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Cambridge Assessment International Education
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/11
Paper 1 Multiple Choice October/November 2017
MARK SCHEME
Maximum Mark: 40
Published
This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the
examination.
Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for
Teachers.
Cambridge International will not enter into discussions about these mark schemes.
Cambridge International is publishing the mark schemes for the October/November 2017 series for most
Cambridge IGCSE®
, Cambridge International A and AS Level components and some Cambridge O Level
components.

9702/11 Cambridge International AS/A Level – Mark Scheme
PUBLISHED
October/November
2017
© UCLES 2017 Page 2 of 3
Question Answer Marks
1 C 1
2 B 1
3 B 1
4 C 1
5 D 1
6 C 1
7 B 1
8 B 1
9 B 1
10 D 1
11 D 1
12 A 1
13 A 1
14 B 1
15 B 1
16 D 1
17 A 1
18 B 1
19 B 1
20 A 1
21 D 1
22 B 1
23 C 1
24 A 1
25 D 1
26 C 1
27 A 1
28 C 1

PUBLISHED
October/November
2017
29 D 1
30 C 1
31 C 1
32 C 1
33 C 1
34 A 1
35 D 1
36 B 1
37 B 1
38 D 1
39 B 1
40 A 1

PHYSICS 9702/12
MARK SCHEME
Maximum Mark: 40
Published
examination.
Teachers.
Cambridge IGCSE®
components.

PUBLISHED
October/November
2017
1 D 1
2 A 1
3 C 1
4 D 1
5 C 1
6 B 1
7 B 1
8 D 1
9 C 1
10 B 1
11 D 1
12 B 1
13 D 1
14 A 1
15 B 1
16 A 1
17 D 1
18 A 1
19 B 1
20 D 1
21 A 1
22 D 1
23 D 1
24 B 1
25 C 1
26 D 1
27 D 1
28 C 1

PUBLISHED
October/November
2017
29 C 1
30 D 1
31 A 1
32 B 1
33 D 1
34 C 1
35 A 1
36 B 1
37 B 1
38 A 1
39 A 1
40 C 1

PHYSICS 9702/13
MARK SCHEME
Maximum Mark: 40
Published
examination.
Teachers.
Cambridge IGCSE®
components.

PUBLISHED
October/November
2017
1 C 1
2 C 1
3 A 1
4 B 1
5 B 1
6 D 1
7 C 1
8 D 1
9 B 1
10 A 1
11 B 1
12 B 1
13 C 1
14 A 1
15 A 1
16 D 1
17 C 1
18 D 1
19 A 1
20 D 1
21 A 1
22 C 1
23 D 1
24 A 1
25 C 1
26 C 1
27 D 1
28 C 1

PUBLISHED
October/November
2017
29 B 1
30 C 1
31 A 1
32 C 1
33 D 1
34 D 1
35 C 1
36 B 1
37 C 1
38 D 1
39 C 1
40 B 1

PHYSICS 9702/21
Paper 2 AS Level Structured Questions October/November 2017
MARK SCHEME
Maximum Mark: 60
Published
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.
Teachers.
Cambridge IGCSE®
components.

PUBLISHED
October/November
2017
1(a) units of F: kg m s–2
C1
units of ρ: kg m–3
and units of v: m s–1
C1
units of K: kg m s–2
/ [kg m–3
(m s–1
)2
]
= m2
A1
1(b)(i) Kρ = 1.5 / 332
C1
= 1.38 × 10–3
FD = 1.38 × 10–3
× 252
or FD / 1.5 = 252
/ 332
FD = 0.86 N
A1
1(b)(ii) a = (1.5 – 0.86) / (1.5 / 9.81) or a = 9.81 – [0.86 / (1.5 / 9.81)] C1
a = 4.2 m s–2
A1
1(c) initial acceleration is g/9.81 (m s–2
)/acceleration of free fall B1
acceleration decreases B1
final acceleration is zero B1

PUBLISHED
October/November
2017
2(a) 30 m s–1
= 108 km h–1
or
100 km h–1
= 28 m s–1
and so exceeds speed limit
B1
2(b) acceleration = gradient or ∆v / (∆)t or (v – u) / t C1
e.g. acceleration = (24 – 20) / 12 [other points on graph line may be used]
= 0.33 m s–2
A1
2(c) distance travelled by Q = ½ × 12 × 30 (= 180 m) C1
distance travelled by P = ½ × (20 + 24) × 12 (= 264 m) C1
distance between cars = 264 – 180
= 84 m
A1
2(d) 30 – 24 = 6 m s–1
‘extra’ time T = 84 / 6 (= 14 s)
or
180 + 30T = 264 + 24T
‘extra’ time T = 84 / 6 (= 14 s)
C1
t = 12 + 14 = 26 s A1

PUBLISHED
October/November
2017
3(a)(i) in a stationary wave energy is not transferred
or
in a progressive wave energy is transferred
B1
3(a)(ii) in a stationary wave (adjacent) particles are in phase
or
in a progressive wave (adjacent) particles are out of phase/have a phase difference/not in phase
B1
3(b)(i) (position where) maximum amplitude B1
3(b)(ii) distance = 0.10 m B1
3(b)(iii) 1. λ = 0.60 / 1.5
= 0.40 m
A1
2. v = fλ C1
f = 340 / 0.40
= 850 Hz
A1
3(b)(iv) λ = 2 × 0.60 or λ = 3 × 0.40 or f = 850 / 3 C1
f = 280 (283) Hz A1

PUBLISHED
October/November
2017
4(a) (strain =) extension / original length B1
4(b)(i) E = σ/ ε C1
maximum stress = 2.1 × 1011
× 4.0 × 10–4
= 8.4 × 107
Pa
A1
4(b)(ii) σ = F / A C1
minimum area = 8.0 × 103
/8.4 × 107
= 9.5 × 10–5
m2
A1
5(a) I1 + I2 = I3 [any subject] B1
5(b) E1 + E3 = I1R1 + I3R3 + I3R4 [any subject] B1
5(c) E1 – E2 = I1R1 – I2R2 [any subject] B1

PUBLISHED
October/November
2017
6(a) force per unit positive charge B1
6(b)(i) EK = ½mv 2
C1
2.4 × 10–16
= ½ × 1.7 × 10–27
× v 2
v = 5.3 × 105
m s–1
A1
6(b)(ii) work done = 2.4 × 10–16
J A1
6(b)(iii) W = Fs C1
F = 2.4 × 10–16
/ 15 × 10–3
= 1.6 × 10–14
N
A1
6(b)(iv) V = Fd / Q
or
V = W / Q
or
E = V / d and E = F / Q
C1
V = (1.6 × 10–14
× 15 × 10–3
) / 1.6 × 10–19
or 2.4 × 10–16
/ 1.6 × 10–19
C1
= 1500 V A1
6(b)(v) straight line with positive gradient starting at the origin and going as far as x = 15 mm B1

PUBLISHED
October/November
2017
7(a) (the ohm is) volt / ampere B1
7(b)(i) R = ρL / A C1
ratio = [ρL / (πd 2
/ 4)] / [0.028ρ × 7.0L / {π(14d)2
/ 4}] = 1000
or
ratio = 142
/ (0.028 × 7) = 1000
A1
7(b)(ii) same current (in connecting and filament wires) and the lamp/filament (wire) has greater resistance B1
7(b)(iii) P = V 2
/ R or P = VI or P = I2
R C1
(for filament wire) R = 122
/ 6.0 or R = 6.0 / 0.502
or R = 12 / 0.50 C1
(for filament wire) R = 24 Ω
(for connecting wire) R = 24 / 1000
= 2.4 × 10–2
Ω
A1
7(b)(iv) resistance of connecting wire increases B1
current in circuit/lamp/filament (wire) decreases
or
potential difference across lamp/filament (wire) decreases
M1
(so) resistance of lamp/filament (wire) decreases A1

PUBLISHED
October/November
2017
8(a) (quark structure is) up, down, down/udd B1
up/u has charge +⅔(e), down/d has charge –⅓(e) C1
+⅔e –⅓e –⅓e = 0 A1
8(b) charge: p +1.6(0) × 10–19
(C) or +e
β–
–1.6(0) × 10–19
(C) or –e
ν zero/0
B1
mass: p 1.67 × 10–27
(kg)/1.7 × 10–27
(kg)
β–
9.1(1) × 10–31
(kg)
ν very small/zero/0
B1

PHYSICS 9702/22
MARK SCHEME
Maximum Mark: 60
Published
Teachers.
Cambridge IGCSE®
components.

PUBLISHED
October/November
2017
1(a)(i) micrometer (screw gauge)/digital calipers B1
1(a)(ii) take several readings (and average) M1
along the wireoraround the circumference A1
1(b)(i) σ = 4 × 25/[π × (0.40 × 10–3
)2
] = 1.99 × 108
Nm–2
or
σ = 25/[π × (0.20 × 10–3
)2
] = 1.99 × 108
Nm–2
A1
1(b)(ii) %F = 2% and %d = 5%
or
∆F/F =
0.5
25
and ∆d/d =
0.02
0.4
C1
%σ = 2% + (2 × 5%)
or
%σ = [0.02 + (2 × 0.05)] × 100
%σ = 12%
A1
1(b)(iii) absolute uncertainty = (12/100) × 1.99 × 108
= 2.4 × 107
C1
σ = 2.0 × 108
± 0.2 × 108
Nm–2
or 2.0 ± 0.2 × 108
Nm–2
A1

PUBLISHED
October/November
2017
2(a) force × perpendicular distance (of line of action of force) to/from a point B1
2(b)(i) 2.4r or (1.2 × 2r) or (1.2r + 1.2r) A1
2(b)(ii) (anticlockwise moment =) 6.0 × r/2 × sinθ C1
6.0 × r/2 × sinθ = 2.4r
θ = 53°
A1
2(b)(iii) 6.0N A1
3(a) p = 1000 × 9.81 × 7.0 × 10–2
or 1000 × 9.81 × 1.9 × 10–2
C1
∆p = 1000 × 9.81 × (7.0 × 10–2
– 1.9 × 10–2
) or 686 – 186
= 500 Pa
A1
3(b) F = pA or (∆)F = ∆p × A C1
upthrust = 500 × (5.1 × 10–2
)2
= 1.3N
or
upthrust = (686 – 186) × (5.1 ×10–2
)2
= 1.3N
or
upthrust = 1000 × 9.81 × 5.1 ×10–2
× (5.1 × 10–2
)2
= 1.3N
A1
3(c) force = 4.0 – 1.3
= 2.7N
A1

PUBLISHED
October/November
2017
3(d) extension/x/e = 2.7/30 C1
= 0.09(m) or 9(cm) C1
height above surface = 9 – 7
= 2cm
A1
3(e)(i) mass = 4.0/9.81 C1
acceleration = 2.7/(4.0/9.81)
= 6.6ms–2
A1
3(e)(ii) viscous force increases (and then becomes constant) M1
(weight and upthrust constant so) acceleration decreases (to zero) A1

PUBLISHED
October/November
2017
4(a) (two) waves travelling (at same speed) in opposite directions overlap B1
waves (are same type and) have same frequency/wavelength B1
4(b)(i) 5 A1
4(b)(ii) T = 1/40 (= 2.5 × 10–2
) C1
time taken = 2.5 × 10–2
/2
= 1.3 × 10–2
s (1.25 × 10–2
s)
A1
4(b)(iii) 180° A1
4(b)(iv) v = fλ C1
λ = 2.0/2.5 (= 0.80m)
v = 0.80 × 40
= 32ms–1
A1

PUBLISHED
October/November
2017
5(a) (coulomb is) ampere second B1
5(b)(i) E = V/d or E = F/Q C1
F = VQ/d
F = (2.0 × 102
× 8.0 × 10–19
)/4.0 × 10–2
= 4.0 × 10–15
N
A1
5(b)(ii) arrow pointing to the left labelled ‘electric force’ and arrow pointing downwards labelled ‘weight’ B1
5(b)(iii) 1. resultant force = √ [(3.9 × 10–15
)2
+ (4.0 × 10–15
)2
] C1
= 5.6 × 10–15
N A1
2. angle = tan–1
(3.9 × 10–15
/4.0 × 10–15
)
= 44°
A1
5(c) downward sloping line from (0, 2.0) M1
magnitude of gradient of line increases with time and line ends at (T, 0) A1

PUBLISHED
October/November
2017
6(a) flow of charge carriers B1
6(b)(i) nALe B1
6(b)(ii) (t is time taken for electrons to move length L)
I = Q/t
B1
I = nALe/t
or
I = nALe/(L/v)
or
I = nAvte/t and I = nAve
B1
6(c)(i) ratio = area at X/area at Y
= [πd 2
/4]/[π(0.69d)2
/4] or d2
/(0.69d)2
or 1/0.692
C1
= 2.1 A1
6(c)(ii) 1. R = ρL/A or R/L ∝ 1/A C1
resistance per unit length = 1.7 × 10–2
× (area at X/area at Y)
= 1.7 × 10–2
× 2.1
= 3.6 × 10–2
Ωm–1
A1
2. P = I2
R or P = V2
/R C1
R = 3.6 × 10–2
× 3.0 × 10–3
(= 1.08 × 10–4
Ω)
P = 0.502
× 1.08 × 10–4
or P = (5.4 × 10–5
)2
/1.08 × 10–4
= 2.7 × 10–5
W
A1

PUBLISHED
October/November
2017
6(c)(iii) (cross-sectional area decreases so) resistance increases M1
(P = I
2
R, so) power increases A1
7(a) lepton(s) B1
7(b) protons: 7 and neutrons: 6 A1
7(c) E = ½mv2
C1
= 0.80 × 106
× 1.60 × 10–19
C1
= 1.28 × 10–13
(J)
v2
= 2 × 1.28 × 10–13
/2.2 × 10–26
v = 3.4 × 106
ms–1
A1
7(d) an (electron) neutrino/ν(e) is also produced (and this has energy) B1

PHYSICS 9702/23
MARK SCHEME
Maximum Mark: 60
Published
Teachers.
Cambridge IGCSE®
components.

PUBLISHED
October/November
2017
1(a)(i) work (done) / time (taken) or energy (transferred) / time (taken) B1
1(a)(ii) Correct substitution of base units of all quantities into any correct equation for power.
Examples:
(P = E / t or W / t gives) kg m2
s–2
/ s = kg m2
s–3
(P = Fs / t or mgh / t gives) kg m s–2
m / s = kg m2
s–3
(P = ½mv2
/ t gives) kg (m s–1
)2
/ s = kg m2
s–3
(P = Fv gives) kg m s–2
m s–1
= kg m2
s–3
(P = VI gives) kg m2
s–2
A–1
s–1
A = kg m2
s–3
A1
1(b)(i) units of A: m2
and units of T: K C1
units of k: kg m2
s–3
/ m2
K4
= kg s–3
K–4
A1
1(b)(ii) curve from the origin with increasing gradient B1
2(a) ρ = m / V or ρ = m / Ah B1
p = F / A or p = W / A B1
p = [ρAhg] / A or p = [ρVg] / [V / h] (so) p = ρgh A1

PUBLISHED
October/November
2017
2(b)(i) 1. weight/gravitational (force)
upthrust (force)/buoyancy (force)
drag/viscous/frictional (force)/fluid resistance/resistance
B1
2. weight = upthrust + viscous (force) B1
2(b)(ii) • decrease in (gravitational) potential energy (of sphere) due to decrease in height (since Ep = mgh)
• increase in thermal energy due to work done against viscous force/drag
• loss/change of (total) Ep equal to gain/change in thermal energy
Any 2 points.
B2
2(c)(i) atmospheric pressure = 9.1(0) × 104
Pa A1
2(c)(ii) (∆)p = ρg(∆)h
(9.15 – 9.10) × 104
= ρ × 9.81 × (0.17 – 0.10)
C1
ρ = 730 (728) kg m–3
A1
3(a) sum/total momentum (of system of bodies) is constant
or
sum/total momentum before = sum/total momentum after
M1
for an isolated system/no (resultant) external force A1
3(b)(i) p = mv C1
(4.0 × 6.0 × sinθ) – (12 × 3.5 × sin 30°) = 0
or
(mAvA × sinθ) – (mBvB × sin 30°) = 0
M1
θ = 61° A1

PUBLISHED
October/November
2017
3(b)(ii) shows the horizontal momentum component of ball A or of ball B as (4.0 × 6.0 × cos θ) or (12 × 3.5 × cos 30°) C1
(4.0 × 6.0 × cos 61°) + (12 × 3.5 × cos 30°) = 4.0v so v = 12 (ms–1
) A1
3(b)(iii) initial EK (= ½ × 4.0 × 122
) = 290 (288) (J) M1
final EK (= ½ × 4.0 × 6.02
+ ½ × 12 × 3.52
) = 150 (145.5) (J) M1
(initial EK > final EK) so inelastic [both M1 marks required to award this mark] A1
4(a) displacement of particles/vibration(s)/oscillation(s) is parallel to/along the direction of energy/propagation B1
4(b) period = 1 / 800 (= 1.25 × 10–3
s) C1
time-base setting = 1.25 × 10–3
/ 2.5 C1
= 5.0 × 10–4
s cm–1
A1
4(c)(i) I ∝ A2
C1
(IX /IY =) [rY / rX] 2
= [AX / AY]2
C1
ratio AY / AX = 120 / 30
= 4.0
A1

PUBLISHED
October/November
2017
4(c)(ii) 1. v = f λ C1
minimum λ = 330 / (800 + 16) = 0.40 m A1
2. fo / fs = v / (v – vs)
816 / 800 = 330 / (330 – vs)
C1
vs = 6.5 m s–1
A1
5(a) force per unit positive charge B1
5(b)(i) s = ½at 2
C1
a = (2 × 0.045) / (1.5 × 10–7
)2
= 4(.0) × 1012
m s–2
A1
5(b)(ii) F = 1.67 × 10–27
× 4.0 × 1012
= 6.7 (6.68) × 10–15
N A1
5(b)(iii) 1. E = F / Q C1
= 6.68 × 10–15
/ 1.6 × 10–19
= 4.2 (4.18) × 104
NC–1
A1
2. E = V / d C1
V = 4.18 × 104
× 0.045
= 1.9 × 103
V
A1

PUBLISHED
October/November
2017
5(c) a = Eq / m
or
F = ma and F = Eq
C1
ratio =
− −
− −
× × × ×
× × × ×
19 27
19 27
(2 1.6 10 ) (1.67 10 )
(1.6 10 ) (4 1.66 10 )
or
×
×
2 1
1 4
= 0.50
A1
6(a)(i) P = VI C1
I = 30 / 120
= 0.25 A
A1
6(a)(ii) Q = 0.25 × 3.0 × 3600 (= 2700) C1
number = (0.25 × 3.0 × 3600) / 1.60 × 10–19
= 1.7 × 1022
A1
6(b) R = V / I or R = P / I 2
or R = V 2
/ P C1
= 120 / 0.25 or = 30 / 0.252
or = 1202
/ 30 = 480 Ω A1

PUBLISHED
October/November
2017
6(c) R = ρl / A C1
A = (6.1 × 10–7
× 580 × 10–3
) / 480 (= 7.37 × 10–10
) C1
d = [(4 × 7.37 × 10–10
) / π]1/2
= 3.1 × 10–5
m
A1
6(d) temperature decreases and so resistance decreases B1
7(a) nucleons = 23 B1
neutrons = 11 B1
7(b) similarity:
same (rest) mass
or
equal (magnitude of) charge
B1
difference:
opposite (sign of) charge
or
one is matter and one is antimatter
or
one is an electron and one is an antielectron
B1

PHYSICS 9702/31
Paper 3 Advanced Practical Skills 1 October/November 2017
MARK SCHEME
Maximum Mark: 40
Published
Teachers.
Cambridge IGCSE®
components.

PUBLISHED
October/November
2017
1(b) Value of θ in range 80–100° with unit. 1
1(c) Value of T in range 0.80–2.00s with unit. 1
1(d) Six sets of readings of θ (different values) and time showing the correct trend (T increases as θ decreases) and without help
from the Supervisor scores 5 marks, five sets scores 4 marks etc.
5
Range: θ ⩾ 120° and θ ⩽ 60°. 1
Column headings:
Each column heading must contain a quantity and a unit where appropriate.
The presentation of the quantity and the unit must conform to accepted scientific convention e.g. T2
/ s2
and θ / °. No unit for
cos (θ / 2).
1
Consistency:
All raw values of time must be given to the nearest 0.1s or all to the nearest 0.01s.
1
Significant figures:
All values of T2
must be given to the same number of s.f. as (or one more than) the number of s.f. in raw values of time.
If raw times recorded to nearest 0.01 s, allow number of s.f. of T2
to be one less than the number of s.f. of the raw times.
1
Values of cos (θ / 2) calculated correctly. 1

PUBLISHED
October/November
2017
1(e)(i) Axes:
Sensible scales must be used, no awkward scales (e.g. 3:10 or fractions).
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.
1
Plotting of points:
All observations 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.
1
Quality:
All points in the table (at least 5) must be plotted on the grid for this mark to be awarded.
It must be possible to draw a straight line that is within 0.05 on the cos (θ / 2) axis (x-axis) of all plotted points.
1
1(e)(ii) Line of best fit:
Judge by balance of all points on the grid about the candidate’s line (at least 5). 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. There must be at least five
points left after the anomalous point is disregarded.
Lines must not be kinked or thicker than half a small square.
1
1(e)(iii) Gradient:
The hypotenuse of the triangle used should be greater than half the length of the drawn line.
Both read-offs must be accurate to half a small square in both the x and y directions.
The method of calculation must be correct.
1
y-intercept:
Check correct read-off from a point on the line and substituted into y = mx + c.
Read-off must be accurate to half a small square in both x and y directions.
or
Intercept read directly from the graph, with read-off at x = 0, accurate to half a small square in the y direction.
1

PUBLISHED
October/November
2017
1(f) Value of P = candidate’s gradient and value of Q =candidate’s intercept.
The values must not be fractions.
1
Units for P and Q correct (s2
). 1
2(a) Value of t in the range 2–9 mm, t to the nearest 0.01cm or 0.001cm. 1
2(b)(i) Value of d to the nearest 0.1cm or better. 1
2(c)(ii) Value of h with unit. 1
2(c)(iii) Correct calculation of V with consistent unit. 1
2(c)(iv) Justification for s.f. in V linked to s.f. in (d – 2t) and h. Allow d, t and h or allow d and h. 1
2(e)(v) Value of y with evidence of repeats. 1
2(f) Percentage uncertainty in y based on absolute uncertainty of 2–8 mm.
If repeated readings have been taken, then the uncertainty can be half the range (but not zero) if the working is clearly shown.
Correct method of calculation to obtain percentage uncertainty.
1
2(g) Second value of h. 1
Second value of y. 1
Quality: second value of y less than first value of y. 1

PUBLISHED
October/November
2017
2(h)(i) Two values of k calculated correctly. 1
2(h)(ii) Valid comment consistent with calculated values of k, testing against a criterion stated by the candidate. 1
2(i)(i) A Two readings/too few readings/only two readings not enough to draw a (valid) conclusion.
B Difficult to measure t with reason e.g. screw thread in way, curved surface, thickness not the same throughout the glass.
C Inaccurate V with reason e.g. non-cylindrical shape of jar/equation gives an approximate value.
D Difficult to judge correct position of nails.
E Difficult to measure y with reason e.g. holding the nail and ruler in position.
1 mark for each point up to a maximum of 4.
4
2(i)(ii) A Take more readings (for different volumes) and plot a graph/take more values of k and compare.
B Improved method of measuring t directly e.g. travelling microscope.
C Improved method of measuring volume e.g. fill with water and use a measuring cylinder/measure circumference with
string to calculate diameter to put into equation for volume.
D Use optical pins/thinner nails.
E Have scale on side of jar/place both nails on lab jacks/use marker pen instead of nails/clamp ruler/use a marker to mark
position of nail.
4

PHYSICS 9702/33
MARK SCHEME
Maximum Mark: 40
Published
Teachers.
Cambridge IGCSE®
components.

PUBLISHED
October/November
2017
1(b)(ii) Value of a with unit and in the range 5.00–60.00s. 1
1(b)(iv) Evidence of repeated readings. 1
1(c) Six sets of readings of m (different values), a and b showing the correct trend (as m increases, a and b also increase) and
without help from the Supervisor scores 5 marks, five sets scores 4 marks etc.
5
Range:
Values of m must include 10 g and 70 g.
1
Column headings:
The presentation of the quantity and unit must conform to accepted scientific convention e.g. a2
/ b / s.
1
Consistency:
All raw values of time must be given to 0.1s or all to 0.01s.
1
All values of a2
/b must be given to the same number of s.f. as (or one more than) the number of s.f. in raw values of time.
If raw times recorded to nearest 0.01s, allow number of significant figures of a2
/ b to be one less than the number of
significant figures of the raw times.
1
Values of a2
/b calculated correctly. 1

PUBLISHED
October/November
2017
1(d)(i) Axes:
1
Plotting of points:
1
Quality:
All points in the table must be plotted on the grid for this mark to be awarded.
It must be possible to draw a straight line that is within 10 g on the mass axis (x-axis) of all plotted points.
1
1(d)(ii) Line of best fit:
Judge by balance of all points on the grid about the candidate’s line (at least 5). There must be an even distribution of points
Allow one anomalous point only if clearly indicated (i.e. circled or labelled) by the candidate. There must be at least five
points left after the anomalous point is disregarded.
1

PUBLISHED
October/November
2017
1(d)(iii) Gradient:
The hypotenuse of the triangle used should be greater than half the length of the drawn line.
1
y-intercept:
Correct read-off from a point on the line substituted into y = mx + c.
or
1
1(e) Value of P = candidate’s gradient and value of Q =candidate’s intercept.
1
Unit for P correct (sg–1
or skg–1
) and unit for Q correct (s). 1

PUBLISHED
October/November
2017
2(a)(ii) Value of c in the range 95.0–100.0 cm. 1
2(c)(ii) Value of x in the range 75.0–85.0 cm. 1
2(c)(iv) Value(s) of raw z to the nearest mm. 1
2(d) Percentage uncertainty in z based on absolute uncertainty of 2–5 mm.
1
2(e)(i) Correct calculation of (x – c / 2). 1
2(e)(ii) Correct calculation of (z – y) / m and consistent unit e.g. cm kg–1
. 1
2(f) Justification for s.f. in (z – y) / m linked to s.f. in z, y and m or (z – y) and m. 1
2(g)(ii) Second value of x. 1
Second value of z. 1
Quality: second value of z greater than first value of z (provided m in (g) > (c)). 1

PUBLISHED
October/November
2017
B Difficult to read x/c on rule owing to thickness of string.
C Difficult to measure y/z/spring with reason e.g. parallax, difficult to judge where end of coiled section is/easy to
knock/difficult to hold ruler still.
D Difficult to judge/adjust rule to be parallel to bench/horizontal (not ‘rule is not parallel to bench’).
E Large percentage uncertainty in (z – y).
4
2(i)(ii) A Take more readings (for different added masses) and plot a graph/take more values of k and compare.
B Use thread/wire/thin(ner) string or any other valid method.
C Use clamped ruler/use pointer(s) on rule or spring/use (vernier) calipers.
D Use a (spirit) level/use ruler and set square with detail.
E Larger difference between masses/larger x value/springs with smaller spring constant.
4

PHYSICS 9702/34
MARK SCHEME
Maximum Mark: 40
Published
Teachers.
Cambridge IGCSE®
components.

PUBLISHED
October/November
2017
1(b)(ii) Value of T with unit in range 1.00–2.00 s. 1
Evidence of repeated readings of nT. 1
1(c) Six sets of readings of m (different values) and T with correct trend and without help from the Supervisor scores 5 marks, five
sets scores 4 marks etc.
5
Range: mmin = 50 g and mmax ⩾ 350 g. 1
Column headings:
Each column heading must contain a quantity and a unit.
The presentation of quantity and unit must conform to accepted scientific convention e.g. T2
/ s2
.
1
Consistency:
All values of time must be given to the nearest 0.1s or all values to the nearest 0.01s.
1
Significant figures of every value of T 2
must be the same as, or one greater than, the s.f. of the raw times as recorded in
table. If raw times recorded to nearest 0.01s, allow number of significant figures of T2
to be one less than the number of
significant figures of the raw times.
1
Values of T2
calculated correctly. 1

PUBLISHED
October/November
2017
1(d)(i) Axes:
1
Plotting of points:
Points must be accurate to within half a small square in both x and y directions.
1
Quality:
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 ±25 g from a straight line in the m direction (x-axis).
1
Judge by balance of all points on the grid (at least 5) about the candidate’s line. There must be an even distribution of points
Allow one anomalous point only if clearly indicated (i.e. circled or labelled) by the candidate.
1
1(d)(iii) Gradient:
The hypotenuse of the triangle used must be greater than half the length of the drawn line.
1
y-intercept:
Correct read-off from a point on the line substituted into y = mx + c or an equivalent expression.
or
Intercept read directly from the graph, with read-off accurate to half a small square in the y direction.
1
1(e) Value of a = candidate’s gradient and value of b = candidate’s intercept.
1
Correct units for a (e.g. s2
g–1
) and b (s2
). 1

PUBLISHED
October/November
2017
2(b)(iii) Value of u with unit in range 30.0–34.0 cm. 1
2(b)(v) (Raw) value(s) of v to nearest 0.1 cm. 1
2(c) Absolute uncertainty in v in range 0.2–0.8 cm and correct method of calculation to obtain percentage uncertainty.
1
2(d) Correct calculation of f. 1
2(e)(iii) Value of vw. 1
Quality: vw > v. 1
2(f) Justification for s.f. in fw linked to s.f. in u and vw. 1
2(g)(ii) Second value of u in range 20.0–24.0 cm. 1
Second value of v. 1
Quality: second value of vw > first value of vw. 1

PUBLISHED
October/November
2017
B Difficult to judge/obtain sharp image/hard to focus.
C Difficult to keep screen steady/vertical (to measure v) or difficult to hold screen and measure distance (at the same time).
D Difficult to measure u (or v) with reason e.g. parallax error/judging front of torch/judging centre of lens.
E LEDs not at the front of torch/u should be measured to the LEDs.
F Difficult to align torch, lens and screen or torch and lens at different heights or lens not vertical.
4
2(i)(ii) A Take more readings and plot a graph/calculate more k values and compare.
B Use dark(ened) room/turn off lights or use improved ‘object’ e.g. cross-hairs/filament lamp/smaller LEDs.
C Mount screen in holder/clamp screen or clamp ruler/fix ruler to bench.
D Make alignment mark on container or use set squares with explanation of use.
E Use LEDs outside the torch/remove glass from torch.
F Use optical bench or draw line/scale/grid on bench or use lens holder to keep lens vertical.
4

PHYSICS 9702/35
MARK SCHEME
Maximum Mark: 40
Published
Teachers.
Cambridge IGCSE®
components.

PUBLISHED
October/November
2017
1(a)(iii) Value of raw V with unit in the range 0.50–2.00 V and to nearest 0.01 V. 1
1(b)(iv) Values of p and q with consistent unit, q > p and both values < 1m. 1
1(c) Six sets of readings of R (different values), p and q with correct trend (as R increases, q decreases) and without help from
Supervisor scores 5 marks, five sets scores 4 marks etc.
5
Range:
Values of R include 10Ω and 33Ω.
1
Column headings:
The presentation of quantity and unit must conform to accepted scientific convention e.g. q/cm, q / R (cm / Ω),
q/ R (cm Ω–1
). No unit given for q / p.
1
Consistency:
All values of p and q must be given to the nearest mm.
1
All values of q / R must be given to 2 or 3 s.f.
1
Calculation: Values of q/ p are correct. 1

PUBLISHED
October/November
2017
1(d)(i) Axes:
1
Plotting of points:
1
Quality:
All points in the table must be plotted on the grid for this mark to be awarded.
It must be possible to draw a straight line that is within ±0.10 on the q/ p axis (normally y-axis) of all plotted points.
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. There must be at least five points
left after the anomalous point is disregarded.
1

PUBLISHED
October/November
2017
1(d)(iii) Gradient:
Gradient sign on answer line matches graph drawn.
Method of calculation must be correct.
1
y-intercept:
Correct read-off from a point on the line substituted correctly into y = mx + c or an equivalent expression.
Read-off accurate to half a small square in both x and y directions.
or
1
1(e) Value of a = candidate’s gradient and value of b = candidate’s intercept.
1
Unit for a correct (e.g. Ω m–1
or Ω cm–1
or Ω mm–1
or Ω / m etc.) and b stated without a unit. 1

PUBLISHED
October/November
2017
2(a) Value of L with unit and L < 50.0 cm. 1
2(b)(ii) All value(s) of raw x to nearest mm with unit. 1
Repeat values of x. 1
2(b)(iii) Percentage uncertainty in x based on absolute uncertainty in x in range 2–8 mm.
1
2(b)(iv) Correct calculation of G where x and L have the same unit. 1
2(b)(v) Justification for s.f. in G linked to s.f. in L and x, or L and (2x – L). 1
2(c)(ii) Value of T with unit in range 0.8–2.0 s. 1
2(d) Second values of L and x. 1
Second value of T. 1
Quality: second value of T < first value of T. 1
2(e)(i) Two values of k calculated correctly. 1
2(e)(ii) Valid comment consistent with calculated values of k, testing against a criterion stated by the candidate. 1

PUBLISHED
October/November
2017
2(f)(i) A Two readings are not enough to draw a conclusion.
B Difficult to measure x with a reason e.g. holding ruler and pulling down on loop at the same time/holding string and
rule/measuring at an angle/holding rule steady/stands move or tilt when loop pulled/judging end points.
C Difficulty linked to the length of the pendulum e.g. knot slips/tying a knot/measuring to the centre of the bob.
D Difficult to measure L with a reason e.g. finger gets in the way/thickness of string.
E Difficulty linked to oscillation e.g. strings on rods move/stands move as the pendulum oscillates or difficult to judge the
end of an oscillation.
4
2(f)(ii) A Take more readings and plot a graph/take more readings and compare k values.
B Improved method to measure x e.g. clamp rule/clamp stands/hang mass on loop.
C Use glue (not knots)/mark the string/attach hook or loop to pendulum/measure string and add radius to length.
D Improved method to measure L e.g. use pins/nails/tape to table.
E Clamp stands/use tape to fix strings to rods/idea of a groove or film/video camera with timer/frame by frame/marker at
centre of oscillation. (Do not award ‘clamp stands’ twice for both B and E.)
4

PHYSICS 9702/36
MARK SCHEME
Maximum Mark: 40
Published
Teachers.
Cambridge IGCSE®
components.

PUBLISHED
October/November
2017
1(c)(ii) T with unit in range 0.1–1.0 s. 1
Evidence of repeated readings of nT where n = 5 or more. 1
1(e) Six sets of readings of x and T showing the correct trend and without help from the Supervisor scores 5 marks, five sets
scores 4 marks etc.
5
Range: 0 ⩽ (x – L/2)min ⩽ 1.0cm. 1
Column headings:
The presentation of quantity and unit must conform to accepted scientific convention. e.g. 1/T2
/ s–2
.
1
Consistency:
All values of x must be given to the nearest mm.
1
Significant figures for every value of 1 /T2
same as, or one greater than, the s.f. of raw time as recorded in table. If raw times
recorded to nearest 0.01s, allow number of significant figures of 1 /T2
to be one less than the number of significant figures of
the raw times.
1
Values of (x – L / 2) calculated correctly. 1

PUBLISHED
October/November
2017
1(f)(i) Axes:
Scales must be chosen so that the plotted points occupy at least half the graph grid in both x and y directions
1
Plotting of points:
Points must be accurate to within half a small square in both x and y directions.
1
Quality:
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 ± 0.25 cm from a straight line in the (x – L / 2) direction.
1
1(f)(ii) Line of best fit:
Judged by balance of all points on the grid (at least 5) about the candidate’s line. There must be an even distribution of points
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.
1
1(f)(iii) Gradient:
1
y-intercept:
Correct read-off from a point on the line substituted into y = mx + c.
or
1
1(g) Value of a = candidate’s gradient and value of b = candidate’s intercept. The values must not be fractions. 1
Unit for a correct (e.g. cm–1
s–2
) and unit for b correct (e.g. s–2
). 1

PUBLISHED
October/November
2017
2(a)(iv) Value of D with unit to nearest mm and in range 10.0–150.0 cm. 1
2(b)(ii) Value of h1 with consistent unit and in range 6.0–10.0 cm. 1
2(c) Absolute uncertainty in h1 of 1 mm and correct method of calculation to obtain percentage uncertainty.
1
2(d)(ii) Correct calculation of d. 1
2(f)(ii) Values of t and w to nearest mm, with unit. 1
2(f)(iii) Correct calculation of E. 1
2(g) Second values of h1 and h2. 1
Quality: Second value of d < first value of d. 1
Second values of h3 and h4. 1
2(h)(ii) Justification for s.f. in k linked to s.f. in d and p, or linked to s.f. in h1, h2, h3 and h4. 1
2(h)(iii) Valid comment relating to the calculated values of k, testing against a criterion stated by the candidate. 1

PUBLISHED
October/November
2017
B Difficult to pile masses in centre, with reason.
C Blocks move/slip/tilt.
D d (or p or t) small so large uncertainty/ large % uncertainty in d (or p or t).
E Permanent deformation of strip.
F Difficult to mark ends of D due to curvature of masses.
4
2(i)(ii) A Take more readings and plot a graph/calculate more k values and compare.
B Improved method for point load, e.g. suspend masses below strip/named method for sticking masses together.
C Named means of stabilising blocks, e.g. stick to bench/stick to strip/stops on bench/clamp blocks to bench.
D Use calipers or travelling microscope or use micrometer for t value or measure stack of MDF pieces.
E Use 8 masses before 10 or turn strip over.
F Use set square with detail/measure length of 10 masses then mark strip.
4

PHYSICS 9702/41
Paper 4 A Level Structured Questions October/November 2017
MARK SCHEME
Maximum Mark: 100
Published
Teachers.
Cambridge IGCSE®
components.

PUBLISHED
October/November
2017
1(a)(i) direction or rate of transfer of (thermal) energy
or
(if different,) not in thermal equilibrium/energy is transferred
B1
1(a)(ii) uses a property (of a substance) that changes with temperature B1
1(b) • temperature scale assumes linear change of property with temperature
• physical properties may not vary linearly with temperature
• agrees only at fixed points
Any 2 points.
B2
1(c)(i) Pt = mc(∆)θ C1
95 × 6 × 60 = 0.670 × 910 × ∆θ M1
∆θ = 56°C so final temperature = 56 + 24 = 80°C A1
or
95 × 6 × 60 = 0.67 × 910 × (θ – 24) (M1)
so final temperature or θ = 80 °C (A1)

PUBLISHED
October/November
2017
1(c)(ii) 1. sketch: straight line from (0,24) to (6,80) B1
2. temperature drop due to energy loss = (80 – 64) = 16°C C1
energy loss = 0.670 × 910 × (80 – 64) = 9800J A1
or
energy to raise temperature to 64°C = 0.670 × 910 × (64 – 24) (C1)
= 24400J
loss = (95 × 6 × 60) – 24400 = 9800J
(A1)
2(a) (angular frequency =) 2π × frequency or 2π / period B1
2(b)(i) 1. displacement = 2.0 cm A1
2. amplitude = 1.5 cm A1
2(b)(ii) reference to displacement of oscillations or displacement from equilibrium position or displacement from 2.0 cm B1
straight line indicates acceleration ∝ displacement B1
negative gradient shows acceleration and displacement are in opposite directions B1

PUBLISHED
October/November
2017
2(b)(iii) ω2
= (–)1/ gradient or ω2
= (–)∆a/ ∆s or a = (–)ω2
x and correct value of x C1
= e.g. (1.8 / 0.03) or (0.9 / 0.015) or (1.2 / 0.02) etc. or 0.9 = ω2
× 0.015
= 60
C1
f = √60 /2π
= 1.2Hz
A1
3(a) force per unit mass B1
3(b) changes in height much less than radius of Earth M1
so (radial) field lines are almost parallel
or
g = GM/R2
≈ GM/(R + h)2
A1

PUBLISHED
October/November
2017
3(c) gravitational force provides/is centripetal force B1
GMm/r2
= mv2
/ r C1
v = (2π × 1.5 × 1011
)/(3600 × 24 × 365) = 2.99 × 104
(ms–1
) C1
6.67 × 10–11
M = 1.5 × 1011
× (2.99 × 104
)2
C1
M = 2.0 × 1030
kg A1
or
GMm/ r2
= mrω2
(C1)
ω = 2π/ (3600 × 24 × 365) = 1.99 × 10–7
(rads–1
) (C1)
6.67 × 10–11
M = (1.5 × 1011
)3
× (1.99 × 10–7
)2
(C1)
M = 2.0 × 1030
kg (A1)
or
T2
= 4π2
r3
/GM (C2)
M = 4π2
× (1.5 × 1011
)3
/ ({3600 × 24 × 365}2
× 6.67 × 10–11
) (C1)
= 2.0 × 1030
kg (A1)

PUBLISHED
October/November
2017
4(a) • acts as ‘return’ (conductor) for signal
• shielding from noise/crosstalk/interference
Two sensible suggestions, 1 mark each.
B2
4(b) • small bandwidth
• (there is) noise/interference/crosstalk
• large attenuation/energy loss
• reflections due to poor impedance matching
B2
4(c) attenuation = 190 × 14 × 10–3
(= 2.66 dB) C1
ratio/ dB = (–)10 lg(P2 / P1) C1
2.66 = –10 lg(POUT / PIN)
POUT / PIN = 0.54
C1
fractional loss = 1 – (POUT / PIN) = 1 – 0.54
= 0.46
A1
or
2.66 = 10 lg(PIN / POUT)
PIN / POUT = 1.85
(C1)
fractional loss = (PIN – POUT) / PIN = (1.85 – 1) / 1.85
= 0.46
(A1)

PUBLISHED
October/November
2017
5(a)(i) force proportional to product of charges and inversely proportional to square of separation A1
5(a)(ii) curve starting at (R, FC) B1
passing through (2R, 0.25FC) B1
5(b) graph: E = 0 when current constant (0 to t1, t2 to t3, t4 to t5) B1
stepped from t1 to t2 and t3 to t4 B1
(steps) in opposite directions B1
later one larger in magnitude B1
6(a)(i) 1 / T = 1 / (2C) + 1 / C C1
T = ⅔C or 0.67C A1
6(a)(ii) same charge on Q as on combination B1
so p.d. is 6.0 V B1
6(b) P: p.d. will decrease (from 3.0V) B1
to zero B1
Q: p.d. will increase (from 6.0 V) B1
to 9.0 V B1

PUBLISHED
October/November
2017
7(a)(i) gain of amplifier is very large B1
V+
is at earth (potential) B1
for amplifier not to saturate M1
difference between V–
and V+
must be very small or V–
must be equal to V+
A1
or
if V–
≠ V+
then feedback voltage (M1)
acts to reduce gap until V–
= V+
when stable (A1)
7(a)(ii) input impedance is infinite B1
(so) current in R1 = current in R2 B1
(VIN – 0) / R1 = (0 – VOUT) / R2 B1
(gain =) VOUT / VIN = – R2 / R1 B1
7(b) graph: correct inverted shape (straight diagonal line from (0,0) to a negative potential, then a horizontal line, then a straight
diagonal line back to the t-axis at the point where VIN = 0)
B1
horizontal line at correct potential of (–)9.0 V B1
both ends of horizontal line occur at correct times (coinciding with when VIN = 2.0V) B1

PUBLISHED
October/November
2017
8(a) DERQ and CFSP B1
8(b)(i) force (on charge) due to magnetic field = force due to electric field
or
Bqv = Eq
or
v = E / B
B1
E = VH /d B1
VH = Bvd B1
8(b)(ii) use of I = nAqv and A = dt M1
algebra clear leading to VH = BI / ntq A1
8(c) (in metal,) n is very large M1
(therefore) VH is small A1

PUBLISHED
October/November
2017
9(a) image of one slice/section (B1)
images (of one slice) taken from different angles (M1)
to give 2D image (of one slice) (A1)
(repeated for) many slices (M1)
to build up 3D image (of whole body/structure) (A1)
Max. 4 marks total 4
9(b) evidence of subtraction of background (–26) C1
evidence of division by three C1
7 11
6 2
A1

PUBLISHED
October/November
2017
10(a) heating depends on current2
/I2
B1
and current2
/I2
is always positive B1
or
a.c. changes direction (every half cycle) (B1)
but heating effect is independent of current direction (B1)
or
voltage and current are always in phase in a resistor (B1)
so V × I is always positive (B1)
or
sketch graph drawn showing power against time (B1)
comment that power is always positive (B1)
10(b)(i) for same power (transmission, higher voltage) → lower current B1
lower current → less power loss in (transmission) cables B1
10(b)(ii) • voltage can be (easily) stepped up/down
• transformers only work with a.c.
• generators produce a.c.
• easier to rectify than invert
B2

PUBLISHED
October/November
2017
11(a) packet/quantum of energy of electromagnetic/EM radiation B1
11(b)(i) E = hf
1.1 × 106
× 1.60 × 10–19
= 6.63 × 10–34
× f
C1
f = 2.7 × 1020
(2.65 × 1020
) Hz A1
11(b)(ii) p = h / λ = hf / c
= (6.63 × 10–34
× 2.65 × 1020
) / (3.00 × 108
)
or
p = E / c
= (1.1 × 1.60 × 10–13
) / (3.00 × 108
)
C1
p = 5.9 × 10–22
(5.87 × 10–22
)Ns A1
11(c) 123 × 1.66 × 10–27
× v = 5.87 × 10–22
C1
v = 2.9 × 103
ms–1
A1

PUBLISHED
October/November
2017
12(a) • emission from radioactive daughter products
• self-absorption in source
• absorption in air before reaching detector
• detector not sensitive to all radiations
• window of detector may absorb some radiation
• dead-time of counter
• background radiation
Any two points.
B2
12(b)(i) curve is not smooth
or
curve fluctuates/curve is jagged
B1
12(b)(ii) clear evidence of allowance for background B1
half-life determined at least twice B1
half-life = 1.5 hours
(1 mark if in range 1.7–2.0; 2 marks if in range 1.4–1.6)
A2
12(c) 1. half-life: no change M1
because decay is spontaneous/independent of environment A1
2. count rate (likely to be or could be) different/is random/cannot be predicted B1

PHYSICS 9702/42
MARK SCHEME
Maximum Mark: 100
Published
Teachers.
Cambridge IGCSE®
components.

PUBLISHED
October/November
2017
1(a) force proportional to product of masses and inversely proportional to square of separation B1
idea of force between point masses B1
1(b) mass of Jupiter (M) = (4/3)πR3
ρ B1
ω = 2π/ T
or
v = 2πnR/ T
B1
(m)ω2
x = GM(m) / x2
or
(m)v2
/ x = GM(m) / x2
M1
substitution and correct algebra leading to ρT2
= 3πn3
/ G A1
1(c)(i) n = (4.32 × 105
) / (7.15 × 104
)
or
n = 6.04
C1
ρ × (42.5 × 3600)2
= (3π × 6.043
) /(6.67 × 10–11
) C1
ρ = 1.33 × 103
kgm–3
A1
1(c)(ii) Jupiter likely to be a gas/liquid (at high pressure) [allow other sensible suggestions] B1

PUBLISHED
October/November
2017
2(a) (thermal) energy per (unit) mass (to cause change of state) B1
(energy required to cause/released in) change of state at constant temperature B1
2(b)(i) 1. (work done on/against) the atmosphere B1
2. water as it turns from liquid to vapour M1
as potential energy of molecules increases A1
or
surroundings as its temperature rises (M1)
as energy is lost/transferred to surroundings (A1)
2(b)(ii) VI – h = M / t × L (where h = power loss)
or
L = (VIt – Q) /M (where Q = energy loss)
C1
(14.2 × 6.4) – (11.5 × 5.2) = (9.1 – 5.0) × L/ 300
or
L = [(14.2 × 6.4) – (11.5 × 5.2)] × 300 /(9.1 – 5.0)
C1
L = 2300 Jg–1
A1

PUBLISHED
October/November
2017
3(a)(i) angle (subtended) where arc (length) is equal to radius M1
(angle subtended) at the centre of a circle A1
3(a)(ii) angular frequency = 2π × frequency or 2π / period B1
3(b)(i) c / ML3
is a constant so acceleration is proportional to displacement B1
minus sign shows that acceleration and displacement are in opposite directions B1
3(b)(ii) c/ ML3
= (2πf )2
C1
c = 4π2
× 3.22
× 0.24 × 0.653
C1
= 27 kgm3
s–2
A1
4(a) quartz/piezo-electric and crystal/transducer B1
p.d. across crystal causes it to distort B1
applying alternating p.d. causes oscillations/vibrations B1
when applied frequency is natural frequency, crystal resonates B1
natural frequency of crystal is in ultrasound range B1
4(b) small(er) structures can be resolved/observed/identified B1

PUBLISHED
October/November
2017
5(a) (0.2 ms) 8.0 (mV) 1000 B1
(0.8 ms) 5.8 (mV) 0101 B1
5(b) series of steps B1
all (step) changes are at 0.2 ms intervals B1
steps with correct levels at correct times
(1 mark if five levels correct; 2 marks if all levels correct)
level 0 8 10 15 5 8
time/ms 0–0.2 0.2–0.4 0.4–0.6 0.6–0.8 0.8–1.0 1.0–1.2
B2
5(c) smaller step heights (possible) B1
smaller changes (in input signal) can be seen/reproduced/represented
or
(allows) more accurate reproduction (of the input signal)
B1

PUBLISHED
October/November
2017
6(a) electric field lines are radial/normal to surface (of sphere) B1
electric field lines appear to originate from centre (of sphere) B1
6(b)(i) tangent drawn at x = 6.0cm and gradient calculation attempted C1
E = 9.0 × 104
N C–1
(1 mark if in range ±1.2; 2 marks if in range ±0.6)
A2
or
correct pair of values of V and x read from curved part of graph and substituted into V = q/ 4πε0x (C1)
to give q = 3.6 × 10–8
C (C1)
(then E = q/ 4πε0x2
and x = 6cm gives) E = 9.0 × 104
N C–1
(A1)
or
(E = q/ 4πε0x2
and V = q / 4πε0x and so) E = V/ x (C1)
giving E = 5.4 × 103
/0.060 (C1)
= 9.0 × 104
NC–1
(A1)
6(b)(ii) (R =) 2.5cm B1
potential inside a conductor is constant
or
field strength inside a conductor zero (so gradient is zero)
B1

PUBLISHED
October/November
2017
7(a)(i) (part of) the output is combined with the input M1
reference to potential/voltage/signal A1
7(a)(ii) • increased (operating) stability
• increased bandwidth/range of frequencies over which gain is constant
• less distortion (of output)
Any 2 points.
B2
7(b)(i) 1. gain = 3.6/ (48 × 10–3
) C1
= 75 A1
2. gain = 1 + RF / R
75 = 1 + (92.5 × 103
) /R
C1
R = 1300 Ω A1
7(b)(ii) for 68 mV, gain × VIN = 5.1(V)
or
output voltage would be greater than the supply voltage
M1
amplifier would saturate (at 5.0 V)
or
output voltage = 5.0 (V)
A1

PUBLISHED
October/November
2017
8(a)(i) DERQ and CFSP B1
8(a)(ii) charge carriers moving normal to (magnetic) field B1
charge carriers experience a force normal to I (and B) B1
charge build-up sets up electric field across the slice
or
build-up of charges results in a p.d. across the slice
B1
charge stops building up/VH becomes constant when FB = FE B1
8(b) VH inversely proportional to n/number density of charge carriers B1
number density of charge carriers (n) lower in semiconductors so VH larger for semiconductor slice B1
or
VH proportional to v/drift velocity (B1)
(for same current) drift velocity (v) higher in semiconductors so VH larger for semiconductor slice (B1)

PUBLISHED
October/November
2017
9(a) region (of space) B1
where an object/particle experiences a force B1
9(b) electric and magnetic fields normal to each other B1
velocity of particle normal to both fields B1
forces (on particle) due to fields are in opposite directions B1
forces are equal for particles with a particular speed/for a selected speed/for speed given by v = E(q)/ B(q) B1
9(c)(i) path labelled Q shown undeviated B1
9(c)(ii) reasonable curve in field and no ‘kink’ on entering, labelled V B1
deviated ‘upwards’ B1
10(a) λ0 marked and graph line passing through EMAX = 0 at λ = λ0 B1
graph line with λ always < λ0 B1
negative gradient with correct concave curvature B1
10(b) curve with negative gradient and correct concave curvature M1
not touching either axis A1

PUBLISHED
October/November
2017
11(a)(i) circles drawn only around the top left and bottom right diodes B1
11(a)(ii) B shown as (+)ve and A shown as (–)ve B1
11(b)(i) Vr.m.s. (= 5.6 / √2) = 4.0V A1
11(b)(ii) 380 = 2πf or f = 60.5Hz C1
number (= 2f ) = 120 A1
11(c)(i) peak values (all) unchanged B1
(all) minima shown at 4.0V B1
three lines from near peak showing concave curves after leaving dotted line
not ‘kinked’ and not cutting the peak
reaching candidate’s minimum at the point where the decay meets the next dotted line
B1
three lines drawn along the dotted lines showing rise in voltage from minima back to peak values B1
11(c)(ii) mean p.d. is higher
or
r.m.s. p.d. is higher
or
capacitor supplies energy to resistor
M1
so (mean) power increases A1

PUBLISHED
October/November
2017
12(a)(i) nucleus emits particles/EM radiation/ionising radiation B1
emission/release from unstable nucleus
or
emission from nucleus is random and/or spontaneous
B1
12(a)(ii) probability of decay (of a nucleus)
or
fraction of (number of undecayed) nuclei that will decay
M1
per unit time A1
12(b) energy is shared with another particle B1
mention of antineutrino B1
12(c)(i) number = [(1.2 × 10–9
) / 131] × 6.02 × 1023
or
number = (1.2 × 10–3
× 10–9
) / (131 × 1.66 × 10–27
)
(= 5.51 × 1012
)
C1
A = λN C1
= [0.086 / (24 × 3600)] × 5.51 × 1012
= 5.5 × 106
Bq
A1
12(c)(ii) 1/ 50 = exp(–0.086t)
or
1/50 = 0.5n
C1
t = 45 days A1

PHYSICS 9702/43
MARK SCHEME
Maximum Mark: 100
Published
Teachers.
Cambridge IGCSE®
components.

PUBLISHED
October/November
2017
1(a)(i) direction or rate of transfer of (thermal) energy
or
(if different,) not in thermal equilibrium/energy is transferred
B1
1(a)(ii) uses a property (of a substance) that changes with temperature B1
1(b) • temperature scale assumes linear change of property with temperature
• physical properties may not vary linearly with temperature
• agrees only at fixed points
Any 2 points.
B2
1(c)(i) Pt = mc(∆)θ C1
95 × 6 × 60 = 0.670 × 910 × ∆θ M1
∆θ = 56°C so final temperature = 56 + 24 = 80°C A1
or
95 × 6 × 60 = 0.67 × 910 × (θ – 24) (M1)
so final temperature or θ = 80 °C (A1)

PUBLISHED
October/November
2017
1(c)(ii) 1. sketch: straight line from (0,24) to (6,80) B1
2. temperature drop due to energy loss = (80 – 64) = 16°C C1
energy loss = 0.670 × 910 × (80 – 64) = 9800J A1
or
energy to raise temperature to 64°C = 0.670 × 910 × (64 – 24) (C1)
= 24400J
loss = (95 × 6 × 60) – 24400 = 9800J
(A1)
2(a) (angular frequency =) 2π × frequency or 2π / period B1
2(b)(i) 1. displacement = 2.0 cm A1
2. amplitude = 1.5 cm A1
2(b)(ii) reference to displacement of oscillations or displacement from equilibrium position or displacement from 2.0 cm B1
straight line indicates acceleration ∝ displacement B1
negative gradient shows acceleration and displacement are in opposite directions B1

PUBLISHED
October/November
2017
2(b)(iii) ω2
= (–)1/ gradient or ω2
= (–)∆a/ ∆s or a = (–)ω2
x and correct value of x C1
= e.g. (1.8 / 0.03) or (0.9 / 0.015) or (1.2 / 0.02) etc. or 0.9 = ω2
× 0.015
= 60
C1
f = √60 /2π
= 1.2Hz
A1
3(a) force per unit mass B1
3(b) changes in height much less than radius of Earth M1
so (radial) field lines are almost parallel
or
g = GM/R2
≈ GM/(R + h)2
A1

PUBLISHED
October/November
2017
3(c) gravitational force provides/is centripetal force B1
GMm/r2
= mv2
/ r C1
v = (2π × 1.5 × 1011
)/(3600 × 24 × 365) = 2.99 × 104
(ms–1
) C1
6.67 × 10–11
M = 1.5 × 1011
× (2.99 × 104
)2
C1
M = 2.0 × 1030
kg A1
or
GMm/ r2
= mrω2
(C1)
ω = 2π/ (3600 × 24 × 365) = 1.99 × 10–7
(rads–1
) (C1)
6.67 × 10–11
M = (1.5 × 1011
)3
× (1.99 × 10–7
)2
(C1)
M = 2.0 × 1030
kg (A1)
or
T2
= 4π2
r3
/GM (C2)
M = 4π2
× (1.5 × 1011
)3
/ ({3600 × 24 × 365}2
× 6.67 × 10–11
) (C1)
= 2.0 × 1030
kg (A1)

PUBLISHED
October/November
2017
4(a) • acts as ‘return’ (conductor) for signal
• shielding from noise/crosstalk/interference
B2
4(b) • small bandwidth
• (there is) noise/interference/crosstalk
• large attenuation/energy loss
• reflections due to poor impedance matching
B2
4(c) attenuation = 190 × 14 × 10–3
(= 2.66 dB) C1
ratio/ dB = (–)10 lg(P2 / P1) C1
2.66 = –10 lg(POUT / PIN)
POUT / PIN = 0.54
C1
fractional loss = 1 – (POUT / PIN) = 1 – 0.54
= 0.46
A1
or
2.66 = 10 lg(PIN / POUT)
PIN / POUT = 1.85
(C1)
fractional loss = (PIN – POUT) / PIN = (1.85 – 1) / 1.85
= 0.46
(A1)

PUBLISHED
October/November
2017
5(a)(i) force proportional to product of charges and inversely proportional to square of separation A1
5(a)(ii) curve starting at (R, FC) B1
5(b) graph: E = 0 when current constant (0 to t1, t2 to t3, t4 to t5) B1
stepped from t1 to t2 and t3 to t4 B1
(steps) in opposite directions B1
later one larger in magnitude B1
6(a)(i) 1 / T = 1 / (2C) + 1 / C C1
T = ⅔C or 0.67C A1
6(a)(ii) same charge on Q as on combination B1
so p.d. is 6.0 V B1
6(b) P: p.d. will decrease (from 3.0V) B1
to zero B1
Q: p.d. will increase (from 6.0 V) B1
to 9.0 V B1

PUBLISHED
October/November
2017
7(a)(i) gain of amplifier is very large B1
V+
is at earth (potential) B1
for amplifier not to saturate M1
difference between V–
and V+
must be very small or V–
must be equal to V+
A1
or
if V–
≠ V+
then feedback voltage (M1)
acts to reduce gap until V–
= V+
when stable (A1)
7(a)(ii) input impedance is infinite B1
(so) current in R1 = current in R2 B1
(VIN – 0) / R1 = (0 – VOUT) / R2 B1
(gain =) VOUT / VIN = – R2 / R1 B1
7(b) graph: correct inverted shape (straight diagonal line from (0,0) to a negative potential, then a horizontal line, then a straight
diagonal line back to the t-axis at the point where VIN = 0)
B1
horizontal line at correct potential of (–)9.0 V B1
both ends of horizontal line occur at correct times (coinciding with when VIN = 2.0V) B1

PUBLISHED
October/November
2017
8(a) DERQ and CFSP B1
8(b)(i) force (on charge) due to magnetic field = force due to electric field
or
Bqv = Eq
or
v = E / B
B1
E = VH /d B1
VH = Bvd B1
8(b)(ii) use of I = nAqv and A = dt M1
algebra clear leading to VH = BI / ntq A1
8(c) (in metal,) n is very large M1
(therefore) VH is small A1

PUBLISHED
October/November
2017
9(a) image of one slice/section (B1)
images (of one slice) taken from different angles (M1)
to give 2D image (of one slice) (A1)
(repeated for) many slices (M1)
to build up 3D image (of whole body/structure) (A1)
Max. 4 marks total 4
9(b) evidence of subtraction of background (–26) C1
evidence of division by three C1
7 11
6 2
A1

PUBLISHED
October/November
2017
10(a) heating depends on current2
/I2
B1
and current2
/I2
is always positive B1
or
a.c. changes direction (every half cycle) (B1)
but heating effect is independent of current direction (B1)
or
voltage and current are always in phase in a resistor (B1)
so V × I is always positive (B1)
or
sketch graph drawn showing power against time (B1)
comment that power is always positive (B1)
10(b)(i) for same power (transmission, higher voltage) → lower current B1
lower current → less power loss in (transmission) cables B1
10(b)(ii) • voltage can be (easily) stepped up/down
• transformers only work with a.c.
• generators produce a.c.
• easier to rectify than invert
B2

PUBLISHED
October/November
2017
11(a) packet/quantum of energy of electromagnetic/EM radiation B1
11(b)(i) E = hf
1.1 × 106
× 1.60 × 10–19
= 6.63 × 10–34
× f
C1
f = 2.7 × 1020
(2.65 × 1020
) Hz A1
11(b)(ii) p = h / λ = hf / c
= (6.63 × 10–34
× 2.65 × 1020
) / (3.00 × 108
)
or
p = E / c
= (1.1 × 1.60 × 10–13
) / (3.00 × 108
)
C1
p = 5.9 × 10–22
(5.87 × 10–22
)Ns A1
11(c) 123 × 1.66 × 10–27
× v = 5.87 × 10–22
C1
v = 2.9 × 103
ms–1
A1

PUBLISHED
October/November
2017
12(a) • emission from radioactive daughter products
• self-absorption in source
• absorption in air before reaching detector
• detector not sensitive to all radiations
• window of detector may absorb some radiation
• dead-time of counter
• background radiation
Any two points.
B2
12(b)(i) curve is not smooth
or
curve fluctuates/curve is jagged
B1
12(b)(ii) clear evidence of allowance for background B1
half-life determined at least twice B1
half-life = 1.5 hours
(1 mark if in range 1.7–2.0; 2 marks if in range 1.4–1.6)
A2
12(c) 1. half-life: no change M1
because decay is spontaneous/independent of environment A1
2. count rate (likely to be or could be) different/is random/cannot be predicted B1

PHYSICS 9702/51
Paper 5 Planning, Analysis and Evaluation October/November 2017
MARK SCHEME
Maximum Mark: 30
Published
Teachers.
Cambridge IGCSE®
components.

PUBLISHED
October/November
2017
1 Defining the problem
d is the independent variable and R is the dependent variable or vary d and measure R 1
keep intensity/power of light source constant 1
Methods of data collection
labelled diagram showing a light source fixed above container of water with the labelled LDR positioned in the beaker 1
correct circuit diagram to measure R, e.g. V and I methods or ohmmeter 1
method to determine R, e.g. =R
p.d. across LDR
current
or read off ohmmeter
1
method to determine d, e.g. use a ruler or drawn labelled vertical ruler adjacent to container with d indicated 1
Method of analysis
plots a graph of R against d2
1
relationship valid if a straight line produced passing through the origin 1
π
=K
4
gradient
1

PUBLISHED
October/November
2017
1 Additional detail including safety considerations Max. 6
D1 dark glasses to prevent damage to eyes due to light source or do not look directly at light source
or
do not touch hot lamp/use gloves to position hot lamp/heat-proof gloves to position lamp
D2 dark room or shielding LDR (so as to avoid light from other sources)
D3 use high intensity lamp or collimated beam or laser
D4 method described to check that current in light source is constant, e.g. use an ammeter and variable
resistor / variable power supply
D5 keep position of light source constant or distance between light source and LDR constant
D6 light source is placed close to water surface to increase intensity/reduce reflections
or
light source is placed further away to make it more directional
D7 use tall container to give a wide range of d or R or to reduce uncertainties
or
use a wide container to reduce reflections
D8 method to position ruler vertically to measure d described e.g. use a set square/spirit level
D9 use of horizontal fiducial mark from ruler to meniscus or middle of LDR, e.g. pin
or
d = reading on rule at surface – reading at top of LDR
D10 ensure that the electrical connections/wire to the LDR are waterproof

PUBLISHED
October/November
2017
2(a)
gradient = −
2mg
s
y-intercept = mg
1
2(b) (T1 –T2)/N
5.5 ± 0.2
4.6 ± 0.2
3.6 ± 0.2
2.8 ± 0.2
1.9 ± 0.2
1.3 ± 0.2
First mark for column heading and values of (T1 –T2)/N.
Second mark for all uncertainties = ±0.2.
2
2(c)(i) Six points plotted correctly.
Must be within half a small square. Diameter of points must be less than half a small square
1
Error bars in P plotted correctly.
All error bars to be plotted. Length of bar must be accurate to less than half a small square and symmetrical.
1

PUBLISHED
October/November
2017
2(c)(ii) Line of best fit drawn.
Must not be drawn from top point to bottom point.
If points are plotted correctly then upper end of line should pass between (0.125, 5.0) and (0.140, 5.0) and lower end of
line should pass between (0.360, 1.5) and (0.380, 1.5).
1
Worst acceptable line drawn correctly (steepest or shallowest possible line).
All error bars must be plotted.
1
2(c)(iii) Gradient determined with a triangle that is at least half the length of the drawn line.
Must be negative.
1
uncertainty = gradient of line of best fit – gradient of worst acceptable line
or
uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)
1
2(c)(iv) y-intercept determined from substitution into y = mx + c. 1
y-intercept determined using gradient of worst acceptable line.
uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line
or
uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)
No ECF from false origin method.
1

PUBLISHED
October/November
2017
2(d)(i) m determined using candidate’s y-intercept and correct units for m and s.
y y
m
g
-intercept -intercept
= =
9.81
1
s determined using candidate’s gradient and m and s given to 2 or 3 significant figures.
Correct substitution of numbers must be seen.
− − ×
= =
2mg y
s
2 -intercept
gradient gradient
1
2(d)(ii) percentage uncertainty in m = percentage uncertainty in y-intercept 1
percentage uncertainty in s = percentage uncertainty in gradient + percentage uncertainty in y-intercept
or
percentage uncertainty in s = percentage uncertainty in gradient + percentage uncertainty m
Maximum/minimum methods:
− × − ×
=
y g m
s
2 max -intercept 2 max
max
min gradient min gradient
or
− × − ×
=
y g m
s
2 min -intercept 2 min
min
max gradient max gradient
or
1

PHYSICS 9702/52
Paper 5 Planning, Analysis and Evalution October/November 2017
MARK SCHEME
Maximum Mark: 30
Published
Teachers.
Cambridge IGCSE®
components.

PUBLISHED
October/November
2017
x is the independent variable and V is the dependent variable or vary x and measure V 1
keep current (in the coil P) constant 1
labelled diagram showing both coils supported 1
two correct circuit diagrams for coil P and coil Q:
power supply connected to one coil and voltmeter/c.r.o. connected to other coil
1
method to determine x, e.g. use a ruler or drawn labelled horizontal ruler adjacent to coils with x indicated 1
method to measure x from centre of coil P to centre of coil Q, e.g. measure width of (each) coil and divide by 2 and add to
separation of coils
1
Method of analysis
plots a graph of ln V against x [or log V against x etc.] 1
relationship valid if a straight line produced 1
k = –gradient 1

PUBLISHED
October/November
2017
Additional detail including safety considerations Max. 6
D1 do not touch hot coil/use gloves to position hot coil/heat-proof gloves to position coil
D2 use large current/number of turns/iron core (to produce large magnetic field/induced e.m.f.)
D3 use high frequency (to produce larger induced e.m.f.)
D4 use an a.c. power supply or signal generator (connected to coil P)
D5 keep the number of turns (on each coil) constant/frequency constant
D6 method described to check that current is constant, e.g. use an ammeter and variable resistor/variable power
supply
D7 repeat measurements of x for different parts of the coil and average
D8 method to position ruler horizontally to measure x described e.g. use a spirit level or same height from bench at
both ends
D9 method to keep coils parallel/co-axial e.g. adjust coil Q until maximum reading or use set square to ensure that
coils are at right angles to the axis
D10 0ln lnV kx V= − +

PUBLISHED
October/November
2017
2(a)
gradient =
2 2
4 L f
g
µ 1
2(b) M / g
2
1
n
850 ± 85 (90) 0.1 or 0.11 or 0.111 or 0.1111
500 ± 50 0.06 or 0.063 or 0.0625
300 ± 30 0.04 or 0.040 or 0.0400
200 ± 20 0.03 or 0.028 or 0.0278
150 ± 15 (20) 0.02 or 0.020 or 0.0204
100 ± 10 0.02 or 0.016 or 0.0156
First mark for uncertainties in first column correct.
Second mark for all second column correct.
2
2(c)(i) Six points plotted correctly.
Must be within half a small square. Diameter of points must be less than half a small square.
1
Error bars in M plotted correctly.
All error bars to be plotted. Length of bar must be accurate to less than half a small square and symmetrical.
1
2(c)(ii) Line of best fit drawn.
Line must not pass through plotted point (0.11, 850) or (0.111, 850).
If points are plotted correctly then lower end of line should pass between (0.032, 250) and (0.036, 250) and upper end of
line should pass between (0.098, 800) and (0.104, 800).
1
Worst acceptable line drawn (steepest or shallowest possible line).
All error bars must be plotted.
1

PUBLISHED
October/November
2017
2(c)(iii) Gradient determined with a triangle that is at least half the length of the drawn line. 1
uncertainty = gradient of line of best fit – gradient of worst acceptable line
or
uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)
1
2(d)(i) µ determined correctly using gradient.
2 2
9.81
gradient
4 120 1.54
µ = ×
× ×
5
7.18123 10 gradientµ −
= × ×
1
µ determined using gradient and given to 2 or 3 significant figures. 1
µ determined using gradient and correct unit gm–1
and in the range 0.560–0.630 (gm–1
). 1

PUBLISHED
October/November
2017
2(d)(ii) Percentage uncertainty in µ.
0.01 5 gradient
% uncertainty 2 2 100
1.54 120 gradient
 ∆
= × + × + × 
 
gradient
% uncertainty 9.63% 100
gradient
∆
= + ×
2 2
9.81 max gradient
max
4 115 1.53
µ
×
=
× ×
2 2
9.81 min gradient
min
4 125 1.55
µ
×
=
× ×
1

PUBLISHED
October/November
2017
2(e) M determined correctly using µ from (d)(i).
2 2
180 1.54
7.833
9.81 1000
M
× ×
= = ×
×
(d)(i)
(d)(i)
1
Absolute uncertainty determined.
0.01 5
% uncertainty 2 2 100 6.9%
1.54 180
 
= × + × × + = + 
 
(d)(ii) (d)(ii)
( )
( )
2 2
4 185 1.55 max
max 8.382 max
4 9.81 1000
M
× × ×
= = ×
× ×
(d)(i)
(d)(i)
( )
( )
2 2
4 175 1.53 min
min 7.308 min
4 9.81 1000
M
× × ×
= = ×
× ×
(d)(i)
(d)(i)
1

PHYSICS 9702/53
Paper 5 Planning, Analysis and Evaluation October/November 2017
MARK SCHEME
Maximum Mark: 30
Published
Teachers.
Cambridge IGCSE®
components.

PUBLISHED
October/November
2017
d is the independent variable and R is the dependent variable or vary d and measure R 1
keep intensity/power of light source constant 1
labelled diagram showing a light source fixed above container of water with the labelled LDR positioned in the beaker 1
correct circuit diagram to measure R, e.g. V and I methods or ohmmeter 1
method to determine R, e.g. =R
p.d. across LDR
current
or read off ohmmeter
1
method to determine d, e.g. use a ruler or drawn labelled vertical ruler adjacent to container with d indicated 1
Method of analysis
plots a graph of R against d2
1
relationship valid if a straight line produced passing through the origin 1
π
=K
4
gradient
1

9702 w17 ms_all

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