9702 w12 ms_all

CAMBRIDGE INTERNATIONAL EXAMINATIONS
GCE Advanced Subsidiary Level and GCE Advanced Level
MARK SCHEME for the October/November 2012 series
9702 PHYSICS
9702/11 Paper 1 (Multiple Choice), maximum raw mark 40
Mark schemes should be read in conjunction with the question paper and the Principal Examiner
Report for Teachers.
Cambridge will not enter into discussions about these mark schemes.
Cambridge is publishing the mark schemes for the October/November 2012 series for most IGCSE,
GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level
components.

Mark Scheme Syllabus Paper
GCE AS/A LEVEL – October/November 2012 9702 11
© Cambridge International Examinations 2012
Question
Number
Key
Question
Number
Key
1 B 21 D
2 C 22 D
3 D 23 A
4 B 24 B
5 B 25 B
6 C 26 D
7 A 27 A
8 D 28 B
9 A 29 C
10 C 30 B
11 C 31 B
12 A 32 A
13 A 33 C
14 A 34 D
15 B 35 C
16 B 36 C
17 C 37 B
18 A 38 C
19 A 39 A
20 C 40 B

9702 PHYSICS
components.

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

9702 PHYSICS
components.

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

1 (a) (i) acceleration = change in velocity / time (taken)
or acceleration = rate of change of velocity B1 [1]
(ii) a body continues at constant velocity unless acted on by a resultant force B1 [1]
(b) (i) distance is represented by the area under graph C1
distance = ½ × 29.5 × 3 = 44.3m (accept 43.5m for 29 to 45m for 30) A1 [2]
(ii) resultant force = weight – frictional force B1
frictional force increases with speed B1
at start frictional force = 0 / at end weight = frictional force B1 [3]
(iii) 1. frictional force increases B1 [1]
2. frictional force (constant) and then decreases B1 [1]
(iv) 1. acceleration = (v2 – v1) / t = (20 – 50) / (17 – 15) C1
= (–) 15m s–2
A1 [2]
2. W – F = ma C1
W = 95 × 9.81 (= 932) C1
F = (95 × 15) + 932 = 2400 (2360) (2357)N A1 [3]
2 (a) resistance = potential difference / current B1 [1]
(b) (i) metal wire in series with power supply and ammeter B1
voltmeter in parallel with metal wire B1
rheostat in series with power supply or potential divider arrangement
or variable power supply B1 [3]
(ii) 1. intercept on graph B1 [1]
2. scatter of readings about the best fit line B1 [1]
(iii) correction for zero error explained B1
use of V and corrected І values from graph C1
resistance = V / І = 22.(2)Ω [e.g. 4.0 / 0.18] A1 [3]
(c) R = 6.8 / 0.64 = 10.625 C1
%R = %V + %І
= (0.1 / 6.8) × 100 + (0.01 / 0.64) × 100 C1
= 1.47% + 1.56%
∆R = 0.0303 × 10.625 = 0.32Ω
R = 10.6 ± 0.3 Ω A1 [3]

3 (a) pressure = force / area B1 [1]
(b) molecules collide with object / surface and rebound B1
molecules have change in momentum hence force acts B1
fewer molecules per unit volume on top of mountain / temperature is less
hence lower speed of molecules B1
hence less pressure A0 [3]
(c) (i) ρ = m / V C1
W = Vρg = 0.25 × 0.45 × 9.81 × 13600 C1
= 15000 (15009)N A1 [3]
(ii) p = W / A (or using p = ρgh) = 15009 / 0.45
= 3.3 × 104
Pa A1 [1]
(iii) pressure will be greater due to the air pressure (acting on the surface of the liquid)
B1 [1]
4 (a) waves pass through the elements / gaps / slits in the grating M1
spread into geometric shadow A1 [2]
(b) (i) 1. displacements add to give resultant displacement B1
each wavelength travels the same path difference or are in phase B1
hence produce a maximum A0 [2]
2. to obtain a maximum the path difference must be λ or phase difference
360° / 2π rad B1
λ of red and blue are different B1
hence maxima at different angles / positions A0 [2]
(ii) nλ = d sinθ C1
N = sin 61° / (2 × 625 × 10–9
) = 7.0 × 105
A1 [2]
(iii) nλ = 2 × 625 is a constant (1250) C1
n = 1 → λ = 1250 outside visible
n = 3 → λ = 417 in visible
n = 4 → λ = 312.5 outside visible
λ = 420 nm A1 [2]
5 (a) when the load is removed then the wire / body object does not return to its original shape /
length B1 [1]
(b) (i) stress = force / area C1
F = 220 × 106
× 1.54 × 10–6
= 340 (338.8)N A1 [2]
(ii) E = (F × l) / (A × e) C1
e = (90 × 106
) × 1.75 / (1.2 × 1011
) = 1.31 × 10–3
m A1 [2]
(c) the stress is no longer proportional to the extension B1 [1]

6 (a) 92 protons in the nucleus and 92 electrons around nucleus B1
143 neutrons (in the nucleus) B1 [2]
(b) (i) α-particle travels short distance in air B1 [1]
(ii) very small proportion in backwards direction / large angles B1
majority pass through with no /small deflections B1
either most of mass is in very small volume (nucleus) and is charged or most of atom is
empty space B1 [3]
(c) I = Q / t C1
n / t = (1.5 × 10–12
) /( 2 × 1.6 × 10–19
) C1
n / t = 4.7 × 106
s–1
A1 [3]

9702 PHYSICS
components.

1 (a) units for D identified as kgm s–2
M1
all other units shown: units for A:m2
units for v2
:m2
s–2
units for ρ:kgm–3
2223
2
smmmkg
smkg
−−
−
=C with cancelling/ simplification to give C no units A1 [2]
(b) (i) straight line from (0,0) to (1,9.8) ± half a square B1 [1]
(ii) ½ mv2
= mgh or using v2
= 2as C1
v = (2 × 9.81 × 1000)1/2
= 140m s–1
A1 [2]
(c) (i) weight = drag (D) ( + upthrust) B1 [1]
Allow mg or W for weight and D or expression for D for drag
(ii) 1. mg = 1.4 ×10–5
× 9.81 C1
1.4 × 10–5
× 9.81 = 0.5 × 0.6 × 1.2 × 7.1 × 10–6
× v2
M1
v = 7.33ms–1
A0 [2]
2. line from (0,0) correct curvature to a horizontal line at velocity of 7ms–1
M1
line reaches 7ms–1
between 1.5s and 3.5s A1 [2]
2 (a) (resultant) force = rate of change of momentum / allow proportional to
or change in momentum / time (taken) B1 [1]
(b) (i) ∆p = (–) 65 × 10–3
(5.2 + 3.7) C1
= (–) 0.58N s A1 [2]
(ii) F = 0.58 / 7.5 × 10–3
= 77(.3) N A1 [1]
(c) (i) 1. force on the wall from the ball is equal to the force on ball from the wall M1
but in the opposite direction A1 [2]
(statement of Newton’s third law can score one mark)
2. momentum change of ball is equal and opposite to momentum change
of the wall / change of momentum of ball and wall is zero B1 [1]
(ii) kinetic energy (of ball and wall) is reduced / not conserved so inelastic B1 [1]
(Allow relative speed of approach does not equal relative speed of separation.)

3 (a) metal: regular / repeated / ordered arrangement / pattern / lattice
or long range order (of atoms / molecules / ions) B1
polymer: tangled chains (of atoms / molecules) or long chains (of
atoms / molecules / ions) B1
amorphous: disordered / irregular arrangement or short range order
(of atoms / molecules / ions) B1 [3]
(b) metal: straight line or straight line then curving with less positive gradient B1
polymer: curve with decreasing gradient with steep increasing gradient at end B1 [2]
4 (a) waves (travels along tube) reflect at closed end / end of tube B1
incident and reflected waves or these two waves are in opposite directions M1
interfere or stationary wave formed if tube length equivalent to
λ / 4, 3λ / 4, etc. A1 [3]
(b) (i) 1. no motion (as node) / zero amplitude B1 [1]
2. vibration backwards and forwards / maximum amplitude
along length B1 [1]
(ii) λ = 330 / 880 (= 0.375m) C1
L = 3λ / 4 C1
L = 3 / 4 × (0.375) = 0.28 (0.281)m A1 [3]
5 (a) (i) І1 = І2 + І3 B1 [1]
(ii) І = V / R or І2 = 12 / 10 (= 1.2A) C1
R = [1/6 + 1 / 10]–1
[total R = 3.75 Ω] or І3 = 12 / 6 (= 2.0A) C1
І1 = 12 / 3.75 = 3.2A or І1 = 1.2 + 2.0 = 3.2 A A1 [3]
(iii) power = VІ or І2
R or V2
/ R
C1
s
2
w
2
3
2
s
2
3
w
2
2
/
/
oror
resistorsseriesinpower
wireinpower
RV
RV
R
R
x
I
I
I
I
V
V
== C1
x = 12 × 1.2 / 12 × 2.0 = 0.6(0) allow 3 / 5 or 3:5 A1 [3]
(b) p.d. BC: 12 – 12 × 0.4 = 7.2 (V) / p.d. AC = 4.8(V) C1
p.d. BD: 12 – 12 × 4 / 6 = 4.0 (V) / p.d. AD = 8.0(V) C1
p.d. = 3.2V A1 [3]
6 (a) extension is proportional to force / load B1 [1]
(b) F = mg C1
x = (mg / k ) = 0.41 × 9.81 / 25 = (4.02 / 25) M1
x = 0.16m A0 [2]

(c) (i) weight and (reaction) force from spring (which is equal to tension in spring) B1 [1]
(ii) F – weight or 0.06 × 25 = ma C1
F = 0.2209 × 25 = 5.52 (N) or 0.22 × 25 = 5.5
a = (5.52 – 0.41× 9.81) / 0.41 or 1.5 / 0.41 and (5.5 – 4.02) C1
a = 3.7 (3.66)m s–2
gives 3.6 ms–2
A1 [3]
(d) elastic potential energy / strain energy to kinetic energy and gravitational
potential energy B1
stretching / extension reduces and velocity increases / height increases B1 [2]
7 (a) He3
2 + He3
2 → He4
2 + 2 p1
1 + Q
A numbers correct (4 and 1) B1
Z numbers correct (2 and 1) B1 [2]
(b) both nuclei have 2 protons B1
the two isotopes have 1 neutron and two neutrons B1 [2]
[allow 1 for ‘same number of protons but different number of neutrons’]
(c) proton number and neutron number B1
energy – mass B1
momentum B1 [2]
(d) (i) γ radiation B1 [1]
(ii) product(s) must have kinetic energy B1 [1]
(e) 13.8MeV = 13.8 × 1.6 × 10–19
× 106
(= 2.208 × 10–12
) C1
60 = n × 13.8 × 1.6 × 10–13
n = 2.7(2) × 1013
s–1
A1 [2]

9702 PHYSICS
components.

1 (a) spacing = 380 or 3.8 × 102
pm B1 [1]
(b) time = 24 × 3600
time = 0.086 (0.0864)Ms B1 [1]
(c) time = distance / speed = 8
11
103
101.5
×
×
C1
= 500(s) = 8.3min A1 [2]
(d) momentum and weight B1 [1]
(e) (i) arrow to the right of plane direction (about 4° to 24°) B1 [1]
(ii) scale diagram drawn
or use of cosine formula v2
= 2502
+ 362
– 2 × 250 × 36 × cos45°
or resolving v = [(36cos45°)2
+ (250 – 36sin45°)2
]1/2
C1
resultant velocity = 226 (220 – 240 for scale diagram)ms–1
allow one mark for values 210 to 219 or 241 to 250ms–1
or use of formula (v2
= 51068) v = 230 (226)ms–1
A1 [2]
2 (a) (i) accelerations (A to B and B to C) are same magnitude B1
accelerations (A to B and B to C) are opposite directions
or both accelerations are toward B B1
(A to B and B to C) the component of the weight down the slope provides
the acceleration B1 [3]
(ii) acceleration = gsin15° C1
s = 0 + ½ at2
s = 0.26 / sin15° = 1.0 C1
°×
×
=
sin1589
2012
.
.
t t = 0.89s A1 [3]
(iii) v = 0 + gsin15t or v2
= 0 + 2gsin15 × 1.0 C1
v = 2.26 ms–1
A1 [2]
(using loss of GPE = gain KE can score full marks)
(b) loss of GPE at A = gain in GPE at C or loss of KE at B = gain in GPE at C B1
h1 = h2 = 0.26m or ½ mv2
= mgh h2 = 0.5 × (2.26)2
/ 9.81 = 0.26m
x = 0.26 / sin 30° = 0.52m A1 [2]
3 (a) power is the rate of doing work or power = work done / time (taken) or
power = energy transferred / time (taken) B1 [1]
(b) (i) as the speed increases drag / air resistance increases B1
resultant force reduces hence acceleration is less B1
constant speed when resultant force is zero B1 [3]
(allow one mark for speed increases and acceleration decreases)

(ii) force from cyclist = drag force / resistive force B1
P = 12 × 48 M1
P = 576W A0 [2]
(iii) tangent drawn at speed = 8.0ms–1
M1
gradient values that show acceleration between 0.44 to 0.48ms–2
A1 [2]
(iv) F – R = ma C1
600 / 8 – R = 80 × 0.5 [using P = 576] 576 / 8 – R = 80 × 0.5 C1
R = 75 – 40 = 35 N R = 72 – 40 = 32N A1 [3]
(v) at 12 ms–1
drag is48N, at 8ms–1
drag is 35 or 32N
R / v calculated as 4 and 4 or 4.4
and consistent response for whether R is proportional to v or not B1 [1]
4 (a) e.m.f. = chemical energy to electrical energy M1
p.d. = electrical energy to thermal energy M1
idea of per unit charge A1 [3]
(b) E = I (R +r) or I = E / (R +r) (any subject) B1 [1]
(c) (i) E = 5.8V B1 [1]
(ii) evidence of gradient calculation or calculation with values from graph
e.g. 5.8 = 4 + 1.0 × r C1
r = 1.8Ω A1 [2]
(d) (i) P = VI C1
P = 2.9 × 1.6 = 4.6 (4.64)W A1 [2]
(ii) power from battery = 1.6 × 5.8 = 9.28 or efficiency = VI / EI C1
efficiency = (4.64 / 9.28) × 100 = 50 % or (2.9 / 5.8) × 100 = 50% A1 [2]
5 (a) travel through a vacuum / free space B1 [1]
(b) (i) B : name: microwaves wavelength: 10– 4
to 10–1
m B1
C : name: ultra-violet / UV wavelength: 10–7
to 10–9
m B1
F : name: X –rays wavelength: 10–9
to 10–12
m B1 [3]
(ii) f = 9
8
10500
103
−
×
×
C1
f = 6(.0) × 1014
Hz A1 [2]

(c) vibrations are in one direction M1
perpendicular to direction of propagation / energy transfer
or good sketch showing this A1 [2]
6 (a) (i) electron B1 [1]
(ii) any two:
can be deflected by electric and magnetic fields or negatively charged /
absorbed by few (1 – 4)mm of aluminum / 0.5 to 2m or metres for range in air /
speed up to 0.99c / range of speeds / energies
B2 [2]
(iii) decay occurs and cannot be affected by external / environmental factors
or two stated factors such as chemical / pressure / temperature / humidity B1 [1]
(b) 3 and 0 for superscript numbers B1
2 and –1 for subscript numbers B1 [2]
(c) energy = 5.7 × 103
× 1.6 × 10–19
(= 9.12 × 10–16
J) C1
v2
= 31
16
10×9.11
10×9.12×2
−
−
C1
v = 4.5 × 107
ms–1
A1 [3]
(d) both have 1 proton and 1 electron B1
1 neutron in hydrogen-2 and 2 neutrons in hydrogen-3 B1 [2]
(special case: for one mark ‘same number of protons / atomic number
different number of neutrons’)

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

1 (b) (ii) Values of raw L in range 2.0cm Y L Y 8.0cm consistent with unit. [1]
(iii) Value of θ < 90° with unit. No raw value greater than 0.5° precision. [1]
(c) Five sets of readings of L, m and θ scores 5 marks, four sets scores 4 marks etc. [5]
Incorrect trend then –1.
Major help from Supervisor –2. Minor help from Supervisor –1.
Range: mmin Y 0.100kg, mmax [ 0.350kg. [1]
Column headings: [1]
Each column heading must contain a quantity and a unit where appropriate.
The unit must conform to accepted scientific convention e.g. m/kg, m sin θ /kg, θ /°.
Consistency: [1]
All values of L must be given to the nearest mm.
Significant figures: [1]
All values of m sin θ must have the same number of significant figures as, or one more
than, the least number of significant figures in m and θ.
Calculation: [1]
Values of m sin θ calculated correctly.
(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 that 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 on the graph grid.
Diameter of plots must be Y half a small square (no blobs).
Check that the points are plotted correctly. Work to an accuracy of half a small square in
both the x and y directions.
Quality: [1]
All points in the table must be plotted (at least 4) for this mark to be scored.
Judge by the scatter of all the points about a straight line.
All points must be within ± 0.01kg in the m sin θ direction of a straight line.
(ii) Line of best fit: [1]
Judge by balance of all the points on the grid (at least 4) 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. Line must not be kinked or thicker than half a small square.

(iii) Gradient: [1]
The sign of the gradient must match the graph. 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 the x and y directions.
The method of calculation must be correct.
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.
Or:
Check the read-off of the intercept directly from the graph.
(e) Value of P = candidate’s gradient. Value of Q = candidate’s intercept. [1]
Do not allow a value presented as a fraction.
Unit for P (mkg–1
or cmkg–1
or mmkg–1
or mg–1
or cmg–1
or mmg–1
) and Q (m or cm or mm)
correct and consistent with value. [1]
[Total: 20]
2 (a) (ii) Value of circumference in range 30.0 – 50.0cm to the nearest mm with unit. [1]
(iii) Absolute uncertainty in circumference is between 2mm – 6mm. [1]
If repeated readings have been taken, then the absolute uncertainty can be half the
range. Correct method used to calculate the percentage uncertainty.
(iv) Value of circumference within 2cm of first value. [1]
(b) (ii) Raw time values to at least 0.1s or 0.01s, value of 0.5s < T < 2.0s. [1]
Evidence of repeats. [1]
(c) (i) Second value of T. [1]
Second value of T > first value of T. [1]
(ii) Third value of T. [1]
(d) (ii) Correct calculation of two values of k. [1]
Correct calculation of third value of k. [1]
(iii) Justification of significant figures in k linked to significant figures in time and m (not just
“raw readings”) [1]
(iv) Sensible comment relating to the calculated values of k, testing against a criterion
specified by the candidate. [1]

(e)
(i) Limitations 4 max. (ii) Improvements 4 max. Do not credit
A three results not enough
/not enough results
take more readings and plot a
graph
two results not
enough
/repeat readings
/few readings
B string too wide for markings
on rule
use thinner string
C rules have different
thicknesses so effective length
of loop changes/
/different lengths so not a fair
test
use rulers of similar thicknesses/
readings/method to take
thickness into account
/use rulers of the same length
D times are small
/large uncertainty in time
use longer strings/improved
method of timing
E difficult to judge start/ end
of/complete oscillation
Position/motion sensor facing the
rule
/video with timer
position sensor at
end or in middle
F swings of 30cm ruler highly
damped
G difficult to make two loops of
the same circumference
method by which this can be
achieved
H large uncertainty in mass method of measuring mass more
precisely
accurate balance
[Total: 20]

9702 PHYSICS
maximum raw mark 40
components.

1 (b) (i) Value of h in range 0.085m Y h Y 0.095m consistent with unit. [1]
(c) Value of T in range 0.6s Y T Y 1.5s consistent with unit. [1]
Evidence of repeats. [1]
(d) Six sets of readings of h and T or raw times scores 4 marks, five sets scores 3 marks etc.
Help from Supervisor –1. [4]
Range: hmax – hmin [ 15.5cm [1]
Column headings:
The unit must conform to accepted scientific convention e.g. T2
h / s2
m (or ms2
) and
h2
/ m2
. [1]
Consistency:
All raw values of h must be given to the nearest mm. [1]
Significant figures:
All values of h2
must have the same number of significant figures as, or one more than, the
number of significant figures in h. [1]
Calculation:
Values of T2
h calculated correctly. [1]
(e) (i) Axes: [1]
Sensible scales must be used, no awkward scales (e.g. 3:10).
Diameter of plots must be Y half a small square (no “blobs”).
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 ± 0.0025m2
(25cm2
) in the h2
direction of a straight line.
candidate. Line must not be kinked or thicker than half a small square.

(iii) Gradient: [1]
The sign of the gradient must match the graph.
The hypotenuse of the triangle should be greater than half the length of the drawn line.
y intercept: [1]
Either:
Correct read-off from a point on the line and substitution into y = mx + c.
Or:
Correct read-off of the intercept directly from the graph.
(f) Value of P = candidate’s gradient. Value of Q = candidate’s intercept. [1]
Unit for P (s2
m–1
or s2
cm–1
or s2
mm–1
) and Q (s2
m or s2
cm or s2
mm) correct and [1]
consistent with value.
[Total: 20]
2 (a) (ii) Value of L in range: 5.0cm Y L Y 15.0cm with unit to nearest mm. [1]
(b) (ii) Value of s in range: 50.0cm Y s Y 70.0cm with unit. [1]
Supervisor’s help –1.
Evidence of repeat measurements. [1]
(iii) Absolute uncertainty in s is between 2cm – 10cm. [1]
(iv) Correct calculation of x. [1]
(c) Raw value(s) of t greater than 1s to a precision of 0.1 or 0.01s with unit. [1]
(d) (i) Correct calculation of v using either value of x with consistent unit. [1]
(ii) Justification of significant figures in v linked to significant figures in t and x or (s – L)
(not just “raw readings”). [1]
(e) (iii) Second value of t. [1]
Second value of s. [1]
Quality: correct trend; If s increases, t increases. [1]
(f) Sensible comment relating to the calculated values of v, testing against a criterion specified
by the candidate. [1]

(g)
A two readings not enough (to draw
a conclusion)
take many readings (for different
masses) and plot a graph
/calculate more v values and
compare
‘repeat readings’
/few readings
/take more
readings and
calculate average
v
B the car does not travel in a
straight line
method of determining the distance
e.g. video + scale/method of
marking a path
/method of guiding trolley in
straight line
C times are short
/large uncertainty in t
use a longer slope
/use a steeper slope
trolley too fast
D difficult to judge when trolley
stopped/
difficult to start the stopwatch
when all wheels on bench/when
trolley at B/when trolley horizontal
improved method of timing eg
video with timer or frame by
frame/motion sensor placed at end
of path/ticker tape timer
light gate(s)
/reaction time
/human error
E there is a drop when the trolley
reaches the end of the board/at B
there is a loss of velocity/kinetic
energy
method to smooth transition e.g.
thinner board/bevelled edge/thin
card placed at transition
F difficult to release without
applying a force/ velocity
/difficult to position head at B after
releasing trolley A
method of releasing trolley e.g.
card/barrier or electromagnet
air resistance
G calculation of x doesn’t take back
of trolley into account
detailed method of measuring from
wheel to the back of the trolley
measuring l
[Total: 20]

9702 PHYSICS
maximum raw mark 40
components.

1 (a) (iv) Value for I1 < 200mA, with consistent unit. [1]
(v) Value for I2 with unit of current. [1]
I2 > I1 [1]
(b) Six sets of readings of I1, I2 and x scores 4 marks, five sets scores 3 marks etc. [4]
Incorrect trend –1.
Major help from Supervisor –2. Minor help from Supervisor –1.
Range: xmax – xmin [ 0.500m. [1]
Column headings: [1]
The unit must conform to accepted scientific convention
e.g. I / A or I(A), 1/x (m–1
), I1/I2
Consistency: [1]
All values of x must be given to the nearest mm.
Significant figures: [1]
All values of I2 / I1 must have the same significant figures as, or one more than, the
least number of significant figures in raw I1 and I2
Calculation: [1]
Values of I2 / I1 calculated correctly.
(c) (i) Axes: [1]
Sensible scales must be used, no awkward scales (e.g. 3:10).
Diameter of plots must be Y 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 ± 0.25m–1
in the 1/x direction of a straight line.
candidate.
Line must not be kinked or thicker than half a small square.

(iii) Gradient: [1]
The sign of the gradient must match the graph.
The hypotenuse of the triangle used must be greater than half the length of the drawn
line.
y intercept: [1]
Either:
Correct read-off from a point on the line and substitution into y = mx + c.
Or:
Correct read-off of the intercept directly from the graph.
(d) Value of P = candidate’s gradient and value of Q = candidate’s intercept. [1]
Unit for P (m or cm or mm, consistent with value) and Q (no unit) correct. [1]
[Total: 20]
2 (a) (i) Value for D in range 10 to 20mm to the nearest mm, with unit. [1]
(ii) Percentage uncertainty in D based on an absolute uncertainty of 0.5, 1, 2 or 3mm. [1]
(b) (ii) Value of x to the nearest mm, in range 1.3 – 1.7cm, with unit. [1]
(iii) Correct calculation of V with consistent unit. [1]
(c) (iv) Raw time values to 0.1s or 0.01s. Value of T in range 0.1 – 1.0s. [1]
Evidence of repeat measurements. [1]
(d) (iv) Second value of x. [1]
(e) Second value of T. [1]
Second value of T < first value of T. [1]
(f) (i) Correct calculation of two values of k. [1]
(ii) Justification of significant figures in k linked to significant figures in D, x and time
(not just “raw readings”). [1]
(iii) Sensible comment relating to the calculated values of k, testing against a specified
criterion. [1]

(g)
* Credit in Bs or Cs, but not both.
** Credit in Es or Fs, but not both.
[Total: 20]
A two results not enough take more readings
and plot a graph/
calculate more k values and
compare
“repeat readings” on its
own
few readings/
only one reading
take more readings and
(calculate) average k
B parallax error in D/
difficult to measure D because
loop is in the way
use Vernier
calipers/micrometer/travelling
microscope to measure D*
use string
C V not accurate because
D not internal diameter
measure thickness/diameter of
wire using micrometer
use travelling microscope/Vernier
calipers to measure D*
D mass swings side-to-side/
horizontal movement/ moves
in more than one plane/non-
uniform oscillation
E times are small/large
uncertainty in T
use bigger mass
improved timing method e.g.
motion/position sensor below
weight/video with timer/video and
view frame-by-frame**
light gates/
human error/reaction
time/
time more cycles/
high frequency
oscillations
F difficult to judge
start of/end of/complete
oscillation
fixed/fiducial marker
improved timing method e.g.
motion/position sensor below
weight/video with timer/video and
view frame-by-frame**
marker fixed to spring/
marker placed at
extreme(s) of oscillation
light gates
G metal strip bends/
not horizontal
use stiffer strip/
thicker strip/support strip at both
ends.
strip not straight/
move spring/use
stronger strip

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

Section A
1 (a) force is proportional to the product of the masses and
inversely proportional to the square of the separation M1
either point masses or separation >> size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1
mv2
/r = GMm/r2
and EK = ½mv2
M1
hence EK = GMm/2r A0 [2]
(ii) 1. ∆EK = ½ × 4.00 × 1014
× 620 × ({7.30 × 106
}–1
– {7.34 × 106
}–1
) C1
= 9.26 × 107
J (ignore any sign in answer) A1 [2]
(allow 1.0 × 108
J if evidence that EK evaluated separately for each r)
2. ∆EP = 4.00 × 1014
× 620 × ({7.30 × 106
}–1
– {7.34 × 106
}–1
) C1
= 1.85 × 108
(allow 1.8 or 1.9 × 108
J)
(iii) either (7.30 × 106
)–1
– (7.34 × 106
)–1
or ∆EK is positive/EK increased M1
speed has increased A1 [2]
2 (a) (i) sum of potential energy and kinetic energy of atoms/molecules/particles M1
reference to random A1 [2]
(ii) no intermolecular forces B1
no potential energy B1
internal energy is kinetic energy (of random motion) of molecules B1 [3]
(reference to random motion here then allow back credit to (i) if M1 scored)
(b) kinetic energy ∝ thermodynamic temperature B1
either temperature in Celsius, not kelvin so incorrect
or temperature in kelvin is not doubled B1 [2]
3 (a) temperature of the spheres is the same B1
no (net) transfer of energy between the spheres B1 [2]
(b) (i) power = m × c × ∆θ where m is mass per second C1
3800 = m × 4.2 × (42 – 18) C1
m = 38g s–1
A1 [3]
(ii) some thermal energy is lost to the surroundings M1
so rate is an overestimate A1 [2]
4 (a) straight line through origin M1
shows acceleration proportional to displacement A1
negative gradient M1
shows acceleration and displacement in opposite directions A1 [4]

(b) (i) 2.8cm A1 [1]
(ii) either gradient = ω2
and ω = 2πf or a = –ω2
x and ω = 2πf C1
gradient = 13.5/(2.8 × 10–2
) = 482
ω = 22rad s–1
C1
frequency = (22/2π =) 3.5Hz A1 [3]
(c) e.g. lower spring may not be extended
e.g. upper spring may exceed limit of proportionality/elastic limit
(any sensible suggestion) B1 [1]
5 (a) (i) ratio of charge and potential (difference)/voltage
(ratio must be clear) B1 [1]
(ii) capacitor has equal magnitudes of (+)ve and (-)ve charge B1
total charge on capacitor is zero (so does not store charge) B1
(+)ve and (-)ve charges to be separated M1
work done to achieve this so stores energy A1 [4]
(b) (i) capacitance of Y and Z together is 24 µF C1
1/C = 1/24 + 1/12
C = 8.0 µF (allow 1 s.f.) A1 [2]
(ii) some discussion as to why all charge of one sign on one plate of X B1
Q = (CV =) 8.0 × 10–6
× 9.0 M1
= 72µC A0 [2]
(iii) 1. V = (72 × 10–6
)/(12 × 10–6
)
= 6.0V (allow 1 s.f.) (allow 72/12) A1 [1]
2. either Q = 12 × 10–6
× 3.0 or charge is shared between Y and Z C1
charge = 36µC A1 [2]
Must have correct voltage in (iii)1 if just quote of 36µC in (iii)2.
6 (a) (i) particle must be moving M1
with component of velocity normal to magnetic field A1 [2]
(ii) F = Bqv sin θ M1
q, v and θ explained A1 [2]
(b) (i) face BCGF shaded A1 [1]
(ii) between face BCGF and face ADHE A1 [1]
(c) potential difference gives rise to an electric field M1
either FE = qE (no need to explain symbols)
or electric field gives rise to force (on an electron) A1 [2]

7 (a) induced e.m.f./current produces effects/acts in such a direction/tends M1
to oppose the change causing it A1 [2]
(b) (i) 1. to reduce flux losses/increase flux linkage/easily magnetised and
demagnetised B1 [1]
2. to reduce energy/heat losses (do not allow ‘to prevent energy losses’) M1
caused by eddy currents A1 [2]
(allow 1 mark for ‘reduce eddy currents’)
(ii) alternating current/voltage B1
gives rise to (changing) flux in core B1
flux links the secondary coil M1
(by Faraday’s law) changing flux induces e.m.f. (in secondary coil) A1 [4]
8 (a) discrete quantity/packet/quantum of energy of electromagnetic radiation B1
energy of photon = Planck constant × frequency B1 [2]
(b) threshold frequency (1)
rate of emission is proportional to intensity (1)
max. kinetic energy of electron dependent on frequency (1)
max. kinetic energy independent of intensity (1)
(any three, 1 each, max 3) B3 [3]
(c) either E = hc/λ or hc/λ = eV C1
λ = 450 nm to give work function of 3.5 eV
energy = 4.4 × 10–19
or 2.8 eV to give λ = 355nm M1
2.8eV < 3.5eV so no emission 355nm < 450nm so no A1 [3]
or work function = 3.5eV
threshold frequency = 8.45×1014
Hz C1
450nm = 6.67×1014
Hz M1
6.67 × 1014
Hz < 8.45 × 1014
Hz A1

Section B
9 (a) e.g. zero output impedance/resistance
infinite input impedance/resistance
infinite (open loop) gain
infinite bandwidth
infinite slew rate
1 each, max. 3 B3 [3]
(b) (i) graph: square wave M1
correct cross-over points where V2 = V1 A1
amplitude 5V A1
correct polarity (positive at t = 0) A1 [4]
(ii) correct symbol for LED M1
diodes connected correctly between VOUT and earth A1
correct polarity consistent with graph in (i) A1 [3]
(R points ‘down’ if (i) correct)
10 X-ray images taken from different angles/X-rays directed from different angles B1
of one section/slice (1)
all images in the same plane (1)
images combined to give image of section/slice B1
images of successive sections/slices combined B1
image formed using a computer B1
image formed is 3D image (1)
that can be rotated/viewed from different angles (1)
(four B-marks plus any two additional marks) B2 [6]
11 (a) e.g. noise can be eliminated/filtered/signal can be regenerated
extra bits can be added to check for errors
multiplexing possible
digital circuits are more reliable/cheaper
data can be encrypted for security
any sensible advantages, 1 each, max. 3 B3 [3]
(b) (i) 1. higher frequencies can be reproduced B1 [1]
2. smaller changes in loudness/amplitude can be detected B1 [1]
(ii) bit rate = 44.1 × 103
× 16 C1
= 7.06 × 105
s–1
number = 7.06 × 106
× 340
= 2.4 × 108
A1 [2]
12 (a) (i) signal in one wire (pair) is picked up by a neighbouring wire (pair) B1 [1]
(ii) outer of coaxial cable is earthed B1
outer shields the core from noise/external signals B1 [2]

(b) attenuation per unit length = 1/L × 10lg(P2/P1) C1
signal power at receiver = 102.5
× 3.8 × 10–8
= 1.2 × 10–5
W C1
attenuation in wire pair = 10lg({3.0 × 10–3
}/{1.2 × 10–5
})
= 24dB C1
attenuation per unit length = 24/1.4
= 17 dBkm–1
A1 [4]
(other correct methods of calculation are possible)

9702 PHYSICS
components.

Section A
1 (a) force is proportional to the product of the masses and
inversely proportional to the square of the separation M1
either point masses or separation >> size of masses A1 [2]
(b) (i) gravitational force provides the centripetal force B1
mv2
/r = GMm/r2
and EK = ½mv2
M1
hence EK = GMm/2r A0 [2]
(ii) 1. ∆EK = ½ × 4.00 × 1014
× 620 × ({7.30 × 106
}–1
– {7.34 × 106
}–1
) C1
= 9.26 × 107
(allow 1.0 × 108
J if evidence that EK evaluated separately for each r)
2. ∆EP = 4.00 × 1014
× 620 × ({7.30 × 106
}–1
– {7.34 × 106
}–1
) C1
= 1.85 × 108
(allow 1.8 or 1.9 × 108
J)
(iii) either (7.30 × 106
)–1
– (7.34 × 106
)–1
or ∆EK is positive/EK increased M1
speed has increased A1 [2]
2 (a) (i) sum of potential energy and kinetic energy of atoms/molecules/particles M1
reference to random A1 [2]
(ii) no intermolecular forces B1
no potential energy B1
internal energy is kinetic energy (of random motion) of molecules B1 [3]
(reference to random motion here then allow back credit to (i) if M1 scored)
(b) kinetic energy ∝ thermodynamic temperature B1
either temperature in Celsius, not kelvin so incorrect
or temperature in kelvin is not doubled B1 [2]
3 (a) temperature of the spheres is the same B1
no (net) transfer of energy between the spheres B1 [2]
(b) (i) power = m × c × ∆θ where m is mass per second C1
3800 = m × 4.2 × (42 – 18) C1
m = 38g s–1
A1 [3]
(ii) some thermal energy is lost to the surroundings M1
so rate is an overestimate A1 [2]
4 (a) straight line through origin M1
shows acceleration proportional to displacement A1
negative gradient M1
shows acceleration and displacement in opposite directions A1 [4]

(b) (i) 2.8cm A1 [1]
(ii) either gradient = ω2
and ω = 2πf or a = –ω2
x and ω = 2πf C1
gradient = 13.5/(2.8 × 10–2
) = 482
ω = 22rad s–1
C1
frequency = (22/2π =) 3.5Hz A1 [3]
(c) e.g. lower spring may not be extended
e.g. upper spring may exceed limit of proportionality/elastic limit
(any sensible suggestion) B1 [1]
5 (a) (i) ratio of charge and potential (difference)/voltage
(ratio must be clear) B1 [1]
(ii) capacitor has equal magnitudes of (+)ve and (-)ve charge B1
total charge on capacitor is zero (so does not store charge) B1
(+)ve and (-)ve charges to be separated M1
work done to achieve this so stores energy A1 [4]
(b) (i) capacitance of Y and Z together is 24 µF C1
1/C = 1/24 + 1/12
C = 8.0 µF (allow 1 s.f.) A1 [2]
(ii) some discussion as to why all charge of one sign on one plate of X B1
Q = (CV =) 8.0 × 10–6
× 9.0 M1
= 72µC A0 [2]
(iii) 1. V = (72 × 10–6
)/(12 × 10–6
)
= 6.0V (allow 1 s.f.) (allow 72/12) A1 [1]
2. either Q = 12 × 10–6
× 3.0 or charge is shared between Y and Z C1
charge = 36µC A1 [2]
Must have correct voltage in (iii)1 if just quote of 36µC in (iii)2.
6 (a) (i) particle must be moving M1
with component of velocity normal to magnetic field A1 [2]
(ii) F = Bqv sin θ M1
q, v and θ explained A1 [2]
(b) (i) face BCGF shaded A1 [1]
(ii) between face BCGF and face ADHE A1 [1]
(c) potential difference gives rise to an electric field M1
either FE = qE (no need to explain symbols)
or electric field gives rise to force (on an electron) A1 [2]

7 (a) induced e.m.f./current produces effects/acts in such a direction/tends M1
to oppose the change causing it A1 [2]
(b) (i) 1. to reduce flux losses/increase flux linkage/easily magnetised and
demagnetised B1 [1]
2. to reduce energy/heat losses (do not allow ‘to prevent energy losses’) M1
caused by eddy currents A1 [2]
(allow 1 mark for ‘reduce eddy currents’)
(ii) alternating current/voltage B1
gives rise to (changing) flux in core B1
flux links the secondary coil M1
(by Faraday’s law) changing flux induces e.m.f. (in secondary coil) A1 [4]
8 (a) discrete quantity/packet/quantum of energy of electromagnetic radiation B1
energy of photon = Planck constant × frequency B1 [2]
(b) threshold frequency (1)
rate of emission is proportional to intensity (1)
max. kinetic energy of electron dependent on frequency (1)
max. kinetic energy independent of intensity (1)
(any three, 1 each, max 3) B3 [3]
(c) either E = hc/λ or hc/λ = eV C1
λ = 450 nm to give work function of 3.5 eV
energy = 4.4 × 10–19
or 2.8 eV to give λ = 355nm M1
2.8eV < 3.5eV so no emission 355nm < 450nm so no A1 [3]
or work function = 3.5eV
threshold frequency = 8.45×1014
Hz C1
450nm = 6.67×1014
Hz M1
6.67 × 1014
Hz < 8.45 × 1014
Hz A1

Section B
9 (a) e.g. zero output impedance/resistance
infinite input impedance/resistance
infinite (open loop) gain
infinite bandwidth
infinite slew rate
1 each, max. 3 B3 [3]
(b) (i) graph: square wave M1
correct cross-over points where V2 = V1 A1
amplitude 5V A1
correct polarity (positive at t = 0) A1 [4]
(ii) correct symbol for LED M1
diodes connected correctly between VOUT and earth A1
correct polarity consistent with graph in (i) A1 [3]
(R points ‘down’ if (i) correct)
10 X-ray images taken from different angles/X-rays directed from different angles B1
of one section/slice (1)
all images in the same plane (1)
images combined to give image of section/slice B1
images of successive sections/slices combined B1
image formed using a computer B1
image formed is 3D image (1)
that can be rotated/viewed from different angles (1)
(four B-marks plus any two additional marks) B2 [6]
11 (a) e.g. noise can be eliminated/filtered/signal can be regenerated
extra bits can be added to check for errors
multiplexing possible
digital circuits are more reliable/cheaper
data can be encrypted for security
any sensible advantages, 1 each, max. 3 B3 [3]
(b) (i) 1. higher frequencies can be reproduced B1 [1]
2. smaller changes in loudness/amplitude can be detected B1 [1]
(ii) bit rate = 44.1 × 103
× 16 C1
= 7.06 × 105
s–1
number = 7.06 × 106
× 340
= 2.4 × 108
A1 [2]
12 (a) (i) signal in one wire (pair) is picked up by a neighbouring wire (pair) B1 [1]
(ii) outer of coaxial cable is earthed B1
outer shields the core from noise/external signals B1 [2]

(b) attenuation per unit length = 1/L × 10lg(P2/P1) C1
signal power at receiver = 102.5
× 3.8 × 10–8
= 1.2 × 10–5
W C1
attenuation in wire pair = 10lg({3.0 × 10–3
}/{1.2 × 10–5
})
= 24dB C1
attenuation per unit length = 24/1.4
= 17 dBkm–1
A1 [4]
(other correct methods of calculation are possible)

9702 PHYSICS
components.

Section A
1 (a) (i) number of molecules B1 [1]
(ii) mean square speed B1 [1]
(b) (i) 1. pV = nRT C1
n = (6.1 × 105
× 2.1 × 104
× 10–6
) / (8.31 × 285) C1
n = 5.4mol A1 [3]
2. either N = nNA
= 5.4 × 6.02 × 1023
C1
= 3.26 × 1024
A1
or
pV = NkT
N = (6.1 × 105
× 2.1 × 104
× 10–6
) / (1.38 × 10–23
× 285) (C1)
N = 3.26 × 1024
(A1) [2]
(ii) either 6.1 × 105
× 2.1 × 10–2
= 1
/3 × 3.25 × 1024
× 4 × 1.66 × 10–27
× <c2
> C1
<c2
> = 1.78 × 106
C1
cRMS = 1.33 × 103
m s–1
A1
or
1
/2 × 4 × 1.66 × 10–27
× <c2
> = 3
/2 × 1.38 × 10–23
× 285 (C1)
<c2
> = 1.78 × 106
(C1)
cRMS = 1.33 × 103
m s–1
(A1) [3]
2 (a) (i) 1. 0.1s, 0.3s, 0.5s, etc (any two) A1 [1]
2. either 0, 0.4s, 0.8s, 1.2s
or
0.2s, 0.6s, 1.0s (any two) A1 [1]
(ii) period = 0.4s C1
frequency = (1/0.4 =) 2.5Hz A1 [2]
(iii) phase difference = 90° or ½ π rad B1 [1]
(b) frequency = 2.4 – 2.5Hz B1 [1]
(c) e.g. attach sheet of card to trolley M1
increases damping / frictional force A1
e.g. reduce oscillator amplitude (M1)
reduces power/energy input to system (A1) [2]

3 (a) (i) (tangent to line gives) direction of force on a (small test) mass B1 [1]
(ii) (tangent to line gives) direction of force on a (small test) charge M1
charge is positive A1 [2]
(b) similarity:
e.g. radial fields
lines normal to surface
greater separation of lines with increased distance from sphere
field strength ∝ 1 / (distance to centre of sphere)2
(allow any sensible answer) B1
difference:
e.g. gravitational force (always) towards sphere B1
electric force direction depends on sign of charge on sphere / towards or
away from sphere B1
e.g. gravitational field/force is attractive (B1)
electric field/force is attractive or repulsive (B1)
(allow any sensible comparison) [3]
(c) gravitational force = 1.67 × 10–27
× 9.81
= 1.6 × 10–26
N A1
electric force = 1.6 × 10–19
× 270 / (1.8 × 10–2
) C1
= 2.4 × 10–15
N A1
electric force very much greater than gravitational force B1 [4]
4 (a) force on proton is normal to velocity and field M1
provides centripetal force (for circular motion) A1 [2]
(b) magnetic force = Bqv B1
centripetal force = mrω2
or mv2
/r B1
v = rω B1
Bqv = Bqrω = mrω2
ω = Bq/m A1 [4]
5 (a) either φ = BA sinθ M1
where A is the area (through which flux passes)
θ is the angle between B and (plane of) A A1
or
φ = BA (M1)
where A is area normal to B (A1) [2]
(b) graph: VH constant and non zero between the poles and zero outside M1
sharp increase/decrease at ends of magnet A1 [2]

(c) (i) (induced) e.m.f. proportional to M1
rate of change of (magnetic) flux (linkage) A1 [2]
(ii) short pulse on entering and on leaving region between poles M1
pulses approximately the same shape but opposite polarities A1
e.m.f. zero between poles and outside A1 [3]
6 (a) (i) connection to ‘top’ of resistor labelled as positive B1 [1]
(ii) diode B and diode D B1 [1]
(b) (i) VP = 4.0V C1
mean power = VP
2
/2R C1
= 42
/ (2 × 2700)
= 2.96 × 10–3
W A1 [3]
(ii) capacitor, correct symbol, connected in parallel with R B1 [1]
(c) graph: half-wave rectification M1
same period and same peak value A1 [2]
7 (a) wavelength associated with a particle M1
that is moving A1 [2]
(b) (i) kinetic energy = 1.6 × 10–19
× 4700 C1
= 7.52 × 10–16
J
either energy = p2
/2m or EK = ½mv2
and p = mv C1
p = √(7.52 × 10–16
× 2 × 9.1 × 10–31
) C1
= 3.7 × 10–23
N s
λ = h/p C1
= (6.63 × 10–34
) / (3.7 × 10–23
)
= 1.8 × 10–11
m A1 [5]
(ii) wavelength is about separation of atoms B1
can be used in (electron) diffraction B1 [2]
8 (a) (i) x = 2 A1 [1]
(ii) either beta particle or electron B1 [1]
(b) (i) mass of separate nucleons = {(92 × 1.007) + (143 × 1.009)} u C1
= 236.931u C1
binding energy = 236.931u – 235.123u
= 1.808u A1 [3]

(ii) E = mc2
C1
energy = 1.808 × 1.66 × 10–27
× (3.0 × 108
)2
= 2.7 × 10–10
J C1
binding energy per nucleon = (2.7 × 10–10
) / (235 × 1.6 × 10–13
) M1
= 7.18MeV A0 [3]
(c) energy released = (95 × 8.09) + (139 × 7.92) – (235 × 7.18) C1
= 1869.43 – 1687.3
= 182MeV A1 [2]
(allow calculation using mass difference between products and reactants)
Section B
9 (a) light-emitting diode (allow LED) B1 [1]
(b) gives a high or a low output / +5V or –5V output M1
dependent on which of the inputs is at a higher potential A1 [2]
(c) (i) provides a reference/constant potential B1 [1]
(ii) determines temperature of ‘switch-over’ B1 [1]
(d) (i) relay A1 [1]
(ii) relay connected correctly for op-amp output and high-voltage circuit B1
diode with correct polarity in output from op-amp B1 [2]
10 (a) background reading = 19 B1 [1]
(b) A = 2 A1
B = 5 A1
C = 9 A1
D = 3 A1 [4]
(Allow 1 mark if only subtracts background reading)
(c) (i) either 5, 14 or 14, 5 (A+D, B+C or v.v.) B1 [1]
(ii) Three numbers and ‘inside’ number is 8 (B+D) B1
Three numbers and ‘outside’ numbers are either 2,9 or 9,2 (A,C or v.v.) B1 [2]
11 (a) high frequency wave B1
the amplitude or the frequency is varied M1
the variation represents the information signal /
in synchrony with (the displacement of) the information signal. A1 [3]

(b) e.g. shorter aerial required
longer transmission range / lower transmitter power / less attenuation
allows more than one station in a region
less distortion
(allow any three sensible suggestions, 1 mark each) B3 [3]
12 (a) (i) e.g. linking a (land) telephone to the (local) exchange B1 [1]
(ii) e.g. connecting an aerial to a television B1 [1]
(iii) e.g. linking a ground station to a satellite B1 [1]
(b) (i) attenuation = 10lg (P2 / P1) C1
total attenuation = 2.1 × 40 (= 84dB) C1
84 = 10lg ({450 × 10–3
} / P)
P = 1.8 × 10–9
W A1 [3]
(answer 1.1 ×108
W scores 1 mark only)
(ii) maximum attenuation = 10lg ({450 × 10–3
} / {7.2 × 10–11
})
= 98dB C1
maximum length = 98/2.1
= 47km A1 [2]

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

1 Planning (15 marks)
Defining the problem (3 marks)
P v is the independent variable or vary v. [1]
P E is the dependent variable or measure E. [1]
P Keep the number of turns on the coil constant. [1]
Methods of data collection (5 marks)
M1 Labelled diagram showing magnet falling vertically through coil. [1]
M2 Voltmeter or c.r.o. connected to the coil. Allow voltage sensor connected to datalogger. [1]
M3 Method to change speed e.g. change height. [1]
M4 Measurements to determine v. Use metre rule to measure distance magnet falls to the
bottom of the coil or metre rule/ruler to measure length of coil or ruler to measure length of
the magnet. [Allow timing instrument to measure the time of the fall from the start to the
bottom of the coil.] [1]
M5 Method of determining v corresponding to appropriate distance e.g. v = √2gh or v=2h/t (for
height method) or v = L/t for length of magnet or coil and by stopwatch, timer or lightgate(s)
connected to datalogger. [Allow v = gt for timing fall to bottom of coil.] [1]
Method of analysis (2 marks)
A Plot a graph of E against v. [Allow lg E against lg v] [1]
A Relationship valid if straight line through origin. [1]
[If lg-lg then straight line with gradient = (+)1 (ignore reference to y-intercept)]
Safety considerations (1 mark)
S Keep away from falling magnet/use sand tray/cushion to catch magnet. [1]
Additional detail (4 marks)
D1/2/3/4 Relevant points might include [4]
Use coil with large number of turns/drop magnet from large heights/strong magnet
1 Detailed use of datalogger/storage oscilloscope to determine maximum E; allow video
camera including slow motion play back
2 Use same magnet or magnet of same strength.
3 Use of short magnet so that v is (nearly) constant
4 Use short/thin coil so that v is (nearly) constant
5 Use a non-metallic vertical guide/tube
6 Method to support vertical coil or guide/tube
7 Repeat experiment for each v 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 Gradient = hc/e
y-intercept = – B/e
Note y-intercept must be negative
(b) T1 1/λ / 106
m–1
Appropriate column heading
T2
1.05 or 1.053
1.14 or 1.143
1.53 or 1.527
1.79 or 1.786
1.98 or 1.980
2.33 or 2.326
Must be values in table.
A mixture of 3 s.f. and 4 s.f. is allowed.
(c) (i) G1 Six points plotted correctly Must be within half a small square.
Penalise ‘blobs’. Ecf allowed from table.
U1 All error bars in V/V plotted
correctly.
Do not allow near misses
(c) (ii) G2 Line of best fit If points are plotted correctly then lower
end of line should pass between (1.12,
0.7) and (1.16,0.7) and upper end of line
should pass between (2.32, 2.25) and
(2.34, 2.25). 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.
(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. Should be about
1.3 × 10–6
.
U2 Uncertainty in gradient Method of determining absolute
uncertainty Difference in worst gradient
and gradient. [± 0.08]
(c) (iv) C2 y-intercept Must be negative
Expect to see point substituted into
y = mx + c
FOX does not score. Do not penalise
POT.
Should be between –0.72 and –0.86
U3 Method of determining uncertainty in
y-intercept
Difference in worst y-intercept and
y-intercept.
[Should be about ± 0.14]. FOX does not
score. Allow ecf from (c)(iv).

(d) (i) C3 h in the range 6.77 × 10–34
to
7.14 × 10–34
and given to 2 or 3
significant figures
Gradient must be used. Penalise 1 s.f. or >3
s.f.
h = gradient × e/c = gradient × 5.33 × 10–28
Allow 6.8 × 10–34
to 7.1 × 10–34
to 2 s.f.
(d) (ii) U4 Percentage uncertainty in h
100×
m
m∆
or 100×
h
h∆
[should be about 6%]
(e) C4 B = –e × y-intercept
and J or CV or VC
Ignore ‘–’ signs. y-intercept must be used
but allow ecf from FOX. Should be between
1.16 × 10–19
J and 1.37 × 10–19
J. If FOX 8.3
× 10–20
J
U5 Absolute uncertainty in B Uncertainty = best B – worst B
= ∆y-intercept × e
[Total: 15]
Uncertainties in Question 2
(c) (iii) Gradient [U2]
Uncertainty = gradient of line of best fit – gradient of worst acceptable line
Uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)
(iv) [U3]
Uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line
Uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)
(d) (ii) [U4]
Percentage uncertainty = 100×
m
m∆
Percentage uncertainty = 100
)min(max
100 2
1
×
−
=×
h
hh
h
h∆
(e) [U5]
Absolute uncertainty = best B – worst B
Absolute uncertainty = ∆y-intercept × e
Absolute uncertainty = B
c
c
×
∆

9702 PHYSICS
maximum raw mark 30
components.

P v is the independent variable or vary v. [1]
P E is the dependent variable or measure E. [1]
P Keep the number of turns on the coil constant. [1]
M1 Labelled diagram showing magnet falling vertically through coil. [1]
M2 Voltmeter or c.r.o. connected to the coil. Allow voltage sensor connected to datalogger. [1]
M3 Method to change speed e.g. change height. [1]
M4 Measurements to determine v. Use metre rule to measure distance magnet falls to the
bottom of the coil or metre rule/ruler to measure length of coil or ruler to measure length of
the magnet. [Allow timing instrument to measure the time of the fall from the start to the
bottom of the coil.] [1]
M5 Method of determining v corresponding to appropriate distance e.g. v = √2gh or v=2h/t (for
height method) or v = L/t for length of magnet or coil and by stopwatch, timer or lightgate(s)
connected to datalogger. [Allow v = gt for timing fall to bottom of coil.] [1]
A Plot a graph of E against v. [Allow lg E against lg v] [1]
S Keep away from falling magnet/use sand tray/cushion to catch magnet. [1]
Use coil with large number of turns/drop magnet from large heights/strong magnet
1 Detailed use of datalogger/storage oscilloscope to determine maximum E; allow video
camera including slow motion play back
2 Use same magnet or magnet of same strength.
3 Use of short magnet so that v is (nearly) constant
4 Use short/thin coil so that v is (nearly) constant
5 Use a non-metallic vertical guide/tube
6 Method to support vertical coil or guide/tube
7 Repeat experiment for each v and average
[Total: 15]

(a) A1 Gradient = hc/e
y-intercept = – B/e
Note y-intercept must be negative
(b) T1 1/λ / 106
m–1
Appropriate column heading
T2
1.05 or 1.053
1.14 or 1.143
1.53 or 1.527
1.79 or 1.786
1.98 or 1.980
2.33 or 2.326
Must be values in table.
A mixture of 3 s.f. and 4 s.f. is allowed.
(c) (i) G1 Six points plotted correctly Must be within half a small square.
Penalise ‘blobs’. Ecf allowed from table.
U1 All error bars in V/V plotted
correctly.
Do not allow near misses
(c) (ii) G2 Line of best fit If points are plotted correctly then lower
end of line should pass between (1.12,
0.7) and (1.16,0.7) and upper end of line
should pass between (2.32, 2.25) and
(2.34, 2.25). Allow ecf from points plotted
incorrectly – examiner judgement.
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.
(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. Should be about
1.3 × 10–6
.
U2 Uncertainty in gradient Method of determining absolute
uncertainty Difference in worst gradient
and gradient. [± 0.08]
(c) (iv) C2 y-intercept Must be negative
Expect to see point substituted into
y = mx + c
FOX does not score. Do not penalise
POT.
Should be between –0.72 and –0.86
U3 Method of determining uncertainty in
y-intercept
Difference in worst y-intercept and
y-intercept.
[Should be about ± 0.14]. FOX does not
score. Allow ecf from (c)(iv).

(d) (i) C3 h in the range 6.77 × 10–34
to
7.14 × 10–34
and given to 2 or 3
significant figures
Gradient must be used. Penalise 1 s.f. or >3
s.f.
h = gradient × e/c = gradient × 5.33 × 10–28
Allow 6.8 × 10–34
to 7.1 × 10–34
to 2 s.f.
(d) (ii) U4 Percentage uncertainty in h
100×
m
m∆
or 100×
h
h∆
[should be about 6%]
(e) C4 B = –e × y-intercept
and J or CV or VC
Ignore ‘–’ signs. y-intercept must be used
but allow ecf from FOX. Should be between
1.16 × 10–19
J and 1.37 × 10–19
J. If FOX 8.3
× 10–20
J
U5 Absolute uncertainty in B Uncertainty = best B – worst B
= ∆y-intercept × e
[Total: 15]
(iv) [U3]
Uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line
Uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)
(d) (ii) [U4]
Percentage uncertainty = 100×
m
m∆
)min(max
100 2
1
×
−
=×
h
hh
h
h∆
(e) [U5]
Absolute uncertainty = best B – worst B
Absolute uncertainty = ∆y-intercept × e
Absolute uncertainty = B
c
c
×
∆

9702 PHYSICS
maximum raw mark 30
components.

P λ is the independent variable or vary λ. [1]
P θ is the dependent variable or measure θ (for each λ). [1]
P Light sources to be of similar intensity/brightness. [1]
M1 Labelled diagram showing observer, light sources with method of producing monochromatic
light e.g. filter/coloured LED. [1]
M2 Method to measure wavelength: record from filter/LED or Young’s slit/diffraction grating
method. [1]
M3 Use a rule to measure the distances. [1]
M4 Method to determine θ, e.g. θ (or sin θ or tan θ) = separation/distance or
( ) distance2
separation
2
tan
×
=θ
Do not allow protractor methods. [1]
M5 Carry out the experiment in a dark room. [1]
A Plot a graph of θ against λ. [Allow lg θ against lg λ]. [1]
S Lamp becomes hot, therefore do not touch/switch off when not in use or use gloves when
moving hot lamp.
OR Light may damage eyes, therefore wear dark glasses or do not look at unprotected lamps.
[1]
1 Use vertical filament lamps. Allow vertical slits.
2 Additional detail on measuring λ e.g. use of equation for Young’s slit/diffraction grating
method.
3 Use of vernier calipers to measure the separation of light sources.
4 Use large distances/separations.
5 θ = sin θ = tan θ for small angles.
6 View with the same eye.
7 Method to ensure distances are perpendicular or observer equidistant from pair of lamps.
8 Repeat experiment for each λ and average.
[Total: 15]

(a) A1 Gradient = kA2
(b) T1
T2 1.3 or 1.33 1.2
0.8(0)(0)(0) 0.74
0.571 or
0.5714
0.54 or 0.55
0.444 or
0.4444
0.41 or 0.411
or 0.410
0.364 or
0.3636
0.34
0.308 or
0.3077
0.29 or 0.30
T1 must be values in 1/M. Ignore row 2.
T2 must be to 2 s.f. or 3 s.f.
U1 From ± 0.2 or ± 0.15 to ± 0.02 or
± 0.03
Allow more than one significant figure.
Do not allow ± 0.1 for row 1.
(c) (i) G1 Six points plotted correctly Must be within half a small square. Penalise ‘blobs’.
Ecf allowed from table.
U2 All error bars in v2
plotted
correctly
Must be accurate within half a small square.
(c) (ii) G2 Line of best fit There must be a balance of points about the line of
best fit – examiner judgement. Allow ecf from points
plotted incorrectly.
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.
(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. Should be about
0.9.
U3 Uncertainty in gradient Method of determining absolute uncertainty.
Difference in worst gradient and gradient.
(d) (i) C2 k = gradient / A2
= gradient / 0.04
Should be about 22.
C3 N m–1
Allow kg s–2
(d) (ii) U4 Percentage uncertainty in k
5%1001002100 +×=××+×
m
m
A
A
m
m ∆∆∆

(e) C4 v in the range 0.534 to 0.559
and given to 2 or 3 s.f.
For 2 s.f. 0.53 to 0.56
U5 Uncertainty in v
[Total: 15]
(d) (ii) [U4]
Percentage uncertainty = 5%1001002100 +×=××+×
m
m
A
A
m
m ∆∆∆
Maximum k =
)(min
max
2
A
m
Minimum k =
)(max
min
2
A
m
)min(max
100 2
1
×
−
=×
k
kk
k
k∆
(e) [U5]
Percentage uncertainty = 100100 2
1
××+×
k
k
A
A ∆∆
Absolute uncertainty = v × percentage uncertainty/100
Maximum v =
0.75
max
max
k
A×
Minimum v =
0.75
min
min
k
A ×
Absolute uncertainty = max v – v or v – min v or )min(max vv −2
1

9702 w12 ms_all

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