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Cambridge Assessment International Education
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/11
Paper 1 Multiple Choice October/November 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 October/November 2018 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
2018
© UCLES 2018 Page 2 of 3
Question Answer Marks
1 D 1
2 D 1
3 B 1
4 B 1
5 C 1
6 D 1
7 A 1
8 C 1
9 D 1
10 A 1
11 C 1
12 A 1
13 C 1
14 A 1
15 B 1
16 B 1
17 C 1
18 B 1
19 B 1
20 C 1
21 D 1
22 C 1
23 A 1
24 D 1
25 A 1
26 A 1
27 C 1
28 C 1

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

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

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

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

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

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

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

PHYSICS 9702/21
Paper 2 AS Level Structured Questions October/November 2018
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.
components.

PUBLISHED
October/November 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
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
1(a)(i) distance in a specified direction (from a point) B1
1(a)(ii) change in velocity/time (taken) B1
1(b)(i) vertical component of velocity = (5.52
– 4.62
)1/2
= 3.0 (ms–1
)
or
5.5cosθ = 4.6 (so θ = 33.2°) and 5.5sin33.2° = 3.0 (ms–1
)
A1
1(b)(ii) s = ut + ½at2
0 = (3.0 × t) – (½ × 9.81 × t 2
)
or
v = u + at
–3.0 = 3.0 – 9.81t
C1
t = 0.61s A1
1(b)(iii) d = 4.6 × 0.61
= 2.8m
A1
1(b)(iv) E = ½mv2
C1
ratio = (½ × m × 4.62
)/(½ × m × 5.52
)
or
ratio = (½ × m × 5.52
– m × 9.81 × 0.459)/(½ × m × 5.52
)
C1
ratio = 0.70 A1
1(c) straight line from positive value of vy at t = 0 to negative value of vy M1
straight line ends at t = T and final magnitude of vy greater than initial magnitude of vy A1

PUBLISHED
2(a) (p =) mv or 4.0 × 45 or 2.0 × 85 or 89v C1
(4.0 × 45) – (2.0 × 85) = 89 v
v = 0.11ms–1
A1
2(b)(i) 1. speed of approach = 47ms–1
and
2. speed of separation = 0
A1
2(b)(ii) speed of separation less than/not equal to speed of approach and so inelastic collision A1
2(c) force is equal to rate of change of momentum B1
force on ball (by block) equal and opposite to force on block (by ball) so rates of change of momentum are equal and
opposite
B1
or
force on ball (by block) equal and opposite to force on block (by ball) (B1)
force is equal to rate of change of momentum so rates of change of momentum are equal and opposite (B1)

PUBLISHED
3(a)(i) work (done)/time (taken) B1
3(a)(ii) energy of a mass due to its position in a gravitational field B1
3(b)(i) P = Fv C1
= 2.0 × 103
× 45
= 9.0 × 104
W
A1
3(b)(ii) 1. W = (2.0 × 103
) × (45 × 3.0 × 60) or W = 9.0 × 104
× 3.0 × 60 C1
W = 1.6 × 107
J A1
2. (∆)EP = mg(∆)h C1
= 1200 × 9.81 × 3.3 × 3.0 × 60
= 7.0 × 106
J
A1
3. W = 1.6 × 107
– 7.0 × 106
= 9.0 × 106
J
A1
3(b)(iii) force = (9.0 × 106
)/(45 × 3.0 × 60)
= 1.1 × 103
N
A1
3(b)(iv) constant velocity so no resultant force B1
no resultant force so in equilibrium B1

PUBLISHED
4(a) when (two or more) waves meet (at a point) B1
(resultant) displacement is the sum of the individual displacements B1
4(b)(i) constant phase difference (between the waves) B1
4(b)(ii) 1. phase difference = 360° or 0 B1
2. path difference = 1.5λ
= 1.5 × 610
= 920nm
A1
4(b)(iii) λ = ax/D C1
x = 22/4 (= 5.5mm) or 22 × 10–3
/4 (= 5.5 × 10–3
m) C1
a = (610 × 10–9
× 2.7)/(5.5 × 10–3
)
= 3.0 × 10–4
m
A1
4(b)(iv) shorter wavelength and (so) separation decreases B1
4(b)(v) • no change to fringe separation/fringe width/number of fringes
• bright fringes are brighter
• dark fringes are unchanged
Any two of the above three points, 1 mark each.
B2

PUBLISHED
5(a) region (of space) where a force acts on a (stationary) charge B1
5(b) E = F/Q B1
F = ma and (so)
Eq
a
m
=
A1
5(c)(i) protons = 96 A1
neutrons = 148 A1
5(c)(ii) mass-energy is conserved/mass change is ‘seen’ as energy B1
energy released as gamma (radiation)/KE of α/KE of Pu B1
5(c)(iii)
9 4
2 4 0
×=
4
2
ratio or 1 9
2 7
2 7
1 9
1 06 0.19 4
1 06 6.12 4 0
1 06 6.14 −
−
−
−
××
××
×
××
××
=
101.602
ratio
C1
ratio = 1.3 A1

PUBLISHED
6(a) sum of e.m.f.(s) equal to sum of p.d.(s) M1
around a loop/around a closed circuit A1
6(b)(i) current in variable resistor = (6.0/2.4) + (6.0/1.2) (= 7.5A) C1
p.d. across variable resistor = 9.0 – 6.0 (= 3.0V) C1
R = 3.0/7.5
= 0.40Ω
A1
or
1 1 1
2.4 1.2TR
= +
RT = 0.80 (Ω)
(C1)
( )
3
9 0.80
R
R
=
+
or
3 6
0.8R
=
(C1)
R = 0.40Ω (A1)
6(b)(ii) P = V2
/R or P = I2
R or P = IV C1
P = 6.02
/24 or 2.52
× 2.4 or 6.0 × 2.5
= 15W
A1

PUBLISHED
6(b)(iii)
1.
L
R
A
ρ
=
C1
ratio = (2.4/1.2) × (3/1)
= 6.0
A1
2. (I = nAvq)
IX /IY = 2.5/5.0 or 1.2/2.4 or 0.5
C1
ratio = (2.5/5.0) × (1/3) or (1.2/2.4) × (1/3)
= 0.17
A1

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

PUBLISHED
marking principles.
descriptors.

PUBLISHED

PUBLISHED
1(a) vX = (6.02
– 4.82
)1/2
= 3.6 (ms–1
)
or
6.0sinθ = 4.8 (so θ = 53.1°) and vx = 6.0cos53.1° = 3.6 (ms–1
)
A1
1(b)(i) straight line from (0, 4.8) to (0.49, 0) M1
straight line continues with same slope to (0.98, –4.8) (labelled Y) A1
1(b)(ii) a horizontal line M1
from (0, 3.6) to (0.98, 3.6) (labelled X) A1
1(c) s = ut + ½at2
= (4.8 × 0.49) + (½ × –9.81 × 0.492
)
or
s = ½(u + v)t or area under graph
= ½ × (4.8 + 0) × 0.49
or
s = vt – ½at2
= ½ × 9.81 × 0.492
or
v2
= u2
+ 2as
s = 4.82
/ (2 × 9.81)
C1
s = 1.2m A1

PUBLISHED
1(d) (∆)E = mg(∆)h C1
E = ½mv2
C1
ratio = (½ × m × 3.62
) / (m × 9.81 × 1.2)
or
ratio = [(½ × m × 6.02
) – (m × 9.81 × 1.2)] / (m × 9.81 × 1.2)
or
ratio = (½ × m × 3.62
) / (½ × m × 4.82
)
C1
ratio = 0.56 A1
1(e) (force due to) air resistance acts in opposite direction to the velocity
or
(with air resistance, average) resultant force is larger (than weight)
B1
2(a) ampere
kelvin
(allow mole, candela)
any two correct answers, 1 mark each
B2
2(b)(i) frictional (force)/friction B1
2(b)(ii) Wcos31° × 3.0 or 90 × 6.0 C1
Wcos31° × 3.0 = 90 × 6.0
W = 210N
A1
2(b)(iii) X = 90sin31°
= 46N
A1

PUBLISHED
3(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
3(b)(i) m = ρV C1
= 1.3 × π × 0.0452
× 1.8 × 2.0 = 0.030 (kg) A1
3(b)(ii) 1. (∆)p = (∆)mv C1
= 0.030 × 1.8
= 0.054Ns
A1
2. F = 0.054 / 2.0 or 0.030 × 1.8 / 2.0
= 0.027N
A1
3(b)(iii) force on air (by propeller) equal to force on propeller (by air) M1
and opposite (in direction) A1
3(b)(iv) resultant force = 0.20 × 0.075 (= 0.015N)
frictional force = 0.027 – 0.015
C1
= 0.012N A1

PUBLISHED
4(a) vibration(s)/oscillation(s) (of particles) parallel to direction of propagation of energy B1
4(b)(i) phase difference = 180° A1
4(b)(ii) v = fλ C1
λ / 2 = 25 (cm) or 0.25 (m) C1
f = 330 / 0.50
= 660Hz
A1
4(b)(iii) (readings from graph =) 2.6 and 4.0 C1
ratio = (2.6 / 4.0)1/2
= 0.81
A1
5(a) nλ = dsinθ C1
λ = 640 × 10–9
(m) C1
2 × 640 × 10–9
= 1.7 × 10–6
× sinθ so θ = 49(°) A1
5(b) 2 × 640 × 10–9
= 3 × λ
or
1.7 × 10–6
× sin 49° = 3 × λ
C1
λ = 4.3 × 10–7
m A1

PUBLISHED
6(a) joule /coulomb B1
6(b)(i) 7.0 = (I × 5.2) + (I × 6.0) + 1.4 C1
I = 0.50A A1
6(b)(ii) R = 1.4 / 0.50
= 2.8Ω
A1
6(b)(iii) P = EI or P = VI or P = I2
R or P = V2
/ R C1
efficiency = [(0.502
×6.0) / (7.0×0.50)] (×100)
or
efficiency = [(0.50×3.0) / (7.0×0.50)] (×100)
or
efficiency = [(3.02
/ 6.0) / (7.0×0.50)] (×100)
C1
efficiency = 43% A1
6(b)(iv) R = ρl / A C1
α = ρ / R
= 3.7 × 10–7
/ 6.0
= 6.2 × 10–8
m
A1

PUBLISHED
7(a) A: (cross-sectional) area (of wire) B1
n: number of free electrons per unit volume or number density of free electrons B1
7(b) line drawn between (X, vx) and (Y, 4vx) M1
line has increasing gradient A1
8(a) antineutrino and positron both underlined (and no other particles) B1
8(b)(i) nucleon number = 27 A1
proton number = 13 A1
8(b)(ii) weak (nuclear force/interaction) B1
8(b)(iii) an (electron) antineutrino / ( )e
ν is produced (and this has energy) B1
X has kinetic energy B1

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

PUBLISHED
marking principles.
descriptors.

PUBLISHED
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
1(a) current
temperature
(allow amount of substance, luminous intensity)
any two correct answers, 1 mark each
B2
1(b)(i) W = 2 × (150 × sin17°) or 2 × (150 × cos73°) C1
W = 88N A1
1(b)(ii) 1. σ = F / A C1
= 150 /(7.5 × 10–6
)
= 2.0 × 107
Pa
A1
2. ε =σ / E C1
= 2.0 × 107
/(2.1 × 1011
)
= 9.5 × 10–5
A1

PUBLISHED
2(a) energy (of a mass/body/object) due to motion/speed/velocity B1
2(b)(i) E = ½mv2
C1
480 = ½ × m × 802
so m = 0.15kg A1
2(b)(ii) 1. E = mgh or ∆E = mg∆h C1
= 0.15 × 9.81 × 210
= 310J
A1
2. work done = 480 – 310
= 170J
A1
2(b)(iii) work done = Fs C1
force = 170/210
= 0.81N
A1
2(b)(iv) curved line from positive value on v-axis to (T, 0) M1
magnitude of gradient decreases A1
2(b)(v) as shell rises force decreases and as shell falls force increases B1
as shell rises force is downward and as shell falls force is upward B1
or
as shell rises the force decreases and is downward (B1)
as shell falls the force increases and is upward (B1)

PUBLISHED
3(a) (resultant) force proportional/equal to rate of change of momentum B1
3(b)(i) ρ = m/V C1
V = π × (7.5 × 10–3
)2
× 13 × 0.2 (= 4.59 × 10–4
m3
)
m = π × (7.5 × 10–3
)2
× 13 × 0.2 × 1000 = 0.46kg
A1
3(b)(ii) 1. (∆)p = (∆m)v C1
(∆)p = 0.46 × 13
= 6.0Ns
A1
2. F = 6.0/0.20
= 30 N
A1
3(b)(iii) force on water (by rocket/nozzle) equal to force on rocket/nozzle (by water) M1
in the opposite direction A1
3(b)(iv) 1. mass = 0.40 + 0.70 – 0.46 = 0.64kg A1
2. acceleration = [30 – (0.64 × 9.81)]/0.64 or 30/0.64 – 9.81 C1
= 37ms–2
A1

PUBLISHED
4(a) graph with x-axis labelled ‘distance’ and wavelength/λ correctly shown B1
graph with x-axis labelled ‘time’ and period/T correctly shown B1
graph with y-axis labelled ‘displacement’ and amplitude/A correctly shown B1
4(b)(i) wave (moves along string and) reflects at fixed point/Y/X/end/wall/boundary B1
the incident and reflected waves interfere/superpose B1
4(b)(ii) 100/40 or 2.5 (cycles/periods/T) C1
1. displacement = 0 B1
2. distance = 130mm A1
4(b)(iii) 1. f = 1/40 × 10–3
= 25Hz
A1
2. v = fλ or λ = vT C1
λ = 30/25 or 30 × 40 × 10–3
(= 1.2m) C1
distance = 1.2 × 1.5
= 1.8m
A1

PUBLISHED
5(a) E = F/Q M1
F = ma and (so) q/m = a/E A1
5(b) m = (4 × 1.60 × 10–19
× 3.5 × 104
)/1.5 × 1012
(= 1.49 × 10–26
kg) B1
= 1.49 × 10–26
/1.66 × 10–27
= 9.0 (u) A1
5(c) protons: 4
and
neutrons: 5
A1
5(d)(i) nuclei have the same charge and so same (magnitudes of) force B1
5(d)(ii) nuclei have different masses and same force and so different (magnitudes of) acceleration B1
6(a) (coulomb is an) ampere second B1
6(b) 8.0 × 10–19
C and 1.6 × 10–19
C both underlined (and no others underlined) B1
6(c) line drawn between (S, 1.00vs) and (T, 0.25vs) M1
line with decreasing magnitude of gradient A1

PUBLISHED
7(a) sum of current(s) in(to) junction = sum of current(s) out of junction
or
(algebraic) sum of current(s) at a junction is zero
B1
7(b)(i) 1. potential difference = 0 A1
2. potential difference = 9.6V A1
7(b)(ii) for resistance in parallel: (1 / RT) = (1 /400) + (1 / 400)
RT = 200(Ω)
C1
V/9.6 = 200/600 C1
V = 3.2V A1

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

PUBLISHED
marking principles.
descriptors.

PUBLISHED

PUBLISHED
1(a) Value of x with unit in the range 23.5–24.5cm. 1
1(b) Value of T with unit in the range 1.0–1.5s and to 0.1s or better. 1
1(c) Six sets of readings of x (different values) and time without help from the Supervisor and showing the correct trend scores 5
marks, five sets scores 4 marks etc.
5
Range: Minimum value of x ˂ 10.5 cm. 1
Column headings:
Each column heading must contain a quantity, a unit and a separating mark where appropriate.
The presentation of the quantity and the unit must conform to accepted scientific convention e.g. T2
x / s2
m.
1
Consistency: All raw values of x must be given to the nearest mm only. 1
Significant figures:
All values of x2
must be given to the same number of significant figures as (or one more than) the number of significant figures
in raw values of x.
1
Calculation: Correct calculation of T2
x. 1
1(d)(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:
General trend of points on graph must be positive.
All points in the table (at least 5) must be plotted on the grid.
It must be possible to draw a straight line that is within ±0.005m2
on the x-axis of all plotted points.
1

PUBLISHED
1(d)(ii) Line of best fit:
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 (i.e. circled or labelled).
Line must not be kinked or thicker than half a small square.
1
1(d)(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 on answer line must match graph.
1
y-intercept:
Correct read-off from a point on the line 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.
1
1(e) Value of A = candidate’s gradient and value of B =candidate’s intercept.
The values must not be fractions.
1
Unit for A correct (s2
m–1
, s2
cm–1
or s2
mm–1
) and unit for B correct (s2
m, s2
cm or s2
mm). 1

PUBLISHED
2(a)(i) Value of M with unit and to the nearest 0.1g or better. 1
2(a)(ii) Correct calculation of m dividing the value in (a)(i) by 25. 1
2(a)(iii) Justification for significant figures in m linked to the s.f. in M. 1
2(b)(i) Value of S in range 1.0g ⩽ S ⩽ 2.0g with consistent unit. 1
2(b)(ii) Correct calculation of c. 1
2(c)(i) Values p and q such that p + q = 25. 1
Evidence of repeated values of p. 1
2(c)(ii) Percentage uncertainty in p based on absolute uncertainty of 1 or 2.
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(d) Second values of p and q. 1
Quality: second value of q less than first value of q. 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
2(f)(i) A Two readings are not enough to draw a (valid) conclusion (not “not enough for accurate results”, “few readings”).
B Difficulty linked to board during movement of clips e.g. paper clips flip over when moving affecting friction/clips move to
the side/clips catch on end of board/board irregular and clips catch on edge/board moves on table as well as clips.
C Difficulty linked to placing or keeping the spheres on the clips e.g. spheres fall off/clips too small to hold the sphere/
difficult to get two neighbouring paper clips level enough to take the two spheres/spheres touch table.
D Difficulty linked to size of clips with a reason e.g. large increments of mass gives large uncertainty in the value of p or
q/the exact force needed to cause movement may not be equal to a whole number of clips.
E Masses of clips or spheres may be different.
1 mark for each point up to a maximum of 4.
4
2(f)(ii) A Take many readings (for different values of n) and plot a graph or take more values of k and compare.
B Workable method e.g. provide a guide/over a pulley wheel/tape or clamp board to bench/sand the end of the board/sand
out irregularities.
C Workable method of adding/holding spheres e.g. use cuboid/disc shaped plasticine/tie paper clips together with thin
cotton to keep flat.
D Improved method to deal with large clips e.g. use more and smaller/lighter clips.
E Check masses using a balance.
4

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

PUBLISHED
marking principles.
descriptors.

PUBLISHED

PUBLISHED
1(a) Value of I with unit and in the range 0.010A ⩽ I ⩽ 0.200A. 1
1(b) Value of y with unit and in the range 20.0–75.0cm. 1
1(c) Six sets of readings of P, Q (different pairs) and y without help from the Supervisor scores 4 marks, five sets scores 3 marks
etc.
4
Range: Resistor values include the pair (15Ω and 12Ω) and the pair (27Ω and 22Ω). 1
Column headings:
Each column heading must contain a quantity, a unit and a separating mark where appropriate.
The presentation of quantity and unit must conform to accepted scientific convention e.g. y/m and PQ / (P + Q) /Ω.
1
Consistency: All values of y must be given to the nearest mm only. 1
Significant figures: Values of PQ/ (P + Q) must be given to 2 or 3 significant figures. 1
Calculation: Values of PQ / (P + Q) are correct. 1
1(d)(i) Axes:
1
Plotting of points:
1
Quality:
General trend of points on graph must be negative.
It must be possible to draw a straight line that is within ±0.050m (to scale) on the y-axis of all plotted points.
1

PUBLISHED
1
1(d)(iii) Gradient:
1
y-intercept:
or
1
1(e) Value of M = –candidate’s gradient and value of N =candidate’s intercept.
1
Units for M (e.g. mΩ–1
or cmΩ–1
or mmΩ–1
) and N (e.g. m or cm or mm) correct. 1
1(f) Correct calculation of E with unit. 1

PUBLISHED
2(a) Value of S with unit in the range 0.5cm ⩽ S ⩽ 2.5cm and to the nearest mm. 1
2(b)(i) Values of S1 and S2 and S1 > S2. 1
Value of raw θ to the nearest degree and value of θ < 45° on answer line. 1
2(b)(ii) Percentage uncertainty in θ based on an absolute uncertainty of 3–10°.
1
2(b)(iii) Correct calculation of e1 cosθ and e2 sinθ. 1
2(b)(iv) Justification for significant figures in e1 cosθ linked to s.f. in:
(S1 – S) and θ
or
e1 and θ
or
S, S1 and θ.
1
2(c) Second values of S1 and S2. 1
Quality: Second values of S1 and S2 greater than first values of S1 and S2. 1
Second value of θ. 1
2(d)(i) Two values of β calculated correctly. 1
2(d)(ii) Valid comment consistent with the calculated values of β, testing against a criterion stated by the candidate. 1
2(e) Correct calculation of k using second value of β with a correct unit. 1

PUBLISHED
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 judge the vertical.
C Difficult to determine θ/angle because of parallax error or cannot see the lines defining the angle.
D Difficult to determine θ/angle because cannot hold the protractor steady.
E Difficult to measure S1 or S2 with reason e.g. springs move or mass moves.
F The stands tilt/topple/slide or the bosses twist on loading.
4
2(f)(ii) A Take many readings (for different values of m) and plot a graph or take more values of β and compare (not “repeat
readings” on its own).
B Method to get vertical reference e.g. plumb line/use of set square on bench with metre rule/use spirit level vertically.
C Use a larger protractor/have a protractor drawn on card at the back/use projection ideas/take a photograph and measure
angle.
D Clamp protractor (valid method).
E Clamp ruler/take photograph with scale or ruler in the frame.
F Use G-clamps/clamp the stands/use heavy weights/lower the bosses.
4

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

PUBLISHED
marking principles.
descriptors.

PUBLISHED

PUBLISHED
1(a) Value of V with unit and in the range 2.00V ⩽ V⩽ 4.00V. 1
1(b) Value of t with unit and in range 1.00–25.00 s. 1
Evidence of repeat readings of t. 1
1(c) Six sets of readings of n and t collected with correct trend (t decreases as n increases) and collected without help from the
Supervisor scores 5 marks, five sets scores 4 marks, etc.
5
Range: nmax ⩾ 7 and nmin = 1. 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. t/s.
There must be no unit for n or
1
n
.
1
Consistency: All values of t must be given to 0.01s or all to 0.1s. 1
Calculation: Values of
1
n
calculated correctly.
1

PUBLISHED
1(d)(i) Axes:
1
Plotting of points:
Diameter of points must be ⩽ half a small square (no “blobs”).
Points must be accurate to within half a small square in both x and y directions.
1
Quality:
All points in the table (at least 5) must be plotted for this mark to be awarded.
It must be possible to draw a straight line that is within ±1.0s from a straight line in the t direction.
1
1
1(d)(iii) Gradient:
1
y-intercept:
or
1

PUBLISHED
1(e) Value of a equal to candidate’s gradient and value of b equal to candidate’s intercept.
1
Unit for a is s and unit for b is s. 1

PUBLISHED
2(a)(i) Raw value(s) for d with unit and to nearest 0.01 mm.
Answer on answer line in range 20.00mm ⩽ d ⩽ 30.00mm.
1
2(a)(ii) Value for w with unit and in range 10.0–15.0 mm. 1
2(b)(i) Value for x to nearest mm.
Answer in range 55.0cm ⩽ x ⩽ 65.0cm.
1
2(b)(ii) Value for t with unit and in range 2.00–4.00s. 1
2(b)(iii) Percentage uncertainty in t based on an absolute uncertainty in the range 0.2–0.5s.
1
2(b)(iv) Correct calculation of v with consistent unit. 1
2(b)(v) Justification linked to significant figures in x and t. 1
2(c) Second value of w. 1
Second value of t. 1
Quality: t greater for greater w. 1
2(d)(i) Two values of k calculated correctly. 1
2(d)(ii) Valid comment consistent with the calculated values of k, testing against a criterion stated by the candidate. 1

PUBLISHED
2(e)(i) A Two readings are not enough to draw a (valid) conclusion (not “not enough for accurate results”, “few readings”).
B Large percentage uncertainty in w or large uncertainty in w because w is small
or
w varies along the length of the track.
C Difficult to align sphere in correct starting position.
D Sphere won’t start without push/force applied when releasing sphere.
E Large percentage uncertainty in t or large uncertainty in t because t is small.
4
2(e)(ii) A Take more readings and plot a graph or calculate more k values and compare (not “repeat readings” on its own).
B Use (vernier/digital) calipers (to measure w).
C Use a stop/card gate.
D Use steeper slope/use steel ball and electromagnet.
E1 Use longer track/use shallower slope.
E2 Improved method of measuring t, e.g. use light gates at top and bottom/use motion sensor in direction of rolling
or video/film/record experiment with timer in view/view frame-by-frame.
4

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

PUBLISHED
marking principles.
descriptors.

PUBLISHED

PUBLISHED
1(a) Answer line value of 0.5cm ⩽ x0 ⩽ 2.5cm with unit. 1
1(b) Value of raw θ to the nearest degree with unit and answer line value of θ < 90°. 1
1(c) Six sets of readings of x and θ (different values) with the correct trend and without help from Supervisor scores 5 marks, five
sets scores 4 marks etc.
5
Range: Values of θ must include one value greater than and one value less than the value of θ in (b). 1
Column headings:
The presentation of quantity and unit must conform to accepted scientific convention e.g. x/ cm, e / cm, e/ cosθ (m), θ /°,
tan θ without a unit.
1
Consistency: All raw values of x must be given to the nearest mm only. 1
All values of e/ cos θ must be given to the same number of significant figures as (or one more than) the number of significant
figures in e or θ, whichever is the smaller.
1
Calculation: Values of e/ cosθ are correct. 1

PUBLISHED
1(d)(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:
Diameter of plotted points must be ⩽ half a small square.
1
Quality:
Trend of points on graph must be negative.
It must be possible to draw a straight line that is within ±0.05 (to scale) on the tanθ axis (normally x-axis) of all plotted points.
1
Allow one anomalous point only if clearly indicated (i.e. circled or labelled). There must be at least 5 points left after the
anomalous point is disregarded.
Lines must not be kinked or thicker than half a small square.
1
1(d)(iii) Gradient:
1
y-intercept:
or
Intercept read directly from the graph, with read-off at tanθ = 0, accurate to half a small square.
1

PUBLISHED
1(e) Value of N = candidate’s gradient and value of M = candidate’s intercept.
1
Units for N and M correct (e.g. m or cm or mm). 1

PUBLISHED
2(a)(i) All values of raw d to the nearest 0.01mm and answer line value of d in the range 0.50mm ⩽ d ⩽ 5.00mm. 1
Evidence of repeats. 1
2(a)(ii) Absolute uncertainty in d in the range 0.02–0.2mm.
1
2(a)(iii) Correct calculation of A with correct unit. 1
2(a)(iv) Justification for significant figures in A linked to significant figures in d. 1
2(b)(i) Value of p. 1
2(b)(ii) Correct calculation of (p + q) / 2. 1
2(c) Second value of d. 1
Second values of p and q. 1
Quality: Second value of (p + q) / 2 less than first value of (p + q) / 2. 1
2(d)(i) Two values of k calculated correctly. 1
2(d)(ii) Valid comment consistent with the calculated values of k, testing against a criterion stated by the candidate. 1

PUBLISHED
2(e)(i) A Two readings are not enough to draw a (valid) conclusion (not “not enough for accurate results”, “few readings”).
B Difficulty with measuring d with a reason e.g. string is soft/string is compressed (by micrometer)/diameter not uniform
along length.
C Difficulty linked to placing paper clips on hanger with a reason e.g. clips slide off/added force from hand/dropping clips
starts movement/difficult not to touch hanger when releasing clips/hanger too small for clips.
D Idea of increments being too large e.g.
large increments of mass (large uncertainty in the value of p or q)
or only a small number of clips are added
or exact force needed to cause movement may not equal a whole number of clips
or mass of each paper clip is too large
E Mass of clips may be different
or
two strings are of different material
4
2(e)(ii) A Take more readings (for different values of d) and plot a graph or take more readings and compare k values (not “repeat
readings” on its own).
B Workable improved method e.g. wrap string round a rod and measure many d/use travelling microscope.
C Workable method of adding or holding paper clips e.g. add a hook to each mass hanger/use a wider mass hanger or
tweezers.
D Improved method of loading e.g. use smaller clips/lighter clips/smaller masses to add to mass hanger or use of
sand/water with detail.
E Method to check clips have the same mass e.g. use a balance.
4

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

PUBLISHED
marking principles.
descriptors.

PUBLISHED

PUBLISHED
1(a) Value of L0 with unit. Value on answer line in the range 1.50—2.50 cm. 1
1(b) Value of θ with unit, to the nearest degree and θ ⩽ 90°. 1
1(c) Six sets of readings of θ and L with the correct trend and without help from the Supervisor scores 5 marks, five sets scores 4
marks, etc.
5
Range: θmax ⩾ 30° and θmin ⩽ 15°. 1
Column headings:
The presentation of quantity and unit must conform to accepted scientific convention e.g. L/mm. There must be no unit for
(sinθ)(cosθ).
1
Consistency: All values of L must be given to the nearest mm only. 1
Number of significant figures for every value of (sinθ)(cosθ) same as, or one greater than, the number of s.f. of θ as recorded
in table.
1
Calculation: Values of (sinθ)(cosθ) calculated correctly. 1

PUBLISHED
1(d)(i) Axes:
1
Plotting of points:
1
Quality:
It must be possible to draw a straight line that is within ±0.05 on the x-axis of all plotted points.
1
1
1(d)(iii) Gradient:
1
y-intercept:
or
1

PUBLISHED
1(e) Value of a equal to candidate’s gradient and given to two or more significant figures.
and
Value of b equal to candidate’s intercept.
1
Unit for a correct and unit for b correct (e.g. m, cm, mm). 1

PUBLISHED
2(a) Value of h0 with unit and to the nearest mm. 1
2(b)(i) Value of h less than h0. 1
2(b)(ii) Correct calculation of y, with unit. 1
2(c) Percentage uncertainty based on an absolute uncertainty in y of 3–6mm.
1
2(d) Value for T on answer line in range 0.2–0.9 s, with unit. 1
Evidence of repeat readings of time. There must be at least two measurements of nT where n ⩾ 5. 1
2(e) Second value of h. 1
Second value of T. 1
Quality: T greater for greater y. 1
2(f)(i) Two values of c calculated correctly. 1
2(f)(ii) Valid comment consistent with the calculated values of c, testing against a criterion specified by the candidate. 1
2(g) Correct calculation of g with consistent unit. 1

PUBLISHED
2(h)(i) A Two readings are not enough to draw a (valid) conclusion (not “not enough for accurate results”, “few readings”).
B Rule may not be vertical (when measuring heights).
C Difficult to measure h0 because blade not horizontal/blade is bent/h0 varies along blade.
D Difficult to measure h because mass is in the way/parallax when measuring h/difficult to judge position of the centre of the
mass.
E Difficult to determine T with a reason e.g. difficult to judge start/end/completion of oscillations or difficult to count
oscillations.
4
2(h)(ii) A Take many readings and plot a graph or take more values of c and compare (not “repeat readings” on its own).
B Method to check rule vertical, e.g. use set square on floor/use plumb line/use spirit level.
C Mark position for centre of the mass on blade then measure h0 at that position.
D Improved method of measuring h, e.g. (clamp) rule and use set square as pointer/hang mass from thread/use a mass
that is narrower than the blade/measure blade height at both sides of mass and average.
E Video/film/record with timer in view (or use frame-by-frame)/use motion sensor above (or below) the blade
or
use larger masses to give longer T/to make counting easier.
4

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

PUBLISHED
marking principles.
descriptors.

PUBLISHED

PUBLISHED
1(a)(i) work done per unit mass B1
work done moving mass from infinity (to the point) B1
1(a)(ii) (near Earth’s surface change in) height ≪ radius or height much less than radius B1
potential inversely proportional to radius and radius approximately constant (so potential approximately constant) B1
1(b) initial kinetic energy = (–) potential energy (at surface)
or
½mv2
= GMm/r
B1
v2
= (2 × 6.67 × 10–11
× 7.4 × 1022
)/ (0.5 × 3.5 × 106
) C1
v = 2.4 × 103
ms–1
A1
2(a) sum of potential and kinetic energies (of molecules/atoms/particles) B1
(energy of) molecules/atoms/particles in random motion B1
2(b)(i) final temperature = initial temperature B1
no change in internal energy B1
2(b)(ii) 1. work done on gas (P→Q): 0 A1
increase in internal energy (P→Q): (+)97.0J A1
2. increase in internal energy (Q→R): –42.5J A1
3. increase in internal energy (R→P): –54.5J A1
thermal energy supplied (R→P): –91.5J A1

PUBLISHED
3(a) ω2
= 2g / L C1
T = 2π /ω C1
ω2
= (2 × 9.81) / 0.19
ω = 10.2 (rad s–1
)
T = 2π/10.2
= 0.62s
A1
3(b)(i) e.g. viscosity of liquid/friction within the liquid/viscous drag/friction between walls of tube and liquid B1
3(b)(ii) (maximum) KE = ½mv0
2
and v0 = ωx0
or
energy = ½mω2
x0
2
C1
change = ½ × 18 × 10–3
× 103 × [(2.0 × 10–2
)2
– (0.95 ×10–2
)2
] C1
= 2.9 × 10–4
J A1

PUBLISHED
4(a) pulses (of ultrasound from generator) B1
reflected at boundaries (between media) B1
time delay (between transmission and receipt) gives information about depth B1
intensity of reflected pulse gives information about nature (of tissues)/type (of tissues)/boundary B1
Any two from:
• (reflected pulses) detected by the (ultrasound) generator
• gel used to minimise reflection at skin/maximise transmission into skin
• degree of reflection depends upon impedances of two media (at boundary)
B2
4(b)(i) product of density and speed M1
speed of ultrasound in medium A1
4(b)(ii) Z1 about equal to Z2 results in negligible/no reflection B1
Z1 ≫ Z2 (or Z1 ≪ Z2) results in mostly reflection B1

PUBLISHED
5(a) Any two reasonable suggestions e.g.:
• noise can be eliminated/(signal/data) can be regenerated
• bits can be added to correct for errors
• data compression/multiplexing (is possible)
• signal can be encrypted/better security
B2
5(b) sketch: series of seven steps B1
each step width 2ms B1
correct levels in correct order (2, 5, 14, 4, 9, 11, 7)
(1 mark for 6 levels correct, 2 marks for 7 levels correct)
A2
5(c)(i) step width reduced
or
higher frequencies can be reproduced
B1
5(c)(ii) step height reduced
or
smaller changes in signal (intensity) can be reproduced
B1
6(a)(i) work done per unit charge B1
work done moving positive charge from infinity (to the point) B1
6(a)(ii) field strength = potential gradient M1
‘–’ sign included or directions discussed A1
6(b)(i) gain in kinetic energy (= loss in potential energy) = charge × p.d. or qV = ½mv2
M1
so v is independent of separation (because separation not in expressions) A1

PUBLISHED
6(b)(ii) (at x = 0.40cm), potential = (–) 75 × 0.40 /1.2
(= (–) 25V)
C1
½mv2
= qV
½ × 4 × 1.66 × 10–27
× v2
= 2 × 1.60 × 10–19
× 25
C1
or
a = Vq/ dm and v2
= 2as (C1)
v2
= (2 × 75 × 2 × 1.60 × 10–19
× 0.40 × 10–2
) / (1.2 × 10–2
× 4 × 1.66 × 10–27
) (C1)
v = 4.9 × 104
ms–1
A1
7(a)(i) gain is constant M1
for all frequencies A1
7(a)(ii) no time delay between input (voltage) and output (voltage) B1
clear reference to change(s) in input and/or output (voltages) B1
7(b) diagram: VIN connected to V+
only B1
midpoint between resistors R1 and R2 connected to V–
only B1
7(c)(i) –3.6V A1
7(c)(ii) (+)5.0V A1

PUBLISHED
8(a) region where there is a force M1
experienced by a current-carrying conductor/moving charge/(permanent) magnet A1
8(b)(i) single path, deflection in ‘upward’ direction B1
acceptable circular arc in whole field B1
no ‘kinks’ at start or end of curvature, and straight outside region of field B1
8(b)(ii) force (on particle) is normal to velocity/direction of motion/direction of speed B1
8(c) magnetic force provides/is the centripetal force B1
Bqv = mv2
/r or r = mv/Bq C1
(if q is doubled), new speed = 2v A1

PUBLISHED
9(a) (induced) e.m.f. proportional/equal to rate M1
of change of (magnetic) flux (linkage) A1
9(b)(i) induced e.m.f. = (∆B)AN/∆t
= (2 × 0.19 × 1.5 × 10–4
× 120) / 0.13
C1
= 0.053V A1
9(b)(ii) reading on voltmeter connected to coil C/V: 0 0.053 0
(all three values required)
A1
reading on voltmeter connected to Hall probe/V: zero in middle column B1
final column correct sign (negative) B1
final column correct magnitude (0.20) B1
10 Any five points from:
• as temperature rises electrons gain energy
• electrons enter conduction band
• (positively charged) holes left in valence band
• more charge carriers (so resistance decreases)
• (as temperature rises,) lattice vibrations increase
• effect of increase in number of electrons or holes or charge carriers outweighs effect of increased lattice vibrations (so
resistance decreases)
B5

PUBLISHED
11(a) discrete amount/quantum/packet of energy M1
of electromagnetic radiation A1
11(b)(i) energy = hc/λ C1
λ = (6.63 × 10–34
× 3.00 × 108
)/(0.51 × 106
× 1.60 × 10–19
)
= 2.4 × 10–12
m
A1
11(b)(ii) p = h/ λ
= (6.63 × 10–34
)/(2.44 × 10–12
)
or
p = E/ c
= (0.51 × 1.60 × 10–13
)/(3.00 × 108
)
C1
p = 2.7 × 10–22
Ns A1
11(c)(i) E = c2
∆m C1
∆m = (0.51 × 1.60 × 10–13
)/(3.00 × 108
)2
= 9.1 × 10–31
kg
A1
11(c)(ii) (momentum is conserved so) nucleus must have momentum in opposite direction to photon B1

PUBLISHED
12(a) unstable nucleus B1
emission of particles/photons B1
emission is spontaneous
or
(particles/radiation) are ionising
B1
12(b)(i) tangent drawn and gradient calculation attempted B1
activity = 1.3 × 106
Bq
(1 mark for answer within ±0.2 × 106
Bq, 2 marks for answer within ±0.1 × 106
Bq)
A2
12(b)(ii) A = λN C1
λ = (1.3 × 106
)/(3.05 × 1010
) = 4.3 × 10–5
s–1
(≈ 4 × 10–5
s–1
) A1
12(c) A = A0e–λt
1.0 × 103
= 4.6 × 103
exp(–5.5 × 10–7
× t)
C1
ln (4.6) = 5.5 × 10–7
× t C1
t = 2.78 × 106
s
= 32 days
A1

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

PUBLISHED
marking principles.
descriptors.

PUBLISHED

PUBLISHED
1(a)(i) force per unit mass B1
1(a)(ii) acceleration = F/m, field strength = F/m, so equal B1
1(b) smooth curve between R and 4R with negative gradient of decreasing magnitude B1
line passing through (R, 1.00g) and (2R, 0.25g) B1
line ending at (4R, 0.0625g) B1
1(c) M = (4/3 × πR3
)ρ C1
g = GM/(2R)2
C1
g = ⅓ × 6.67 × 10–11
× π × 3.4 × 106
× 4.0 × 103
= 0.95ms–2
A1

PUBLISHED
2(a) gas that obeys equation pV = constant × T M1
symbols p,V and T explained A1
2(b)(i) pV = ⅓ Nm<c2
> and M = Nm
(and so) p = ⅓ρ<c2
>
C1
2.12 × 107
= ⅓ × [3.20/(1.84 × 10–2
)]×<c2
> C1
cr.m.s. = 605ms–1
A1
2(b)(ii) 1. pV = nRT and T = (22 + 273)K C1
n = (2.12 × 107
× 1.84 × 10–2
)/(8.31 × 295)
= 159 mol
A1
2. mass = 3.20/(159 × 6.02 × 1023
)
or
mass = [2 × (3/2) × 1.38 × 10–23
× 295]/6052
C1
mass = 3.34 × 10–26
kg A1
2(c) A = (3.34 × 10–26
)/(1.66 × 10–27
)
= 20
A1

PUBLISHED
3(a) (thermal) energy per unit mass (to cause change of state) B1
(energy transfer during) change of state between solid and liquid at constant temperature B1
3(b)(i) Any one from:
• rate of increase in mass (of beaker and water) is constant
• level of water rises at a constant rate
• volume of water (in beaker) increases at a constant rate
• constant time between drops
• constant rate of dripping
B1
3(b)(ii) (electrical power supplied =) 12.8 × 4.60
(= 58.9W)
C1
(rate of transfer to ice =) [(185.0 – 121.5) × 332]/[5.00 × 60]
(= 70.3W)
C1
1. rate = 70.3W A1
2. rate = 70.3 – 58.9
= 11.4W
A1

PUBLISHED
4(a) (defining equation of s.h.m. is) a = –kx where k is a constant or a ∝ –x B1
g and L are constant (so a ∝ –x and hence s.h.m.) B1
4(b) T = 0.50s and T = 2π/ω C1
ω2
= 2g/L C1
L = (2 × 9.81 × 0.502
)/4π2
= 0.12m
A1
4(c)(i) Any one from:
• viscosity of liquid
• friction within the liquid
• viscous drag
• friction/resistance between walls of tube and liquid
B1
4(c)(ii) (maximum) KE = ½mv0
2
and v0 = ωx0
or
energy = ½mω2
x0
2
C1
ratio = (1.3 / 2.0)2
= 0.42
A1

PUBLISHED
5(a) amplitude of carrier (wave) varies B1
variation in synchrony with displacement of information signal B1
5(b)(i) wavelength = (3.0 × 108
)/(900 × 103
)
= 3.3 × 102
m
A1
5(b)(ii) amplitude varies (continuously) between a maximum and a minimum B1
variations repeat 5000 times each second
or
variations repeat every 0.2ms
or
variations above and below 4.0V
B1
5(b)(iii) 10000Hz A1
5(c)(i) Any two from:
• (orbit is) above the Equator
• (orbit is) from west to east/same direction as Earth’s rotation
• orbit is circular/orbit has a particular radius
B2
5(c)(ii) 1. minimal reflection/absorption/attenuation by atmosphere
or
maximum penetration of/transmission through atmosphere
B1
2. uplink signal is greatly attenuated/must be greatly amplified B1
prevents downlink signal swamping the uplink signal B1

PUBLISHED
6(a)(i) work done per unit charge B1
work done moving positive charge from infinity (to the point) B1
6(a)(ii) field strength = potential gradient M1
negative sign included or directions discussed A1
6(b) horizontal straight lines, at non-zero potential, within the spheres B1
magnitude of potential greater at surface of sphere A than at surface of sphere B B1
concave curve between A and B, with a minimum nearer to B B1
lines show V positive all the way from 0 to D B1
7(a) R/RT = 2.4/1.8
or
at 4.0°C, RT = 3.2kΩ
C1
hence R / 3.2 = 2.4/1.8
R = 4.3kΩ
A1
7(b) RT = 3.37kΩ
or
RT is greater (than 3.2kΩ)
B1
V+
> V–
M1
hence output is +5.0V A1

PUBLISHED
7(c) correct LED symbol B1
two diodes shown connected, in parallel and with opposite polarities, between VOUT and earth M1
diodes labelled to show correct polarities consistent with (b)
(G pointing from VOUT to earth and B pointing from earth to VOUT if (b) correct)
A1
8(a) force per unit current B1
force per unit length (of wire) B1
current normal to (magnetic) field B1
8(b)(i) forces (on PQ and RS) are horizontal B1
(hence they create) no moment about the pivot B1
or
forces (on PQ and RS) are equal and opposite (B1)
(hence there is) no net force (on the two sections) (B1)
8(b)(ii) realisation of the need to apply moments C1
BILx = mgy
B × 2.7 × 1.2 × 10–2
× 7.5 = 45 × 10–6
× 9.81 × 8.8
C1
B = 1.6 × 10–2
T A1

PUBLISHED
9(a) 0 → t1 horizontal straight line at non-zero value of VH
and
t3 → t4 horizontal straight line at different non-zero VH
B1
t1 → t3 straight diagonal line with negative gradient
and
graph line starts at (0, V0) and ends at (t4, –2V0)
B1
9(b) E = 0 for 0 → t1 and t3 → t4 B1
E is non-zero at all points between t1 → t3 M1
E has constant magnitude between t1 → t3 A1

PUBLISHED
10(a) V0 = √2 × Vr.m.s. = √2 × 9.9 (= 14V)
and
ω = 2πf = 2π × 50 (= 314rads–1
)
C1
V = 14 sin 314t A1
10(b) enables (resonating) nuclei to be located B1
resonant frequency depends on magnetic field strength B1
Any one from:
• non-uniform field is (accurately) calibrated
• (non-uniform) field may be varied to enable detection in different positions
• unique (magnetic) field strength/frequency at each point
B1
10(c) I = I0 exp(–µx) C1
I = I0 [exp(–µx)bone × exp(–µx)soft tissue]
I = I0 [exp(–2.9 × 0.40) × exp(–0.92 × 1.4)]
C1
I/I0 = 0.0865 C1
ratio/dB = 10 lg 0.0865
= –11dB
A1

PUBLISHED
11(a) discrete amount/quantum/packet of energy M1
of electromagnetic radiation A1
11(b) mostly dark/dark background B1
coloured lines B1
11(c)(i) 6 A1
11(c)(ii) 1. maximum photon energy = 13.6 – 0.85
(= 12.75eV)
C1
maximum kinetic energy = (13.6 – 0.85) – 5.6
= 7.2eV
A1
2. energy = hc/λ C1
λ = (6.63 × 10–34
× 3.00 × 108
) / [(13.6 – 0.85) × 1.60 × 10–19
] C1
= 9.8 × 10–8
m A1

PUBLISHED
12(a) fusion: two nuclei combine to form a (single) nucleus B1
fission: a (single) large nucleus divides to form (smaller) nuclei B1
Any one from:
• fusion is initiated by (very) high temperatures
• fission is initiated by neutron bombardment
• resulting nuclei in fission are of similar size
• (both processes) release energy
• binding energy per nucleon increases
• total binding energy increases
• fission involves release of neutrons
B1
12(b)(i) neutron B1
12(b)(ii) 1. zero A1
2. (4 × 11.3290 × 10–13
) – (2 × 1.7813 × 10–13
) – (3 × 4.5285 × 10–13
) C1
energy change = 45.316 × 10–13
– 17.148 × 10–13
= 2.82 × 10–12
J
A1
12(b)(iii) 1.0mol or NA nuclei of each
energy = 2.817 × 10–12
× 6.02 × 1023
= 1.7 × 1012
J
A1

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

PUBLISHED
marking principles.
descriptors.

PUBLISHED

9702 w18 ms_all

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