9702 w10 ms_all

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
GCE Advanced Subsidiary Level and GCE Advanced Level
MARK SCHEME for the October/November 2010 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.
• CIE will not enter into discussions or correspondence in connection with these mark schemes.
CIE is publishing the mark schemes for the October/November 2010 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 A/AS LEVEL – October/November 2010 9702 11
© UCLES 2010
Question
Number
Key
Question
Number
Key
1 D 21 A
2 C 22 B
3 B 23 A
4 A 24 C
5 B 25 A
6 C 26 A
7 B 27 D
8 A 28 A
9 C 29 D
10 B 30 B
11 B 31 C
12 B 32 A
13 A 33 B
14 B 34 C
15 D 35 C
16 B 36 B
17 D 37 A
18 D 38 C
19 D 39 D
20 D 40 C

9702 PHYSICS
examination.
syllabuses.

© UCLES 2010
Question
Number
Key
Question
Number
Key
1 A 21 A
2 B 22 A
3 A 23 C
4 C 24 C
5 D 25 D
6 D 26 C
7 D 27 B
8 A 28 A
9 C 29 A
10 B 30 D
11 B 31 A
12 C 32 C
13 D 33 D
14 D 34 B
15 D 35 B
16 B 36 B
17 C 37 B
18 C 38 D
19 C 39 C
20 D 40 B

9702 PHYSICS
examination.
syllabuses.

© UCLES 2010
Question
Number
Key
Question
Number
Key
1 B 21 B
2 B 22 D
3 C 23 A
4 D 24 A
5 A 25 C
6 A 26 D
7 C 27 A
8 B 28 D
9 B 29 B
10 B 30 A
11 B 31 C
12 C 32 C
13 D 33 C
14 A 34 B
15 B 35 A
16 D 36 A
17 D 37 B
18 B 38 C
19 A 39 C
20 D 40 D

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.

GCE AS/A LEVEL – October/November 2010 9702 21
© UCLES 2010
1 (a) length, current, temperature, amount of substance, (luminous intensity)
any three, 1 each B3 [3]
(b) (i) F: kg m s–2
B1
ρ: kg m–3
B1
v: m s–1
B1 [3]
(ii) some working e.g. kg m s–2
= m2
kg m–3
(m s–1
)k
M1
hence k = 2 A1 [2]
2 (a) (i) horizontal speed constant at 8.2 m s–1
C1
vertical component of speed = 8.2 tan 60° M1
= 14.2 m s–1
A0 [2]
(ii) 14.22
= 2 × 9.8 × h (using g = 10 then –1) C1
vertical distance = 10.3 m A1 [2]
(iii) time of descent = 14.2 / 9.8 = 1.45 s C1
x = 1.45 × 8.2
= 11.9 m A1 [2]
(b) (i) smooth path curved and above given path M1
hits ground at more acute angle A1 [2]
(ii) smooth path curved and below given path M1
hits ground at steeper angle A1 [2]
3 (a) force = rate of change of momentum (allow symbols if defined) B1 [1]
(b) (i) ∆ρ = 140 × 10–3
× (5.5 + 4.0) C1
= 1.33 kg m s–1
A1 [2]
(ii) force = 1.33 / 0.04 M1
= 33.3 N A0 [1]
(c) (i) taking moments about B C1
(33 × 75) + (0.45 × g × 25) = FA × 20 C1
FA = 129 N A1 [3]
(ii) FB = 33 + 129 + 0.45g C1
= 166 N A1 [2]

© UCLES 2010
4 (a) (i) F / A B1 [1]
(ii) ∆L / L B1 [1]
(iii) allow FL / A∆L B1 [1]
(iv) allow ρL / A or ρ (L + ∆L) / A B1 [1]
(b) (i) ∆L = FL / EA
= (30 × 2.6) / (7.0 × 1010
× 3.8 × 10–7
) M1
= 2.93 × 10–3
m = 2.93 mm A0 [1]
(ii) ∆R = ρ∆L / A C1
= (2.6 × 10–8
× 2.93 × 10–3
) / (3.8 × 10–7
)
= 2.0 × 10–4
Ω A1 [2]
(c) change in resistance is (very) small M1
so method is not appropriate A1 [2]
5 (a) when a wave passes through a slit / by an edge M1
the wave spreads out / changes direction A1 [2]
(b) diagram: wavelength unchanged M1
wavefront flat at centre, curving into geometrical shadow A1 [2]
(c) d sin θ = nλ C1
for θ = 90°
1 / (650 × 103
) = n × 590 × 10–9
M1
n = 2.6
number of orders is 2 A1 [3]
(d) intensity / brightness decreases (as order increases) B1 [1]
6 (a) (i) either P = V 2
/ R or P = VI and V = IR C1
R = 4.0 Ω A1 [2]
(ii) sketch vertical axis labelled appropriately B1
(straight) line from origin then curved in correct direction B1
line passes through 12 V, 3.0 A B1 [3]
(b) (i) 2.0 kW A1 [1]
(ii) 0.5 kW A1 [1]
(iii) total resistance = 3R / 2 C1
power = 0.67 kW A1 [2]

© UCLES 2010
7 (a) either different forms of same element
or nuclei have same number of protons M1
different numbers of neutrons (in the nucleus) A1 [2]
(b) (i) proton number conserved B1
nucleon number conserved B1
mass-energy conserved B1 [3]
(ii) 1. Z = 36 A1 [1]
2. x = 3 A1 [1]

9702 PHYSICS
examination.
syllabuses.

GCE A LEVEL – October/November 2010 9702 22
© UCLES 2010
1 (a) (i) scalar quantity has magnitude (allow size) B1
vector quantity has magnitude and direction B1 [2]
(ii) 1. temperature: scalar B1 [1]
2. acceleration: vector B1 [1]
3. resistance: scalar B1 [1]
(b) either triangle / parallelogram with correct shape C1
tension = 14 .3N (allow ± 0.5 N) A2 [3]
(if > ±0.5N but ≤ ±1N, allow 1 mark)
or R = 25 cos 35° (C1)
T = R tan 35° (C1)
T = 14.3N (A1)
or T = 25 sin 35° (C2)
T = 14.3N (A1)
or R and T resolved vertically and horizontally (C2)
leading to T = 14.3N (A1)
2 (a) (i) VH = 12.4 cos 36° (= 10.0 m s–1
) C1
distance= 10.0 × 0.17
= 1.7m A1 [2]
(ii) VV = 12.4 sin 36° (= 7.29m s–1
) C1
h = 7.29 × 0.17 – ½ × 9.81 × 0.172
C1
= 1.1m A1 [3]
(b) smooth curve with ball hitting wall below original B1
smooth curve showing rebound to ground with correct reflection at wall B1 [2]
3 (a) point at which (whole) weight (of body) (allow mass for weight) M1
appears / seems to act ... (for mass need ‘appears to be concentrated’) A1 [2]
(b) (i) point C shown at centre of rectangle ± 5mm B1 [1]
(ii) arrow vertically downwards, from C with arrow starting from the same
margin of error as in (b)(i) B1 [1]
(c) (i) reaction / upwards / supporting / normal reaction force M1
friction M1
force(s) at the rod A1 [3]
(ii) comes to rest with (line of action of) weight acting through rod
allow C vertically below the rod B1
so that weight does not have a moment about the pivot / rod B1 [2]

© UCLES 2010
4 (a) energy = average force × extension B1
= ½ × F × x B1
(Hooke’s law) extension proportional to (applied) force B1
hence F = kx B1
so E = ½kx2
A0 [4]
(b) (i) correct area shaded B1 [1]
(ii) 1.0 cm2
represents 1.0mJ or correct units used in calculation C1
ES = 6.4 ± 0.2mJ A2 [3]
(for answer > ±0.2mJ but ≤ ±0.4mJ, then allow 2/3 marks)
(iii) arrangement of atoms / molecules is changed B1 [1]
5 (a) (i) distance (of point on wave) from rest / equilibrium position B1 [1]
(ii) distance moved by wave energy / wavefront during one cycle of the source
or minimum distance between two points with the same phase or between
adjacent crests or troughs B1 [1]
(b) (i) T = 0.60s B1 [1]
(ii) λ = 4.0cm B1 [1]
(iii) either v = λ/T or v = fλ and f = 1/T C1
v = 6.7cms–1
A1 [2]
(c) (i) amplitude is decreasing M1
so, it is losing power A1 [2]
(ii) intensity ~ (amplitude)2
C1
ratio = 2.02
/ 1.12
C1
= 3.3 A1 [3]
6 (a) (i) at 22.5 °C, RT = 1600Ω or 1.6kΩ C1
total resistance = 800Ω A1 [2]
(ii) either use of potential divider formula or current = 9 / 2000 (4.5mA) C1
V = (0.8/2.0) × 9 V = (9/2000) × 800
= 3.6V = 3.6V A1 [2]
(b) (i) total resistance = 4/5 × 1200 C1
= 960Ω A1 [2]
(ii) for parallel combination, 1/960 = 1/1600 + 1/RT
RT = 2400Ω / 2.4kΩ C1
temperature = 11°C A1 [2]

© UCLES 2010
(c) e.g. only small part of scale used / small sensitivity B1
non-linear B1 [2]
(any two sensible suggestions, 1 each, max 2)
7 (a) (i) most α-particles were deviated through small angles B2 [2]
(allow 1 mark for ‘straight through’ / undeviated)
(ii) small fraction of α-particles deviated through large angles M1
greater than 90° (allow rebound back) A1 [2]
(b) e.g. β-particles have a range of energies
β-particles deviated by (orbital) electrons
β-particle has (very) small mass
(any two sensible suggestions, 1 each, max 2) B2 [2]
Do not allow β-particles have negative charge or β-particles have high speed

9702 PHYSICS
examination.
syllabuses.

© UCLES 2010
1 (a) allow 0.05mm → 0.15mm B1 [1]
(b) allow 0.25s → 0.5s B1 [1]
(c) allow 8N → 12N B1 [1]
ignore number of significant figures
2 crystalline: atoms / ions / particles in a regular arrangement / lattice
long range order / orderly pattern B1
(lattice) repeats itself (1)
polymer: long chain molecules / chains of monomers B1
some cross-linking between chains / tangled chains (1)
amorphous: disordered arrangement of molecules / atoms / particles B1
any ordering is short-range (1)
(three ‘B’ marks plus any other 2 marks) B2 [5]
3 connect microphone / (terminals of) loudspeaker to Y-plates of c.r.o. B1
adjust c.r.o. to produce steady wave of 1 (or 2) cycles / wavelengths on screen B1
measure length of cycle / wavelength λ and note time-base b M1
frequency = 1 / λb A1 [4]
(assume b is measured as s cm–1
, unless otherwise stated)
(if statement is ‘measure T, f = 1/T’ then last two marks are lost)
4 (a) acceptable straight line drawn (touching every point) B1 [1]
(b) the distance fallen is not d C1
d is the distance fallen plus the diameter of the ball A1 [2]
(‘d is not measured to the bottom of the ball’ scores 2/2)
(c) (i) diameter: allow 1.5 ± 0.5cm (accept one SF) A1 [1]
no ecf from (a)
(ii) gradient = 4.76, ± 0.1 with evidence that origin has not been used C1
gradient = g / 2 C1
g = 9.5m s–2
A1 [3]

© UCLES 2010
5 (a) (i) Fig. 5.2 B1 [1]
(ii) Fig. 5.3 B1 [1]
(b) kinetic energy increases from zero then decreases to zero B1 [1]
(c) (i) ∆EP = mg∆h / mgh C1
= 94 × 10–3
× 9.8 × 2.6 × 10–2
using g = 10 then –1
= 0.024J A1 [2]
(ii) either 0.024= ½ k × (2.6 × 10–2
)2
or ½ kd2
= ½k × (2.6 × 10–2
)2
– ½kd2
C1
0.012= ½k × d2
kd2
= ½k × (2.6 × 10–2
)2
C1
d = 0.018m d = 0.018m
= 1.8cm = 1.8cm A1 [3]
6 (a) when two (or more) waves meet (at a point) B1
(resultant) displacement is (vector) sum of individual displacements B1 [2]
(b) (i) λ = ax / D (if no formula given and substitution is incorrect then 0/3) C1
590 × 10–9
= (1.4 × 10–3
× x) / 2.6 C1
x = 1.1mm A1 [3]
(ii) 1. 180° (allow π if rad stated) A1 [1]
2. at maximum, amplitude is 3.4 units and at minimum, 0.6 units C1
intensity ~ amplitude2
allow I ~ a2
C1
ratio = 3.42
/ 0.62
= 32 A1 [3]
7 (a) (i) path: reasonable curve upwards between plates B1
straight and at a tangent to the curve beyond the plates B1 [2]
(ii) 1. (F =) E.g B1 [1]
2. (t =) L / v B1 [1]
(b) (i) total momentum of a system remains constant or total momentum of a
system before a collision equals total momentum after collision M1
provided no external force acts on the system A1 [2]
(do not accept ‘conserved’ but otherwise correct statement gets 1/2)
(ii) (∆p =) EqL / v allow ecf from (a)(ii) B1 [1]
(iii) either charged particle is not an isolated system M1
so law does not apply A1 [2]
or system is particle and ‘plates’ (M1)
equal and opposite ∆p on plates / so law applies (A1)

© UCLES 2010
8 (a) (i) either P = V2
/ R or I = 1200 / 230 or 5.22 C1
R = (230 × 230) / 1200
R = 2302
/ 1200 or R = 230 / 5.22 M1
= 44.1Ω = 44.1Ω A0 [2]
(ii) R = ρL / A C1
= (1.7 × 10–8
× 9.2 × 2) / (π × {0.45 × 10–3
}2
) M1
= 0.492Ω A0 [2]
(b) current = 230 /44.6 C1
power = (230 /44.6)2
× 44.1 C1
= 1170W A1 [3]
(allow full credit for solution based on potential divider)
(c) e.g. less power dissipated in the heater / smaller p.d. across heater /
more power loss in cable / current lower B1
cable becomes heated / melts B1 [2]
(any two sensible suggestions, 1 each, max 2)
9 (a) nucleus emits α-particles or β-particles and/or γ-radiation B1
to form a different / more stable nucleus B1 [2]
(b) (i) fluctuations in count rate (not ‘count rate is not constant’) B1 [1]
(ii) no effect B1 [1]
(iii) if the source is an α-emitter B1
either α-particles stopped within source (and gain electrons)
or α-particles are helium nuclei B1 [2]
allow 1/2 for ‘parent nucleus gives off radiation to form daughter nucleus’

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

© UCLES 2010
1 (a) (i) No help from Supervisor. [1]
(ii) Values of a and b with consistent units to the nearest mm. [1]
(b) Six sets of readings of a, b and R scores 5 marks, five sets scores 4 marks etc. [5]
Incorrect trend then –1. Correct trend b/a increases, R increases.
Major help from supervisor –1.
Range: used R = 8000 Ω or 7000 Ω. [1]
Column headings (R/Ω, a/m, b/m, b/a). [1]
Must have R and either b/a or a and b columns.
Each column heading must contain a quantity and a unit where appropriate.
Ignore any units in the body of the table.
There must be some distinguishing mark between the quantity and the unit (solidus is
expected but accept, for example, R (Ω).
Consistency of presentation of readings. [1]
All values of raw a and b must be given to the nearest mm.
Significant figures. [1]
Significant figures for b/a must be the same as, or one more than, the least number of
s.f. used in a or b.
Correct calculation of b/a. [1]
(c) (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. Ignore units.
Scale markings should be no more than three large squares apart.
All observations must be plotted. Ignore any plot off the grid. [1]
Write a ringed total of plotted points.
Ring and check a suspect point.
Work to an accuracy of half a small square.
Do not accept blobs (points with diameter > 0.5 small square).
(ii) Line of best fit. [1]
Judge by balance of at least 5 trend points about candidate’s line.
There must be an even distribution of points either side of the line along the full
length.
Line must not be kinked. Do not allow lines thicker than half a small square.
Quality. [1]
Scatter of points must be less than ± 200 Ω in the R – axis about a straight line.
All points in the table must be plotted (at least 5) for this mark to be awarded.
(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.

© UCLES 2010
(d) Gradient =
X
1
[1]
Value of X in range 3000–3600 Ω with unit.
(e)
a
b
= 1 [1]
Correct reading off graph. [1]
[Total: 20]
2 (c) (ii) Measurement of h to nearest mm with consistent unit. 0.900 m < h < 1.100 m [1]
(d) (ii) Value of mA – mB = 20 g with consistent unit. [1]
(iii) Value of t with unit. t < 5 seconds [1]
Evidence of repeated measurements of t. [1]
(e) Absolute uncertainty in t in range 0.1–0.6 s. [1]
If repeated readings have been taken, then the uncertainty can be half the range.
Correct method of calculation to get percentage uncertainty. [1]
(f) Second value of mA – mB = 40 g [1]
Second value of t. [1]
Quality: second value of t < first value of t. [1]
(g) (i) Values of k calculated correctly. [1]
(ii) Justification of sf in k linked t and (mA – mB) or mA and mB or masses. [1]
(iii) Valid conclusion based on the calculated values of k. [1]
Candidate must test against a stated criterion.

© UCLES 2010
(h) Identifying limitations marks and suggesting improvements
(i) Limitations [4] (ii) Improvements [4] Do not credit
Ap Two readings are not
enough (to draw a
conclusion)
As Take more readings and plot a
graph/calculate more values of k.
One reading/few
readings/take
more readings
and average.
Bp Masses hit each other/
masses slipping off.
Bs Use larger pulley/method of
securing masses to hanger.
Cp Uncertain starting position Cs Method of fixing rule e.g. clamp
rule/electromagnetic release
mechanism
Dp Difficult to measure time
as time short/reaction time
large compared with time.
Ds Drop through greater height/
expand on trap door mechanism/
light gate with timer/motion sensor
with data logger/video timer with
timer.
Ep Friction at pulley Es Lubricate pulley Friction between
pulley and string
Fp Retort stand moves Fs Method of fixing to the bench
e.g. clamp/add weights
Gp Mass (values) not
accurate
Gs Use balance/method of measuring
mass
Do not credit parallax error.
[Total: 20]

9702 PHYSICS
maximum raw mark 40
examination.
syllabuses.

© UCLES 2010
1 (a) (ii) Value of raw h to the nearest mm (unit needed). h > 20 cm. [1]
(b) Evidence of repeat times: of one swing repeated several times or the time for a number
of swings recorded at least once (not fixed time and count n). [1]
Value of 0.5 < T < 3 s.
(c) Six sets of readings of x and T scores 5 marks, five sets scores 4 marks etc. [5]
Incorrect or no trend then –1 (Correct trend x increases, T2
decreases). SH –1.
Write a ringed total next to the table.
Maximum value of x at least h/2. [1]
Column headings (x / m, x / mm, T / s, T2
/s2
). [1]
Must have x and T2
columns.
Each column heading must contain a quantity and a unit.
Ignore any units in the body of the table.
There must be some distinguishing mark between the quantity and the unit (solidus is
expected but accept, for example, x (m)).
Consistency of presentation of raw readings. [1]
All values of raw x must be given to the nearest mm and all values of raw time to the
same number of d.p. (either 1 or 2).
Significant figures. [1]
Significant figures for T2
must be the same as, or one more than, the least number of
significant figures used in the raw time data. Also if raw time is given to the nearest
hundredth of a second accept one less significant figure in T2
.
Correct calculation of T2
. Do not allow t2
. [1]

© UCLES 2010
(d) (i) Axes: [1]
Sensible scales must be used. No awkward scales (e.g. 3:10).
All observations must be plotted on the grid. [1]
Write a ringed total of plotted points.
Ring and check a suspect plot.
(ii) Line of best fit. [1]
Judge by balance of at least 5 points about the candidate’s line.
length.
Line must not be kinked. Do not allow lines thicker than half a small square.
Quality. [1]
Scatter of points must be less than ± 1 cm (to scale) in the x (cm) direction of a
straight line. All points in table must be plotted (at least 5) for this mark to be
awarded.
(iii) Gradient. [1]
Negative sign must be seen on answer line consistent with graph.
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. Allow ecf of gradient value.
Or:
Check read-off of intercept directly from the graph.
(e) Value of
gradient
intercept
−
−
=
y
B
A
(Expect value to be approximately equal to h). [1]
Unit for A/B correct (e.g. m) consistent with value. [1]
Allow candidate’s value 0.5 h < A/B < 1.5 h.
[Total: 20]

© UCLES 2010
2 Measurement of dA in range 0.20 mm < dA < 0.40 mm to nearest 0.01 mm or 0.001 mm with
consistent unit. If OOR allow SV ± 0.10 mm. [1]
Evidence of repeated measurements of d (or in (e)). [1]
(c) (i) Measurement of L to nearest mm with consistent unit. [1]
(ii) Absolute uncertainty in L is 2 mm–10 mm. [1]
Correct method of calculation to get percentage uncertainty. [1]
(d) (ii) Measurement of VA. Any supervisor’s help –1. [1]
(e) Value of dB. Major help from supervisor –1. [1]
(f) (ii) Measurement of VB to at least nearest 0.1 V with unit. V < 2 V. If > 2 V check SV. [1]
Quality: VB < VA. [1]
(g) (i) Values of k calculated correctly. [1]
(ii) Justification of sf in k linked to L and d and V. [1]

© UCLES 2010
(h)
Ap Two readings are not
enough (to draw a
conclusion.
As Take more readings and plot a
graph/calculate more values of k.
One reading/
few readings/
take more
readings and
average.
Bp Difficult to measure length
because (give a reason)
e.g. clips have a width/
clip slips. Difficult to make
L the same (for both
experiments).
Bs Use sliding jockeys/narrower clips/
solder contacts/use longer wire (to
reduce % error).
Cp Voltmeter scale not
sensitive enough/not
precise enough/only reads
to 0.1 or 0.05 V.
Cs Use digital voltmeter/use a
voltmeter that reads to 0.01 V.
Voltmeter not
accurate
enough. More
accurate
voltmeter.
Dp Wires kinked/Wires not
straight/Difficult to keep
wire straight/difficult to
prevent short circuiting.
Ds Method of keeping wire (during
experiment) straight e.g. tape to
ruler, hang weights off end, clamp
wire.
Parallax error.
Ep Difficult to make I the
same (for both
experiments).
Es Method to obtain continuous
variation in the current e.g. (slide
wire) potentiometer/potential
divider/finer wire rheostat/longer
rheostat.
Fp Contact resistance/
fluctuating ammeter or
voltmeter readings.
Fs Method of cleaning contacts e.g.
sand clips. Tighten clips.
Ignore reference to parallax error, zero error on meters, heating effects of wire, cell runs
down, video the experiment.
[Total: 20]

9702 PHYSICS
maximum raw mark 40
examination.
syllabuses.

© UCLES 2010
1 (c) Measurements for h1 and h2 to nearest mm [1]
Check raw values if readings are repeated.
The difference between h1 and h2 is < 2 mm. [1]
(d) (iii) Six sets of readings of n, h1 and h2 scores 5 marks, five sets scores [5]
4 marks etc.
Incorrect trend then –1.
Help from supervisor then –1.
Range – [1]
n values must include 10 or greater.
Column headings – [1]
There must be some distinguishing mark between the quantity and the unit.
E.g. h1/cm or h1(cm) but not 1/((h1 – h1)/cm).
Consistency of presentation of raw readings – [1]
All values of h1 and h2 must be given to the same precision.
Significant figures – [1]
S.f. for 1/(h1 – h2) must be the same as, or one more than, the s.f. in the difference
(h1 – h2).
Calculation – [1]
1/(h1 – h2) calculated correctly.
(Graph) Axes – [1]
Sensible scales must be used, no awkward scales (e.g. 3:10). Scales must be
chosen so that the plotted points must occupy at least half the graph grid in both x
and y directions.
Scale markings must be no more than 3 large squares apart.
Plotting of points – [1]
All observations must be plotted.
Do not accept blobs (points with diameter > half a small square).
Ring and check a suspect plot. Tick if correct. Re-plot if incorrect.
Line of best fit – [1]
Judge by balance of at least 5 trend points about the candidate's line. There must
be an even distribution of points either side of the line along the full length.
Line must not be kinked.
Quality – [1]
Scatter of points must be less than ±0.02 on the 1/n axis about the examiner’s line.
All points must be plotted (at least 5) for this mark to be scored.

© UCLES 2010
(e) (iii) Gradient
The hypotenuse must be at least half the length of the drawn line. [1]
Intercept [1]
Check that the read-off or the method of calculation is correct.
(f) Value of a = value of gradient and value of b = value of intercept. [1]
Do not allow a value presented as a fraction.
Units for a and b are correct. [1]
E.g. cm–1
or m–1
but must be consistent with the values.
Allow no unit for b if b = 0.
[Total: 20]
2 (a) (i) Value of d in range 5 cm to 15 cm. [1]
Evidence of repeated measurements of d. [1]
(ii) Correct calculation of A. [1]
Do not allow a value in terms of π.
(b) (i) Measurement for x in range 0.8 cm < x < 1.0 cm to nearest mm. [1]
(ii) Absolute uncertainty 1 or 2 mm (or half the range of repeats), and correct method
of calculation. [1]
(c) (ii) Measurement for h to nearest mm. [1]
(d) (iii) Value for t > 1 s and given to 0.1 s or 0.01 s. [1]
Check raw data if there are repeats.
(iv) Correct calculation of R, with consistent unit (e.g. cm3
s–1
). [1]
(e) (i) Values for x, V and h. [1]
(ii) Correct trend (R increases with h). [1]
(f) (i) Values of k calculated correctly. [1]
(ii) Valid conclusion based on the calculated values of k. Candidate must test against
a stated criterion. [1]

© UCLES 2010
(g)
(i) Problems 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.
More readings and
calculate the average/
only one reading.
B Bottle not circular/
diameter at P different to
that at Q.
Collect water and measure
volume/remeasure
diameter at P.
C Bottle deforms when
measuring d.
Use vernier callipers to
measure d.
Use string to measure
d.
D Difficult to see water
level/meniscus
problems/refraction
problems.
Use coloured water/liquid. Use oil.
E Labels get wet/ink runs Use waterproof labels/ink
F Difficult to judge when to
start/stop timing.
Use video, with timing
method.
Human reaction time
error.
G Large uncertainty in x. Use travelling microscope
to measure x.
X Another valid point
E.g. Flowrate calculated
is not the flowrate at h.
E.g. Measure h to point
midway between marks.
Move marks closer
together.
Ignore ‘parallax problems’ unless there is a convincing diagram.
Ignore ‘use assistant’.
Ignore ‘use distance sensor’ unless there is a convincing diagram.
Ignore ‘use a computer/datalogger/light gates’.
Ignore ‘bottle not vertical’.
[Total: 20]

9702 PHYSICS
maximum raw mark 40
examination.
syllabuses.

© UCLES 2010
1 (a) (i) Value of d to the nearest 0.01 mm or 0.001 mm with consistent unit. [1]
0.20 < d < 0.60 mm.
(b) (iii) Value of x in range 40 cm–60 cm with consistent unit. [1]
Value of I with units.
(c) Six sets of readings of x and I scores 5 marks, five sets scores 4 marks etc. Incorrect
trend then –1. Minor help from supervisor –1 ; major help from supervisor –2 [5]
Range [1]
xmax > 70 cm; xmin < 30 cm
Column headings [1]
Each column heading must contain a quantity and a unit.
There must be some distinguishing mark between the quantity and the unit
(solidus is expected but accept, for example, 1/I (A–1
). Do not allow 1/I (A))
Consistency of presentation of raw readings. [1]
All values of x must be given to the nearest mm.
Significant figures [1]
S.F. in 1/I must be the same as, or one more than, the least number of significant
figures used in raw I.
Calculation [1]
Correct calculation of 1/I.

© UCLES 2010
(d) (i) Axes [1]
Sensible scales must be used. Awkward scales (e.g. 3:10) are not allowed.
Plotting of points [1]
All observations must be plotted on the grid.
Ring and check a suspect plot.
(ii) Line of best fit [1]
Judge by the balance of at least 5 points about the candidate’s line.
length.
Lines must not be kinked. Do not accept lines thicker than half a small square.
Quality [1]
All points in the table (minimum 5) must be plotted for this mark to be scored. All
points must be within 2 cm (to scale) in x direction of a straight line.
(iii) Gradient [1]
Intercept [1]
Either:
Check correct read-off from a point on the line, and substitution into y = mx + c.
Read-off must be accurate to half a small square. Allow ecf of gradient value.
Or:
Check read-off of intercept directly from graph.
(e) Values obtained in (a)(ii) and (d)(iii) substituted correctly into equation:
ARN
M ρ
= [1]
Do not allow substitution methods to find M or N
Value for ρ in range: 1 × 10–7
Ω m – 5 × 10–6
Ω m with consistent unit. [1]
[Total: 20]

© UCLES 2010
2 (a) (ii) Measurement of x to nearest mm. x < 15.0 cm with consistent unit. [1]
–1 for supervisor’s help.
(b) (iii) Measurement of θ (less than 90°) with unit. [1]
(iv) Absolute uncertainty in θ in the range 2°–10°. [1]
Correct method of calculation of percentage uncertainty.
(v) m = 50 g with consistent unit [1]
M = 60 g with consistent unit [1]
(vi) Correct calculation of m/M (0.83 or 0.833). No units. [1]
(c) Measurement of θ [1]
m = 40 g; M = 70 g [1]
Quality: θ2 > θ1 [1]
(d) (i) Correct calculation of two values of k. [1]
(ii) Justification of sf in k linked to θ, m and M [1]

© UCLES 2010
(e) Identifying limitations (4 marks) and suggesting improvements (4 marks)
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. Only one reading.
B Difficult to balance with
reason e.g. unstable or
effect of fans/draughts/a.c.
Drill hole higher up/switch off
fans/a.c./close windows.
Closed room.
C Difficult to judge when
wooden strip
horizontal/parallel (to the
bench).
Method of ensuring strip is
horizontal/parallel to bench
e.g. use a spirit level or
metre rule(s) to measure
height of both ends/sight
against window. Allow
detailed use of set square.
Strip not straight/parallel/
horizontal.
Use set square.
D Difficult keeping x constant/
weights move.
Method of fixing cotton loop
to rule e.g. tape, glue.
E Difficult to measure θ
because hard to judge
vertical/movement of hand.
Use a plumb line/clamped
ruler/clamp protractor.
Bigger protractor. Paper
behind protractor.
F Friction at pulley/between
nail and wooden strip.
Use lubricant/method of
reducing friction.
Friction. Better pulley/
smooth(er) string/thin(ner)
string. Friction between
string and pulley.
Lubrication between string
and pulley.
G Mass (values) not accurate. Use balance/method of
weighing mass.
Weigh masses.
Do not credit ‘parallax problems’, ‘use assistant’ or references to sensors, computers or
videos.
[Total: 20]

9702 PHYSICS
maximum raw mark 40
examination.
syllabuses.

© UCLES 2010
1 (c) Measurements for all raw l in range 19.5 to 20.5 cm. [1]
(e) (i) Measurements for all raw h1 and h2 to nearest mm. [1]
(iii) Measurement for raw d to nearest mm, with unit, in range 1.5 to 2.5 cm. [1]
(f) Five sets of readings of h1, h2 and d scores 4 marks, four sets scores 3 marks etc. [4]
Incorrect trend then –1.
Range – [1]
d values used must include dmin ≤ 3 cm and dmax ≥ 8 cm
Column headings – [1]
There must be some distinguishing mark between the quantity and the unit.
e.g. θ /o
, 1/tan θ, sin θ, sin (θ /o
) not sin θ /o
, not (1/tan θ)/o
Consistency of presentation of raw readings – [1]
All h values in the table must be given to the same precision.
Significant figures – [1]
S.f. for 1/tan θ must be the same as, or one more than, the minimum s.f. given
for (h1 – h2) and l.
Calculation – [1]
1/tan θ calculated correctly.
(Graph) Axes – [1]
Sensible scales must be used, no awkward scales (e.g. 3:10).
Scales must be chosen so that the plotted points must occupy at least half the graph
Scales must be correctly labelled with the quantity that is being plotted. Ignore units.
Scale markings must be no more than three large squares apart.
Plots – [1]
All observations must be plotted.
Ring and check a suspect plot. Tick if correct.
Re-plot if incorrect. Work to an accuracy of half a small square.
Diameter of plots must be ≤ half a small square (no blobs).
Line of best fit – [1]
Judge by balance of all plots, at least 4 trend points, about the candidate's line.
There must be an even distribution of points either side of the line along the full length.
Line must not be kinked.
Quality – [1]
Scatter of points must be less than ± 0.25 cm in the d direction about the examiner’s line.
All points in table must be plotted (at least 4) for this mark to be scored.

© UCLES 2010
(g) (iii) Gradient [1]
Intercept [1]
Check that the read-off from graph or the method of calculation (substitution of
correct read offs into y = mx + c) is correct.
(h) Value of a = gradient and value of b = intercept. [1]
Unit for a (m–1
cm–1
or mm–1
) consistent with value and b (no unit). [1]
[Total: 20]
2 (b) (i) Raw length and width to nearest mm with unit. Help from supervisor –1 [1]
Values of length and width in range 1 cm to 10 cm. [1]
Correct calculation of A, with consistent unit. [1]
(ii) S.f. in A same as/one more than the (smallest) s.f. in length and width [1]
(not just “raw readings”).
(d) (i) Measurement of F, with unit, F < 10 N. [1]
Evidence of repeated measurements of F. [1]
(ii) Uncertainty in measurements of F stated, in range 0.1 to 0.5 N. [1]
(e) Values of second length and second width. [1]
Correct calculation of A. [1]
Measurement of F. [1]
Second F = first F (within 1 N). [1]
(f) Justification of a valid conclusion based on two values of F being within (or outside)
the uncertainty in (d)(ii). [1]

© UCLES 2010
(g)
(i) Limitations 4 max (ii) Improvements 4 max No credit/not enough
A Two readings not enough
(to draw a conclusion)/too
few readings/only two
readings.
Take many readings for
different areas and plot a
graph/compare more F
values
Repeat readings
Few readings
One reading
NOT average F
B Maximum force reached
without warning(suddenly)/
reading over quickly, link
to short time
Method of recording
maximum reading e.g.
force sensor + data
logger/video recording to
find force/meter which
retains max reading/ use
masses and pulley system
Position sensors
/parallax/computer
methods/bald human
reaction time error/
increase force slowly/fast
paper/high speed camera/
slow camera
C Reason for the problem of
detecting paper movement/
difficult to look at meter
and paper at same time.
Method to indicate
movement e.g. contrasting
colours of paper/drawing a
reference mark
Difficult to know when
paper moves.
Fast movement
D Position of eraser (and
weights) not fixed/
Mass(weight) of eraser
changes/irregularity of
rubber shape (not
rectangular)
Method to ensure same
position e.g. mark position
on top paper/method to
ensure constant mass e.g.
use malleable strip which
can be bent to change A/
change total masses to
account for change in
mass of rubber/pile up
unused rubber pieces on
top/improved method to
measure rubber e.g.
vernier caliper
Keep mass constant
E Variation in direction of
force/misalignment of
paper strips (which affects
F).
Method to ensure direction
is constant e.g. align strips
along straight edge/draw a
line to follow/method to
equalise levels
F Uneven bench surface
(leading to contact area
being less than A).
Method to ensure
smoother surface e.g. use
named surface e.g. glass
or melamine/sand the
surface
Use smoother surface
X: Increase mass so increase the force (reducing % uncertainty in force).
Do not credit references to zero error/accuracy/digital meter friction between papers/re-
zeroing after each experiment/2 people/paper tearing/clip deforming.
[Total: 20]

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

© UCLES 2010
Section A
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) graph: correct curvature M1
from (R,1.0gS) & at least one other correct point A1 [2]
(c) (i) fields of Earth and Moon are in opposite directions M1
either resultant field found by subtraction of the field strength
or any other sensible comment A1
so there is a point where it is zero A0 [2]
(allow FE = –FM for 2 marks)
(ii) GME / x2
= GMM / (D – x)2
C1
(6.0 × 1024
) / (7.4 × 1022
) = x2
/ (60RE – x)2
C1
x = 54RE A1 [3]
(iii) graph: g = 0 at least ⅔ distance to Moon B1
gE and gM in opposite directions M1
correct curvature (by eye) and gE > gM at surface A1 [3]
2 (a) (i) no forces (of attraction or repulsion) between atoms / molecules / particles B1 [1]
(ii) sum of kinetic and potential energy of atoms / molecules M1
due to random motion A1 [2]
(iii) (random) kinetic energy increases with temperature M1
no potential energy
(so increase in temperature increases internal energy) A1 [2]
(b) (i) zero A1 [1]
(ii) work done = p∆V C1
= 4.0 × 105
× 6 × 10–4
= 240J (ignore any sign) A1 [2]
(iii)
change work done / J heating / J increase in internal
energy / J
P → Q
Q → R
R → P
240
0
–560
0
320
0
240
320
–560
(correct signs essential)
(each horizontal line correct, 1 mark – max 3) B3 [3]

© UCLES 2010
3 (a) (i) resonance B1 [1]
(ii) amplitude 16mm and frequency 4.6Hz A1 [1]
(b) (i) a = (–)ω2
x and ω = 2πf C1
a = 4π2
× 4.62
× 16 × 10–3
C1
= 13.4ms–2
A1 [3]
(ii) F = ma C1
= 50 × 10–3
× 13.4
= 2.0N A1 [2]
(c) line always ‘below’ given line and never zero M1
peak is at 4.6Hz (or slightly less) and flatter A1 [2]
4 (a) charge / potential (difference) (ratio must be clear) B1 [1]
(b) (i) V = Q / 4πε0r B1 [1]
(ii) C = Q / V = 4πε0r and 4πε0 is constant M1
so C ~ r A0 [1]
(c) (i) r = C / 4πε0r C1
r = (6.8 × 10–12
) / (4π × 8.85 × 10–12
) C1
= 6.1 × 10–2
m A1 [3]
(ii) Q = CV = 6.8 × 10–12
× 220
= 1.5 × 10–9
C A1 [1]
(d) (i) V = Q/C = (1.5 × 10–9
) / (18 × 10–12
)
= 83V A1 [1]
(ii) either energy = ½CV 2
C1
∆E = ½ × 6.8 × 10–12
× 2202
– ½ × 18 × 10–12
× 832
C1
= 1.65 × 10–7
– 6.2 × 10–8
= 1.03 × 10–7
J A1 [3]
or energy = ½QV (C1)
∆E = ½ × 1.5 × 10–9
× 220 – ½ × 1.5 × 10–9
× 83 (C1)
= 1.03 × 10–7
J (A1)

© UCLES 2010
5 (a) field into (the plane of) the paper B1 [1]
(b) force due to magnetic field provides the centripetal force B1
mv2
/ r = Bqv C1
B = (20 × 1.66 × 10–27
× 1.40 × 105
) / (1.6) × 10–19
× 6.4 × 10–2
) B1
= 0.454T A0 [3]
(c) (i) semicircle with diameter greater than 12.8cm B1 [1]
(ii) new flux density =
20
22
× 0.454 C1
B = 0.499T A1 [2]
6 (a) (i) e.g. prevent flux losses / improve flux linkage B1 [1]
(ii) flux in core is changing B1
e.m.f. / current (induced) in core B1
induced current in core causes heating B1 [3]
(b) (i) that value of the direct current producing same (mean) power / heating M1
in a resistor A1 [2]
(ii) power in primary = power in secondary M1
VP IP = VS IS A1 [2]
7 (a) (i) e.g. electron / particle diffraction B1 [1]
(ii) e.g. photoelectric effect B1 [1]
(b) (i) 6 A1 [1]
(ii) change in energy = 4.57 × 10–19
J
λ = hc / E C1
= (6.63 × 10–34
× 3.0 × 108
) / (4.57 × 10–19
)
= 4.4 × 10–7
m A1 [2]
8 (a) splitting of a heavy nucleus (not atom/nuclide) M1
into two (lighter) nuclei of approximately same mass A1 [2]
(b) n1
0
He4
2 (allow α4
2 ) M2
Li7
3 A1 [3]
(c) emitted particles have kinetic energy B1
range of particles in the control rods is short / particles stopped in rods /
lose kinetic energy in rods B1
kinetic energy of particles converted to thermal energy B1 [3]

© UCLES 2010
Section B
9 (a) (i) non-inverting (amplifier) B1 [1]
(ii) (G =) 1 + R2 / R1 B1 [1]
(b) (i) gain = 1 + 100 / 820 C1
output = 17 mV A1 [2]
(ii) 9V A1 [1]
(R2 / R1 scores 0 in (a)(ii) but possible 1 mark in each of (b)(i) and (b)(ii)
(1 + R1 / R2) scores 0 in (a)(ii), no mark in (b)(i), possible 1 mark in (b)(ii)
(1 – R2 / R1) or R1 / R2 scores 0 in (a)(ii), (b)(i) and (b)(ii))
10 (a) (i) density × speed of wave (in the medium) B1 [1]
(ii) ρ = (7.0 × 106
) / 4100
= 1700kgm–3
A1 [1]
(b) (i) I = IT + IR B1 [1]
(ii) 1. α = (0.1 × 106
)2
/ (3.1 × 106
)2
C1
= 0.001 A1 [2]
2. α ≈ 1 A1 [1]
(c) either very little transmission at an air-skin boundary M1
(almost) complete transmission at a gel-skin boundary M1
when wave travels in or out of the body A1 [3]
or no gel, majority reflection (M1)
with gel, little reflection (M1)
when wave travels in or out of the body (A1)
11 (a) (i) unwanted random power / signal / energy B1 [1]
(ii) loss of (signal) power / energy B1 [1]
(b) (i) either signal-to-noise ratio at mic. = 10lg (P2 / P1) C1
= 10lg ({2.9 × 10–6
} / {3.4 × 10–9
})
= 29dB A1
maximum length = (29 – 24) / 12 C1
= 0.42km = 420m A1 [4]
or signal-to-noise ratio at receiver = 10 lg (P2 / P1) (C1)
at receiver, 24 = 10 lg(P / {3.4 × 10–9
})
P = 8.54 × 10–7
W (A1)
power loss in cables = 10lg({2.9 × 10–6
} / {8.54 × 10–7
}) (C1)
= 5.3dB
length = 5.3 / 12km
= 440m (A1)

© UCLES 2010
(ii) use an amplifier M1
coupled to the microphone A1 [2]
(repeater amplifiers scores no mark)
12 (a) (carrier wave) transmitted from Earth to satellite (1)
satellite receives greatly attenuated signal (1)
signal amplified and transmitted back to Earth B1
at a different (carrier) frequency B1
different frequencies prevent swamping of uplink signal (1)
e.g. of frequencies used (6/4GHz, 14/11GHz, 30/20GHz) (1)
(two B1 marks plus any two other for additional physics) B2 [4]
(b) advantage: e.g. much shorter time delay M1
because orbits are much lower A1
e.g. whole Earth may be covered (M1)
in several orbits / with network (A1)
disadvantage: e.g. either must be tracked
or limited use in any one orbit M1
more satellites required for continuous operation A1 [4]

9702 PHYSICS
examination.
syllabuses.

© UCLES 2010
Section A
1 (a) force per unit mass (ratio idea essential) B1 [1]
(b) graph: correct curvature M1
from (R,1.0gS) & at least one other correct point A1 [2]
(c) (i) fields of Earth and Moon are in opposite directions M1
either resultant field found by subtraction of the field strength
or any other sensible comment A1
so there is a point where it is zero A0 [2]
(allow FE = –FM for 2 marks)
(ii) GME / x2
= GMM / (D – x)2
C1
(6.0 × 1024
) / (7.4 × 1022
) = x2
/ (60RE – x)2
C1
x = 54RE A1 [3]
(iii) graph: g = 0 at least ⅔ distance to Moon B1
gE and gM in opposite directions M1
correct curvature (by eye) and gE > gM at surface A1 [3]
2 (a) (i) no forces (of attraction or repulsion) between atoms / molecules / particles B1 [1]
(ii) sum of kinetic and potential energy of atoms / molecules M1
due to random motion A1 [2]
(iii) (random) kinetic energy increases with temperature M1
no potential energy
(so increase in temperature increases internal energy) A1 [2]
(b) (i) zero A1 [1]
(ii) work done = p∆V C1
= 4.0 × 105
× 6 × 10–4
= 240J (ignore any sign) A1 [2]
(iii)
change work done / J heating / J increase in internal
energy / J
P → Q
Q → R
R → P
240
0
–560
0
320
0
240
320
–560
(correct signs essential)
(each horizontal line correct, 1 mark – max 3) B3 [3]

© UCLES 2010
3 (a) (i) resonance B1 [1]
(ii) amplitude 16mm and frequency 4.6Hz A1 [1]
(b) (i) a = (–)ω2
x and ω = 2πf C1
a = 4π2
× 4.62
× 16 × 10–3
C1
= 13.4ms–2
A1 [3]
(ii) F = ma C1
= 50 × 10–3
× 13.4
= 2.0N A1 [2]
(c) line always ‘below’ given line and never zero M1
peak is at 4.6Hz (or slightly less) and flatter A1 [2]
4 (a) charge / potential (difference) (ratio must be clear) B1 [1]
(b) (i) V = Q / 4πε0r B1 [1]
(ii) C = Q / V = 4πε0r and 4πε0 is constant M1
so C ~ r A0 [1]
(c) (i) r = C / 4πε0r C1
r = (6.8 × 10–12
) / (4π × 8.85 × 10–12
) C1
= 6.1 × 10–2
m A1 [3]
(ii) Q = CV = 6.8 × 10–12
× 220
= 1.5 × 10–9
C A1 [1]
(d) (i) V = Q/C = (1.5 × 10–9
) / (18 × 10–12
)
= 83V A1 [1]
C1
∆E = ½ × 6.8 × 10–12
× 2202
– ½ × 18 × 10–12
× 832
C1
= 1.65 × 10–7
– 6.2 × 10–8
= 1.03 × 10–7
J A1 [3]
or energy = ½QV (C1)
∆E = ½ × 1.5 × 10–9
× 220 – ½ × 1.5 × 10–9
× 83 (C1)
= 1.03 × 10–7
J (A1)

© UCLES 2010
5 (a) field into (the plane of) the paper B1 [1]
(b) force due to magnetic field provides the centripetal force B1
mv2
/ r = Bqv C1
B = (20 × 1.66 × 10–27
× 1.40 × 105
) / (1.6) × 10–19
× 6.4 × 10–2
) B1
= 0.454T A0 [3]
(c) (i) semicircle with diameter greater than 12.8cm B1 [1]
(ii) new flux density =
20
22
× 0.454 C1
B = 0.499T A1 [2]
6 (a) (i) e.g. prevent flux losses / improve flux linkage B1 [1]
(ii) flux in core is changing B1
e.m.f. / current (induced) in core B1
induced current in core causes heating B1 [3]
(b) (i) that value of the direct current producing same (mean) power / heating M1
in a resistor A1 [2]
(ii) power in primary = power in secondary M1
VP IP = VS IS A1 [2]
7 (a) (i) e.g. electron / particle diffraction B1 [1]
(ii) e.g. photoelectric effect B1 [1]
(b) (i) 6 A1 [1]
(ii) change in energy = 4.57 × 10–19
J
λ = hc / E C1
= (6.63 × 10–34
× 3.0 × 108
) / (4.57 × 10–19
)
= 4.4 × 10–7
m A1 [2]
8 (a) splitting of a heavy nucleus (not atom/nuclide) M1
into two (lighter) nuclei of approximately same mass A1 [2]
(b) n1
0
He4
2 (allow α4
2 ) M2
Li7
3 A1 [3]
(c) emitted particles have kinetic energy B1
range of particles in the control rods is short / particles stopped in rods /
lose kinetic energy in rods B1
kinetic energy of particles converted to thermal energy B1 [3]

© UCLES 2010
Section B
9 (a) (i) non-inverting (amplifier) B1 [1]
(ii) (G =) 1 + R2 / R1 B1 [1]
(b) (i) gain = 1 + 100 / 820 C1
output = 17 mV A1 [2]
(ii) 9V A1 [1]
(R2 / R1 scores 0 in (a)(ii) but possible 1 mark in each of (b)(i) and (b)(ii)
(1 + R1 / R2) scores 0 in (a)(ii), no mark in (b)(i), possible 1 mark in (b)(ii)
(1 – R2 / R1) or R1 / R2 scores 0 in (a)(ii), (b)(i) and (b)(ii))
10 (a) (i) density × speed of wave (in the medium) B1 [1]
(ii) ρ = (7.0 × 106
) / 4100
= 1700kgm–3
A1 [1]
(b) (i) I = IT + IR B1 [1]
(ii) 1. α = (0.1 × 106
)2
/ (3.1 × 106
)2
C1
= 0.001 A1 [2]
2. α ≈ 1 A1 [1]
(c) either very little transmission at an air-skin boundary M1
(almost) complete transmission at a gel-skin boundary M1
when wave travels in or out of the body A1 [3]
or no gel, majority reflection (M1)
with gel, little reflection (M1)
when wave travels in or out of the body (A1)
11 (a) (i) unwanted random power / signal / energy B1 [1]
(ii) loss of (signal) power / energy B1 [1]
(b) (i) either signal-to-noise ratio at mic. = 10lg (P2 / P1) C1
= 10lg ({2.9 × 10–6
} / {3.4 × 10–9
})
= 29dB A1
maximum length = (29 – 24) / 12 C1
= 0.42km = 420m A1 [4]
or signal-to-noise ratio at receiver = 10 lg (P2 / P1) (C1)
at receiver, 24 = 10 lg(P / {3.4 × 10–9
})
P = 8.54 × 10–7
W (A1)
power loss in cables = 10lg({2.9 × 10–6
} / {8.54 × 10–7
}) (C1)
= 5.3dB
length = 5.3 / 12km
= 440m (A1)

© UCLES 2010
(ii) use an amplifier M1
coupled to the microphone A1 [2]
(repeater amplifiers scores no mark)
12 (a) (carrier wave) transmitted from Earth to satellite (1)
satellite receives greatly attenuated signal (1)
signal amplified and transmitted back to Earth B1
at a different (carrier) frequency B1
different frequencies prevent swamping of uplink signal (1)
e.g. of frequencies used (6/4GHz, 14/11GHz, 30/20GHz) (1)
(two B1 marks plus any two other for additional physics) B2 [4]
(b) advantage: e.g. much shorter time delay M1
because orbits are much lower A1
e.g. whole Earth may be covered (M1)
in several orbits / with network (A1)
disadvantage: e.g. either must be tracked
or limited use in any one orbit M1
more satellites required for continuous operation A1 [4]

9702 PHYSICS
examination.
syllabuses.

© UCLES 2010
Section A
1 (a) (i) rate of change of angle / angular displacement M1
swept out by radius A1 [2]
(ii) ω × T = 2π B1 [1]
(b) centripetal force is provided by the gravitational force B1
either mr(2π/T)2
= GMm/r 2
or mrω 2
= GMm/r 2
M1
r 3
× 4π2
= GM × T 2
A1
GM/4π2
is a constant (c) A1
T 2
= cr 3
A0 [4]
(c) (i) either T 2
= (45/1.08)3
× 0.6152
or T 2
= 0.30 × 453
C1
T = 165 years A1 [2]
(ii) speed = (2π × 1.08 × 108
) / (0.615 × 365 × 24 × 3600) C1
= 35 km s–1
A1 [2]
2 (a) atoms / molecules / particles behave as elastic (identical) spheres (1)
volume of atoms / molecules negligible compared to volume of containing vessel (1)
time of collision negligible to time between collisions (1)
no forces of attraction or repulsion between atoms / molecules (1)
atoms / molecules / particles are in (continuous) random motion (1)
(any four, 1 each) B4 [4]
(b) pV = 3
1
Nm<c2
> and pV = nRT or pV = NkT B1
3
1
Nm<c2
> = nRT or = NkT and <EK> = ½m<c2
> B1
n = N/NA or k = R/NA B1
<EK> =
2
3
× R/NA × T A0 [3]
(c) (i) reaction represents either build-up of nucleus from light nuclei
or build-up of heavy nucleus from nuclei M1
so fusion reaction A1 [2]
(ii) proton and deuterium nucleus will have equal kinetic energies B1
1.2 × 10–14
= 2
3
× 8.31 / (6.02 × 1023
) × T C1
T = 5.8 × 108
K A1 [3]
(use of E = 2.4 × 10–14
giving 1.16 × 109
K scores 1 mark)
(iii) either inter-molecular / atomic / nuclear forces exist
or proton and deuterium nucleus are positively charged / repel B1 [1]

© UCLES 2010
3 (a) (i) 8.0 cm A1 [1]
(ii) 2πf = 220 C1
f = 35 (condone unit) A1 [2]
(iii) line drawn mid-way between AB and CD (allow ±2 mm) B1 [1]
(iv) v = ωa C1
= 220 × 4.0
= 880 cm s–1
A1 [2]
(b) (i) 1. line drawn 2 cm above AB (allow ±2 mm) B1 [1]
2. arrow pointing upwards B1 [1]
(ii) 1. line drawn 2 cm above AB (allow ±2 mm) B1 [1]
2. arrow pointing downwards B1 [1]
(iii) v = ω√(a2
– x2
)
= 220 × √(4.02
– 2.02
) C1
= 760 cm s–1
A1 [2]
(incorrect value for x, 0/2 marks)
4 (a) (i) work done moving unit positive charge M1
from infinity to the point A1 [2]
(ii) charge / potential (difference) (ratio must be clear) B1 [1]
(b) (i) capacitance = (2.7 × 10–6
) / (150 × 103
) C1
(allow any appropriate values)
capacitance = 1.8 × 10–11
(allow 1.8 ±0.05) A1 [2]
or energy = ½QV and Q = CV C1
energy = ½ × 1.8 × 10–11
× (150 × 103
)2
or ½ × 2.7 × 10–6
× 150 × 103
= 0.20 J A1 [2]
(c) either since energy ~ V 2
, capacitor has (½)2
of its energy left
or full formula treatment C1
energy lost = 0.15 J A1 [2]

© UCLES 2010
5 (a) magnetic flux = BA
= 89 × 10–3
× 5.0 × 10–2
× 2.4 × 10–2
C1
= 1.07 × 10–4
Wb A1 [2]
(b) (i) e.m.f. = ∆φ / ∆t C1
(for ∆φ = 1.07 × 10–4
Wb), ∆t = 2.4 × 10–2
/ 1.8 = 1.33 × 10–2
s C1
e.m.f. = (1.07 × 10–4
) / (1.33 × 10–2
)
= 8.0 × 10–3
V A1 [3]
(ii) current = 8.0 × 10–3
/ 0.12 M1
≈ 70 mA A0 [1]
(c) force on wire = BIL
= 89 × 10–3
× 70 × 10–3
× 6.0 × 10–2
C1
≈ 4 × 10–4
(N) M1
suitable comment e.g. this force is too / very small (to be felt) A1 [3]
6 (a) power / heating depends on I2
M1
so independent of current direction A1 [2]
(b) either maximum power = I0
2
R or average power = IRMS
2
R M1
I0 = √2 × IRMS M1
maximum power = 2 × average power
ratio = 0.5 A1 [3]
7 (a) force due to E-field is equal and opposite to force due to B-field B1
Eq = Bqv B1
v = E/B B1 [3]
(b) either charge and mass are not involved in the equation in (a)
or FE and FB are both doubled
or E, B and v do not change M1
so no deviation A1 [2]
8 (a) minimum frequency for electron to be emitted (from surface) M1
of electromagnetic radiation / light / photons A1 [2]
(b) E = hc / λ or E = hf and c = fλ C1
either threshold wavelength = (6.63 × 10–34
× 3.0 × 108
) / (5.8 × 10–19
)
= 340 nm
or energy of 340 nm photon = 4.4 × 10–19
J
or threshold frequency = 8.7 × 1014
Hz
or 450 nm → 6.7 × 1014
Hz A1
appropriate comment comparing wavelengths / energies / frequencires B1
so no effect on photo-electric current B1 [4]

© UCLES 2010
Section B
9 (a) (i) edges can be (clearly) distinguished B1 [1]
(ii) e.g. size of X-ray source / anode / target / aperture
scattering of X-ray beam
pixel size
(any two, 1 each) B2
further detail e.g. use of lead grid B1 [3]
(b) X-ray image involves a single exposure B1
CT scan: exposure of a slice from many different angles M1
repeated for different slices A1
CT scan involves a (much) greater exposure B1 [4]
10 (a) e.g. infinite input impedance / resistance
zero output impedance / resistance
infinite gain
infinite bandwidth
infinite slew rate
(any three, 1 each) B3 [3]
(b) (i) with switch open, V –
is less (positive) than V +
M1
output is negative A1
with switch closed, V –
is more (positive) than V +
so output is positive A1 [3]
(allow similar scheme if V –
more positive than V +
treated first)
(ii) 1. diodes connected correctly between output and earth M1
2. green identified correctly A1 [2]
(do not allow this mark if not argued in (i))
11 (a) (i) I / I0 = exp(–1.5 × 2.9) C1
= 0.013 A1 [2]
(ii) I / I0 = exp(–4.6 × 0.95)
= 0.013 A1 [1]
(b) attenuation (coefficients) in muscle and in fat are similar B1
attenuation (coefficients) in bone and muscle / fat are different B1
contrast depends on difference in attenuation B1 [3]

© UCLES 2010
12 (a) (i) 1. signal has same variation (with time) as the data B1
2. consists of (a series of) ‘highs’ and ‘lows’ B1
either analogue is continuously variable (between limits)
or digital has no intermediate values B1 [3]
(ii) e.g. can be regenerated / noise can be eliminated
extra data can be added to check / correct transmitted signal
(any two reasonable suggestions, 1 each) B2 [2]
(b) (i) analogue signal is sampled at (regular time) intervals B1
sampled signal is converted into a binary number B1 [2]
(ii) one channel is required for each bit (of the digital number) B1 [1]

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

© UCLES 2010
1 Planning (15 marks)
Defining the problem (3 marks)
P1 f is the independent variable and V is the dependent variable or vary f and measure V [1]
P2 Keep the current in coil X constant [1]
P3 Keep the number of turns on coil (Y)/area of coil Y constant
Do not credit reference to coil X only. [1]
Methods of data collection (5 marks)
M1 Two independent coils labelled X and Y. [1]
M2 Alternating power supply/signal generator connected to coil X in a workable circuit. [1]
M3 Coil Y connected to voltmeter/c.r.o. in a workable circuit. [1]
M4 Use c.r.o. to determine period/frequency or read off signal generator. [1]
M5 Method to keep current constant in coil X: adjust signal generator/use of rheostat. [1]
Method of analysis (2 marks)
A1 Plot a graph of V against f. [1]
A2 Relationship valid if straight line through origin [1]
Safety considerations (1 mark)
S1 Reference to hot coils – switch off when not in use/use gloves/do not touch coils. Must refer
to hot coils. [1]
Additional detail (4 marks)
D1/2/3/4 Relevant points might include [4]
1. Use large current in coil X/large number of coils on coil Y (to increase emf).
2. Use iron core (to increase emf).
3. Detail on measuring emf e.g. height × y-gain.
4. Avoid other alternating magnetic fields.
5. Detail on measuring frequency from c.r.o. to determine period and hence f.
6. Use of ammeter/c.r.o. and resistor to check current is constant
7. Use insulated wire for coils.
8. Keep coil Y and coil X in the same relative positions.
Do not allow vague computer methods.
[Total: 15]

© UCLES 2010
2 Analysis, conclusions and evaluation (15 marks)
Part Mark Expected Answer Additional Guidance
(a) A1 Gradient = b
y-intercept = lg a
Allow log a but not ln a
(b) T1
T2 1.9777 0.292 or 0.2923
1.9294 0.265 or 0.2648
1.8751 0.241 or 0.2405
1.8129 0.210 or 0.2095
1.7404 0.170 or 0.1703
1.6532 0.127 or 0.1271
T1 for lg l column – ignore rounding errors; min
2 dp.
T2 for lg T column – must be values given
A mixture is allowed
U1 From ± 0.004 or ± 0.005 to ± 0.006
or ± 0.007
Allow more than one significant figure.
(c) (i) G1 Six points plotted correctly Must be within half a small square; penalise ≥
half a small square. Penalise ‘blobs’ ≥ half a
small square. Ecf allowed from table.
U2 Error bars in lg (T/s) plotted
correctly.
All error bars must be plotted. Check first and
last point. Must be accurate within half a small
square; penalise ≥ half a small square.
(ii) G2 Line of best fit If points are plotted correctly then lower end of
line should pass between (1.65, 0.124) and
(1.65, 0.128) and upper end of line should
pass between (2.00, 0.300) and (2.00, 0.306).
Allow ecf from points plotted incorrectly; five
trend plots needed – 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 all
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; penalise ≥ half a
small square.
U3 Uncertainty in gradient Method of determining absolute uncertainty
Difference in worst gradient and gradient.
(iv) C2 y-intercept Must be negative. Check substitution of point
from line into c = y – mx.
Allow ecf from (c)(iii).

© UCLES 2010
U4 Uncertainty in y-intercept Method of determining absolute uncertainty
Difference in worst y-intercept and y-intercept.
Do not allow ecf from false origin read-off
(FOX). Allow ecf from (c)(iv).
(d) C3 a = 10y-intercept
y-intercept must be used. Expect an answer of
about 0.19. If FOX expect answer of about 1.3.
C4 b = gradient and in the range 0.495
to 0.520 and to 2 or 3 sf
Allow 0.50 to 0.52 to 2 sf
Penalise 1 sf or ≥4 sf
U5 Absolute uncertainty in a and b Difference in a and worst a.
Uncertainty in b should be the same as the
uncertainty in the gradient.
[Total: 15]
Uncertainties in Question 2
(c) (iii) Gradient [U3]
1. Uncertainty = gradient of line of best fit – gradient of worst acceptable line
2. Uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)
(c) (iv) [U4]
1. Uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line
2. Uncertainty = ½ (y-intercept of steepest worst line – y-intercept of shallowest worst
line)
(d) [U5]
1. Uncertainty = 10 best y-intercept
- 10 worst y-intercept

9702 PHYSICS
maximum raw mark 30
examination.
syllabuses.

© UCLES 2010
P1 f is the independent variable and V is the dependent variable or vary f and measure V [1]
P2 Keep the current in coil X constant [1]
P3 Keep the number of turns on coil (Y)/area of coil Y constant
Do not credit reference to coil X only. [1]
M1 Two independent coils labelled X and Y. [1]
M2 Alternating power supply/signal generator connected to coil X in a workable circuit. [1]
M3 Coil Y connected to voltmeter/c.r.o. in a workable circuit. [1]
M4 Use c.r.o. to determine period/frequency or read off signal generator. [1]
M5 Method to keep current constant in coil X: adjust signal generator/use of rheostat. [1]
A1 Plot a graph of V against f. [1]
A2 Relationship valid if straight line through origin [1]
S1 Reference to hot coils – switch off when not in use/use gloves/do not touch coils. Must refer
to hot coils. [1]
1. Use large current in coil X/large number of coils on coil Y (to increase emf).
2. Use iron core (to increase emf).
3. Detail on measuring emf e.g. height × y-gain.
4. Avoid other alternating magnetic fields.
5. Detail on measuring frequency from c.r.o. to determine period and hence f.
6. Use of ammeter/c.r.o. and resistor to check current is constant
7. Use insulated wire for coils.
8. Keep coil Y and coil X in the same relative positions.
[Total: 15]

© UCLES 2010
(a) A1 Gradient = b
y-intercept = lg a
Allow log a but not ln a
(b) T1
T2 1.9777 0.292 or 0.2923
1.9294 0.265 or 0.2648
1.8751 0.241 or 0.2405
1.8129 0.210 or 0.2095
1.7404 0.170 or 0.1703
1.6532 0.127 or 0.1271
T1 for lg l column – ignore rounding errors; min
2 dp.
T2 for lg T column – must be values given
U1 From ± 0.004 or ± 0.005 to ± 0.006
or ± 0.007
(c) (i) G1 Six points plotted correctly Must be within half a small square; penalise ≥
half a small square. Penalise ‘blobs’ ≥ half a
small square. Ecf allowed from table.
U2 Error bars in lg (T/s) plotted
correctly.
All error bars must be plotted. Check first and
last point. Must be accurate within half a small
square; penalise ≥ half a small square.
(ii) G2 Line of best fit If points are plotted correctly then lower end of
line should pass between (1.65, 0.124) and
(1.65, 0.128) and upper end of line should
pass between (2.00, 0.300) and (2.00, 0.306).
Allow ecf from points plotted incorrectly; five
trend plots needed – examiner judgement.
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 all
error bars are plotted.
(iii) C1 Gradient of best fit line The triangle used should be at least half the
small square.
U3 Uncertainty in gradient Method of determining absolute uncertainty
(iv) C2 y-intercept Must be negative. Check substitution of point
from line into c = y – mx.
Allow ecf from (c)(iii).

© UCLES 2010
U4 Uncertainty in y-intercept Method of determining absolute uncertainty
Difference in worst y-intercept and y-intercept.
Do not allow ecf from false origin read-off
(FOX). Allow ecf from (c)(iv).
(d) C3 a = 10y-intercept
y-intercept must be used. Expect an answer of
about 0.19. If FOX expect answer of about 1.3.
C4 b = gradient and in the range 0.495
to 0.520 and to 2 or 3 sf
Allow 0.50 to 0.52 to 2 sf
Penalise 1 sf or ≥4 sf
U5 Absolute uncertainty in a and b Difference in a and worst a.
Uncertainty in b should be the same as the
uncertainty in the gradient.
[Total: 15]
(c) (iv) [U4]
1. Uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line
2. Uncertainty = ½ (y-intercept of steepest worst line – y-intercept of shallowest worst
line)
(d) [U5]
1. Uncertainty = 10 best y-intercept
- 10 worst y-intercept

9702 PHYSICS
maximum raw mark 30
examination.
syllabuses.

© UCLES 2010
P1 c, d or A is the independent variable and R is the dependent variable or vary c, d or A and
measure R. [1]
P2 If c varied then (t and) d or A kept constant, if d varied then (t and) c or A kept constant, if A
varied then c or d kept constant. [1]
P3 Keep temperature constant. [1]
M1 Circuit diagram to measure resistance. [1]
M2 Use micrometer screw gauge to measure d or t. (Allow digital or vernier callipers) [1]
M3 Measure c with a ruler/metre rule. [1]
M4 Method of making contact with the strip e.g. use electrodes of at least same dimension as c
or d or t or conducting paint methods. Do not allow crocodile clips, unless it is clear that the
whole area of the end of the strip is covered. [1]
M5 Method to determine resistance. [1]
A1 Plot a graph of R against c, 1/d or 1/A depending on orientation. Other alternatives possible,
e.g. R against 1/c depending on orientation [1]
A2 Must be consistent with A1: ρ = A × gradient or t × gradient/c [1]
Other alternatives possible, e.g. ρ = d × gradient/t
S1 Reference sharp edges or cutting metals, e.g. wear gloves. [1]
1. Insulate aluminium strip
2. Take many readings of t or d and average
3. Use a protective resistor/circuit designed to reduce current
4. Rearrange equation to determine graph using c, d and t or A
5. Determine typical resistance of aluminium strip
6. Likely meter range of ammeter/voltmeter/ohmmeter
7. Detail on cutting strip e.g. mark using set square
[Total: 15]

© UCLES 2010
(a) A1
C
t
− Must be negative. Allow
C
15
− .
(b) T1
T2 150 1.28 or 1.281
100 1.61 or 1.609
66.7 1.86 or 1.856
50.0 1.97 or 1.974
33.3 2.08 or 2.079
T1 for 1/R column – ignore sf and rounding
errors
T2 for ln (V/V) column – must be values given
U1 From ± 0.05 or ± 0.06 to ± 0.02 or
± 0.03
(c) (i) G1 Five points plotted correctly Must be within half a small square; penalise ≥
half a small square. Ecf allowed from table.
Penalise ‘blobs’ ≥ half a small square.
U2 Error bars in ln(V/V) plotted
correctly.
All plots to have error bars; penalise ≥ half a
small square. Check first and last point. Must
be accurate within half a small square.
(ii) G2 Line of best fit If points are plotted correctly then upper end of
line should pass between (20, 2.16) and
(20, 2.18) and lower end of line should pass
between (160, 1.20) and (160, 1.225). Allow
ecf from points plotted incorrectly – examiner
judgement.
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 all error bars are plotted.
(iii) C1 Gradient of best fit line
Must be negative
The triangle used should be at least half the
small square. Do not penalise POT.
U3 Uncertainty in gradient Method of determining absolute uncertainty.
(d) (i) C2 C = –15/gradient Gradient must be used.
Allow ecf from (c)(iii). Do not penalise POT.
C3 2.14 × 10–3
F to 2.24 × 10–3
F
and to 2 or 3 sf
Must be in range – penalise POT.
Allow equivalent unit including s Ω–1
, C V–1
,
A s V–1

© UCLES 2010
(ii) U4 Determines % uncertainty in C Uses worst gradient or worst calculated C
value.
Do not check calculation.
(e) C4 Determines R correctly Expect to see an answer about 3000 Ω.
R = 6.514/candidate’s C; allow ecf from (d)(i)
U5 Determines absolute uncertainty Determines worst value of R or (d)(ii) × R
[Total: 15]
(d) (ii) [U4]
1. Works out worst C then determines % uncertainty
2. Works out percentage uncertainty in gradient
(e) [U5]
1. Works out worst R then determines difference
2. R
C
C
RR 




 ∆
=




 ∆
=∆
gradient
gradient

9702 w10 ms_all

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