9702 w15 ms_all

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CAMBRIDGE INTERNATIONAL EXAMINATIONS
Cambridge International Advanced Subsidiary and Advanced Level
MARK SCHEME for the October/November 2015 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 2015 series for most
Cambridge IGCSE®
, Cambridge International A and AS Level components and some
Cambridge O Level components.

Mark Scheme Syllabus Paper
Cambridge International AS/A Level – October/November 2015 9702 11
© Cambridge International Examinations 2015
Question
Number
Key
Question
Number
Key
1 B 21 B
2 B 22 A
3 A 23 B
4 A 24 C
5 B 25 A
6 C 26 A
7 C 27 D
8 B 28 A
9 D 29 B
10 D 30 A
11 A 31 C
12 D 32 A
13 C 33 B
14 A 34 C
15 B 35 A
16 D 36 A
17 B 37 D
18 D 38 D
19 D 39 D
20 B 40 B

9702 PHYSICS
Cambridge IGCSE®

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

9702 PHYSICS
Cambridge IGCSE®

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

1 (a) temperature B1
current B1 [2]
(allow amount of substance, luminous intensity)
(b) (i) 1. E = (stress/strain =) [force/area] / [extension/original length]
units of stress: kgms–2
/m2
and no units for strain B1
units of E: kgm–1
s–2
A0 [1]
2. units for T: s, l: m and M: kg
K2
= T2
E/Ml3
hence units: s2
kgm–1
s–2
/kg3
(= m–4
) C1
units of K: m–2
A1 [2]
(ii) % uncertainty in E = 4% (for T2
) + 0.6% (for l3
) + 0.1% (for M) + 3% (for K2
)
= 7.7% B1
E = [(1.48 × 105
)2
× 0.2068 × (0.892)3
]/(0.45)2
= 1.588 × 1010
C1
7.7% of E = 1.22 × 109
C1
E = (1.6 ± 0.1) × 1010
kgm–1
s–2
A1 [4]
2 (a) ps = 10–12
(s) or T = 4 × 50 × 10–12
(s) B1
v = fλ or v = λ/T C1
λ = 3.0 × 108
× 4 × 50 × 10–12
C1
= 0.06(0)m A1 [4]
(b) 1500 = 3.0 × 108
× 4 × time-base setting or T = 5 × 10–6
s C1
time-base setting = 1.3 (1.25)µscm–1
A1 [2]
3 (a) work done is force × distance moved in direction of force
or
no work done along PQ as no displacement/distance moved in direction of force B1
work done is same in vertical direction as same distance moved in direction of
force B1 [2]

(b) (i) at maximum height t = 1.5(s) or s = ½(u + v)t, s = 11m and t = 1.5s C1
Vv = 0 + 9.81 × 1.5 Vv = (11 × 2) / 1.5
= 15 (14.7)ms–1
A1 [2]
(ii) straight line from (0,0) to (3.00, 25.5) B1 [1]
(iii) at maximum height Vh = 25.5/3 (= 8.5ms–1
) B1
ratio = mgh/½mv2
C1
= (2 × 9.81 × 11.0)/(8.5)2
= 3.0 (2.99) A1 [3]
(iv) deceleration is greater/resultant force (weight and friction force) is greater M1
time is less A1 [2]
4 (a) density = mass/volume C1
mass = 7900 × 4.5 × 24 × 10–6
= 0.85 (0.853)kg M1 [2]
(b) pressure = force/area C1
force = Wcos40° C1
pressure = (0.85 × 9.81cos 40°)/24 × 10–4
= 2.7 (2.66) × 103
Pa A1 [3]
(c) F = ma C1
W sin40° – f = ma C1
0.85 × 9.81 × sin40° – f = 0.85 × 3.8
f (= 5.36 – 3.23) = 2.1N [5.38 – 3.242 if 0.8532kg is used for the mass] A1 [3]

5 (a) progressive: all particles have same amplitude
stationary: no nodes or antinodes or maximum to minimum/zero amplitude B1
progressive: adjacent particles are not in phase
stationary: waves particles are in phase (between adjacent nodes) B1 [2]
(b) (i) wavelength 1.2m (zero displacement at 0.0, 0.60m, 1.2m, 1.8m, 2.4m)
either peaks at 0.30m and 1.5m and troughs at 0.90m and 2.1m
or vice versa (but not both) B1
maximum amplitude 5.0mm B1 [2]
(ii) 180° or π rad A1 [1]
(iii) at t = 0 particle has kinetic energy as particle is moving B1
at t = 5.0ms no kinetic energy as particle is stationary
so decrease in kinetic energy (between t = 0 and t = 5.0ms) B1 [2]
6 (a) energy converted from chemical to electrical per unit charge B1 [1]
(b) (i) current = E/(R + r) C1
= 6.0/(16 + 0.5)
= 0.36 (0.364)A A1 [2]
(ii) terminal p.d. = (0.36 × 16) = 5.8V or (6 – 0.36 × 0.5)
= 5.8V A1 [1]
(c) (i) use of R = ρl/A or proportionality with length and inverse
proportionality with area or d2
C1
d/2 and l/2 gives resistance of Z = 2RY = 24(Ω) C1
R = resistance of parallel combination = [1/24 + 1/12]–1
= 8(.0)(Ω) A1 [3]
(ii) resistance of circuit less therefore current larger B1
lost volts greater therefore terminal p.d. less B1 [2]
(d) power = I 2
R or VI or V2
/R C1
current in second circuit (= 6.0/12.5) = 0.48(A) B1
ratio = [(0.36)2
× 16] /[(0.48)2
× 12] = 0.75 [0.77 if full s.f. used] B1 [3]

7 (a) (i) curved path towards negative (–) plate (right-hand side) B1 [1]
(ii) range of α-particle is only few cm in air/loss of energy of the α-particles due
to collision with air molecules/ionisation of the air molecules B1 [1]
(iii) V = E × d C1
= 140 × 106
× 12 × 10–3
= 1.7 (1.68)MV A1 [2]
(b) β have opposite charge to α therefore deflection in opposite direction B1
β has a range of velocities/energies hence number of different deflections B1
β have less mass or q/m is larger hence deflection is greater
or
β with (very) high speed (may) have less deflection B1 [3]
(c)
emitted particle change in Z change in A
α-particle –2 –4
β-particle +1 0
A1 [1]

9702 PHYSICS
Cambridge IGCSE®

1 (a) v = fλ C1
λ = (3.0 ×108
)/(4.6 ×1020
) C1
(= 6.52 ×10–13
=) 0.65(2)pm A1 [3]
(b) t = (8.5 ×1016
)/(3.0 ×108
) C1
(= 2.83 ×108
=) 0.28(3)Gs A1 [2]
(c) mass, power and temperature all underlined and no others B1 [1]
(d) (i) arrow in the direction 30° to 40° south of east B1 [1]
(ii) triangle of velocities completed (i.e. correct scale diagram) or correct working
given C1
e.g. [142
+ 8.02
– 2(14)(8.0) cos 60°]1/2
or [(14 – 8.0 cos 60°)2
+ (8.0 sin 60°)2
]1/2
resultant velocity = 12(.2) (or 12.0 to 12.4 from scale diagram)ms–1
A1 [2]
2 (a) (i) v = u + at C1
0 = 3.6 – 3.0t
t (= 3.6/3.0) = 1.2s A1 [2]
(ii) (distance to rest from P = (3.6 × 1.2)/2 =) 2.2 (2.16)m A1 [1]
or
[0 – (3.6)2
]/[2 × (–3.0)] = 2.2 (2.16)m
or
3.6 × 1.2 – ½ × 3.0 × (1.2)2
= 2.2 (2.16)m
or
0 + ½ × 3.0 × (1.2)2
= 2.2 (2.16)m
(b) distance = 6.0 – 2.16 (= 3.84) C1
v2
= u2
+ 2as = 2 × 3.0 × 3.84 (= 23.04) M1
or
x + 2 × 2.16 = 6.0 gives x = 1.68(m) (C1)
v2
= 3.62
+ 2 × 1.68 × 3.0 (= 23.04) (M1)
or correct method with intermediate time calculated (t = 1.6s from Q to R)
v = 4.8ms–1
A0 [2]

(c) straight line from v = 3.6 ms–1
to v = 0 at t = 1.2s B1
straight line continues with the same gradient as v changes sign B1
straight line from v = 0 intercept to v = –4.8ms–1
B1 [3]
(d) difference in KE = ½m(v2
– u2
)
= 0.5 × 0.45 (4.82
– 3.62
) [= 5.184 – 2.916] C1
= 2.3 (2.27)J A1 [2]
3 (a) (i) k = F/x or 1/gradient C1
(k = 4.4/(5.4 × 10–2
) =) 81 (81.48)Nm–1
A1 [2]
(ii) work done = area under line or ½Fx or ½kx2
C1
(= 0.5 × 4.4 × 5.4 × 10–2
=) 0.12 (0.119)J A1 [2]
(b) (i) kinetic energy/Ek of trolley/T (and block) changes to EPE/strain
energy/elastic energy of spring B1
EPE changes to KE of trolley/T and KE of block or to give lower KE to trolley B1 [2]
(ii) change in momentum = m(v + u) C1
= 0.25 (0.75 + 1.2) = 0.49 (0.488)Ns A1 [2]
4 (a) product of the force and the perpendicular distance to/from a point/pivot B1 [1]
(b) (i) 4000 × 2.8 × sin 30° or 500 × 1.4 × sin 30° or T × 2.8
or 4000 × 1.4 or 500 × 0.7 B1
4000 × 2.8 × sin 30° + 500 × 1.4 × sin 30° = T × 2.8 M1
hence T = 2100 (2125)N A0 [2]
(ii) (Tv = 2100 cos 60° =) 1100 (1050)N A1 [1]
(iii) there is an upward (vertical component of) force at A B1
upward force at A + Tv = sum of downward forces/weight+load/4500N B1 [2]

5 (a) (i) I = V/R C1
(= 240/1500 =) 0.16A A1 [2]
(ii) I2 = 0.40 – 0.16 (= 0.24) C1
0.24(350 + R) = 240
R = 650Ω A1 [2]
(iii) power = IV or I2
R or V2
/R C1
ratio = (84 × 0.24)/(88 × 0.16)
or [(0.24)2
× 350]/[(0.16)2
× 550]
or (842
/350)/(882
/550)
or 20.16/14.08
= 1.4(3) A1 [2]
(b) (i) p.d. across 350Ω resistor = 0.24 × 350
or p.d. across 550Ω resistor = 0.16 × 550 C1
V350 = 84 (V) and V550 = 88 (V) gives VAB = 4.0V
or V950 = 152 (V) and VR = 156 V gives VAB = 4.0V A1 [2]
(ii) p.d. across R increases or potential at B increases or V350 decreases hence
VAB increases B1 [1]
6 (a) internal resistance causes lost volts B1
p.d. across lamp is less than 12V, power is less than 48W B1 [2]
(b) (i) greater lost volts or p.d. across cell/lamp reduced, less current in lamp B1 [1]
(ii) p.d. across lamp/current in lamp decreases, hence resistance decreases B1 [1]
7 (a) (i) 3.2mm A1 [1]
(ii) 20mm A1 [1]
(b) (i) energy is transferred/propagated (through the water) or wave
profile/wavefronts move (outwards from dipper) so progressive B1 [1]
(ii) to produce waves with constant/zero phase difference/coherent waves B1 [1]

(c) (i) path difference is λ B1
water vibrates/oscillates with amplitude about 2 × 3.2mm B1 [2]
(ii) path difference is λ/2 so little/no motion/displacement/amplitude B1 [1]
8 (a) result: majority/most (of the α-particles) went straight through/were deviated by
small angles M1
conclusion: most of the atom is (empty) space or size/volume of nucleus very
small compared with atom A1
result: a small proportion were deflected through large angles or >90° or came
straight back M1
conclusion: the mass or majority of mass is in a (very) small charged
volume/region/nucleus A1 [4]
(b) ρ = m/V C1
mass of atom and mass of nucleus (approx.) equal stated or cancelled or values
given e.g. 63u or 63 × 1.66 × 10–27
C1
ratio = (rA)3
/(rN)3
= (1.15 × 10–10
)3
/(1.4 × 10–14
)3
or
ratio = (dA)3
/(dN)3
= (2.3 × 10–10
)3
/(2.8 × 10–14
)3
= 5.5 × 1011
A1 [3]

9702 PHYSICS
Cambridge IGCSE®

1 (a) energy or W: kgm2
s–2
or
power or P: kgm2
s–3
M1
intensity or I: kgm2
s–2
m–2
s–1
(from use of energy expression)
or
kgm2
s–3
m–2
(from use of power expression)
indication of simplification to kgs–3
A1 [2]
(b) (i) ρ: kgm–3
, c: ms–1
, f: s–1
, x0: m M1
substitution of terms in an appropriate equation and simplification to show K
has no units A1 [2]
(ii) I = 20 × 1.2 × 330 × (260)2
× (0.24 × 10–9
)2
C1
= 3.1 × 10–11
(Wm–2
) C1
= 31 (30.8)pWm–2
A1 [3]
2 (a) (i) (the loudspeakers) are connected to the same signal generator B1 [1]
(ii) 1. the waves (that overlap) have phase difference of zero or path difference
of zero and so
either constructive interference
or displacement larger B1 [1]
2. the waves (that overlap) have phase difference of (n + ½) × 360° or
(n + ½) × 2π rad or path difference of (n + ½)λ and so
either destructive interference
or displacements cancel/smaller B1 [1]
3. the waves (that overlap) are in phase or have phase difference of n360°
or 2πn rad or path difference of nλ and so
either constructive interference
or displacement larger B1 [1]
(b) time period = 0.002s or 2ms C1
wave drawn is half time period B1
amplitude 1.0cm (same as Fig. 2.2) B1 [3]

3 (a) (i) 1. s = ut + ½at2
192 = ½ × 9.81 × t2
C1
t = 6.3 (6.26)s A1 [2]
2. max Ek (= mgh) = 0.27 × 9.81 × 192 C1
or
calculation of v (= 61.4) and use of EK (= ½ mv2
) = ½ × 0.27 × (61.4)2
(C1)
max Ek = 510 (509)J A1 [2]
(ii) velocity is proportional to time or velocity increases at a constant rate
as acceleration is constant or resultant force is constant B1 [1]
(iii) use of v = at or v2
= 2as or E = ½mv2
to give v = 61(.4)ms–1
B1 [1]
(b) (i) R increases with velocity B1
resultant force is mg – R or resultant force decreases B1
acceleration decreases B1 [3]
(ii) at v = 40ms–1
, R = 0.6(N) C1
0.27 × 9.8 – 0.6 = 0.27 × a
a = 7.6 (7.58) ms–2
A1 [2]
(iii) R = weight for terminal velocity B1
either weight requires velocity to be about 80ms–1
or at 60ms–1
, R is less than weight
so does not reach terminal velocity B1 [2]
4 (a) (i) reaction/vertical force = weight – P cos 60° C1
= 180 – 35 cos 60°
= 160 (163)N A1 [2]
(ii) work done = 35 sin 60° × 20 C1
= 610 (606) J A1 [2]

(b) (i) work done by force P = work done against frictional force B1 [1]
(ii) horizontal component of P is equal and opposite to frictional force B1
vertical component of P + normal reaction force equal and opposite to weight B1 [2]
5 (a) (i) resistance = V/I B1
very high/infinite resistance at low voltages B1
resistance decreases as V increases B1 [3]
(ii) p.d. from graph 0.50(V) C1
resistance = 0.5/(4.4 × 10–3
)
= 110 (114) Ω A1 [2]
(b) (i) current (= 1.2/375) = 3.2 × 10–3
A A1 [1]
(ii) current in diode = 4.4 × 10–3
(A)
total resistance = 1.2/4.4 × 10–3
= 272.7(Ω) C1
resistance of R1 = 272.7 – 113.6 = 160 (159)Ω A1
or
p.d. across diode = 0.5V and p.d. across R1 = 0.7V (C1)
resistance of R1 = 0.7/4.4 × 10–3
= 160 (159)Ω (A1) [2]
(iii) power = IV or I2
R or V2
/R C1
ratio = (4.4 × 0.5)/(3.2 × 1.2)
or [(4.4)2
× 114]/[(3.2)2
× 375]
or [(0.5)2
× 375]/[114 × (1.2)2
]
= 0.57 A1 [2]
6 (a) waves from loudspeaker (travel down tube and) are reflected at closed end B1
two waves (travelling) in opposite directions with same frequency/wavelength
overlap B1 [2]
(b) (i) 0.51m A1
0.85m A1 [2]
(ii) A at open end, N at closed end, with an N and A in between, equally spaced
(by eye) B1 [1]

7 (a) stress or σ = F/A C1
max. tension = UTS × A = 4.5 × 108
× 15 × 10–6
= 6800 (6750)N A1 [2]
(b) ρ = m/V C1
weight = mg = ρVg = ρALg
6750 = 7.8 × 103
× 15 × 10–6
× L × 9.81 C1
L = 5.9 (5.88) × 103
m A1
or
maximum mass = 6750/9.81 = 688kg (C1)
mass per unit length = ρA = 0.117kgm–1
(C1)
L = 688/0.117 = 5.9 × 103
m (A1)
or
maximum mass = 6750/9.81 = 688kg (C1)
volume = m/ρ = 0.0882m3
= LA (C1)
L = 0.0882/15 × 10–6
= 5.9 × 103
m (A1) [3]
8 (a) mass-energy
proton number or charge
nucleon number B2 [2]
(b) (i) Ek = ½ mv2
and p = mv with working leading to
[via Ek = ½m2
v2
/m or ½m(p/m)2
]
to Ek =
m
p
2
2
B1 [1]
(ii) p = (2Ekm)½
hence (2[Ekm]α)½
= (2[Ekm]Th)½
C1
2 × [Ek]Th × 234 = 2 × 6.69 × 10–13
× 4 C1
[Ek]Th = 1.14 × 10–14
J
= 71(.5) keV A1
or
calculation of speed of α-particle = 1.42 × 107
ms–1
calculation of momentum of α-particle/nucleus = 9.43 × 10–20
Ns (C1)
[Ek]Th = 1.14 × 10–14
J (C1)
= 71(.5)keV (A1) [3]

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

1 (a) (iv) Value of d in range 19.5cm to 20.5cm with unit. [1]
(c) (ii) Value of N with evidence of repeat readings. [1]
(e) Six sets of readings of d and N scores 5 marks, five sets scores 4 marks etc. [5]
Incorrect trend –1. Help from Supervisor –1.
Range: [1]
Smallest value of d < 9.5cm.
Column headings: [1]
Each column heading must contain a quantity and a unit where appropriate. The
presentation of quantity and unit must conform to accepted scientific convention
e.g. d –1
/m–1
.
Consistency: [1]
All values of d must be given to the nearest mm.
Significant figures: [1]
Every value of 1/d must be given to the same number of significant figures as (or
one more than) the number of significant figures in the corresponding value of d.
Calculation: [1]
√N calculated correctly to the number of significant figures given by the candidate.
(f) (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 should be no more than three large squares apart.
Plotting: [1]
All observations in the table must be plotted on the grid.
Diameter of plotted points must be ⩽ half a small square (no “blobs”).
Points must be plotted to an accuracy of half a small square.
Quality: [1]
All points in the table must be plotted (at least 5) for this mark to be awarded.
Scatter of points must be no more than ±0.5 in the √N direction from a
straight line.
(ii) Line of best fit: [1]
Judge by balance of all points on the grid about the candidate’s line (at least
5 points). There must be an even distribution of points either side of the line
along the full length.
Allow one anomalous point only if clearly indicated (i.e. circled or labelled) by
the candidate.
Lines must not be kinked or thicker than half a square.

(iii) Gradient: [1]
The hypotenuse of the triangle must be greater than half the length of the drawn
line. Do not allow ∆x/∆y. Sign of gradient must match graph drawn.
Both read-offs must be accurate to half a small square in both the x and y
directions.
y-intercept: [1]
Either:
Correct read-offs from a point on the line substituted into y = mx + c or an
equivalent expression. Read-offs must be accurate to half a small square in
both x and y directions.
Or:
Intercept read directly from the graph, with read-off accurate to half a small
square.
(g) Value of A=candidate’s gradient and value of B=candidate’s intercept. [1]
Unit for A correct (e.g. m or cm or mm) and consistent with value.
No unit given for B. [1]
2 (a) (ii) Value for x to the nearest mm. [1]
x in the range 0.155m to 0.165m. [1]
(b) (i) Correct calculation of C in m (correct to 2 s.f.). [1]
(ii) Justification for significant figures in C linked to significant figures in x and h. [1]
(c) (ii) Value for R with unit. [1]
Evidence of repeat readings. [1]
(iii) Absolute uncertainty in R in range 5mm to 20mm and correct method of
calculation to obtain percentage uncertainty. If repeated readings have been
taken, then the uncertainty can be half the range (but not zero) if the working
is clearly shown. [1]
(d) Second value of x. [1]
Second value of R. [1]
Second value of R < first value of R. [1]
(e) (i) Two values of k calculated correctly. [1]
(ii) Valid comment consistent with the calculated values of k, testing against a
criterion specified by the candidate. [1]

(f) (i) Limitations (4 max.) (ii) Improvements (4 max.) Do not credit
A Not enough readings to draw a
conclusion
Take many readings for different
holes and plot a graph/
obtain more k values and
compare
Few readings/
only one reading/
not enough readings for
an accurate result/
“repeat readings”
on its own/
take more readings
and (calculate)
average k
B Ball rolls off block/ball does not
move along straight line from the
wood/rod does not hit marble
square on each time/rod hits
marble at an angle
Small groove in the wood to
place the marble
C Difficult to measure distance rod is
pulled back/difficult to hold rod still
before release
Use another stand or stop
D Difficult to measure R with reason
e.g. marble skids in sand leaving
elongated hole/can’t fit ruler in
sand tray/parallax error
Improved method for measuring
R e.g. video with scale/use
carbon paper/ink on marble/put
scale on the sand
E Difficult to flatten sand/know when
sand is horizontal
Use a straight edge/
use a spirit level
F Difficult to measure x with reason
e.g. wooden rod moves
Method of measuring x/
clamp rule close by/
draw scale on rod

9702 PHYSICS
maximum raw mark 40
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.
Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for
Teachers.
Cambridge is publishing the mark schemes for the October/November 2015 series for most Cambridge IGCSE®
,
Cambridge International A and AS Level components and some Cambridge O Level components.

1 (b) (i) Value for L to the nearest mm, with unit. [1]
(c) Second value of h > first value of h. [1]
(d) (ii) Six sets of readings of m, h and θ scores 5 marks, five sets scores 4 marks etc. [5]
Help from Supervisor –1.
Incorrect trend –1. Correct trend is h increases as m increases.
Range: [1]
Range of values to include mmin < 60g and mmax > 80g.
Each column heading must contain a quantity and a unit.
The presentation of quantity and unit must conform to accepted scientific
convention e.g. h/cosθ (cm).
Consistency: [1]
All values of h must be given to the nearest mm.
Every value of h/cosθ must be given to 2 or 3 significant figures only.
Calculation: [1]
Values of h/cosθ calculated correctly to the number of significant figures
given by the candidate.
(e) (i) Axes: [1]
Plotting: [1]
Quality: [1]
Scatter of points must be no more than 10g in the m direction of a straight
line.
Allow one anomalous point only if clearly indicated (i.e. circled or labelled) by the
candidate.

(iii) Gradient: [1]
The hypotenuse of the triangle must be greater than half the length of the
drawn line.
Do not allow ∆x/∆y. Sign of gradient must match graph drawn.
directions.
y-intercept: [1]
Either:
equivalent expression.
Read-offs must be accurate to half a small square in both x and y directions.
Or:
Intercept read directly from the graph, with read-off accurate to half a small square.
(f) Value of A = candidate’s gradient and value of B = candidate’s intercept. [1]
Unit for A correct (e.g. mkg–1
or cmg–1
) and unit for B correct (m or cm or mm). [1]
2 (a) (iii) Values of x, y and z to the nearest mm with unit. [1]
Value of z > value of x. [1]
(iv) Absolute uncertainty in y of 1mm to 4mm and correct method of calculation to
obtain percentage uncertainty. If repeated readings have been taken, then the
uncertainty can be half the range (but not zero) if working is clearly shown. [1]
(b) Correct calculation of C with consistent unit. [1]
(c) (ii) Value for T with unit in range 5.0s > T > 0.5s. [1]
Evidence of repeat readings for T. [1]
(iv) Justification for significant figures in T2
linked to significant figures in the (raw) times. [1]
(d) Second values of x, y and z. Value of y within 5mm of value in (a)(iii). [1]
Second value of T. [1]
Second value of T < first value of T. [1]
(ii) Valid comment consistent with the calculated values of k, testing against
a criterion specified by the candidate. [1]

conclusion
Take many readings (for different
masses) and plot a graph/
obtain more k values and compare
“Repeat readings” on its
own/
few readings/
only one reading/
not enough readings for
an accurate result/
take more readings and
(calculate) average k
B Rod is bent when loaded Use smaller masses/
rigid/stiff/thick rod
Just “rod is bent”/
shorter rod
C Difficult to get horizontal Use a spirit level or named
instrument.
D Difficult to measure distances with
reason e.g. rod unstable/awkward
with metre rule/rod moves/holding
ruler mid-air
Add a scale on rod/
use travelling microscope/
clamp ruler
Parallax
Do not award if reason
given is bent rod.
E y not constant with a reason e.g.
spring/loop moves around during
oscillations
Cut groove or drill hole in wooden
rod/
tape to wooden rod
F Difficult to judge the start/end of an
oscillation
or
Difficult to judge when to start/stop
the stopwatch
(Fiducial) marker at centre/
video and timer/view frame by
frame/
motion sensor placed below/above
More oscillations/
high speed camera/
reaction time/
human error
G Oscillation in more than one
plane/irregular oscillations
Wind/draughts/
switch off air
conditioning/
close doors and
windows

9702 PHYSICS
maximum raw mark 40
Cambridge IGCSE®

1 (b) (i) Value of θ to the nearest degree and in the range 135° to 165°. [1]
(ii) Value of L in range 5.0 to 10.0cm, with unit. [1]
(d) Six sets of readings of θ and L scores 5 marks, five sets scores 4 marks etc. [5]
Range: [1]
θmax ⩾ 160° and θmin ⩽ 140°.
Each column heading must contain a quantity and a unit. The presentation of
quantity and unit must conform to accepted scientific convention e.g. θ/°.
1/sin(θ –90°) must have no unit.
Consistency: [1]
All values of L must be given to the nearest mm.
Every value of 1/sin(θ –90°) must be given to 2 or 3 significant figures only.
Calculation: [1]
Values of 1/sin(θ –90°) calculated correctly to the number of significant figures
given by the candidate.
(e) (i) Axes: [1]
Plotting: [1]
Quality: [1]
Scatter of points must be no more than ±0.3 (to scale) cm in the L direction
from a straight line.
the candidate.

(iii) Gradient: [1]
The hypotenuse of the triangle must be greater than half the length of the drawn line.
Do not allow ∆x/∆y. Sign of gradient must match graph drawn.
directions.
y-intercept: [1]
Either:
Or:
Intercept read directly from the graph, with read-off accurate to half a small square.
(f) Value of a = candidate’s gradient and value of b = candidate’s intercept. [1]
Unit for a is correct (e.g. cm–1
) and no unit for b. [1]
2 (a) (i) All values of d to nearest mm, with unit, in range 5 to 30mm. [1]
Value of l greater than value of d. [1]
(ii) Correct calculation of V with consistent unit. [1]
(b) Justification for significant figures in V linked to significant figures in d and l. [1]
(c) (iii) t in range 5.00s to 30.00s, with unit. [1]
Evidence of repeat readings of t. [1]
(d) Absolute uncertainty in t in range 0.5s to 5.0s and correct method of calculation to
absolute uncertainty can be half the range (but not zero) if working is clearly shown. [1]
(e) Second values of d and l. [1]
Second value of t. [1]
Second value of t < first value of t. [1]
(f) (i) Two values of k calculated correctly. [1]

(g) (i) Limitations (4 max.) (ii) Improvements (4 max.) Do not credit
conclusion
Take more readings and plot a
graph/
compare
Few readings/
only one reading/
not enough readings
for an accurate result/
on its own/
take more readings
and (calculate)
average k
B d is small/large uncertainty in d Improved method of measuring
d e.g. micrometer/vernier
calipers/digital calipers/travelling
microscope
Difficult to measure d/
parallax error/
“calipers” on its own/
use bigger/larger
components
C Volume of component not accurate,
with reason e.g. component not
cylindrical/has groove.
Method to find volume of
component e.g. use liquid
displacement method
D Difficult to judge/know/see when
LED goes out.
Use dark(ened) room/
light meter/
light sensor/
cardboard tube over LED/
voltmeter to measure time for
p.d. to fall below specific value
Use video
E Poor/dirty/loose contacts Method of cleaning contacts e.g.
iron wool

9702 PHYSICS
maximum raw mark 40
Cambridge IGCSE®

1 (b) (i) Value of a with consistent unit and in the range 0 to 50.0cm. [1]
(v) Value of V with unit to nearest 0.001V and in range 0 to 2V. [1]
(c) Six sets of readings of a, b and V scores 5 marks, five sets scores 4 marks etc. [5]
Incorrect trend –1. Minor help from Supervisor –1. Major help from Supervisor –2.
Range: [1]
amax – amin ⩾ 30.0cm.
Each column heading must contain a quantity and a unit. The presentation of
quantity and unit must conform to accepted scientific convention. e.g. a/m or
a(m), 1/V/V–1
.
Consistency: [1]
All values of a and b must be given to the nearest mm.
Every value of 1/V must be given to the same number of significant figures as (or
one more than) the number of significant figures in the corresponding value of V.
Calculation: [1]
Values of 1/V calculated correctly to the number of significant figures given by
the candidate.
(d) (i) Axes: [1]
Plotting: [1]
Quality: [1]
Scatter of points must be no more than ± 0.050m (5.0cm) to scale in the b
direction from a straight line.
the candidate.

(iii) Gradient: [1]
The hypotenuse of the triangle must be greater than half the length of the
drawn line. Do not allow ∆x/∆y. Sign of gradient must match graph drawn.
directions.
y-intercept: [1]
Either:
Correct read-off from a point on the line substituted into y = mx + c or an
Or:
Intercept read directly from the graph, with read-off accurate to half a small
square.
(e) Value of P = candidate’s gradient and value of Q = candidate’s intercept. [1]
Unit for P is correct (e.g. m–1
V–1
) and unit for Q is correct (e.g. V–1
). [1]
2 (b) (iv) All values of x with unit to nearest mm. Average x ⩽ 20.0cm. [1]
(c) (i) Value of θ to the nearest degree in the range 25° to 35°. [1]
(ii) Absolute uncertainty in θ in range 2° to 5° and correct method of calculation
to obtain percentage uncertainty. If repeated readings have been taken, then
the absolute uncertainty can be half the range (but not zero) if working is
clearly shown. [1]
(iii) Correct calculation of cos2
(θ/2). No unit. [1]
(iv) Justification for significant figures in cos2
(θ/2) linked to significant figures in θ. [1]
(d) (ii) T in range 0.5s to 2.5s, with unit. [1]
Evidence of repeat readings of T. [1]
(f) Second value of θ. [1]
Second values of T1 and T2. [1]
Second value of T1 /T2 < first value of T1 /T2 when rounded to 2 s.f. [1]
(g) (i) Two values of k calculated correctly. [1]

(h) (i) Limitations (4 max.) (ii) Improvements (4 max.) Do not credit
conclusion
graph/
compare
Few readings/
only one reading/
not enough readings
for an accurate result/
on its own/
take more readings
and (calculate)
average k
B Difficult to measure θ or read
protractor with reason e.g. rod is
above protractor/rod obscures view
Improved method of measuring
θ e.g. shadow projection with
light above/thinner rod/rod with
smaller diameter/plumb-lines
hung from rod/larger
protractor/360° protractor
Just “difficult to
measure θ”/
smaller rod/
“protractor reads to
1°”
C Oscillations orT affected by...
e.g. air movement
force on release
different forces
angle of release
unwanted modes of oscillation
Improved method of release e.g.
card gate
or
switch off air conditioning/close
windows/closed room
Release by
electromagnet/
cutting string/
damping/
air resistance/
friction
D Large percentage uncertainty in
time/period T is short
Valid method to improve timing
e.g. use video with timer/frame
by frame/motion sensor and
position at side of cradle
or
Increase T by... e.g. heavier
nail/longer nail/string
Oscillations too fast/
high/low speed
camera/
video on its own/
human reaction time/
just “difficult to
determine time”/
fiducial marker
E Valid problem linked to magnetism
e.g. nail weakly magnetised/metal
stand attracts nail/interference by
Earth’s magnetic field.
Valid method to overcome
problem linked with magnetism
e.g. stroke nail more times/use
of coil
Nail loses magnetism
F Method of fixing
paper/protractor/magnets e.g.
tape/Blu-tack/draw protractor on
paper
Just “stick paper”
without method

9702 PHYSICS
maximum raw mark 40
Cambridge IGCSE®

1 (a) Value of H in the range 13.0cm to 17.0cm. [1]
(c) (iii) Value of F to nearest 0.1N. [1]
(iv) Value of x correct and in range 3.5cm to 4.5cm. [1]
(d) Six sets of readings of hw, hb and F scores 4 marks, five sets scores 3 marks etc. [4]
Range: [1]
xmax – xmin ⩾ 6.0cm.
Each column heading must contain a quantity and a unit.
The presentation of quantity and unit must conform to accepted scientific convention,
e.g. (H–x)3
/cm3
.
Consistency: [1]
All values of hw and hb must be given to the nearest mm.
Significant figures for (H–x)3
must be the same as, or one greater than, the
number of s.f. for (H–x).
Calculation: [1]
Values of (H–x)3
calculated correctly.
(e) (i) Axes: [1]
Scales must be chosen so that the plotted points occupy at least half the graph
grid in both x and y directions.
Plotting: [1]
Quality: [1]
Scatter of points must be less than ± 0.2N from a straight line in the F direction.
Judged by balance of all points on the grid about the candidate’s line (at least 5
points). There must be an even distribution of points either side of the line along
the full length.
Allow one anomalous point only if clearly indicated (i.e. circled or labelled) by the
candidate.

(iii) Gradient: [1]
The hypotenuse of the triangle used must be greater than half the length of the
drawn line.
The method of calculation must be correct.
Both read-offs must be accurate to half a small square in both the x and y directions.
y-intercept: [1]
Either:
Or:
Intercept read directly from the graph, with read-off accurate to half a small square
in the y direction.
(f) Value of a = candidate’s gradient and value of b = candidate’s intercept. [1]
Unit for a is correct (e.g. Ncm–3
) and unit for b is correct (e.g. N). [1]
2 (a) (i) All values of d to nearest 0.001cm and in range 0.100cm to 0.500cm. [1]
(ii) All values of D to nearest 0.1cm. [1]
(b) Values of l, h and s present.
Value of h to nearest 0.1cm and in range 7.5cm to 8.5cm, with unit. [1]
(c) (ii) t on answer line in range 1.00s to 20.00s, with unit. [1]
Evidence of repeated readings for t. [1]
(iii) Absolute uncertainty in t in range 0.2s to 0.5s and correct method of calculation to
absolute uncertainty can be half the range (but not zero) if the working is clearly
shown. [1]
(iv) Calculation of g correct to the second s.f., with consistent unit (e.g. ms–2
). [1]
(d) (ii) Second values of d and t. [1]
t greater for smaller d. [1]
(ii) Justification for significant figures in k linked to significant figures in t and d. [1]
(iii) Valid comment consistent with the calculated values of k, tested against a

conclusion
graph/
compare
“Repeat readings”
on its own/
few readings/
only one reading/
take more readings
and (calculate)
average k/
two readings not
enough for accurate
results
B Parallax error when measuring
h or D
Use calipers/
use set square as pointer
Rule not vertical
when measuring h
C Plastic deforms when measuring
larger d/diameter with tubing
Use larger diameter axle instead
of tubing
D Push force to start flywheel may
vary/push force may affect time
Release mechanism, with detail/
use steeper ramp
“Force may be too
large”/
start before top line
E Flywheel doesn’t travel straight/
flywheel hits (sides of) track
Sensible method of preventing
collision, e.g. level the track
sideways/widen the track
Reduce friction/
mark both sides of
track
F Use video with timer/
view frame by frame/
use light gates at start and end/
use longer track
Reaction time

9702 PHYSICS
9702/41 Paper 4 (A2 Structured Questions), maximum raw mark 100
Cambridge IGCSE®

Section A
1 (a) (i) gravitational force provides/is the centripetal force B1
GMmS /x2
= mSv2
/x (allow x or r; allow m or mS) M1
EK = ½mSv2
and clear algebra leading to EK = GMmS /2x A1 [3]
(ii) EP = – GMmS /x (sign essential) B1 [1]
(iii) ET = EK + EP
= GMmS /2x – GMmS /x C1
= – GMmS /2x (allow ECF from (a)(ii)) A1 [2]
(b) (i) decreases B1 [1]
(ii) decreases B1 [1]
(iii) decreases B1 [1]
(iv) increases B1 [1]
(for answers in (b) allow ECF from (a)(iii))
2 (a) obeys the equation pV = nRT or pV/T = constant M1
all symbols explained; T in kelvin/thermodynamic temperature A1 [2]
(b) (i) temperature rise = 48K A1 [1]
(ii) <c2
> ∝ T or equivalent C1
<c2
> = (353/305) × 1.9 × 106
C1
cr.m.s. = 1480ms–1
A1 [3]
3 (a) heat/thermal energy gained by system or energy transferred to system by heating B1
plus work done on the system or minus work done by the system B1 [2]
(b) (i) either volume decreases so work done on the system
or small volume change so work done on system negligible M1
(thermal) energy absorbed to break lattice structure M1
internal energy increases A1 [3]
(ii) gas expands so work done by gas (against atmosphere) M1
no time for thermal energy to enter or leave the gas M1
internal energy decreases A1 [3]
4 (a) free: (body oscillates) without any loss of energy/no resistive forces/no external
forces applied B1
forced: continuous energy input (required)/body is made to vibrate by an
(external) periodic force/driving oscillator B1 [2]

(b) (i) idea of resonance B1
maximum amplitude at natural frequency B1
frequency = 2.1Hz (allow 2.08 to 2.12Hz) B1 [3]
(ii) peak not very sharp/amplitude not infinite so frictional forces are present B1 [1]
(c) v = ωx0
= 2π × 2.1 × 4.7 × 10–2
(allow ECF from (b)(i)) C1
= 0.62ms–1
A1 [2]
5 (a) (i) force proportional to the product of the two/point charges B1
and inversely proportional to the square of their separation B1 [2]
(ii) 1. force radially away from sphere/to right/to east B1 [1]
2. (maximum) at/on surface of sphere or x = r B1 [1]
3. F ∝ 1/x2
or F = q1q2/(4πε0x2
) C1
ratio = 16 A1 [2]
(b) E = q/(4πε0x2
) or E ∝ q C1
maximum charge = (2.0/1.5) × 6.0 × 10–7
C1
= 8.0 × 10–7
C
additional charge = 2.0 × 10–7
C A1 [3]
6 (a) (i) force = mg M1
along the direction of the field/of the motion A1 [2]
(ii) no force B1 [1]
(b) (i) force due to E-field downwards so force due to B-field upwards B1
into the plane of the paper B1 [2]
(ii) force due to magnetic field = Bqv B1
force due to electric field = Eq B1
(use of FB and FE not explained, allow 1/2)
forces are equal (and opposite) so Bv = E or Eq = Bqv so E = Bv B1 [3]
(c) sketch: smooth curved path M1
in ‘upward’ direction A1 [2]
7 (a) minimum frequency of e.m. radiation/a photon (not “light”) M1
for emission of electrons from a surface A1 [2]
(reference to light/UV rather than e.m. radiation, allow 1/2)

(b) EMAX corresponds to electron emitted from surface B1
electron (below surface) requires energy to bring it to surface, so less than EMAX B1 [2]
(c) (i) 1/λ0 = 1.85 × 106
(allow 1.82 to 1.88) C1
f0 = c/λ0
= 3.00 × 108
× 1.85 × 106
= 5.55 × 1014
Hz A1 [2]
(ii) Φ = hf0
= 6.63 × 10–34
× 5.55 × 1014
(allow ECF from (c)(i)) C1
= 3.68 × 10–19
J A1 [2]
(d) sketch: straight line with same gradient M1
intercept between 1.0 and 1.5 A1 [2]
8 (a) nucleus: small central part/core of an atom B1
nucleon: proton or a neutron B1
particle contained within a nucleus B1 [3]
(b) (i) 1. decay constant = ln2/(3.8 × 24 × 3600) C1
= 2.1 × 10–6
s–1
A1 [2]
2. A = λN
97 = 2.1 × 10–6
× N C1
N = 4.6 × 107
A1 [2]
(ii) 1.0m3
contains (6.02 × 1023
)/(2.5 × 10–2
) air molecules C1
ratio = (4.6 × 107
× 2.5 × 10–2
)/(6.02 × 1023
)
= 1.9 × 10–18
A1 [2]

Section B
9 (a) (i) (+) 3.0V B1 [1]
(ii) potential = 6.0 × {2.0 / (2.0 + 2.8)} C1
= 2.5V A1 [2]
(iii) potential = 6.0 × {2.0 / (2.0 + 1.8)}
= 3.2V A1 [1]
(b) at 10°C, VA > VB M1
VOUT is –9.0V (allow “negative saturation”) A1
at 20°C, VOUT is +9.0V B1
(if 20°C considered initially, mark as M1,A1,B1)
sudden switch (from –9V to +9V) when VA = VB B1 [4]
10 (a) sharpness: clarity of edges/resolution (of image) B1
contrast: difference in degree of blackening (of structures) B1 [2]
(b) (i) X-rays produced when (high speed) electrons hit target/anode B1
either electrons have been accelerated through 80kV
or electrons have (kinetic) energy of 80keV B1 [2]
(ii) IT /I = e–3.0×1.4
C1
= 0.015 A1 [2]
(c) for good contrast, µx or eµx
or e–µx
must be very different B1
µx or eµx
or e–µx
for bone and muscle will be different than that for muscle M1
so good contrast A1 [3]
11 (a) frequency of carrier wave varies M1
in synchrony with the displacement of the signal/information wave A1 [2]
(b) (i) 5.0V A1 [1]
(ii) 720kHz A1 [1]
(iii) 780kHz A1 [1]
(iv) 7500 A1 [1]

12 (a) (i) (gradual) loss of power/intensity/amplitude (not “signal”) B1 [1]
(ii) e.g. noise can be eliminated (not “there is no noise”) M1
because pulses can be regenerated A1
e.g. much greater data handling/carrying capacity M1
because many messages can be carried at the same time/greater
bandwidth A1
e.g. more secure (M1)
because it can be encrypted (A1)
e.g. error checking (M1)
because extra information/parity bit can be added (A1) [4]
(allow any two sensible suggestions with ‘state’ M1 and ‘explain’ A1)
(b) attenuation = 10 lg(145/29) (= 7.0) C1
attenuation per unit length = 7.0/36
= 0.19dBkm–1
A1 [2]

9702 PHYSICS
Cambridge IGCSE®

Section A
1 (a) (i) gravitational force provides/is the centripetal force B1
GMmS /x2
= mSv2
/x (allow x or r; allow m or mS) M1
EK = ½mSv2
and clear algebra leading to EK = GMmS /2x A1 [3]
(ii) EP = – GMmS /x (sign essential) B1 [1]
(iii) ET = EK + EP
= GMmS /2x – GMmS /x C1
= – GMmS /2x (allow ECF from (a)(ii)) A1 [2]
(b) (i) decreases B1 [1]
(ii) decreases B1 [1]
(iii) decreases B1 [1]
(iv) increases B1 [1]
(for answers in (b) allow ECF from (a)(iii))
2 (a) obeys the equation pV = nRT or pV/T = constant M1
all symbols explained; T in kelvin/thermodynamic temperature A1 [2]
(b) (i) temperature rise = 48K A1 [1]
(ii) <c2
> ∝ T or equivalent C1
<c2
> = (353/305) × 1.9 × 106
C1
cr.m.s. = 1480ms–1
A1 [3]
3 (a) heat/thermal energy gained by system or energy transferred to system by heating B1
plus work done on the system or minus work done by the system B1 [2]
(b) (i) either volume decreases so work done on the system
or small volume change so work done on system negligible M1
(thermal) energy absorbed to break lattice structure M1
internal energy increases A1 [3]
(ii) gas expands so work done by gas (against atmosphere) M1
no time for thermal energy to enter or leave the gas M1
internal energy decreases A1 [3]
4 (a) free: (body oscillates) without any loss of energy/no resistive forces/no external
forces applied B1
forced: continuous energy input (required)/body is made to vibrate by an
(external) periodic force/driving oscillator B1 [2]

(b) (i) idea of resonance B1
maximum amplitude at natural frequency B1
frequency = 2.1Hz (allow 2.08 to 2.12Hz) B1 [3]
(ii) peak not very sharp/amplitude not infinite so frictional forces are present B1 [1]
(c) v = ωx0
= 2π × 2.1 × 4.7 × 10–2
(allow ECF from (b)(i)) C1
= 0.62ms–1
A1 [2]
5 (a) (i) force proportional to the product of the two/point charges B1
and inversely proportional to the square of their separation B1 [2]
(ii) 1. force radially away from sphere/to right/to east B1 [1]
2. (maximum) at/on surface of sphere or x = r B1 [1]
3. F ∝ 1/x2
or F = q1q2/(4πε0x2
) C1
ratio = 16 A1 [2]
(b) E = q/(4πε0x2
) or E ∝ q C1
maximum charge = (2.0/1.5) × 6.0 × 10–7
C1
= 8.0 × 10–7
C
additional charge = 2.0 × 10–7
C A1 [3]
6 (a) (i) force = mg M1
along the direction of the field/of the motion A1 [2]
(ii) no force B1 [1]
(b) (i) force due to E-field downwards so force due to B-field upwards B1
into the plane of the paper B1 [2]
(ii) force due to magnetic field = Bqv B1
force due to electric field = Eq B1
(use of FB and FE not explained, allow 1/2)
forces are equal (and opposite) so Bv = E or Eq = Bqv so E = Bv B1 [3]
(c) sketch: smooth curved path M1
in ‘upward’ direction A1 [2]
7 (a) minimum frequency of e.m. radiation/a photon (not “light”) M1
for emission of electrons from a surface A1 [2]
(reference to light/UV rather than e.m. radiation, allow 1/2)

(b) EMAX corresponds to electron emitted from surface B1
electron (below surface) requires energy to bring it to surface, so less than EMAX B1 [2]
(c) (i) 1/λ0 = 1.85 × 106
(allow 1.82 to 1.88) C1
f0 = c/λ0
= 3.00 × 108
× 1.85 × 106
= 5.55 × 1014
Hz A1 [2]
(ii) Φ = hf0
= 6.63 × 10–34
× 5.55 × 1014
(allow ECF from (c)(i)) C1
= 3.68 × 10–19
J A1 [2]
(d) sketch: straight line with same gradient M1
intercept between 1.0 and 1.5 A1 [2]
8 (a) nucleus: small central part/core of an atom B1
nucleon: proton or a neutron B1
particle contained within a nucleus B1 [3]
(b) (i) 1. decay constant = ln2/(3.8 × 24 × 3600) C1
= 2.1 × 10–6
s–1
A1 [2]
2. A = λN
97 = 2.1 × 10–6
× N C1
N = 4.6 × 107
A1 [2]
(ii) 1.0m3
contains (6.02 × 1023
)/(2.5 × 10–2
) air molecules C1
ratio = (4.6 × 107
× 2.5 × 10–2
)/(6.02 × 1023
)
= 1.9 × 10–18
A1 [2]

Section B
9 (a) (i) (+) 3.0V B1 [1]
(ii) potential = 6.0 × {2.0 / (2.0 + 2.8)} C1
= 2.5V A1 [2]
(iii) potential = 6.0 × {2.0 / (2.0 + 1.8)}
= 3.2V A1 [1]
(b) at 10°C, VA > VB M1
VOUT is –9.0V (allow “negative saturation”) A1
at 20°C, VOUT is +9.0V B1
(if 20°C considered initially, mark as M1,A1,B1)
sudden switch (from –9V to +9V) when VA = VB B1 [4]
10 (a) sharpness: clarity of edges/resolution (of image) B1
contrast: difference in degree of blackening (of structures) B1 [2]
(b) (i) X-rays produced when (high speed) electrons hit target/anode B1
either electrons have been accelerated through 80kV
or electrons have (kinetic) energy of 80keV B1 [2]
(ii) IT /I = e–3.0×1.4
C1
= 0.015 A1 [2]
(c) for good contrast, µx or eµx
or e–µx
must be very different B1
µx or eµx
or e–µx
for bone and muscle will be different than that for muscle M1
so good contrast A1 [3]
11 (a) frequency of carrier wave varies M1
in synchrony with the displacement of the signal/information wave A1 [2]
(b) (i) 5.0V A1 [1]
(ii) 720kHz A1 [1]
(iii) 780kHz A1 [1]
(iv) 7500 A1 [1]

12 (a) (i) (gradual) loss of power/intensity/amplitude (not “signal”) B1 [1]
(ii) e.g. noise can be eliminated (not “there is no noise”) M1
because pulses can be regenerated A1
e.g. much greater data handling/carrying capacity M1
because many messages can be carried at the same time/greater
bandwidth A1
e.g. more secure (M1)
because it can be encrypted (A1)
e.g. error checking (M1)
because extra information/parity bit can be added (A1) [4]
(allow any two sensible suggestions with ‘state’ M1 and ‘explain’ A1)
(b) attenuation = 10 lg(145/29) (= 7.0) C1
attenuation per unit length = 7.0/36
= 0.19dBkm–1
A1 [2]

9702 PHYSICS
Cambridge IGCSE®

1 (a) (gravitational) force proportional to product of masses
and inversely proportional to square of separation M1
either point masses or particles or ‘size’ ≪ separation A1 [2]
(b) gravitational force provides the centripetal force B1
either GMm/x2
= mxω2
or mv2
/x M1
either ω = 2π/T or v = 2πx/T and working to GM = 4π2
x 3
/T2
A1 [3]
(c) either use of gradient of graph or line through origin so can use single point
or line shown extrapolated to origin B1
gradient = (4.5 × 1014
)/0.35
6.67 × 10–11
× M = 4π2
× (4.5 × 1014
× 109
)/(0.35 × {24 × 3600}2
)
correct conversion for km3
and power of 10 C1
correct conversion for day2
C1
M = 1.02 × 1026
kg A1 [4]
2 (a) total volume of molecules negligible compared to that of containing vessel
no intermolecular forces
molecules in random motion
time of collision small compared with the time between collisions
large number of molecules
any two B2 [2]
(b) in a real gas there is a range of velocities or must take the average of v2
B1 [1]
(c) (i) either p =
3
1
ρ<c2
>
or 1.0 × 105
=
3
1
× 1.2 × <c2
> C1
<c2
> = 2.5 × 105
C1
cr.m.s. = 500ms–1
A1 [3]
(ii) T ∝ <c2
> C1
<c2
> = 2.5 × 105
× 480/300
= 4.0 × 105
m2
s–2
(allow ECF from (c)(i)) A1 [2]
3 (a) same temperature B1
no (net) transfer of thermal energy (between the bodies) B1 [2]
(b) (i) 41.3 K B1 [1]
(ii) 330.4 K B1 [1]

(c) ∆EK =
2
3
× 1.9 × 60
= 171 J C1
work done = p∆V
= 1.2 × 105
× 950 × 10–6
C1
= 114 J C1
thermal energy = 114 + 171
= 285 (290) J A1 [4]
4 (a) acceleration/force proportional to distance from a fixed point or displacement M1
either acceleration/force and displacement in opposite directions
or acceleration/force (always) directed towards a fixed point/mean
position/equilibrium position A1 [2]
(b) hρg = Mg/A B1
h × 790 × 4.9 × 10–4
= 70 × 10–3
leading to h = 0.18m or 18cm A1 [2]
(c) (i) 1. ω2
= (790 × 4.9 × 10–4
× 9.81)/(70 × 10–3
) C1
= 54.25
ω = 7.37 (rads–1
)
period (= 2π/ω) = 0.85s C1
t1 = 0.43 s A1 [3]
2. t3 = 1.28 s (allow 2 s.f.) A1 [1]
(ii) energy of peak = ½Mω2
x0
2
B1
change = ½ × 70 × 10–3
× 54.25 {(2.2 × 10–2
)2
– (1.0 × 10–2
)2
} C1
= 7.3 × 10–4
J A1 [3]

5 (a) charges in metal do not move B1
no (resultant) force on charges so no (electric) field B1 [2]
(allow 1/2 for “no field inside sphere”)
(b) either average field strength = ½ (28 + 54) NC–1
C1
average force = 8.5 × 10–9
× ½ (28 + 54) C1
= 3.49 × 10–7
N
change in potential energy = 3.49 × 10–7
× 2.0 × 10–2
= 7.0 × 10–9
J (allow 1 s.f.) A1
(allow range 54 ± 1)
or (for a point charge) V = Ex (C1)
∆V = (54 × 5.0 × 10–2
) – (28 × 7.0 × 10–2
) (C1)
× (2.70 – 1.96)
= 6.3 × 10–9
J (allow 1 s.f.) (A1)
(allow range 54 ± 1)
or ∆V is area under curve (C1)
∆V = 0.74 V (C1)
× 0.74
= 6.3 × 10–9
J (allow 1 s.f.) (A1) [3]
(allow range 0.70 to 0.84)
6 (a) magnetic fields are equal in magnitude/strength/flux density M1
magnetic fields are opposite in direction M1
fields superpose/add/cancel to give zero/negligible resultant field A1 [3]
(b) core causes increase in magnetic flux in the solenoid/induced poles in core
or field induced in core B1
changing flux threads/cuts the turns on the solenoid M1
(by Faraday’s law) an e.m.f. is induced in the solenoid A1
by Lenz’s law, this e.m.f. opposes the battery e.m.f. A1 [4]
7 (a) (i) V0 (= 14 2 ) = 19.8 (20) V A1 [1]
(ii) ω (= 2π × 750) = 4700rads–1
A1 [1]
(b) large amount of charge required to charge capacitor M1
capacitor would charge and discharge rapidly/in a very short time
or capacitor would charge and discharge 750/1500 times per second M1
I = Q/t, so large current A1 [3]

8 (a) hc/λ = Φ + EMAX M1
h = Planck constant, c = speed of light/e.m. radiation A1 [2]
(b) (i) gradient of line is hc M1
h and c are both constants A1 [2]
(ii) Φ = 2.28 × 1.6 × 10–19
C1
= 3.65 × 10–19
(J)
hc/λ0 = 3.65 × 10–19
λ0 = (6.63 × 10–34
× 3.0 × 108
)/(3.65 × 10–19
) C1
= 5.45 × 10–7
m A1 [3]
9 (a) energy required to separate the nucleons (in a nucleus)
or energy required to separate the protons and neutrons in a nucleus M1
(or energy released when nucleons combine (to form a nucleus)/energy released
when protons and neutrons combine to form a nucleus)
either completely or to infinity A1 [2]
(either free protons and neutrons or from infinity)
(b) (i) either different forms of same element or nuclei having same number of M1
protons with different numbers of neutrons A1 [2]
(ii) 1784 MeV (accept min. 3 s.f.) A1
7.57 MeV A1 [2]
(c) (i) λ = ln2/(7.1 × 108
× 365 × 24 × 3600) = 3.1 × 10–17
s–1
B1 [1]
(ii) A = λN
5000 = 3.1 × 10–17
× N C1
N = 1.61 × 1020
mass = 235 × (1.61 × 1020
)/(6.02 × 1023
) C1
= 0.063 g (accept min. 2 s.f.) A1 [3]

Section B
10 (a) correct LED symbol B1
separately connected between VOUT and earth with opposite polarities M1
diode B ‘pointing’ from VOUT to earth A1 [3]
(ignore protective resistors)
(b) diode in VOUT line M1
diode ‘pointing’ towards VOUT from earth A1
relay coil connected between VOUT and earth M1
switch connected across lamp A1 [4]
(if a diode is placed across the relay it must point down otherwise max. 2/4;
one diode but wrong direction max. 3/4)
11 (a) e.g. scattering (in metal)
non-parallel beam (not just “A closer than B”)
reflection (from metal)
diffraction in the metal/lattice
any two B2 [2]
(b) (i) 1. ratio = eµx
= exp(0.27 × 4.0) C1
= 2.94 (2.9) A1 [2]
2. ratio = exp(0.27 × 2.5) × exp(3.0 × 1.5) C1
= 1.96 × 90
= 177 (180) A1 [2]
(do not penalise unit error more than once)
(ii) each ratio gives measure of transmission B1
ratios (in (i)) very different so good contrast B1 [2]
12 (a) (i) serial-to-parallel converter B1 [1]
(ii) digital-to-analogue converter or DAC B1 [1]
(iii) (audio) amplifier or AF amplifier B1 [1]
(b) (i) 4 A1 [1]
(ii) 1011 A1 [1]
(c) correct levels at 0.25ms intervals
0, 8, 11, 10, 15 A1
and 7, 4 A1
series of steps, each of depth 0.25 ms M1
voltage levels shown in correct intervals A1 [4]

13 (a) advantage: e.g. shorter time delay
greater coverage over a long time B1
disadvantage: e.g. satellite needs to be tracked
more satellites for (continuous) coverage/communication
(any sensible suggestions) B1 [2]
(b) (i) frequencies linking Earth with satellite B1
6 GHz is uplink frequency }
4 GHz is downlink frequency } (allow vice versa) B1 [2]
(ii) either signal from Earth to satellite is attenuated greatly
or downlink must be amplified greatly before transmission B1
downlink would swamp uplink unless frequencies are different B1 [2]

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

1 Planning (15 marks)
Defining the problem (3 marks)
P m is the independent variable, or vary m. [1]
P (tan)φ is the dependent variable, or measure (tan) φ. [1]
P Keep the temperature of the oil constant. [1]
Methods of data collection (5 marks)
M Labelled diagram showing labelled protractor positioned to determine φ for tilted
cylinder.
Allow distances marked to determine φ and use of a rule. [1]
M Use of balance/scales to measure the mass of the oil/cylinder. [1]
M Mass of oil = mass of (oil + cylinder) – mass of cylinder. [1]
M Use of vernier calipers/micrometer/rule to measure d. [1]
M Repeat each experiment for the same value of m and average φ. [1]
Method of analysis (2 marks)
A Plot a graph of
φtan
1
against m.
(Allow 3d
m or 3d
m
ρ or ρ
m
. Do not allow log-log graphs.) [1]
A a = gradient ×ρd3
and b = y-intercept; must be consistent with suggested graph. [1]
Safety considerations (1 mark)
S Precaution linked to preventing spilling oil, e.g. use a tray/lid/cloth to absorb oil (do not
allow just wiping or mopping)
or precaution linked to preventing glass cylinder breaking, e.g. padding/cushion
or use of gloves to prevent skin irritation (do not allow “because oil is slippery”). [1]
Additional detail (4 marks)
D Relevant points might include [4]
1 Repeat measurements of d in different directions and average
2 Use of video with slow motion/frame by frame playback to determine φ
3 Use large protractor to reduce percentage uncertainty or trigonometry
relationship related to measurements to be taken
4 Use the same (diameter) cylinder (not “same size” but allow “same size and shape”)
5 Slowly/gently/gradually tilt cylinder of oil/use of rough surface (to prevent sliding)
6 Experimental method to determine density of oil and ρ = m/V
7 Relationship is valid if the graph is a straight line that does NOT pass through the
origin/has an intercept; must be consistent with suggested graph
Do not allow vague computer methods.

2 Analysis, conclusions and evaluation (15 marks)
Mark Expected Answer Additional Guidance
(a) A1
gradient = 2
Edπ
ρ4
y-intercept =
E
r
(b) T1
I
1
/A–1
Allow
I
1
(A–1
) or 





A
11
I
.
T2
4.2 or 4.17
5.0 or 5.00
5.9 or 5.88
6.7 or 6.67
7.7 or 7.69
8.3 or 8.33
Allow a mixture of significant figures.
Must be table values.
U1 ± 0.2 to ± 0.6 or ± 0.7 or ± 0.8 Allow more than one significant figure.
(c) (i) G1 Six points plotted correctly Must be within half a small square.
Do not allow “blobs”.
ECF allowed from table.
U2 Error bars in 1/I plotted
correctly
All error bars to be plotted. Must be accurate to
less than half a small square. Length of bar
must be accurate to less than half a small
square. Do not allow less than 0.05.
(ii) G2 Line of best fit If points are plotted correctly then lower end of
line should pass between (41, 4.5) and (44, 4.5)
and upper end of line should pass between
(83, 8.0) and (88, 8.0).
Line should not go from bottom to top points.
G3 Worst acceptable straight line.
Steepest or shallowest
possible line that passes
through all the error bars.
Line should be clearly labelled or dashed.
Examiner judgement on worst acceptable line.
Lines must cross. Mark scored only if error bars
are plotted.
(iii) C1 Gradient of line of best fit 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 8.)
U3 Absolute uncertainty in
gradient
Method of determining absolute uncertainty:
difference in worst gradient and gradient.
(iv) C2 y-intercept Check substitution into y = mx + c.
Allow ECF from (c)(iii).
(Should be about 0.7–1.5.)

U4 Absolute uncertainty in y-
intercept
Uses worst gradient and point on WAL.
Do not check calculation.
(d) (i) C3 ρ = 2.415 × 10–7
× gradient
Must be in the range
1.80 × 10–6
to 2.10 × 10–6
and
given to 2 or 3 s.f.
Must use gradient.
ρ =
4
2
Edπ
× gradient
[2 × 10–6
Ωm = 2 × 10–4
Ωcm = 2 × 10–3
Ωmm]
C4 r = E × y-intercept
= 3.2 × y-intercept
and Ωm and Ω given
Must include units for ρ and r.
Allow VA–1
or kgm2
A–2
s–3
for Ω.
(ii) U5 Percentage uncertainty in ρ Must be greater than 9.6%.
Uncertainties in Question 2
(c) (iii) Gradient [U3]
uncertainty = gradient of line of best fit – gradient of worst acceptable line
uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)
(iv) [U4]
uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line
uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)
(d) (ii) [U5]
percentage uncertainty = 





×++
∆
0.31
0.01
2
3.2
0.1
m
m
× 100
= 





×
∆
100
m
m
+ 3.125 + 2 × 3.226
max. p =
( )
4
100.323.3
23−
×××π
× max. gradient
min. p =
( )
4
100.303.1
23−
×××π
× min. gradient

9702 PHYSICS
maximum raw mark 30
Cambridge IGCSE®

P m is the independent variable, or vary m. [1]
P (tan)φ is the dependent variable, or measure (tan) φ. [1]
P Keep the temperature of the oil constant. [1]
M Labelled diagram showing labelled protractor positioned to determine φ for tilted
cylinder.
Allow distances marked to determine φ and use of a rule. [1]
M Use of balance/scales to measure the mass of the oil/cylinder. [1]
M Mass of oil = mass of (oil + cylinder) – mass of cylinder. [1]
M Use of vernier calipers/micrometer/rule to measure d. [1]
M Repeat each experiment for the same value of m and average φ. [1]
A Plot a graph of
φtan
1
against m.
(Allow 3d
m or 3d
m
ρ or ρ
m
. Do not allow log-log graphs.) [1]
A a = gradient ×ρd3
and b = y-intercept; must be consistent with suggested graph. [1]
S Precaution linked to preventing spilling oil, e.g. use a tray/lid/cloth to absorb oil (do not
allow just wiping or mopping)
or precaution linked to preventing glass cylinder breaking, e.g. padding/cushion
or use of gloves to prevent skin irritation (do not allow “because oil is slippery”). [1]
1 Repeat measurements of d in different directions and average
2 Use of video with slow motion/frame by frame playback to determine φ
3 Use large protractor to reduce percentage uncertainty or trigonometry
relationship related to measurements to be taken
4 Use the same (diameter) cylinder (not “same size” but allow “same size and shape”)
5 Slowly/gently/gradually tilt cylinder of oil/use of rough surface (to prevent sliding)
6 Experimental method to determine density of oil and ρ = m/V
7 Relationship is valid if the graph is a straight line that does NOT pass through the
origin/has an intercept; must be consistent with suggested graph

(a) A1
gradient = 2
Edπ
ρ4
y-intercept =
E
r
(b) T1
I
1
/A–1
Allow
I
1
(A–1
) or 





A
11
I
.
T2
4.2 or 4.17
5.0 or 5.00
5.9 or 5.88
6.7 or 6.67
7.7 or 7.69
8.3 or 8.33
Must be table values.
U1 ± 0.2 to ± 0.6 or ± 0.7 or ± 0.8 Allow more than one significant figure.
U2 Error bars in 1/I plotted
correctly
square. Do not allow less than 0.05.
line should pass between (41, 4.5) and (44, 4.5)
and upper end of line should pass between
(83, 8.0) and (88, 8.0).
Line should not go from bottom to top points.
are plotted.
gradient
(iv) C2 y-intercept Check substitution into y = mx + c.
Allow ECF from (c)(iii).
(Should be about 0.7–1.5.)

U4 Absolute uncertainty in y-
intercept
Uses worst gradient and point on WAL.
Do not check calculation.
(d) (i) C3 ρ = 2.415 × 10–7
× gradient
Must be in the range
1.80 × 10–6
to 2.10 × 10–6
and
Must use gradient.
ρ =
4
2
Edπ
× gradient
[2 × 10–6
Ωm = 2 × 10–4
Ωcm = 2 × 10–3
Ωmm]
C4 r = E × y-intercept
= 3.2 × y-intercept
and Ωm and Ω given
Must include units for ρ and r.
Allow VA–1
or kgm2
A–2
s–3
for Ω.
(ii) U5 Percentage uncertainty in ρ Must be greater than 9.6%.
(iv) [U4]
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) [U5]
percentage uncertainty = 





×++
∆
0.31
0.01
2
3.2
0.1
m
m
× 100
= 





×
∆
100
m
m
+ 3.125 + 2 × 3.226
max. p =
( )
4
100.323.3
23−
×××π
× max. gradient
min. p =
( )
4
100.303.1
23−
×××π
× min. gradient

9702 PHYSICS
maximum raw mark 30
Cambridge IGCSE®

P m is the independent variable and E is the dependent variable or vary m and measure E.
Do not allow time. [1]
P Keep the temperature change of water constant. Allow two specified temperatures.
Do not allow “keep temperature constant”. [1]
P Keep the mass or volume of water constant. [1]
M Labelled diagram including labelled thermometer with bulb in water and at least one
other label. [1]
M Workable circuit diagram to determine E: power supply, heater and ammeter and
voltmeter, or joulemeter or wattmeter. [1]
M Method to determine change in temperature: measure initial temperature, measure final
temperature and subtract, or measure initial temperature and specific temperature
change. [1]
M Use balance/scales to measure mass of blocks. [1]
M Stir water (so that metal is in thermal equilibrium). [1]
A Plot a graph of E against m.
Do not allow log–log graphs. [1]
A a = gradient and b = y-intercept; must be consistent with suggested graph. [1]
S Precaution linked to hot heater/water, e.g. use gloves or use tongs for hot blocks.
Do not allow goggles. [1]
1 Method to ensure that e.m.f. of the power supply is constant/current in heater is constant,
e.g. adjust variable power supply/variable resistor to ensure p.d./current is constant
2 Keep the starting temperature of water/metal constant
3 Wait for water and metal temperatures to equalise
4 Add insulation to sides of beaker/lid (to prevent energy losses)
5 Use of timer and equation, e.g. E = Pt = ItV for candidate’s method
6 Use large temperature change to reduce percentage uncertainty
7 Relationship is valid if the graph is a straight line that does not pass through the origin

(a) A1
gradient =
m
Pg
(b) T1 T/s, v/ms–1
and v2
/m2
s–2
Allow T(s), v(ms–1
) and v2
(m2
s–2
).
T2
8.7 or 8.74
19 or 19.3
27 or 27.4
37 or 36.6
45 or 44.9
52 or 52.3
Must be values of v2
in table (if v not rounded).
All values of v2
must be 2 s.f. or 3 s.f.
U1 From ± 0.9 or ± 1 to ± 3 Allow more than one significant figure.
U2 Error bars in v2
plotted
correctly
square.
line should pass between (0.16, 10) and
(0.18, 10) and upper end of line should pass
between (0.70, 50) and (0.72, 50).
Line should not go from top to bottom points.
are plotted.
gradient
(d) (i) C2
P =
g
m
× gradient
= 2.55 × 10–3
× gradient
Must use gradient. Should be about 0.19.
C3 kg
(ii) U4 Percentage uncertainty in P Must be greater than 4%.

(e) (i) C4 v in the range 4.70 to 4.90 and
(ii) U5 Percentage uncertainty in v Allow credit if absolute uncertainty in mass used
correctly.
(d) (ii) [U4]
percentage uncertainty = 10004.0
gradient
gradient
100
0.025
0.001
gradient
gradient
×





+
∆
=×





+
∆
max. P =
9.81
0.026
× max. gradient
max. P =
9.81
0.024
× min. gradient
(e) (ii) [U5]
percentage uncertainty = 100
0.5
0.005
2
1
×





+
∆
×
P
P
max. v =
0.040
0.5059.81max. ××P
min. v =
0.040
0.4959.81min. ××P

9702 w15 ms_all

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9702 w15 ms_all