0cm (0.100m) or less
Column headings: [1]
Each column heading must contain a quantity and a unit
The unit must conform to accepted scientific convention e.g. d /m, d(m) or d in m,
(h – h0)/m, (L/2 – d)2
/m2
Consistency: [1]
All values of d and h must be given to the nearest mm.
All values of (L/2 – d)2
to 2 or 3 s.f.
Calculation: [1]
Values of (L/2 – d)2
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 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.
All observations in the table must be plotted.
Diameter of plots must be < half a small square (no blobs).
Plots must be accurate to half a small square.
Quality: [1]
All points in the table must be plotted (at least 5) for this mark to be awarded. Scatter of
points must be less than 0.5cm (0.005m) of (h – h0) of a straight line.
Judge by balance of all points on the grid about the candidate’s line (at least 5 points).
Allow one anomalous point only if clearly indicated by the candidate.

(iii) Gradient: [1]
Both read-offs must be accurate to half a small square in both x and y directions.
Do not allow ∆x / ∆y.
y-intercept: [1]
Either:
Check correct read off from a point on the line and substituted into y = mx + c.
Read off must be accurate to half a small square in both x and y directions.
Or:
Check read-off of the intercept directly from the graph.
(e) Value of a = candidate’s gradient. Value of b = candidate’s intercept. [1]
Unit for a (e.g. m) and b (e.g. m2
) consistent with values. [1]
[Total: 20]
2 (b) (i) Value of ball diameter or d to the nearest 0.1mm (or 0.01mm). [1]
Values of ball diameter and d in range 5mm < d < 25mm. [1]
(ii) Absolute uncertainty is between 2mm and 5mm. [1]
If repeated readings have been taken, then the absolute uncertainty can be half the
range. Correct method shown to find the percentage uncertainty.
(iii) Correct calculation of A with consistent unit. [1]
(c) (ii) Value of F, with unit. [1]
Evidence of repeat measurements of F here or in (d)(ii). [1]
(d) (ii) Second value of d. [1]
Second value of A is given to the same number of s.f. (or one more s.f.) than d2. [1]
Second value of F. [1]
Quality: When d increases (second d value is larger than first d value) F also increases
(second F value is larger than first F value) and vice versa. [1]
(e) (i) Two values of k calculated correctly. [1]
(ii) Sensible comment relating to the calculated values of k, testing against a criterion
specified by the candidate. [1]

(f)
A two results not enough take more readings and plot a
graph/
compare
‘repeat readings’ on its own/
few readings/
(calculate) average k/
only one reading
B difficult to form a perfect
sphere or disc/diameter of
sphere or disc varied
method to make uniform
spheres/discs e.g. moulds
pre-sized spheres/
repeat diameter
measurement and average
C reason for difficulty in
measuring d e.g. viewed
through ruler/parallax error in
d
method to improve
measurement of d e.g.
travelling microscope
eyes in line
D difficult to pull newton-meter
parallel to ruler/bench
method to ensure force is
parallel to ruler e.g. use a
long string/pulley and
weights*
E difficult to judge reading on
newton-meter when detaches
with reason e.g. ruler moves
suddenly/without warning (so
difficult to read newton-meter
at the instant the ruler starts
to move)/force drops to zero
immediately after detachment
method to read force at
detachment e.g. newton
meter with a ‘max hold’
facility/video and playback or
freeze frame/ use system of
pulley and weights or sand to
measure F*/ use force sensor
and datalogger or computer*
video to take reading/
digital (electronic) newton
meter/
parallax related to newton
meter/
difficult to measure force/
issue of viewing ruler and
meter simultaneously
F contact area less than
calculated disc area/bulging
disc
G difficult to zero newton-meter
when used horizontally
improved method to measure
F: e.g. use system of pulley
and weights or sand*/use
force sensor with datalogger
or computer*
zero error in newton-meter/
just a pulley
Do not allow: reaction time/human error/using vernier caliper/helpers/use of micrometer
screw gauge/effect of temperature/change in stickiness of Blu-Tack.
*This answer can be credited as D, E or G (but not more than once).
[Total: 20]

9702 PHYSICS
maximum raw mark 40
examination.
syllabuses.

1 (a) (ii) Value of h0 in range 0.70m > h0 > 0.50m. Consistent with unit. [1]
(b) (iii) Value of h, less than h0 in (a)(ii), with unit. [1]
(c) Five sets of readings of h and m scores 5 marks, four sets scores 4 marks etc.
Major help from Supervisor –2 (setting up apparatus). Minor help from Supervisor –1. [5]
Range of m: [1]
To include 0.350kg.
Each column heading must contain a quantity and a unit.
The unit must conform to accepted scientific convention e.g. m / kg, m(kg) or m in kg,
(h0 – h)/m / m kg–1
, 1/m / kg–1
Consistency: [1]
All values of h must be given to the nearest mm.
Significant figures for every row of values of 1/m same as or one greater than m as recorded
in the table.
Calculation: [1]
Values of (h0 – h) /m calculated correctly.
(d) (i) Axes: [1]
Scales must be chosen so that the plotted points occupy at least half the graph grid in
both x and y directions.
Diameter of plots must be ≤ half a small square (no ‘blobs’).
Quality: [1]
All points in the table must be plotted (at least 4) for this mark to be awarded. Scatter of
points must be less than 0.5 kg–1
(0.0005 g–1
) of 1/m of a straight line.
(iii) Gradient: [1]

y-intercept: [1]
Either:
Or:
(e) Value of P = candidate’s gradient. Value of Q = candidate’s intercept. [1]
Unit for P (e.g. m) consistent with value, and Q (m kg–1
) [1]
[Total: 20]
2 (b) (ii) Value of θ0 to the nearest degree or 0.5° in range 70° # θ # 80° [1]
(iii) Value of θ with unit, θ < θ0 [1]
(iv) Correct calculation of (θ0 – θ) [1]
(c) (i) Value of raw d with unit to nearest mm. [1]
(ii) Absolute uncertainty in 2mm < d < 5mm. [1]
range. Correct method shown to find the percentage uncertainty.
(d) Second value of θ0 within 1ºC of first value of θ0. [1]
Second value of θ. [1]
Second value of ∆θ > first value of ∆θ (check second value of d > first value of d). [1]
Evidence of repeat readings of d here or in (c)(i). [1]
(e) (i) Two values of k calculated correctly. [1]
(ii) Justification of s.f. in k linked to significant figures in d and ∆θ. [1]
(iii) Sensible comment relating to the calculated values of k, testing against a criterion

(f)
A two results not enough take more readings and plot a
graph/
compare
few readings/
only one reading
B heat lost through sides and /or
bottom
method to reduce heat loss/
lag/
insulate/
polystyrene container
use of lid/
heat loss in warming bowl/cup/
draughts/
heat loss to surroundings
C temperature change is small/
∆θ values too close
time for longer/
higher starting temperature/
greater range of surface areas
D large (percentage) uncertainty
in ∆θ
use thermometer with greater
sensitivity or precision/
use thermometer that can read
to 0.1ºC
use more accurate
thermometer/
thermometer not precise
enough/
not just ‘digital thermometer’
E water in bowl barely covers
(bulb of) thermometer
use larger volume of water/
use of thermocouple/
other small temperature sensor
(e.g. probe)
not just ‘digital thermometer’
any reference to stirrer/
non-uniform temperature/
thermometer touching base
F parallax error in measuring d /
reason for difficulty in access
in measuring d
use dividers/calipers string measurements to
measure d
G difficult to mark level with
reason
method of making mark stay
e.g. depth gauge/ calibrated
marks/ marker on outside
Do not allow: use of coloured ink/reaction time/fans/draughts/water left behind/beakers not
accurate/ helpers.
[Total: 20]

9702 PHYSICS
maximum raw mark 40
examination.
syllabuses.

1 (a) (iii) Value for I0 in range 2.0 to 4.0mA, with unit. [1]
(b) (ii) First value of I (greater than I0). [1]
(c) Six sets of readings of R and I scores 5 marks, five sets scores 4 marks etc. [5]
Major help from Supervisor –2. Minor help from Supervisor –1. Incorrect trend then –1.
Range: [1]
Values of R must include 0.22kΩ or 0.33kΩ and 3.3kΩ or 4.7kΩ.
Each column heading must contain a quantity and a unit.
There must be some distinguishing mark between the quantity and the unit.
Consistency: [1]
Values of I must be given either all to the nearest 0.1mA or all to the nearest 0.01mA.
Every value of 1/R must be given to either 2 or 3 significant figures.
Calculated values: [1]
1/R calculated correctly.
(d) (i) Axes: [1]
Sensible scales must be used (no awkward scales such as 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.
Diameter of plots must be < half a small square (no blobs). Plotting must be accurate to
half a small square.
Quality: [1]
Range of I must be at least 2mA, and all points must be within 0.5mA of a straight line.
All points in the table must be plotted (at least 5) for this mark to be scored.
Judge by balance of all points on the grid about the candidate’s line (at least
5 points).There must be an even distribution of points either side of the line along the full
length.
One anomalous point is allowed only if clearly indicated (i.e. circled or labelled) by the
candidate.

(d) (iii) Gradient: [1]
The hypotenuse must be at least half the length of the drawn line.
Do not allow ∆x/∆y.
y-intercept: [1]
Either:
Correct read off from a point on the line is substituted into y = mx + c.
Or:
(e) Calculation of b is correct, [1]
i.e. b = (candidate’s gradient value)/(candidate’s intercept value).
Value for b in range 0.8kΩ to 1.2kΩ, with unit. [1]
[Total: 20]
2 (b) Value of d in range 0.80 to 0.99mm, to nearest 0.01mm, with unit. [1]
Evidence of repeated measurements for d. [1]
(c) Percentage uncertainty in d based on absolute uncertainty of 0.01mm. [1]
Correct calculation to get percentage uncertainty.
(d) (iv) Value of θ in range 91° to 180° to nearest degree, with unit. [1]
Evidence of repeated measurements for θ. [1]
(v) Correct calculation of sin(180° – θ ). [1]
sin(180° – θ ) given to 2 or 3 s.f. [1]
(e) Second value of d. [1]
Second value of θ. [1]
Quality: θ larger for smaller d. [1]
(f) (i) Correct calculation of two values of k. [1]
(ii) Valid conclusion based on the calculated values of k. Candidate must test correctly
against a stated criterion. [1]

(g)
A two results not enough take more readings and plot
a graph/
compare
few readings/
only one reading
B θ (or angle, or scale reading,
or protractor reading, or
pointer reading) is difficult to
measure, with reason linked
to rapid motion or short time
video and view playback/
slow motion camera/
video to read angle/
add a ‘max hold’ pointer/
angle sensor with data
logger (or computer)
just ‘use a computer’/
‘reading’ difficult to measure
C parallax error in θ
measurement
use mirror scale/
description of method to
reduce error
view at right angles/
trial and improvement
D θ (or reading) is difficult (or
inaccurate, or imprecise)
because pointer is thick
– use thinner pointer/
use larger scale
E pointer attachment moves description of secure method
of attachment
F – description of method of
fixing block to bench
[Total: 20]

9702 PHYSICS
maximum raw mark 40
examination.
syllabuses.

1 (b) (ii) Ammeter reading with unit, in range 1mA < I < 1A. Must see n = 3. [1]
(c) Six sets of readings of I and n scores 5 marks, five sets scores 4 marks etc.
Incorrect trend then –1. Correct trend is I decreases as n increases.
Major help from Supervisor –2. Minor help from Supervisor –1. [5]
Range of 6 or 7. [1]
Column heading: [1]
Each column heading must contain a quantity and a unit where appropriate.
The unit must conform to accepted scientific convention e.g. I / A, I (A), I in A, n + 1 /I / A–1
.
Consistency: [1]
All values of I must be given to the nearest 0.1mA or better.
Significant figures for every row of values of (n + 1) / I same as or one greater than s.f. in I,
as recorded in the table.
Calculation: [1]
Values of (n + 1) / I calculated correctly.
(d) (i) Axes: [1]
Scales must be chosen so that the plotted points must occupy at least half the graph grid
in both x and y directions.
Diameter of plots must be ≤ half a small square (no ‘blobs’).
Quality: [1]
Judge by scatter of all points about best fit line. All points in the table must be plotted for
this mark to be scored. At least 5 plots needed.
All points must be within 0.2 of n from a best line.
(iii) Gradient: [1]

y-intercept: [1]
Either:
Or:
Check read-off of intercept directly from the graph.
(e) Value of P = candidate’s gradient. Value of Q = candidate’s intercept. [1]
Do not allow fractions.
(f) Value of V in range 1V ≤ V ≤ 2V. [1]
(g) R with appropriate unit Ω or VA–1
. Expect 50Ω or 0.05VmA–1
or 0.05kΩ [1]
[Total: 20]
2 (b) (ii) Value of x with unit to the nearest mm in range: 40.0 cm ≤ x ≤ 60.0 cm. [1]
(c) (ii) Value of x1 with consistent unit. [1]
(iii) Correct calculation of d1 with unit. [1]
(iv) Absolute uncertainty in d1 in range 2 – 5 mm. [1]
range. Correct method shown to find the percentage uncertainty
(d) (ii) Value of x2. [1]
(e) (iii) Value of 1 s < T < 4 s. [1]
(f) Second value of T. [1]
Second value of T < first value of T. [1]
(g) (i) Two values of k calculated correctly. [1]
(ii) Justification of sf in k linked to significant figures in d and T. [1]
(iii) Sensible comment relating to the calculated values of k, testing against a criterion

(h)
A two results not enough take more readings with discs
of other materials / mass and
plot a graph/
compare
repeat readings
few readings
B reason why difficult to record/
measure x2/x1 directly
use a taller /narrower shape
take measurement to each end
and average/ hole in middle to
see x1/x2/ hang masses with
string
C difficult to get circular shape/flat
top/ same shape/
two shapes not the same because
of groove in 100g mass
use a mould/ use a plane
surface to press down on
plasticine
use rubber masses
D pivot/100g mass moved while x2
being determined
method of securing 100g mass
to rule/ rubber pivot
fix pivot and ruler
E oscillation not in one plane only
F difficult to determine end/start of
oscillation/ difficult to turn through
90° each time
use of (fiducial) marker(s)/
video with timer
use a protractor
[Total: 20]

9702 PHYSICS
9702/41 Paper 4 (A2 Structured Questions), maximum raw mark 100
examination.
syllabuses.

Section A
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1
either as r decreases, object/mass/body does work
or work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fall
/gravitational field strength) B1
g = GM/r2
B1
if r @ h, g is constant B1
∆EP = force × distance moved M1
= mgh A0
or ∆EP = m∆φ (C1)
= GMm(1/r1 – 1/r2) = GMm(r2 – r1)/r1r2 (B1)
if r2 ≈ r1, then (r2 – r1) = h and r1r2 = r2
(B1)
g = GM/r2
(B1)
∆EP = mgh (A0) [4]
(d) ½mv2
= m∆φ
v2
= 2 × GM/r C1
= (2 × 4.3 × 1013
) / (3.4 × 106
) C1
v = 5.0 × 103
ms–1
A1 [3]
(Use of diameter instead of radius to give v = 3.6 × 103
ms–1
scores 2 marks)
2 (a) (i) either random motion
or constant velocity until hits wall/other molecule B1 [1]
(ii) (total) volume of molecules is negligible M1
compared to volume of containing vessel A1
or
radius/diameter of a molecule is negligible (M1)
compared to the average intermolecular distance (A1) [2]
(b) either molecule has component of velocity in three directions
or c2
= cX
2
+ cY
2
+ cZ
2
M1
random motion and averaging, so <cX
2
> = <cY
2
> = <cZ
2
> M1
<c2
> = 3<cX
2
> A1
so, pV = ⅓Nm<c2
> A0 [3]
(c) <c2
> ∝ T or crms ∝ T C1
temperatures are 300K and 373K C1
crms = 580ms–1
A1 [3]
(Do not allow any marks for use of temperature in units of ºC instead of K)

3 (a) (numerically equal to) quantity of (thermal) energy required to change
the state of unit mass of a substance M1
without any change of temperature A1 [2]
(Allow 1 mark for definition of specific latent heat of fusion/vaporisation)
(b) either energy supplied = 2400 × 2 × 60 = 288000J C1
energy required for evaporation = 106 × 2260 = 240000J C1
difference = 48000J
rate of loss = 48000 / 120 = 400W A1
or energy required for evaporation = 106 × 2260 = 240000J (C1)
power required for evaporation = 240000 / (2 × 60) = 2000W (C1)
rate of loss = 2400 – 2000 = 400W (A1) [3]
4 (a) a = (–)ω2
x and ω = 2π/T C1
T = 0.60s C1
a = (4π2
× 2.0 × 10–2
) / (0.6)2
= 2.2ms–2
A1 [3]
(b) sinusoidal wave with all values positive B1
all values positive, all peaks at EK and energy = 0 at t = 0 B1
period = 0.30s B1 [3]
5 (a) force per unit positive charge acting on a stationary charge B1 [1]
(b) (i) E = Q / 4πε0r2
C1
Q = 1.8 × 104
× 102
× 4π × 8.85 × 10–12
× (25 × 10–2
)2
M1
Q = 1.25 × 10–5
C = 12.5µC A0 [2]
(ii) V = Q / 4πε0r
= (1.25 × 10–5
) / (4π × 8.85 × 10–12
× 25 × 10–2
) C1
= 4.5 × 105
V A1 [2]
(Do not allow use of V = Er unless explained)

6 (a) (i) peak voltage = 4.0V A1 [1]
(ii) r.m.s. voltage (= 4.0/√2) = 2.8V A1 [1]
(iii) period T = 20ms M1
frequency = 1 / (20 × 10–3
) M1
frequency = 50Hz A0 [2]
(b) (i) change = 4.0 – 2.4 = 1.6V A1 [1]
(ii) ∆Q = C∆V or Q = CV C1
= 5.0 × 10–6
× 1.6 = 8.0 × 10–6
C A1 [2]
(iii) discharge time = 7ms C1
current = (8.0 × 10–6
) / (7.0 × 10–3
) M1
= 1.1(4) × 10–3
A A0 [2]
(c) average p.d. = 3.2V C1
resistance = 3.2 / (1.1 × 10–3
)
= 2900Ω (allow 2800Ω) A1 [2]
7 (a) sketch: concentric circles (minimum of 3 circles) M1
separation increasing with distance from wire A1
correct direction B1 [3]
(b) (i) arrow direction from wire B towards wire A B1 [1]
(ii) either reference to Newton’s third law
or force on each wire proportional to product of the two currents M1
so forces are equal A1 [2]
(c) force always towards wire A/always in same direction B1
varies from zero (to a maximum value) (1)
variation is sinusoidal / sin2
(1)
(at) twice frequency of current (1)
(any two, one each) B2 [3]
8 (a) packet/quantum/discrete amount of energy M1
of electromagnetic radiation A1
(allow 1 mark for ‘packet of electromagnetic radiation’)
energy = Planck constant × frequency (seen here or in b) B1 [3]
(b) each (coloured) line corresponds to one wavelength/frequency B1
energy = Planck constant × frequency
implies specific energy change between energy levels B1
so discrete levels A0 [2]

9 (a) (i) either probability of decay (of a nucleus) M1
per unit time A1 [2]
or λ = (–)(dN/dt) / N (M1)
(–)dN/dt and N explained (A1)
(ii) in time t½, number of nuclei changes from N0 to ½N0 B1
½ = exp(–λ t½) or 2 = exp (λ t½) B1
ln (½) = –λ t½ and ln (½) = –0.693 or ln 2 = λ t½ and ln 2 = 0.693 B1
0.693 = λ t½ A0 [3]
(b) 228 = 538exp(–8λ) C1
λ = 0.107 (hours–1
) C1
t½ = 6.5hours (do not allow 3 or more SF) A1 [3]
(c) e.g. random nature of decay
background radiation
daughter product is radioactive
(any two sensible suggestions, 1 each) B2 [2]

Section B
10 (a) light-dependent resistor (allow LDR) B1 [1]
(b) (i) two resistors in series between +5V line and earth M1
midpoint connected to inverting input of op-amp A1 [2]
(ii) relay coil between diode and earth M1
switch between lamp and earth A1 [2]
(c) (i) switch on/off mains supply using a low voltage/current output B1 [1]
(allow ‘isolates circuit from mains supply’)
(ii) relay will switch on for one polarity of output (voltage) C1
switches on when output (voltage) is negative A1 [2]
11 (a) (i) e.m. radiation produced whenever charged particle is accelerated M1
electrons hitting target have distribution of accelerations A1 [2]
(ii) either wavelength shorter/shortest for greater/greatest acceleration
or λmin = hc/ Emax
or minimum wavelength for maximum energy B1
all electron energy given up in one collision/converted to single photon B1 [2]
(b) (i) hardness measures the penetration of the beam C1
greater hardness, greater penetration A1 [2]
(ii) controlled by changing the anode voltage C1
higher anode voltage, greater penetration/hardness A1 [2]
(c) (i) long-wavelength radiation more likely to be absorbed in the body/less
likely to penetrate through body B1 [1]
(ii) (aluminium) filter/metal foil placed in the X-ray beam B1 [1]
12 (a) strong uniform (magnetic) field M1
either aligns nuclei
or gives rise to Larmor/resonant frequency in r.f. region A1
non-uniform (magnetic) field M1
either enables nuclei to be located
or changes the Larmor/resonant frequency A1 [4]
(b) (i) difference in flux density = 2.0 × 10–2
× 3.0 × 10–3
= 6.0 × 10–5
T A1 [1]
(ii) ∆f = 2 × c × ∆B C1
= 2 × 1.34 × 108
× 6.0 × 10–5
= 1.6 × 104
Hz A1 [2]

13 (a) (i) no interference (between signals) near boundaries (of cells) B1 [1]
(ii) for large area, signal strength would have to be greater and this could
be hazardous to health B1 [1]
(b) mobile phone is sending out an (identifying) signal M1
computer/cellular exchange continuously selects cell/base station
with strongest signal A1
computer/cellular exchange allocates (carrier) frequency (and slot) A1 [3]

9702 PHYSICS
examination.
syllabuses.

Section A
1 (a) force proportional to product of masses and inversely proportional to
square of separation (do not allow square of distance/radius) M1
either point masses or separation @ size of masses A1 [2]
(b) (i) ω = 2π / (27.3 × 24 × 3600) or 2π / (2.36 x 106
) M1
= 2.66 × 10–6
rads–1
A0 [1]
(ii) GM = r3
ω2
or GM = v2
r C1
M = (3.84 × 105
× 103
)3
× (2.66 × 10–6
)2
/ (6.67 × 10–11
) M1
= 6.0 × 1024
kg A0 [2]
(special case: uses g = GM/r2
with g = 9.81, r = 6.4 × 106
scores max 1 mark)
(c) (i) grav. force = (6.0 × 1024
) × (7.4 × 1022
) × (6.67 × 10–11
)/(3.84 × 108
)2
C1
= 2.0 × 1020
N (allow 1SF) A1 [2]
(ii) either ∆EP = Fx because F constant as x ! radius of orbit B1
∆EP = 2.0 × 1020
× 4.0 × 10–2
C1
= 8.0 × 1018
J (allow 1SF) A1 [3]
or ∆EP = GMm/r1 – GMm/r2 C1
Correct substitution B1
8.0 × 1018
J A1
(∆EP = GMm/r1 + GMm/r2 is incorrect physics so 0/3)
2 (a) energy = ½mω2
a2
and ω = 2πf C1
= ½ × 37 × 10–3
× (2π × 3.5)2
× (2.8 × 10–2
)2
M1
= 7.0 × 10–3
J A0 [2]
(allow 2π × 3.5 shown as 7π)
Energy = ½ mv2
and v = rω (C1)
Correct substitution (M1)
Energy = 7.0 × 10–3
J (A0)
(b) EK = EP
½mω2
(a2
– x2
) = ½mω2
x2
or EK or EP = 3.5mJ C1
x = a/√2 = 2.8 /√2 or EK = ½mω2
(a2
– x2
) or EP = ½mω2
x2
C1
= 2.0cm A1 [3]
(EK or EP = 7.0mJ scores 0/3)
Allow: k = 17.9 (C1)
E = ½ kx2
(C1)
x = 2.0cm (A1)

(c) (i) graph: horizontal line, y-intercept = 7.0mJ with end-points of line at
+2.8cm and –2.8cm B1 [1]
(ii) graph: reasonable curve B1
with maximum at (0,7.0) end-points of line at (–2.8, 0)
and (+2.8, 0) B1 [2]
(iii) graph: inverted version of (ii) M1
with intersections at (–2.0, 3.5) and (+2.0, 3.5) A1 [2]
(Allow marks in (iii), but not in (ii), if graphs K & P are not labelled)
(d) gravitational potential energy B1 [1]
3 (a) sum of potential energy and kinetic energy of atoms/molecules/particles M1
reference to random (distribution) A1 [2]
(b) (i) as lattice structure is ‘broken’/bonds broken/forces between
molecules reduced (not molecules separate) B1
no change in kinetic energy, potential energy increases M1
internal energy increases A1 [3]
(ii) either molecules/atoms/particles move faster/ <c2
> is increasing
or kinetic energy increases with temperature (increases) B1
no change in potential energy, kinetic energy increases M1
internal energy increases A1 [3]
4 (a) (i) as r decreases, energy decreases/work got out (due to) M1
attraction so point mass is negatively charged A1 [2]
(ii) electric potential energy = charge × electric potential B1
electric field strength is potential gradient B1
field strength = gradient of potential energy graph/charge A0 [2]
(b) tangent drawn at (4.0, 14.5) B1
gradient = 3.6 × 10–24
A2
(for < ±0.3 allow 2 marks, for < ±0.6 allow 1 mark)
field strength= (3.6 × 10–24
) / (1.6 × 10–19
)
= 2.3 × 10–5
Vm–1
(allow ecf from gradient value) A1 [4]
(one point solution for gradient leading to 2.3 × 10–5
Vm–1
scores 1 mark only)

5 (a) (long) straight conductor carrying current of 1A M1
current/wire normal to magnetic field M1
(for flux density 1T,) force per unit length is 1Nm–1
A1 [3]
(b) (i) (originally) downward force on magnet (due to current) B1
by Newton’s third law (allow “N3”) M1
upward force on wire A1 [3]
(ii) F = BIL
2.4 × 10–3
× 9.8 = B × 5.6 × 6.4 × 10–2
C1
B = 0.066T (need 2SF) A1 [2]
(g missing scores 0/2, but g = 10 leading to 0.067T scores 1/2)
(c) new reading is 2.4√2g C1
either changes between +3.4g and –3.4g
or total change is 6.8g A1 [2]
6 (a) oil drop charged by friction/beta source B1
between parallel metal plates B1
plates are horizontal (1)
adjustable potential difference/field between plates B1
until oil drop is stationary B1
mg = q × V/d B1
symbols explained (1)
oil drop viewed through microscope (1)
m determined from terminal speed of drop (when p.d. is zero) (1)
(any two extras, 1 each) B2 [7]
(b) 3.2 × 10–19
C A1 [1]
7 (a) minimum energy to remove an electron from the metal/surface B1 [1]
(b) gradient = 4.17 × 10–15
(allow 4.1 → 4.3) C1
h = 4.15 × 10–15
× 1.6 × 10–19
or h = 4.1 to 4.3 × 10–15
eVs A1
= 6.6 × 10–34
Js A0 [2]
(c) graph: straight line parallel to given line
with intercept at any higher frequency B1
intercept at between 6.9 × 1014
Hz and 7.1 × 1014
Hz B1 [3]

8 (a) nuclei having same number of protons/proton (atomic) number B1
different numbers of neutrons/neutron number B1 [2]
(allow second mark for nucleons/nucleon number/mass number/atomic
mass if made clear that same number of protons/proton number)
(b) probability of decay per unit time is the decay constant C1
λ = ln 2 / t½
= 0.693 / (52 × 24 × 3600) C1
= 1.54 × 10–7
s–1
A1 [3]
(c) (i) A = A0 exp(–λt)
7.4 × 106
= A0 exp(–1.54 × 10–7
× 21 × 24 × 3600) C1
A0 = 9.8 × 106
Bq A1 [2]
(alternative method uses 21 days as 0.404 half-lives)
(ii) A = λN and mass = N × 89 / NA C1
mass = (9.8 × 106
× 89) / (1.54 × 10–7
× 6.02 × 1023
)
= 9.4 × 10–9
g A1 [2]

Section B
9 (a) e.g. infinite input impedance/resistance
zero output impedance/resistance
infinite (open loop) gain
infinite bandwidth
infinite slew rate
(any four, one mark each) B4 [4]
(b) graph: square wave M1
180° phase change A1
amplitude 5.0 V A1 [3]
(c) correct symbol for LED M1
diodes connected correctly between VOUT and earth A1
diodes identified correctly A1 [3]
(special case: if diode symbol, not LED symbol, allow 2nd
and 3rd
marks to be scored)
10 (a) e.g. beam is divergent/obeys inverse square law
absorption (in block)
scattering (of beam in block)
reflection (at boundaries)
(b) (i) I = I0exp(–µx) C1
I0/I = exp(0.27 × 2.4)
= 1.9 A1 [2]
(ii) I0/I = exp(0.27 × 1.3) × exp(3.0 × 1.1) C1
= 1.42 × 27.1
= 38.5 A1 [2]
(c) either much greater absorption in bone than in soft tissue
or Io / I much greater for bone than soft tissue B1 [1]
11 (a) (i) loss of (signal) power B1 [1]
(ii) unwanted power (on signal) M1
that is random A1 [2]
(b) for digital, only the ‘high’ and the ‘low’ / 1 and 0 are necessary M1
variation between ‘highs’ and ‘lows’ caused by noise not required A1 [2]
(c) attenuation = 10lg(P2 / P1) C1
either 195 = 10lg({2.4 × 103
} / P)
or –195 = 10lg(P / 2.4 × 103
) C1
P = 7.6 × 10–17
W A1 [3]

12 (a) (i) modulator B1 [1]
(ii) serial-to-parallel converter (accept series-to-parallel converter) B1 [1]
(b) (i) enables one aerial to be used for transmission and receipt of signals A1 [1]
(ii) all bits for one number arrive at one time B1
bits are sent out one after another B1 [2]

9702 PHYSICS
examination.
syllabuses.

Section A
1 (a) work done in bringing unit mass from infinity (to the point) B1 [1]
(b) gravitational force is (always) attractive B1
either as r decreases, object/mass/body does work
or work is done by masses as they come together B1 [2]
(c) either force on mass = mg (where g is the acceleration of free fall
/gravitational field strength) B1
g = GM/r2
B1
if r @ h, g is constant B1
∆EP = force × distance moved M1
= mgh A0
or ∆EP = m∆φ (C1)
= GMm(1/r1 – 1/r2) = GMm(r2 – r1)/r1r2 (B1)
if r2 ≈ r1, then (r2 – r1) = h and r1r2 = r2
(B1)
g = GM/r2
(B1)
∆EP = mgh (A0) [4]
(d) ½mv2
= m∆φ
v2
= 2 × GM/r C1
= (2 × 4.3 × 1013
) / (3.4 × 106
) C1
v = 5.0 × 103
ms–1
A1 [3]
(Use of diameter instead of radius to give v = 3.6 × 103
ms–1
scores 2 marks)
2 (a) (i) either random motion
or constant velocity until hits wall/other molecule B1 [1]
(ii) (total) volume of molecules is negligible M1
compared to volume of containing vessel A1
or
radius/diameter of a molecule is negligible (M1)
compared to the average intermolecular distance (A1) [2]
(b) either molecule has component of velocity in three directions
or c2
= cX
2
+ cY
2
+ cZ
2
M1
random motion and averaging, so <cX
2
> = <cY
2
> = <cZ
2
> M1
<c2
> = 3<cX
2
> A1
so, pV = ⅓Nm<c2
> A0 [3]
(c) <c2
> ∝ T or crms ∝ T C1
temperatures are 300K and 373K C1
crms = 580ms–1
A1 [3]
(Do not allow any marks for use of temperature in units of ºC instead of K)

3 (a) (numerically equal to) quantity of (thermal) energy required to change
the state of unit mass of a substance M1
without any change of temperature A1 [2]
(Allow 1 mark for definition of specific latent heat of fusion/vaporisation)
(b) either energy supplied = 2400 × 2 × 60 = 288000J C1
energy required for evaporation = 106 × 2260 = 240000J C1
difference = 48000J
rate of loss = 48000 / 120 = 400W A1
or energy required for evaporation = 106 × 2260 = 240000J (C1)
power required for evaporation = 240000 / (2 × 60) = 2000W (C1)
rate of loss = 2400 – 2000 = 400W (A1) [3]
4 (a) a = (–)ω2
x and ω = 2π/T C1
T = 0.60s C1
a = (4π2
× 2.0 × 10–2
) / (0.6)2
= 2.2ms–2
A1 [3]
(b) sinusoidal wave with all values positive B1
all values positive, all peaks at EK and energy = 0 at t = 0 B1
period = 0.30s B1 [3]
5 (a) force per unit positive charge acting on a stationary charge B1 [1]
(b) (i) E = Q / 4πε0r2
C1
Q = 1.8 × 104
× 102
× 4π × 8.85 × 10–12
× (25 × 10–2
)2
M1
Q = 1.25 × 10–5
C = 12.5µC A0 [2]
(ii) V = Q / 4πε0r
= (1.25 × 10–5
) / (4π × 8.85 × 10–12
× 25 × 10–2
) C1
= 4.5 × 105
V A1 [2]
(Do not allow use of V = Er unless explained)

6 (a) (i) peak voltage = 4.0V A1 [1]
(ii) r.m.s. voltage (= 4.0/√2) = 2.8V A1 [1]
(iii) period T = 20ms M1
frequency = 1 / (20 × 10–3
) M1
frequency = 50Hz A0 [2]
(b) (i) change = 4.0 – 2.4 = 1.6V A1 [1]
(ii) ∆Q = C∆V or Q = CV C1
= 5.0 × 10–6
× 1.6 = 8.0 × 10–6
C A1 [2]
(iii) discharge time = 7ms C1
current = (8.0 × 10–6
) / (7.0 × 10–3
) M1
= 1.1(4) × 10–3
A A0 [2]
(c) average p.d. = 3.2V C1
resistance = 3.2 / (1.1 × 10–3
)
= 2900Ω (allow 2800Ω) A1 [2]
7 (a) sketch: concentric circles (minimum of 3 circles) M1
separation increasing with distance from wire A1
correct direction B1 [3]
(b) (i) arrow direction from wire B towards wire A B1 [1]
(ii) either reference to Newton’s third law
or force on each wire proportional to product of the two currents M1
so forces are equal A1 [2]
(c) force always towards wire A/always in same direction B1
varies from zero (to a maximum value) (1)
variation is sinusoidal / sin2
(1)
(at) twice frequency of current (1)
(any two, one each) B2 [3]
8 (a) packet/quantum/discrete amount of energy M1
of electromagnetic radiation A1
(allow 1 mark for ‘packet of electromagnetic radiation’)
energy = Planck constant × frequency (seen here or in b) B1 [3]
(b) each (coloured) line corresponds to one wavelength/frequency B1
energy = Planck constant × frequency
implies specific energy change between energy levels B1
so discrete levels A0 [2]

9 (a) (i) either probability of decay (of a nucleus) M1
per unit time A1 [2]
or λ = (–)(dN/dt) / N (M1)
(–)dN/dt and N explained (A1)
(ii) in time t½, number of nuclei changes from N0 to ½N0 B1
½ = exp(–λ t½) or 2 = exp (λ t½) B1
ln (½) = –λ t½ and ln (½) = –0.693 or ln 2 = λ t½ and ln 2 = 0.693 B1
0.693 = λ t½ A0 [3]
(b) 228 = 538exp(–8λ) C1
λ = 0.107 (hours–1
) C1
t½ = 6.5hours (do not allow 3 or more SF) A1 [3]
(c) e.g. random nature of decay
background radiation
daughter product is radioactive

Section B
10 (a) light-dependent resistor (allow LDR) B1 [1]
(b) (i) two resistors in series between +5V line and earth M1
midpoint connected to inverting input of op-amp A1 [2]
(ii) relay coil between diode and earth M1
switch between lamp and earth A1 [2]
(c) (i) switch on/off mains supply using a low voltage/current output B1 [1]
(allow ‘isolates circuit from mains supply’)
(ii) relay will switch on for one polarity of output (voltage) C1
switches on when output (voltage) is negative A1 [2]
11 (a) (i) e.m. radiation produced whenever charged particle is accelerated M1
electrons hitting target have distribution of accelerations A1 [2]
(ii) either wavelength shorter/shortest for greater/greatest acceleration
or λmin = hc/ Emax
or minimum wavelength for maximum energy B1
all electron energy given up in one collision/converted to single photon B1 [2]
(b) (i) hardness measures the penetration of the beam C1
greater hardness, greater penetration A1 [2]
(ii) controlled by changing the anode voltage C1
higher anode voltage, greater penetration/hardness A1 [2]
(c) (i) long-wavelength radiation more likely to be absorbed in the body/less
likely to penetrate through body B1 [1]
(ii) (aluminium) filter/metal foil placed in the X-ray beam B1 [1]
12 (a) strong uniform (magnetic) field M1
either aligns nuclei
or gives rise to Larmor/resonant frequency in r.f. region A1
non-uniform (magnetic) field M1
either enables nuclei to be located
or changes the Larmor/resonant frequency A1 [4]
(b) (i) difference in flux density = 2.0 × 10–2
× 3.0 × 10–3
= 6.0 × 10–5
T A1 [1]
(ii) ∆f = 2 × c × ∆B C1
= 2 × 1.34 × 108
× 6.0 × 10–5
= 1.6 × 104
Hz A1 [2]

13 (a) (i) no interference (between signals) near boundaries (of cells) B1 [1]
(ii) for large area, signal strength would have to be greater and this could
be hazardous to health B1 [1]
(b) mobile phone is sending out an (identifying) signal M1
computer/cellular exchange continuously selects cell/base station
with strongest signal A1
computer/cellular exchange allocates (carrier) frequency (and slot) A1 [3]

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)
P1 Frequency or period of rotation or ω is the independent variable and θ is the dependent variable
or vary f or T or ω and measure θ. [1]
P2 ω = 2πf = 2π/T [1]
P3 Keep the length of the rigid rod constant; ignore reference to mass. [1]
Methods of data collection (5 marks)
M1 Labelled diagram of apparatus: small object, pole attached to a rotating device
(motor, turntable). [1]
M2 Method to change the speed of the rotating device. [1]
M3 Method to determine frequency or time period (e.g. stop watch to time a number of rotations,
rev counter/tachometer, light gates connected to a timer/frequency meter). [1]
M4 Use fiducial mark or light gates perpendicular to motion of object. [1]
M5 Method to measure angle – use protractor or rule for measurements for trigonometry methods.
This must be shown correctly on diagram or explained in text. [1]
Method of analysis (2 marks)
A1 Plot a graph of cos θ against 1/ω 2
. [1]
A2 Relationship is valid if straight line through the origin [1]
Safety considerations (1 mark)
S1 Use a protective screen in case mass detaches from the pole. Do not use goggles. [1]
Additional detail (4 marks)
Relevant points might include [4]
1 Large motor speed to produce measurable θ.
2 Additional detail on measuring angle e.g. large protractor fixed to pole.
3 Projection method, slow motion freeze frame video, camera with detail.
4 cos θ = h/l or equivalent.
5 Method of checking pole is vertical – use a set square.
6 Additional detail on measuring angular velocity, e.g. time at least 10 rotations.
7 Wait for motion to become stable.
Do not allow vague computer methods.
[Total: 15]

2 Analysis, conclusions and evaluation (15 marks)
Part Mark Expected Answer Additional Guidance
(a) A1 Gradient = r
y-intercept = lg s
Allow log or ln
(b) T1
T2
1.70 or 1.699 0.41 or 0.415
1.78 or 1.778 0.53 or 0.531
1.85 or 1.845 0.64 or 0.643
1.90 or 1.903 0.73 or 0.732
1.95 or 1.954 0.82 or 0.820
1.98 or 1.978 0.86 or 0.857
Ignore significant figures. A mixture is allowed.
U1 From ± 0.03 or ± 0.04, to
± 0.01 (±0.012)
Allow more than one significant figure.
(c) (i) G1 Six points plotted correctly Must be within half a small square. Penalise
‘blobs’ (more than half a small square). Ecf
allowed from table.
U2 Error bars in lg (y / mm)
plotted correctly.
Must be accurate within half a small square.
(ii) G2 Line of best fit If points are plotted correctly then lower end of
line should pass between (1.655, 0.35) and
(1.665, 0.35) and upper end of line should pass
between (2.00, 0.89) and (2.00, 0.90). 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.
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.
(iii) C1 Gradient of best fit line The triangle used should be at least half the
length of the drawn line. Check the read offs.
Work to half a small square. Do not penalise
POT. (Should be about 1.6)
U3 Uncertainty in gradient Method of determining absolute uncertainty.
Difference in worst gradient and gradient.
(iv) C2 Negative y-intercept Must be negative. FOX does not score.
Expect to see point substituted 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) C3 r = gradient and is given to 2
or 3 s.f. and in the range 1.57
to 1.64
Allow 1.6 to 2 s.f.
Penalise 1 s.f. or >3 s.f.
C4 s = 10y-intercept
y-intercept must be used.
(Should be about 0.005 or 5 × 10–3
)
Allow ecf for method from (c)(iv).
U5 Absolute uncertainty in r and
s
Uncertainty in r should be the same as the
uncertainty in the gradient.
Difference in worst s and s.
[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)
(iv) [U4]
Uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line
(d) [U5]
Uncertainty = best s –worst s

9702 PHYSICS
maximum raw mark 30
examination.
syllabuses.

P1 v is the independent variable and θ is the dependent variable or vary v and measure θ. [1]
P2 Keep the (shape and) size/volume/surface area/mass of balloon/helium constant [1]
Do not credit ‘same balloon’.
P3 Keep the temperature (air/helium/balloon) constant. [1]
M1 Labelled diagram of apparatus: balloon, string fixed and method of producing wind.
Method of producing wind to be approximately horizontal to balloon. [1]
M2 Suspend mass from balloon. [1]
M3 Method to change wind speed, e.g. change setting, variable power supply/resistor/change
distance from fan. [1]
M4 Method to measure wind speed, e.g. wind speed indicator/detector, anemometer [1]
A1 Plot a graph of tan θ against 1/v2
. [1]
A2 Relationship valid if straight line through origin [1]
S1 Avoid the moving blades of the fan (safety screen, switch off when changing experiment);
goggles to avoid air stream into eye. [1]
D1/2/3/4Relevant points might include [4]
1 Large wind speed to produce measurable deflection/large cross-sectional area of balloon.
2 Additional detail on measuring angle e.g. use a large protractor, projection method.
3 tan θ = h/l.
4 Measuring air speed at point where balloon is positioned.
5 Adjust height of fan so that air flow is horizontally aligned to the balloon.
6 Reason for adding mass to increase stability/deflection.
7 Keep windows shut/air conditioning switched off/use of wind tunnel to avoid draughts.
8 Wait for the balloon to become stable.
[Total: 15]

(a) A1
Gradient =
µ
T
2
1
Allow equivalent, e.g.
µ4
T
(b) T1 1/L / m–1
or (1/L) / m–1
Allow 1/L (m–1
), 1/L / 1/m, 1/L (1/m)
T2 1.83 or 1.835
2.08 or 2.083
2.35 or 2.353
2.50 or 2.500
2.82 or 2.817
3.13 or 3.125
Values must correspond to table. A mixture of
3 s.f. and 4 s.f. is allowed
U1 From ± 0.01 or ± 0.02, to ± 0.05 Allow more than one significant figure.
allowed from table.
U2 All Error bars in 1/L/m–1
plotted
correctly.
Check second and last point for accuracy.
(ii) G2 Line of best fit If points are plotted correctly then lower end
of line should pass between (1.8, 250) and
(1.8, 254) and upper end of line should pass
between (3.18, 450) and (3.2, 448). Allow ecf
judgement.
G3 Worst acceptable straight line.
Steepest or shallowest possible
line that passes through all the
error bars.
Should pass from left of top error bar to right
of bottom error bar or right of top error bar to
left of bottom error bar. Mark scored only if all
error bars are plotted.
POT. (Should be about 140)
U3 Uncertainty in gradient Method of determining absolute uncertainty
(d) (i) C2 Value of µ using gradient µ = 7.5/gradient2
Gradient must be used.
(Should be about 0.00037 or 3.7 × 10–4
)
C3 kg m–1
or N Hz–2
m–2
Allow other correct units e.g. N s2
m–2
or Pa s2
or N (Hz m)–2
(ii) U4 10% + 2 × percentage
uncertainty in gradient
Check working. Must be larger than 10%.

(e) C4 r given to 2 or 3 s.f. and
in the range 1.15 × 10–4
to
1.18 × 10–4
Allow 1.2 to 2 s.f.
U5 (d)(ii) / 2 Check working if not (d)(ii) / 2
[Total: 15]
(d) (ii) [U4]
Percentage uncertainty = 10 + 2
m
m∆
Percentage uncertainty = 100min4
max
2
×
−
×
µ
µ
m
T
Percentage uncertainty = 100max4
min
2
×
−
×
µ
µ
m
T
(e) (ii) [U5]
Percentage uncertainty = 100
max
×
−
r
rr
Percentage uncertainty = 100
min
×
−
r
rr

9702 PHYSICS
maximum raw mark 30
examination.
syllabuses.

P1 Frequency or period of rotation or ω is the independent variable and θ is the dependent variable
or vary f or T or ω and measure θ. [1]
P2 ω = 2πf = 2π/T [1]
P3 Keep the length of the rigid rod constant; ignore reference to mass. [1]
M1 Labelled diagram of apparatus: small object, pole attached to a rotating device
(motor, turntable). [1]
M2 Method to change the speed of the rotating device. [1]
M3 Method to determine frequency or time period (e.g. stop watch to time a number of rotations,
rev counter/tachometer, light gates connected to a timer/frequency meter). [1]
M4 Use fiducial mark or light gates perpendicular to motion of object. [1]
A1 Plot a graph of cos θ against 1/ω 2
. [1]
A2 Relationship is valid if straight line through the origin [1]
S1 Use a protective screen in case mass detaches from the pole. Do not use goggles. [1]
Relevant points might include [4]
1 Large motor speed to produce measurable θ.
2 Additional detail on measuring angle e.g. large protractor fixed to pole.
3 Projection method, slow motion freeze frame video, camera with detail.
4 cos θ = h/l or equivalent.
5 Method of checking pole is vertical – use a set square.
6 Additional detail on measuring angular velocity, e.g. time at least 10 rotations.
7 Wait for motion to become stable.
[Total: 15]

(a) A1 Gradient = r
y-intercept = lg s
Allow log or ln
(b) T1
T2
1.70 or 1.699 0.41 or 0.415
1.78 or 1.778 0.53 or 0.531
1.85 or 1.845 0.64 or 0.643
1.90 or 1.903 0.73 or 0.732
1.95 or 1.954 0.82 or 0.820
1.98 or 1.978 0.86 or 0.857
Ignore significant figures. A mixture is allowed.
U1 From ± 0.03 or ± 0.04, to
± 0.01 (±0.012)
Allow more than one significant figure.
allowed from table.
U2 Error bars in lg (y / mm)
plotted correctly.
(ii) G2 Line of best fit If points are plotted correctly then lower end of
line should pass between (1.655, 0.35) and
(1.665, 0.35) and upper end of line should pass
between (2.00, 0.89) and (2.00, 0.90). Allow ecf
judgement.
G3 Worst acceptable straight
line.
Steepest or shallowest
possible line that passes
through all the error bars.
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.
POT. (Should be about 1.6)
U3 Uncertainty in gradient Method of determining absolute uncertainty.
(iv) C2 Negative y-intercept Must be negative. FOX does not score.
Expect to see point substituted 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) C3 r = gradient and is given to 2
or 3 s.f. and in the range 1.57
to 1.64
Allow 1.6 to 2 s.f.
C4 s = 10y-intercept
y-intercept must be used.
(Should be about 0.005 or 5 × 10–3
)
Allow ecf for method from (c)(iv).
U5 Absolute uncertainty in r and
s
Uncertainty in r should be the same as the
uncertainty in the gradient.
Difference in worst s and s.
[Total: 15]
(iv) [U4]
Uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line
(d) [U5]
Uncertainty = best s –worst s

9702 s12 ms_all

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