9702 s18 ms_all

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Cambridge Assessment International Education
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/11
Paper 1 Multiple Choice May/June 2018
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 May/June 2018 series for most
Cambridge IGCSE™, Cambridge International A and AS Level and Cambridge Pre-U components, and
some Cambridge O Level components.

9702/11 Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2018
© UCLES 2018 Page 2 of 3
Question Answer Marks
1 A 1
2 C 1
3 C 1
4 A 1
5 C 1
6 A 1
7 B 1
8 A 1
9 B 1
10 A 1
11 D 1
12 D 1
13 C 1
14 C 1
15 D 1
16 B 1
17 C 1
18 B 1
19 B 1
20 B 1
21 A 1
22 B 1
23 D 1
24 C 1
25 B 1
26 A 1
27 B 1
28 D 1

PUBLISHED
May/June 2018
29 B 1
30 C 1
31 A 1
32 D 1
33 A 1
34 C 1
35 B 1
36 B 1
37 D 1
38 C 1
39 D 1
40 D 1

PHYSICS 9702/12
MARK SCHEME
Maximum Mark: 40
Published
examination.
Teachers.

PUBLISHED
May/June 2018
1 C 1
2 B 1
3 D 1
4 D 1
5 A 1
6 C 1
7 C 1
8 D 1
9 C 1
10 D 1
11 C 1
12 B 1
13 A 1
14 C 1
15 D 1
16 B 1
17 C 1
18 C 1
19 B 1
20 D 1
21 D 1
22 B 1
23 D 1
24 A 1
25 D 1
26 C 1
27 D 1
28 A 1

PUBLISHED
May/June 2018
29 C 1
30 D 1
31 A 1
32 A 1
33 B 1
34 B 1
35 A 1
36 A 1
37 C 1
38 C 1
39 B 1
40 B 1

PHYSICS 9702/13
MARK SCHEME
Maximum Mark: 40
Published
examination.
Teachers.

PUBLISHED
May/June 2018
1 B 1
2 B 1
3 A 1
4 A 1
5 D 1
6 B 1
7 C 1
8 B 1
9 C 1
10 A 1
11 D 1
12 D 1
13 B 1
14 A 1
15 D 1
16 B 1
17 B 1
18 D 1
19 C 1
20 C 1
21 D 1
22 B 1
23 D 1
24 B 1
25 D 1
26 A 1
27 A 1
28 B 1

PUBLISHED
May/June 2018
29 C 1
30 A 1
31 B 1
32 D 1
33 B 1
34 A 1
35 C 1
36 D 1
37 D 1
38 D 1
39 C 1
40 A 1

PHYSICS 9702/21
Paper 2 AS Level Structured Questions May/June 2018
MARK SCHEME
Maximum Mark: 60
Published
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 IGCSE™, Cambridge International A and AS Level and Cambridge Pre-U components,
and some Cambridge O Level components.

PUBLISHED
May/June 2018
Generic Marking Principles
These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the
specific content of the mark scheme or generic level descriptors for a question. Each question paper and mark scheme will also comply with these
marking principles.
GENERIC MARKING PRINCIPLE 1:
Marks must be awarded in line with:
• the specific content of the mark scheme or the generic level descriptors for the question
• the specific skills defined in the mark scheme or in the generic level descriptors for the question
• the standard of response required by a candidate as exemplified by the standardisation scripts.
Marks awarded are always whole marks (not half marks, or other fractions).
Marks must be awarded positively:
• marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond the
scope of the syllabus and mark scheme, referring to your Team Leader as appropriate
• marks are awarded when candidates clearly demonstrate what they know and can do
• marks are not deducted for errors
• marks are not deducted for omissions
• answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assessed by the
question as indicated by the mark scheme. The meaning, however, should be unambiguous.
Rules must be applied consistently e.g. in situations where candidates have not followed instructions or in the application of generic level
descriptors.

PUBLISHED
May/June 2018
Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may
be limited according to the quality of the candidate responses seen).
Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or
grade descriptors in mind.

PUBLISHED
May/June 2018
1(a) a scalar has magnitude (only) B1
a vector has magnitude and direction B1
1(b) power: scalar
temperature: scalar
momentum: vector
(two correct 1 mark, all three correct 2 marks)
B2
1(c)(i) arrow labelled R in a direction from 5° to 20° north of west B1
1(c)(ii) v2
= 282
+ 952
– (2 × 28 × 95 × cos 115°)
or
v2
= [(95 + 28cos65°)2
+ (28sin65°)2
]
C1
v = 110 ms–1
(109.8 ms–1
) A1
or (scale diagram method)
triangle of velocities drawn (C1)
v = 110 ms–1
(allow 108–112 ms–1
) (A1)

PUBLISHED
May/June 2018
2(a) a body continues at (rest or) constant velocity unless acted upon by a resultant force B1
2(b)(i)1. from 0–2s, distance = ½ × 2 × 6.8 (= 6.8m)
and
from 2–3s, distance = ½ × 1 × 3.4 (= 1.7 m)
C1
magnitude of displacement = 5.1 m A1
direction of displacement is down(wards) B1
2(b)(i)2. (∆E) = mg∆h or (E) = mgh or (E) = Wh C1
(∆)E = 15 × 5.1
= (–) 77 J
A1
2(b)(ii) a = (v – u) / t or a = gradient or a = dv /dt C1
a = 3.4 ms–2
A1
2(b)(iii) T – W = ma or T – mg = ma C1
T = 15 + (15/ 9.81) × 3.4 = 20 N or 20.2N A1
2(b)(iv) E = F / Aε or E = σ / ε and σ = F / A C1
ε = 20 / (2.8 × 10–5
× 1.7 × 1011
) C1
= 4.2 × 10–6
A1
2(b)(v) block is in equilibrium/has no resultant force B1
block could be stationary (or have constant velocity/speed)
(so no, not possible to deduce)
B1

PUBLISHED
May/June 2018
3(a) mass is the property (of a body/object) resisting changes in motion
or
mass is the quantity of matter (in a body)
B1
3(b)(i) force on A (by B) equal and opposite to force on B (by A) or both A and B exert equal and opposite forces on each other B1
force is rate of change of momentum and time (of contact) is same B1
3(b)(ii) p = mv or 3M × 0.40 or M × 0.25 or 3M × 0.2 or Mv C1
(3M × 0.40) – (M × 0.25) = (3M × 0.2) + Mv C1
v = (3 × 0.40) – 0.25 – (3 × 0.2)
= 0.35 ms–1
A1
3(b)(iii) 1. relative speed of approach = 0.40 + 0.25
= 0.65 ms–1
A1
2. relative speed of separation = 0.35 – 0.20
= 0.15ms–1
A1
3(b)(iv) (relative) speed of separation not equal to/less than (relative) speed of approach or answers (to (b)(iii) are) not equal
and so inelastic collision
B1

PUBLISHED
May/June 2018
4(a)(i) time for one oscillation/one vibration/one cycle
or
time between adjacent wavefronts/points in phase
or
shortest time between two wavefronts/points in phase
B1
4(a)(ii) distance moved by wavefront/energy during one cycle/oscillation/period (of source)
or
minimum distance between two wavefronts
or
distance between two adjacent wavefronts
or
minimum distance between two points having the same displacement and moving in the same direction
B1
4(b)(i) v = λ / T or v = fλ and f = 1/ T C1
λ = 20 × 0.60
= 12 cm
A1
4(b)(ii) phase difference = 360° × (0.20 / 0.60) or 360° × (0.40 / 0.60)
= 120° or 240°
A1
4(b)(iii) I ∝ A2
C1
IQ / IP =AQ
2
/ AP
2
= 2.02
/
3.02
= 0.44
A1
4(b)(iv) displacement = 1.00 – 3.00
= –2.00mm
A1

PUBLISHED
May/June 2018
5(a)(i) waves spread at (each) slit/gap B1
5(a)(ii) constant phase difference (between (each of) the waves) B1
5(b)(i) nλ = dsinθ B1
dsinθ is the same and 3λ1 = 4λ2 so λ2 / λ1 = 0.75 A1
5(b)(ii) λ2 / λ1 = 0.75 and λ1 – λ2 = 170
λ1 = 680nm
A1
6(a) joule / coulomb B1
6(b)(i) lamps have same p.d./lamps have p.d. of 2.7V B1
current = 0.15 + 0.090
= 0.24 A
A1
6(b)(ii) R = (4.5 – 2.7) / 0.24
or
RP = 18(Ω) and RQ = 30(Ω)
I / RT = 1/18 + 1 /30 and so RT = 11.25
4.5 = 0.24 × (R + 11.25)
C1
R = 7.5Ω A1

PUBLISHED
May/June 2018
6(b)(iii) R = ρl /A C1
RP /RQ = [(2.7 / 0.15) / (2.7 / 0.09)] (= 0.60) C1
ratio = 0.60 × 22
= 2.4
A1
6(b)(iv) less p.d. across resistor/greater p.d. across P B1
greater current through P and so resistance (of P) increases B1
7(a) arrow pointing vertically down the page B1
7(b) E = ½mv2
C1
E = 460 × 1.60 × 10–19
(= 7.36 × 10–17
(J)) C1
v = [(2 × 460 × 1.60 × 10–19
) /(9.11 × 10–31
)]½
= 1.3 × 107
m s–1
A1
7(c) β–
particles have range of/different/various speeds/velocities/momenta/energies M1
so they follow different paths A1

PHYSICS 9702/22
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.

PUBLISHED
May/June 2018
marking principles.
descriptors.

PUBLISHED
May/June 2018

PUBLISHED
May/June 2018
1(a) rate of change of momentum B1
1(b) kg m s–2
A1
1(c) units for Q: As and for r: m C1
units for ε = (As × As) / (kg m s–2
× m2
)
= A2
kg–1
m–3
s4
A1

PUBLISHED
May/June 2018
2(a) sum/total momentum (of a system of bodies) is constant
or
sum/total momentum before = sum/total momentum after
M1
for an isolated system or no (resultant) external force A1
2(b)(i) (p =) mv or (3.0M × 7.0) or (2.0M × 6.0) or (1.5M × 8.0) C1
3.0M × 7.0 = 2.0M × 6.0sinθ + 1.5M × 8.0sinθ C1
θ = 61° A1
or (vector triangle method)
momentum vector triangle drawn (C1)
θ = 61° (2 marks for ±1°, 1 mark for ±2°) (A2)
or (use of cosine rule)
p = mv or (3.0M × 7.0) or (2.0M × 6.0) or (1.5M × 8.0) (C1)
(21M)2
= (12M)2
+ (12M)2
– (2 × 12M × 12M × cos 2θ ) (C1)
θ = 61° (A1)
2(b)(ii) (E =) ½mv2
C1
ratio = (½ × 2.0M × 6.02
) / (½ × 1.5M × 8.02
)
= 0.75
A1

PUBLISHED
May/June 2018
3(a) time = 12 s A1
3(b) distance (up slope) = ½ × 12 × 18 (= 108) C1
distance (down slope) = ½ × 12 × 6 (= 36) C1
displacement from A = 108 – 36
= 72 m
A1
3(c) v = u + at or a = gradient or a = ∆v / (∆)t C1
a = 6 / 12 = 0.50 (ms–2
) (other points from the line may be used) A1
or
v2
= u2
+ 2as and u = 0
or
v2
= 2as
(C1)
a = 6.02
/ (2 × 36) = 0.50 (m s–2
) (A1)
or
s = ut + ½at2
and u = 0
or
s = ½at2
(C1)
a = 2 × 36 / 122
= 0.50 (m s–2
) (A1)
or
s = vt – ½at2
(C1)
a = 2 × (6 × 12 – 36)/ 122
= 0.50 (ms–2
) (A1)

PUBLISHED
May/June 2018
3(d)(i) F = 70 × 0.50 (= 35) C1
frictional force = 80 – 35
= 45N
A1
3(d)(ii) sin θ = 80/ (70 × 9.81) C1
θ = 6.7° A1
3(e)(i) f0 = (900 × 340)/ (340 + 12) C1
= 870Hz A1
3(e)(ii) speed/velocity (of sledge) decreases and (so) frequency increases B1

PUBLISHED
May/June 2018
4(a)(i) distance moved by wavefront/energy during one cycle/oscillation/period (of source)
or
minimum distance between two wavefronts
or
distance between two adjacent wavefronts
B1
4(a)(ii) (position where) maximum amplitude B1
4(b)(i) λ = 4 × 0.045
( = 0.18 (m) or 18 (cm))
C1
v = fλ C1
f = 340 / 0.18
= 1900 Hz
A1
4(b)(ii) distance = λ / 2
(= 0.09 (m) or 9 (cm))
C1
time = 0.09 / 0.0075
= 12 s
A1
or
t = 4.5 / 0.75 and t = 13.5 / 0.75 (C1)
time = 18 – 6
= 12 s
(A1)

PUBLISHED
May/June 2018
5(a) ρ = m / V C1
= (560 / 9.81)/ (1.2 × 0.018)
= 2600 kg m–3
A1
5(b) (∆)p = 940 × 9.81 × 1.2 C1
(upthrust =) 940 × 9.81 × 1.2 × 0.018 = 200 N A1
5(c)(i) tension = 560 – 200
= 360 N
A1
5(c)(ii) P= Fv C1
= 360 × 0.020
= 7.2 W
A1
5(d)(i) upthrust decreases B1
tension (in wire) increases M1
power (output of motor) increases A1
5(d)(ii) there is work done (on the cylinder) by the upthrust
or
GPE of oil decreases (as it fills the space left by cylinder and so total energy is conserved)
B1

PUBLISHED
May/June 2018
6(a)(i) sum of current(s) into junction = sum of current(s) out of junction
or
(algebraic) sum of current(s) at a junction is zero
B1
6(a)(ii) charge B1
6(b)(i)1. E = I2
Rt or E = VIt or E = (V2
/ R)t C1
E = 2.52
× 2.0 × 5.0 × 60 or 5.0 × 2.5 × 5.0 × 60 or (5.02
/ 2.0) × 5.0 × 60
= 3800 J
A1
6(b)(i)2. p.d. = 8.0 – (2.0 × 2.5)
= 3.0 V
A1
6(b)(ii) IX = 3.0 /15 = 0.20 (A) C1
IY = 2.5 – 0.20 = 2.3 (A) C1
RY = 3.0 /2.3
= 1.3 Ω
A1
or
RT = 3.0 / 2.5 = 1.2(Ω) or (8.0 / 2.5) – 2.0 = 1.2(Ω) (C1)
1/ 1.2 = 1 / 15 + 1 / RY (C1)
RY = 1.3 Ω (A1)

PUBLISHED
May/June 2018
6(b)(iii)1. Z has larger radius/diameter/(cross-sectional) area B1
Z has (material of) smaller resistivity/greater conductivity B1
6(b)(iii)2. current/I (in battery) increases M1
(P = EI so) power/P (produced by battery) increases A1
7(a) circle(s) drawn only around β–
and ν symbols B1
7(b) (electron) antineutrino B1
7(c) kinetic (energy) B1
7(d) Y has one more proton (and one less neutron)/X has one less proton (and one more neutron)
or
Y has more protons (and fewer neutrons)/X has fewer protons (and more neutrons)
or
a neutron changes to a proton
or
the number of protons increases
M1
(so) not isotopes A1
7(e) up down down changes to up up down or udd → uud
or
down changes to up or d → u
B1

PHYSICS 9702/23
MARK SCHEME
Maximum Mark: 60
Published
Teachers.

PUBLISHED
May/June 2018
marking principles.
descriptors.

PUBLISHED
May/June 2018

PUBLISHED
May/June 2018
Question Answer Mark
1(a)(i) zero error or wrongly calibrated scale B1
1(a)(ii) reading scale from different angles or wrongly interpolating between scale readings/divisions B1
1(b)(i) P = V2
/R or P = VI and V = IR C1
P = 5.02
/ 125 or 5.0 × 0.04 or (0.04)2
× 125
= 0.20 W
A1
1(b)(ii) %V = 2% or ∆V / V = 0.02 C1
%P = (2 × 2%) + 3% or %P = (2 × 0.02 + 0.03) × 100
= 7%
A1
1(b)(iii) absolute uncertainty in P = (7 / 100) × 0.20
= 0.014
C1
power = 0.20 ± 0.01 W or (2.0 ± 0.1) × 10–1
W A1

PUBLISHED
May/June 2018
2(a)(i) (work =) force × distance moved in the direction of the force B1
2(b)(i) ρ = m / V C1
= (20 / 9.81) / (4/3 × π × 0.163
)

= 120 kg m–3
A1
2(b)(ii) the pressure on the lower surface (of sphere) is greater than the pressure on the upper surface (of sphere) B1
2(b)(iii) a = (170 – 20) / (20 / 9.81) C1
= 74 m s–2
A1
2(b)(iv) D = 170 – 20 (= 150) C1
810 × (0.162
) × v2
= 150 C1
v = 2.7 m s–1
A1
2(b)(v) 4870 = (4850 × v) / (v – 6.30) C1
v = 1530 m s–1
A1

PUBLISHED
May/June 2018
3(a) v = u + at
v = 9.6 – (9.81 × 0.37) = 6.0 ms–1
A1
3(b) s = ½ × (9.6 + 6.0) × 0.37
or
6.02
= 9.62
– (2 × 9.81 × s)
or
s = (9.6 × 0.37) – (½ × 9.81 × 0.372
)
or
s = (6.0 × 0.37) + (½ × 9.81 × 0.372
)
C1
s = 2.9 m A1
3(c)(i) (∆)E = mg(∆)h C1
∆E = 0.056 × 9.81 × 2.9
= 1.6 J
A1
3(c)(ii) E = ½mv2
C1
∆E = ½ × 0.056 × (6.02
– 3.82
)
= 0.60 J
A1
3(d) force on ball (by ceiling) equal to force on ceiling (by ball) M1
and opposite (in direction) A1
3(e) (p =) mv or 0.056 × 6.0 or 0.056 × 3.8 C1
change in momentum = 0.056 × (6.0 + 3.8)
= 0.55 Ns
A1

PUBLISHED
May/June 2018
3(f) resultant force = 0.55/ 0.085 (= 6.47 N) C1
force by ceiling = 6.47 – (0.056 × 9.81)
= 5.9 N
A1
4(a) (Young modulus =) stress/ strain B1
4(b)(i) k = F / ∆L or 1 / gradient C1
= 90 × 103
/ (2 × 10–3
) (or other point on line)
= 4.5 × 107
N m–1
A1
4(b)(ii) E = ½F∆L or E = ½k(∆L)2
C1
= ½ × 90 × 103
× 2 × 10–3
or ½ × 4.5 × 107
× (2 × 10–3
)2
C1
= 90 J A1
4(c) straight line starting from (0, 150) and below original line M1
line ends at (90, 147) A1

PUBLISHED
May/June 2018
5(a) intensity ∝ (amplitude)2
B1
5(b)(i) v = fλ or c = fλ C1
f = 3.00 × 108
/ 0.060
= 5.0 × 109
Hz
A1
5(b)(ii) (at X path) difference = 3λ M1
(at X phase) difference = 0 or 1080° M1
so intensity is at a maximum/it is an intensity maximum A1
5(b)(iii) 1. decrease in the distance between (adjacent intensity) maxima/minima B1
2. (intensity) maxima and minima exchange places B1

PUBLISHED
May/June 2018
6(a) R = ρL / A C1
3.0 = ρ / [π × (0.48 × 10–3
/ 2)2
] C1
ρ = 5.4 × 10–7
Ω m A1
6(b)(i) p.d. = 5.0 – (2.0 × 1.6)
= 1.8 V
A1
6(b)(ii)1. current in resistor = 1.8 / 4.5 (= 0.40A) C1
current in wire = 1.6 – 0.40 (= 1.2A) C1
RX = 1.8 / 1.2
= 1.5 Ω
A1
or
RT = 1.8 / 1.6 or (5.0 / 1.6) – 2.0 (= 1.125Ω) (C1)
(1 / 1.125) = (1 / 4.5) + (1 / RX) (C1)
RX = 1.5 Ω (A1)
6(b)(ii)2. length = 1.5 / 3.0 or 1.5 × 1.8 × 10–7
/ (5.4 × 10–7
)
= 0.50 m
A1

PUBLISHED
May/June 2018
7(a)(i) Q plotted at (82, 210) A1
7(a)(ii) R plotted at (83, 210) A1
7(b) lepton(s) B1
7(c) up down down changes to up up down or udd → uud
or
down changes to up or d → u
B1

PHYSICS 9702/31
Paper 3 Advanced Practical Skills 1 May/June 2018
MARK SCHEME
Maximum Mark: 40
Published
Teachers.

PUBLISHED
May/June 2018
marking principles.
descriptors.

PUBLISHED
May/June 2018

PUBLISHED
May/June 2018
1(a) Value of raw d to nearest mm with unit and in the range 29.0–31.0cm. 1
1(b) Value of T in the range 0.8–2.0s with unit and evidence of at least two sets of nT where n ⩾ 5. 1
1(c) Second set of values of s and T. 1
1(d) Five sets of readings of s and time (different values) with the correct trend and without help from the Supervisor scores 4
marks, four sets scores 3 marks etc.
4
Range: at least one value of s ⩽ 70.0cm and at least one value of s ⩾ 85.0cm 1
Column headings:
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. s/ cm, T / s, T2
/ s2
or T2
(s2
).
1
Consistency:
All raw time values must be given to 0.1s or all given to 0.01 s.
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 the raw value(s) of
time.
If raw times are given to 0.01s, allow T2
to be recorded to 1 s.f. less than the raw times.
1
Calculation: Values of T2
are correct. 1

PUBLISHED
May/June 2018
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 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.
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 ± 2.5 cm (to scale) on the s axis (normally 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 4 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 four
points left after the anomalous point is disregarded.
Line must not be kinked or thicker than half a small square.
1
1(e)(iii) Gradient:
The hypotenuse of the triangle used must be greater than half the length of the drawn line.
The method of calculation must be correct. Do not allow ∆x/∆y.
Both read-offs must be accurate to half a small square in both the x and y directions.
Sign of gradient must match graph.
1
y-intercept:
Correct read-off from a point on the line substituted correctly into y = mx + c or an equivalent expression.
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 s = 0, accurate to half a small square in y direction.
1

PUBLISHED
May/June 2018
1(f) Value of P = candidate’s gradient and value of Q = candidate’s intercept.
The values must not be fractions.
1
Unit for P correct (e.g. s2
m–1
or s2
cm–1
or s2
mm–1
).
and
Unit for Q is s2
.
1
2(a) Value of x with unit and in the range 1.0–5.0mm. 1
2(b) All value(s) of raw y to nearest mm with unit. 1
Evidence of repeat values of y. 1
2(c) Absolute uncertainty in y in the range 2–4 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 find percentage uncertainty.
1
2(d) Second value of x. 1
Second value of y. 1
Quality: second value of y < first value of y. 1
2(e)(i) Two values of k calculated correctly. 1
2(e)(ii) Valid comment consistent with the calculated values of k, testing against a criterion stated by the candidate. 1
2(f)(i) Value of y. 1
2(f)(ii) Correct calculation of x. 1
2(f)(iii) Justification for s.f. in x linked to s.f. in y. 1

PUBLISHED
May/June 2018
2(g)(i) A Two readings are not enough to draw a (valid) conclusion (not “not enough for accurate results”, “few readings”).
B Difficulty with alignment at A with a reason e.g. string twists/magnet will not settle/magnet continually moves/draughts
or air conditioning disturb magnet.
C Difficulty with finding position C or position after 30 oscillations or finding y with a reason e.g. magnet only stays at
position C for a short time/difficult to estimate where magnet will stop/difficult to judge the end of an
oscillation/parallax.
D Difficulty linked to x with a reason e.g. metre rule (too large or not precise enough/large percentage uncertainty in
x/value of x is small/ adjustment difficult or clumsy with clamp and cork/angled lower surface of magnet.
E Difficulty with the oscillation e.g. magnet rotates while swinging/wrong mode of oscillation/magnetic stand attracts the
magnet affecting its motion.
1 mark for each point up to a maximum of 4.
4
2(g)(ii) A Take many readings and plot a graph or take more readings and compare k values (not “repeat readings” on its
own).
B Improved method of alignment with A e.g. valid improvement to suspension e.g. nylon thread/wire/double point of
suspension/switch off AC/windshield.
C Improved method of locating C e.g. use a scale/use of a pointer with detail of how pointer is used e.g. attached to
magnet/use to refine position of C.
D1 Improved method to adjust x e.g. scissor jack/rearrange suspension to use screw thread on clamp.
D2 Improved method to measure x e.g. travelling microscope/vernier or digital calipers/use of thin spacers e.g. sheets of
paper/shim.
E Use a wooden or plastic stand i.e. a named non-magnetic material.
4

PHYSICS 9702/32
MARK SCHEME
Maximum Mark: 40
Published
Teachers.

PUBLISHED
May/June 2018
marking principles.
descriptors.

PUBLISHED
May/June 2018

PUBLISHED
May/June 2018
1(a)(i) Value of h to nearest mm, with unit. 1
1(a)(ii) Value of t in the range 1.0–5.0s, with unit. 1
At least two readings of t. 1
1(b) Six sets of readings of h and t showing the correct trend and without help from the Supervisor scores 4 marks, five sets
scores 3 marks etc.
4
Range: tmax ⩾ 2.0s and tmin ⩽ 1.5s. 1
Column headings:
The presentation of quantity and unit must conform to accepted scientific convention e.g. (1 / t2
)/ s–2
.
1
Consistency:
All raw values of t must be given to 0.01s or all must be given to 0.1s.
1
Significant figures for every value of 1 / t 2
same as, or one greater than, the number of s.f. of t as recorded in table.
1
Calculation:
Values of 1/ t 2
calculated correctly to the number of s.f. given by the candidate.
1

PUBLISHED
May/June 2018
1(c)(i) Axes:
1
Plotting of points:
Plots 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 ± 0.05 s–2
from a straight line in the 1 / t2
direction.
1
1(c)(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 only if clearly indicated (i.e. circled or labelled) by the candidate.
Lines must not be kinked or thicker than half a small square.
1
1(c)(iii) Gradient:
The hypotenuse of the triangle used must be greater than half the length of the drawn line.
The method of calculation must be correct. Do not allow ∆x / ∆y.
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 at x = 0 accurate to half a small square in y direction.
1
1(d) Value of a equal to candidate’s gradient and value of b equal to candidate’s intercept. 1
Unit for a correct (e.g. mm–1
s–2
or cm–1
s–2
) and unit for b is s–2
. 1

PUBLISHED
May/June 2018
2(a) Value for L to nearest mm with unit and in range 50.0–75.0cm. 1
2(b) Value for x with unit. 1
Raw value(s) of x to nearest mm and value on answer line in range 100–200 mm. 1
2(c) Absolute uncertainty in x value of 0.2–0.5cm 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 the working is
clearly shown.
1
2(d) Correct calculation of λ to the number of significant figures used by the candidate. 1
2(e) Justification linking s.f. in λ to s.f. in L. 1
2(f) Correct calculation of f. 1
2(g) Second values of L and x. 1
Quality: x increases as µ increases. 1
Second values of λ and f. 1
2(h)(i) Two values of k calculated correctly. 1
2(h)(ii) Valid comment consistent with the calculated values of k, testing against a numerical criterion specified by the candidate. 1

PUBLISHED
May/June 2018
2(i)(i) A Two readings are not enough to draw a (valid) conclusion (not “not enough for accurate results”, “few readings”).
B Difficult to see string pattern.
C String pattern dies away quickly.
D Difficult to produce exact pattern required.
E Difficulty with wooden block e.g. doesn’t grip blade/hard to hold down block and oscillate blade/view pattern at same
time.
F Hacksaw/mass not at a node in the string pattern.
4
2(i)(ii) A Take many readings and plot a graph or take more values of k and compare (not “repeat readings” on its own).
B Put contrasting/black/white background behind string.
C Use photo/video to assess pattern or description of workable method of producing continuous oscillations.
D Add a scale/marks to hacksaw blade.
E Clamp/put weight on the movable wooden block.
F Workable solution to allow a node e.g. move excitation point away from node/use heavier mass.
4

PHYSICS 9702/33
MARK SCHEME
Maximum Mark: 40
Published
Teachers.

PUBLISHED
May/June 2018
marking principles.
descriptors.

PUBLISHED
May/June 2018

PUBLISHED
May/June 2018
1(a) Values of V1 and V2 with units and to the nearest 0.01V. 1
|V2 | > |V1|. 1
1(b) Six sets of readings of x, V1 and V2 (different values) showing the correct trend and without help from the Supervisor scores 5
marks, five sets scores 4 marks etc.
5
Range: xmin ⩽ 10.0 cm and xmax ⩾ 70.0cm. 1
Column headings:
The presentation of quantity and unit must conform to accepted scientific convention e.g. (V1 / x) / Vm–1
.
1
Consistency:
All values of x must be given to the nearest mm.
1
All values of V1 / x must be given to 2 or 3 s.f.
1
Calculation: Values of V1 / x are correct. 1
1(c)(i) Axes:
1
Plotting of points:
Points must be plotted to an accuracy of half a small square.
1
Quality:
It must be possible to draw a straight line that is within 0.040 on the (V2 – V1) axis of all plotted points.
1

PUBLISHED
May/June 2018
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 by the candidate.
1
1(c)(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
Intercept read directly from the graph with read-off at x = 0, accurate to half a small square.
1
1(d) Value of P = candidate’s gradient and value of Q = candidate’s intercept.
1
Unit for P correct (e.g. m or cm or mm) and unit of Q correct (e.g. V) unless zero. 1

PUBLISHED
May/June 2018
2(a) Value of L to the nearest mm with unit and in range 50.0–60.0cm. 1
2(b)(i) Value of A to the nearest degree with unit and in the range 11–17°. 1
2(b)(ii) Percentage uncertainty in A based on absolute uncertainty of 2°–5°.
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(c)(i) Value of T in the range 1.20–1.60s with unit. 1
Evidence of at least two sets of nT where n ⩾ 5. 1
2(c)(ii) Correct calculation of d. 1
2(c)(iii) Justification for s.f. in d linked to s.f. in A (or measured angle). 1
2(d) Second value of A. 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 specified by the candidate. 1

PUBLISHED
May/June 2018
2(f)(i) A Two readings are not enough to draw a (valid) conclusion (not “not enough for accurate results”, “few readings”).
B Difficult to measure L with reason e.g. locating centre of bob/parallax/bob or cork gets in the way of the ruler.
C Difficult to measure A or 45° or angle with a reason e.g. no indication of vertical/hard to hold and measure from
protractor/hard to hold protractor steady/string too thick/parallax error.
D Problem with bob moving between setting and releasing.
E Difficulty setting rod at L / 2 with reason e.g. setting rod and angle at the same time/rod moves when tightening the
clamp/rod moves when pendulum hits.
F Difficulty in judging start/end of oscillation.
4
2(f)(ii) A Take many readings (for different values of A) and plot a graph or take more values of k and compare (not “repeat
readings” on its own).
B Improved method of measuring L e.g. measure diameter and add on (or take away) radius from length of string from cork
to top (bottom) of bob, or e.g. clear use of pointer(s).
C Improved method of measuring angle e.g. use a plumb line/clamp protractor/use thinner string, or e.g. shadow projection
ideas/photograph and measure angle.
D Improved mechanism of releasing the bob e.g. clamp bob/use a card gate/add a stop.
E Mark string.
F Method of improving timing e.g. put a marker anywhere except at the ends/video with timer (or replay frame by frame).
4

PHYSICS 9702/34
MARK SCHEME
Maximum Mark: 40
Published
Teachers.

PUBLISHED
May/June 2018
marking principles.
descriptors.

PUBLISHED
May/June 2018

PUBLISHED
May/June 2018
1(a) Value of VS in range 2.00–4.00V, with unit. 1
1(b) Value of V less than VS. 1
Evidence of repeat readings of V. 1
1(c) Six sets of readings of n and V with the correct trend and without help from the Supervisor scores 4 marks, five sets scores
3 marks, etc.
4
Range: nmin = 1 or 0 and nmax ⩾ 7. 1
Column headings:
The presentation of quantity and unit must conform to accepted scientific convention, e.g. (1/V) / V–1
or 1/V (V–1
).
1
Consistency:
All raw values of V must be given to 0.01V, without trailing zeros.
1
Significant figures for every value of 1/V the same as, or one greater than, the s.f. of V as recorded in the table.
1
Calculation: Values of 1/V calculated correctly. 1
1(d)(i) Axes:
1
Plotting of points:
Points must be plotted to an accuracy of 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 ± 0.025V–1
from a straight line in the 1/V direction.
1

PUBLISHED
May/June 2018
1(d)(ii) Line of best fit:
Judge by balance of all points on the grid about the candidate’s line. There must be an even distribution of points either side
of the line along the full length.
Allow one anomalous point only if clearly indicated by the candidate. There must be at least 5 points left after the anomalous
point is disregarded.
1
1(d)(iii) Gradient:
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 at n = 0, accurate to half a small square in the y direction.
1
1(e) Value of a equal to candidate’s gradient and value of b equal to candidate’s intercept. 1
Unit for a is V–1
and unit for b is V–1
. 1

PUBLISHED
May/June 2018
2(a) Correct calculation of M. 1
2(b)(i) Value for T in range 0.50–1.00s, with unit. 1
Evidence of repeat readings of time, with at least two sets of nT where n ⩾ 5. 1
2(b)(ii) Absolute uncertainty in time measurement of 0.20–0.50s 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 the working is
clearly shown.
1
2(c)(i) Second value of M. 1
2(c)(ii) Second value of T. 1
Quality: T greater for greater M. 1
2(d)(i) Two values of k calculated correctly. 1
2(d)(ii) Valid comment relating to the calculated values of k, testing against a criterion specified by the candidate. 1
2(e)(i) Value(s) for D to nearest mm and value for D (on answer line) in range 0.050–0.200m. 1
2(e)(ii) Correct calculation of A. 1
2(e)(iii) Correct calculation of ρ using second value of k. 1

PUBLISHED
May/June 2018
2(f)(i) A Two readings are not enough to draw a (valid) conclusion (not “not enough for accurate results”, “few readings”).
B Reason for M not being accurate, e.g. mass of bottle ignored/uncertainty in the volume of water linked to precision of
beaker.
C Difficulty with oscillation with reason, e.g. masses move/bottle hits side of bucket/does not oscillate vertically/other modes
of oscillation.
D Difficult to measure D with reason, e.g. bottle flexes when measuring D/non-uniform D/bottle not circular/D varies with
depth.
E Difficult to judge when an oscillation starts/ends/is completed.
4
2(f)(ii) A Take more readings and plot a graph or take more values of k and compare (not “repeat readings” on its own).
B Use electronic balance/top-pan balance or use measuring cylinder.
C Method of fixing mass hanger in position in bottle, e.g. Blu-Tack, glue, tape
or use suitable alternative to slotted masses, e.g. sand/lead shot/single mass
or use wider bucket/larger diameter.
D Use vernier/digital calipers
or details of alternative method to find D, e.g use two set squares/two wooden blocks
or measure D in different directions/positions and find average.
E Use video and timer/video and view frame by frame or position/motion sensor above bucket.
4

PHYSICS 9702/35
MARK SCHEME
Maximum Mark: 40
Published
Teachers.

PUBLISHED
May/June 2018
marking principles.
descriptors.

PUBLISHED
May/June 2018

PUBLISHED
May/June 2018
1(a)(i) Value of z in the range 29.0–31.0cm to the nearest mm with unit. 1
1(a)(ii) Value of y with unit and y ⩽ 35.0cm. 1
1(b) Six sets of readings of x and y (different values) showing the correct trend and without help from the Supervisor scores 5
marks, five sets scores 4 marks etc.
5
Range: values of x must include at least one negative value. 1
Column headings:
The presentation of the quantity and unit must conform to accepted scientific convention e.g. x / m.
1
Consistency:
All raw values of x and y must be given to the nearest mm.
1
1(c)(i) Axes:
1
Plotting of points:
Diameter of plotted point must be ⩽ half a small square (no “blobs”).
All points must be plotted to an accuracy of half a small square.
1
Quality:
It must be possible to draw a straight line that is within 1.0 cm (to scale) on the 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 by the candidate.
1

PUBLISHED
May/June 2018
1(c)(iii) Gradient:
1
y-intercept:
Correct read-off from a point on the line and substituted into y = mx + c.
or
Intercept read directly from the graph with read-off at x = 0, accurate to half a small square.
1
1(d) Value of A = candidate’s gradient and value of B=candidate’s intercept.
1
No unit for A and unit for B correct (m, cm, mm). 1
1(e) Correct calculation of p to the number of s.f. given by the candidate. 1
1(f) Line W drawn with the same gradient but lower value of y-intercept. 1

PUBLISHED
May/June 2018
2(a)(i) Value of T less than 1.0s with unit and evidence of at least two sets of nT where n ⩾ 5. 1
2(a)(ii) Correct calculation of f to the number of s.f. given by the candidate. 1
2(a)(iii) Justification for s.f. in f linked to s.f. in time or period. 1
2(b)(i) Value of n. 1
2(b)(ii) Value of I. 1
2(b)(iii) Percentage uncertainty in I based on absolute uncertainty ⩾ 0.2µA.
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(c)(i) Second value of T. 1
Second value of T > first value of T. 1
2(c)(ii) Values of n and I. 1
Quality: second I < first I. 1
2(d)(i) Two values of k calculated correctly. 1
2(d)(ii) Valid comment consistent with calculated values of k, testing a criterion specified by the candidate. 1

PUBLISHED
May/June 2018
2(e)(i) A Two readings are not enough to draw a (valid) conclusion (not “not enough for accurate results”, “few readings”).
B Reason for difficulty with oscillation e.g. magnet struck top of tube/magnet not passing through all turns/some turns are
not near magnet/magnet oscillates outside of coil.
C Difficulty in judging end/start of (complete) oscillation.
D Difficulty with the current readings with reason e.g. because of positive and negative values/meter does not refresh quickly
enough/current small/resistance of connecting leads high/maximum current lasts for short time.
E Difficulty with the practical setup e.g. tube fell over/magnet fell off/spring moving along rod.
F n is not a whole number.
4
2(e)(ii) A Take many readings (for different values of n or f) and plot a graph or take more values of k and compare (not “repeat
readings” on its own).
B Method of improving difficulty with oscillation e.g. use wider tube/use shorter tube/bunch up coils/longer magnet.
C Method of improving timing e.g. put a marker with position (except at the ends)/video with timer (or replay frame by
frame)/position or motion sensor placed below.
D Method to reduce difficulties with current e.g. use (centre-zero) analogue meter/use c.r.o./galvanometer/use more turns of
wire/smaller masses/stiffer spring/stronger magnet/sand contacts/video ammeter and replay to find maximum current.
E Named method of attaching to table/holder e.g. clamp tube/tape magnet to weight/tape spring to rod/tape tube to table.
F Calculate n from the length of the wire and the diameter of the tube.
4

PHYSICS 9702/41
Paper 4 A Level Structured Questions May/June 2018
MARK SCHEME
Maximum Mark: 100
Published
Teachers.

PUBLISHED
May/June 2018
marking principles.
descriptors.

PUBLISHED
May/June 2018

PUBLISHED
May/June 2018
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)(i) velocity changes/direction of motion changes/there is an acceleration/there is a resultant force
so not in equilibrium
B1
1(b)(ii)1. gravitational force equals/is centripetal force C1
GMm/R2
= mRω2
and ω = 2π/T
or
Gm/R2
= mv2
/R and v = 2πr/T
or
GMm / R2
= mR (2π / T)2
M1
convincing algebra leading to k = GM/ 4π2 A1
1(b)(ii)2. correct use of R3
/T2
for one planet
(c gives 3.54 × 1021
; e and g both give 3.56 × 1021
)
C1
3.5(5) × 1021
= (6.67 × 10–11
×
M) / 4π2
M = 2.1 × 1033
kg
A1
two or three values of R3
/T2
correctly calculated and used in a valid way to find a value for M based on more than one k B1

PUBLISHED
May/June 2018
2(a)(i) straight line through origin indicates acceleration ∝ displacement B1
negative gradient shows acceleration and displacement are in opposite directions B1
2(a)(ii) a = –ω2
y and ω = 2πf
4.5 = (2π × f)2
× 8.0 × 10–3
(or other valid read-off)
C1
f = 3.8 Hz A1
2(b)(i) maximum displacement upwards/above rest/above the equilibrium position B1
2(b)(ii) (just leaves plate when) acceleration = 9.81m s–2
C1
9.81 = (2π × 3.8)2
× y0
or
9.81 = 563 × y0
C1
amplitude = 17 mm A1

PUBLISHED
May/June 2018
3(a)(i) sum of potential and kinetic energies (of molecules/atoms/particles) B1
(energy of) molecules/atoms/particles in random motion B1
3(a)(ii) (in ideal gas) no intermolecular forces so no potential energy B1
internal energy is (solely) kinetic energy (of particles) B1
(mean) kinetic energy (of particles) proportional to (thermodynamic) temperature of gas B1
3(b) pV = NkT C1
6.4 × 106
× 1.8 × 104
× 10–6
= N × 1.38 × 10–23
× 298 C1
or
pV = nRT and N = n × NA (C1)
6.4 × 106
× 1.8 × 104
× 10–6
= n × 8.31 × 298
n = 46.5 (mol)
N = 46.5 × 6.02 × 1023
(C1)
N = 2.8 × 1025
A1

PUBLISHED
May/June 2018
4(a) e.g. microphone
weighing scales/pressure sensor
lighters/spark generation
watches/clocks/regulation of time
B1
4(b) pulses (of ultrasound) B1
reflected at boundaries (between media) B1
(reflected pulses) detected by (ultrasound) generator B1
Any three from:
• time delay (between transmission and receipt) gives information about depth (of boundary)
• intensity of reflected pulse gives information about (nature of) boundary
• gel used to minimise reflection at skin/maximise transmission into skin
• degree of reflection depends upon impedances of two media (at boundary)
B3

PUBLISHED
May/June 2018
5(a)(i) west to east B1
5(a)(ii) above the Equator B1
5(a)(iii) value in range (1–300) × 109
Hz A1
5(b)(i) gain/ dB = 10 lg(P2 / P1) C1
–195 = 10 lg(P / 3000)
or
195 = 10 lg(3000 / P)
C1
power = 9.5 × 10–17
W A1
5(b)(ii) up-link has been (greatly) attenuated (before reaching satellite)
or
down-link signal must be (greatly) amplified (before transmission back to Earth)
or
up-link has (much) smaller intensity/power than down-link
B1
(different frequency) prevents down-link (signal) swamping up-link (signal) B1
6(a) force per unit charge B1
6(b) E = Q/ (4πε0r2
) C1
2.0 × 104
= Q / (4π × 8.85 × 10–12
× 0.262
)
charge = 1.5 × 10–7
C
A1
6(c) charge (= Q [52 / 26]2
) = 4Q C1
additional charge = 3Q A1

PUBLISHED
May/June 2018
7(a) (capacitance =) charge/potential M1
charge is (numerically equal to) charge on one plate A1
potential is potential difference between plates A1
7(b)(i) 4.5 × 10–6
C A1
7(b)(ii) 9.0 × 10–8
C A1
7(b)(iii) capacitance = (9.0 × 10–8
) / 120 C1
= 7.5 × 10–10
F A1
7(c) total capacitance is halved B1
current is halved B1

PUBLISHED
May/June 2018
8(a)(i) (fraction of) output is combined with the input M1
output (fraction) subtracted/deducted from input A1
8(a)(ii) any two valid points e.g.:
• greater bandwidth/gain constant over a larger range of frequencies/greater bandwidth
• smaller gain
B2
8(b)(i) gain = (–)9600/ 800 C1
= –12 A1
8(b)(ii) 1. 1.2 V B1
2. –6 V B1
8(b)(iii) replace the 9600Ω resistor with an LDR B1

PUBLISHED
May/June 2018
9(a) using Fleming’s left-hand rule force on wire is upwards B1
by Newton’s third law, force on magnet is downwards B1
9(b)(i) F = BIL C1
= 3.7 × 10–3
× 5.1 × 8.5 × 10–2
= 1.6 × 10–3
N
A1
9(b)(ii) F = 1.6 × 10–3
N A1
9(c) sketch: sinusoidal wave with two cycles B1
amplitude 2.3 × 10–3
N B1
period 0.05s B1
10(a) induced e.m.f. proportional to rate M1
of change of (magnetic) flux (linkage)
or
of cutting (magnetic) flux
A1
10(b) current in coil produces flux B1
(by Faraday’s law) changing flux induces e.m.f. in ring B1
current in ring causes field (around ring) B1
(by Lenz’s law) field around ring opposes field around coil B1

PUBLISHED
May/June 2018
11(a)(i) packet/quantum/discrete amount of energy M1
of electromagnetic radiation A1
11(a)(ii) (maximum) energy of emitted electrons is independent of intensity
or
no emission of electrons below the threshold frequency regardless of intensity
or
no emission of electrons when photon energy is less than work function (energy) regardless of intensity
B1
11(b) in darkness: conduction band empty so high resistance B1
in daylight: electrons in valence band absorb photons B1
in daylight: electrons ‘jump’ to conduction band B1
this leaves holes in valence band B1
more charge carriers in daylight so resistance decreases B1

PUBLISHED
May/June 2018
12(a)(i) I = I0 e–µx
C1
= I0 exp(–0.90 × 2.8)
= 0.080I0
A1
12(a)(ii) I = I0 exp[(–0.90 × 1.5) × (–3.0 × 1.3)] C1
= I0 (0.259 × 0.20)
= 0.0052I0
A1
12(b)(i) difference in degrees of blackening M1
between structures A1
12(b)(ii) large difference in intensities so good contrast B1

PUBLISHED
May/June 2018
13(a) emission of particles/radiation by unstable nucleus B1
spontaneous emission B1
13(b)(i) use of graph to determine half-life = 14 minutes B1
hence λ = ln 2 / (14 × 60) (s–1
) C1
N at 14 minutes = 4.4 × 107
and A = λN C1
activity = 4.4 × 107
× ln2 / (14 × 60)
= 3.6 × 104
Bq
A1
or
correct tangent drawn at time t = 14 minutes (B1)
magnitude of gradient of tangent identified as activity (C1)
correct working for gradient leading to activity (C1)
activity = 3.6 × 104
Bq (A1)
13(b)(ii) 3.6 × 104
= λ × 4.4 × 107
or
λ = ln2 / (14.0 × 60)
C1
λ = 8.2 × 10–4
s–1
A1

PHYSICS 9702/42
Paper 4 A Level Structured Questions May/June 2018
MARK SCHEME
Maximum Mark: 100
Published
Teachers.

PUBLISHED
May/June 2018
marking principles.
descriptors.

PUBLISHED
May/June 2018

PUBLISHED
May/June 2018
1(a)(i) direction of force on a (small test) mass
or
path in which a (small test) mass will move
B1
1(a)(ii) (at surface,) lines (of force) are radial B1
Earth has large radius/height above surface is small
so lines are (approximately) parallel
B1
parallel lines → constant field strength B1
1(b) (change in) KE of rock = (change in) PE
or
½mv2
= GMm / R
C1
(m)v2
= (m)(2 × 6.67 × 10–11
× 7.4 × 1022
) / (1.7 × 103
× 103
) C1
v = 2.4 × 103
ms–1
A1
correct conclusion based on comparison of v with 2.8 km s–1
B1
or
(change in) KE of rock = (change in) PE (C1)
(at infinity) EP = (6.67 × 10–11
× 7.4 × 1022
× m) / (1.7 × 103
× 103
)
= 2.9 × 106
m
(C1)
EK of rock = ½ × m × (2.8 × 103
)2
= 3.9 × 106
m (A1)
correct conclusion based on comparison of EK and EP values (B1)
or

PUBLISHED
May/June 2018
(change in) KE of rock = (change in) PE
or
½mv2
= GMm /R
(C1)
(m) (2800)2
= (m) (2 × 6.67 × 10–11
× 7.4 × 1022
) / R (C1)
R = 1.3 × 103
km (A1)
correct conclusion based on comparison of R with 1.7 × 103
km (B1)
or
(change in) KE of rock = (change in) PE
or
½mv 2
= GMm / R
(C1)
(m) (2800)2
= (m) (2 × 6.67 × 10–11
× M) / (1.7 × 106
) (C1)
M = 1.0 × 1023
kg (A1)
correct conclusion based on comparison of M with 7.4 × 1022
kg (B1)

PUBLISHED
May/June 2018
2(a) no intermolecular forces (so no potential energy) B1
2(b)(i) mean square speed (of molecule(s)) B1
2(b)(ii) kelvin/thermodynamic/absolute temperature B1
2(c)(i)1. pV = NkT C1
4.7 × 10–2
× 2.6 × 105
= N × 1.38 × 10–23
× 446 C1
or
pV = nRT and N = nNA
4.7 × 10–2
× 2.6 × 105
= n × 8.31 × 446
n = 3.3 (mol)
(C1)
N = 3.3 × 6.02 × 1023
(C1)
N = 2.0 × 1024
A1
2(c)(i)2. average increase = 2900 / (2.0 × 1024
)
= 1.5 × 10–21
J
A1
2(c)(ii) ∆EK = (3/2)k (∆)T
1.5 × 10–21
= (3/2) × 1.38 × 10–23
× (∆)T
C1
(∆)T in range 70–72 K C1
T = 173 + 273 + 70
= 520 K
A1

PUBLISHED
May/June 2018
3(a) (during melting,) bonds between atoms/molecules are broken B1
potential energy of atoms/molecules is increased B1
no/little work done so required input of energy is thermal B1
3(b)(i) (∆Q =) mc∆θ C1
loss = (160 × 0.910 × 15) + (330 × 4.18 × 15)
= 2.3 × 104
J
A1
3(b)(ii) 2.3 × 104
= (48 × 2.10 × 18) + 48L + (48 × 4.18 × 23) C1
48L = 1.66 × 104
L = 350 J g–1
A1

PUBLISHED
May/June 2018
4(a) acceleration proportional to displacement B1
acceleration directed towards fixed point
or
displacement and acceleration in opposite directions
B1
4(b)(i) 1. amplitude decreases gradually so light damping
or
oscillations continue so light damping
B1
2. loss of energy B1
due to friction in wheels
or
due to friction between wheels and surface (during slipping)
or
due to air resistance (on trolley)
B1
4(b)(ii)1. ω2
= 2k / m C1
= (2 × 230) / 0.950 C1
ω = 22 rad s–1
A1
4(b)(ii)2. T = 2π / ω C1
T = (2π / 22) = 0.286 s
time = 1.5T
= 0.43 s
A1

PUBLISHED
May/June 2018
5(a)(i) range of frequencies (of signal) B1
5(a)(ii) advantage: e.g. better quality (of reproduction)
greater rate of transfer of data
less distortion
B1
disadvantage: e.g. fewer stations (in any frequency range) B1
5(b)(i) 5.0V A1
5(b)(ii) maximum: 674 kHz A1
minimum: 626 kHz A1
5(b)(iii) T = 1 / (10 × 103
) = 1.0 × 10–4
s
minimum time = T / 2
= 5.0 × 10–5
s
A1

PUBLISHED
May/June 2018
6(a) capacitance = charge / potential M1
charge is (numerically equal to) charge on one plate A1
potential is potential difference between plates A1
6(b)(i) two in series, in parallel with the other (correct symbols) A1
6(b)(ii) two in parallel connected to one in series (correct symbols) A1
6(c)(i) capacitance = 1.2 µF A1
6(c)(ii) 1. Q = CV C1
= 1.2 × 8.0
= 9.6 µC
A1
2. E = ½QV and V = Q / C
or
E = ½CV2
and V = Q / C
or
E = ½Q2
/ C
C1
E = ½ (9.6 × 10–6
)2
/ (3.0 × 10–6
)
= 1.5 × 10–5
J
A1

PUBLISHED
May/June 2018
7(a)(i) (fraction of) output is combined with the input M1
output (fraction) subtracted/deducted from input A1
7(a)(ii) Any two valid points e.g.
• greater bandwidth/gain constant over a larger range of frequencies
• smaller gain
B2
7(b)(i) gain = 1 + (6400 / 800)
= 9.0
A1
7(b)(ii) 1. (+)5.4 V A1
2. –9.0 V A1
7(b)(iii) replace the 6400 Ω resistor with a thermistor B1

PUBLISHED
May/June 2018
8(a) electric and magnetic fields at right-angles to one another (may be shown on a clearly labelled diagram) B1
particle enters fields (with velocity) normal to the (two) fields (may be shown on a clearly labelled diagram) B1
no deviation for particles with selected velocity B1
8(b) magnetic force equals/is the centripetal force C1
Bqv = mv2
/ r C1
M = Bqr / v
= (94 × 10–3
× 1.6 × 10–19
× 0.075) / (3.4 × 104
)
M1
division by 1.66 × 10–27
shown, to give m = 20 u A1
8(c) sketch: semicircle clear (in same direction) B1
with larger radius B1

PUBLISHED
May/June 2018
9(a) (magnetic) flux density × area B1
magnetic flux density normal to area
or
reference to cross-sectional area
or
× sin (angle between B and A)
B1
× number of turns on coil B1
9(b) e.m.f. = BAN / t
or
e.m.f = rate of change of flux linkage
C1
= (7.5 × 10–3
× π × {1.2 × 10–2
}2
× 160) / 0.15
= 3.6 × 10–3
V
A1
9(c) sketch: zero for 0–0.10 s, 0.25–0.35 s, and 0.425–0.55 s, and non-zero outside these ranges B1
two horizontal steps, with zero voltage either side B1
with same polarity B1
correct values (1st step 3.6 mV and 2nd step 7.2 mV) B1

PUBLISHED
May/June 2018
10(a) emission of electron B1
when electromagnetic radiation incident (on surface) B1
10(b)(i) packet/quantum/discrete amount of energy M1
of electromagnetic radiation A1
10(b)(ii) E = hc /λ C1
= (6.63 × 10–34
× 3.00 × 108
) / (420 × 10–9
)
= 4.7 × 10–19
J
A1
10(b)(iii) sodium: yes
zinc: no
B1

PUBLISHED
May/June 2018
11(a) X-ray image(s) taken of one slice M1
(many images) taken from different angles A1
(computer) produces 2D image of slice B1
(this is) repeated for (many) slices M1
to build up a 3D image (of structure) A1
11(b)(i) combining of images involves (very) large number of calculations B1
11(b)(ii) CT scan consists of (very) many (single X-ray) images B1

PUBLISHED
May/June 2018
12(a) emission of particles/radiation by unstable nucleus B1
spontaneous emission B1
12(b)(i) P – the curve that starts with a high number
D – the curve with the peak
S – the curve that increases from zero throughout
(one correct 1 mark, all three correct 2 marks)
B2
12(b)(ii) λt½ = 0.693
λ = 0.693 / (60.0 × 60)
C1
= 1.93 × 10–4
s–1
A1
12(c) half-life of F is much shorter than half-life of E B1
nuclei of F decay (almost) as soon as they are produced B1

9702 s18 ms_all

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