9702 m18 ms_all

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Cambridge Assessment International Education
Cambridge International Advanced Subsidiary and Advanced Level
PHYSICS 9702/12
Paper 1 Multiple Choice March 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 March 2018 series for most
Cambridge IGCSE®
, Cambridge International A and AS Level components and some Cambridge O Level
components.

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

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

PHYSICS 9702/22
Paper 2 AS Level Structured Questions March 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.
Cambridge IGCSE®
components.

PUBLISHED
March 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
March 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
March 2018
1(a) acceleration: vector
speed: scalar
power: scalar
All three correct scores 2 marks. Only two correct scores 1 mark.
B2
1(b)(i) time = 0.43 / 1.1
= 0.39 s
A1
1(b)(ii) s = ut + ½at 2
= ½ × 9.81 × 0.392
C1
= 0.75 m A1
1(b)(iii) 1 horizontal line at a non-zero value of a. B1
2 curved line from origin with increasing gradient. B1
1(c) acceleration (of free fall) is unchanged / not dependent on mass
and so no effect (on time taken).
A1
Question Answer Mark
2(a)(i) force × distance moved in the direction of the force B1
2(a)(ii) energy (of a mass/body) due to motion / speed / velocity B1
2(b)(i) 1 E = ½mv 2
C1
(∆)E = ½ × 580 × (222
– 122
) = 9.9 × 104
J A1
2 (∆)E = mg(∆)h
∆E = 580 × 9.81 × 13
C1
= 7.4 × 104
J A1

PUBLISHED
March 2018
2(b)(ii) length = (2π×13) / 4 or (π×26) / 4 or (π×13) / 2 = 20 m A1
2(b)(iii) work done against resistive force = 9.9 × 104
– 7.4 × 104
average resistive force = (9.9 × 104
– 7.4 × 104
) / 20
C1
= 1300 N A1
2(b)(iv) from horizontal/right to vertical / up or 90° A1
2(b)(v) p = mv or (580 × 22) or (580 × 12) C1
∆p = [ (580×12)2
+ (580×22)2
]0.5
C1
= 1.5 × 104
N s A1
3(a)(i) force / (cross-sectional) area B1
3(a)(ii) extension / original length B1
3(b)(i) measure / determine / find diameter B1
using a micrometer / digital calipers B1
several measurements in different places / along the wire / around the circumference (and average them) B1
3(b)(ii) E = σ / ε or E = FL / Ax or E = gradient × (L / A)
E = (4 × 2.5) / (0.8 × 10–3
) × (9.4 × 10–8
)
C1
= 1.3 × 1011
Pa A1

PUBLISHED
March 2018
3(b)(iii) E = ½Fx or E = ½kx 2
or E = area under graph
E = ½ × (2+4) × 0.4 × 10–3
or E = (½ × 4 × 0.8×10–3
) – (½ × 2 × 0.4×10–3
)
or E = [ ½ × 5000 × (0.8×10–3
)2
] – [ ½ × 5000 × (0.4×10–3
)2
]
C1
E = 1.2 × 10–3
J A1
3(c) straight line from the origin and above the original line M1
straight line passes through (0.80, 8.0) A1
4(a) (two) waves (travelling at same speed) in opposite directions overlap B1
(waves are same type and) have same frequency / wavelength B1
4(b)(i) v = fλ
f = 330 / 0.18
C1
= 1800 Hz (1830 Hz) A1
4(b)(ii) T = 1 / 1800 (= 5.5 × 10–4
)
time-base setting = (1.5 × 5.5 ×10–4
) / 8.0 or 1 / (1800 × 5.3)
C1
= 1.0 × 10–4
s cm–1
A1
4(b)(iii) waveform drawn with same period as original waveform B1
waveform drawn with amplitude of 1.7 cm B1
4(c)(i) distance = λ / 2 = 0.18 / 2
= 0.090 m
A1

PUBLISHED
March 2018
4(c)(ii) letter N shown at level B and at level A and not anywhere else. B1
4(c)(iii) m = ρAx
= 0.79 × 13 × 9.0 (=92.4) or 790 × 13×10–4
× 0.090 (=0.0924)
t = 92.4 / 6.7 or 0.0924 / 0.0067
C1
= 14 s A1
5(a) sum of e.m.f.(s) = sum of p.d.(s) M1
around a loop / around a closed circuit A1
5(b)(i) 1 6.0 – 4.0I = 0
I = 1.5 A
A1
2 6.0 + 6.0 = I (4.0 + R + 1.5)
12 = 1.5 (4.0 + R + 1.5)
C1
R = 2.5 Ω A1
or 6.0 = I (R + 1.5)
6.0 = 1.5 (R + 1.5)
(C1)
R = 2.5 Ω (A1)
or combines 6 = 4I and 6 = I(R + 1.5) to give
4 = R + 1.5
(C1)
R = 2.5 Ω (A1)

PUBLISHED
March 2018
5(b)(ii) I = Anvq
ratio = 12
/ 22
C1
= 0.25 A1
5(b)(iii) total (circuit) resistance increases B1
current / I decreases or P ∝ I or P ∝ 1 / (total resistance) M1
power (transformed) decreases A1
6(a) –1 / decreases by 1 A1
6(b) I = Q / t or Ne / t C1
= (9.8×1010
× 1.6×10–19
) / (2.0 × 60)
= 1.3 × 10–10
(A)
C1
= 130 pA A1
6(c) antineutrino(s) (emitted) / other particle(s) (emitted) C1
energy / momentum shared with antineutrino(s) A1

PHYSICS 9702/33
Paper 3 Advanced Practical Skills 1 March 2018
MARK SCHEME
Maximum Mark: 40
Published
Teachers.
Cambridge IGCSE®
components.

PUBLISHED
March 2018
marking principles.
descriptors.

PUBLISHED
March 2018

PUBLISHED
March 2018
1(a) Value of L0 in the range 59 cm to 61 cm, with unit. 1
1(b) Value of L less than L0 1
1(c) Six sets of readings of n and L (with the correct trend and without help from supervisor) scores 5 marks, five sets scores 4
marks etc.
5
Range of values:
Values of n must include at least 2 and 8.
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. 1 / L / cm–1
.
1
Consistency:
Values of raw L must all be given to the nearest mm.
1
Significant figures:
Significant figures for every value of 1 / L same as, or one greater than, the s.f. of L as recorded in table.
1
Calculation:
Correct calculation of 1 / L with no rounding error.
1

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March 2018
1(d)(i) Axes:
Sensible scales must be used, no awkward scales (e.g. 3:10).
Scales must be chosen so that the plotted points occupy at least half the graph grid in both x and y directions
Scales must be labelled with the quantity which is being plotted.
Scale markings should be no more than 3 large squares apart.
1
Plotting of points:
All observations must be plotted on the grid.
Diameter of plotted points must be ⩽ half a small square (no blobs).
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
±0.005 cm–1
(±0.5 m–1
) from a straight line in the 1 / L direction.
1
1(d)(ii) Line of best fit:
Judged by balance of all points on the grid (at least 5) about the candidate’s line. There must be an even distribution of points
either side of the line along the full length.
One anomalous point is allowed 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(d)(iii) Gradient:
The hypotenuse of the triangle used must be greater than half the length of the drawn line.
Method of calculation must be correct.
Both read-offs must be accurate to half a small square in both the x and y directions.
1
y-intercept:
Either
Correct read-off from a point on the line substituted into y = mx + c or an equivalent expression, with read-off accurate to half
a small square in both x and y directions.
Or
Intercept read directly from the graph, with read-off at 1 / n = zero accurate to half a small square in y direction.
1

PUBLISHED
March 2018
1(e) Value of a equal to candidate’s gradient. Value of b equal to candidate’s intercept. 1
Unit for a and unit for b are dimensionally correct (e.g. a = 16.9 m–1
and b = 1.7 m–1
) 1
2(a) Values for dA and dB, with unit, both <1 mm 1
Raw values for dA and dB recorded to nearest 0.01 mm. 1
2(b) Value for WB, with unit, >2 mm 1
2(c) Absolute uncertainty in WB value of 0.2 cm to 0.5 cm and correct method of calculation to obtain percentage uncertainty.
If several readings have been taken, then the absolute uncertainty can be half the range (but not zero if values are equal).
1
2(d) Value for WA. 1
Evidence of several raw measurements used to find the average, either here or in 2(b). 1
2(e) Value for second WB. 1
Quality:
Second WB greater than first WB.
1
Value for second WA. 1
2(f)(i) Two values of k calculated correctly. 1
2(f)(ii) Justification based on s.f. in dA, dB, WB and WA. 1
2(f)(iii) Sensible comment relating to the calculated values of k, testing against a criterion specified by the candidate. 1

PUBLISHED
March 2018
2(g)(i) Two k values are not enough to draw a valid conclusion 4 max
Difficult to: place glass plate / remove (B) wires / press down, without moving A wires
(Water drop) irregular shape so difficult to find average width
Parallax when measuring water drop
Graph grid is 2 mm (or just ‘grid is large’) / large % uncertainty in w
Outline of drop not clear because colourless / drop is colourless so difficult to see edge / drop is colourless so difficult to see
width
Force not constant
2(g)(ii) Take more readings and plot a graph / calculate more k values and compare 4 max
Tape / clamp / weight / stick the wires
Measure with ruler / calipers
Grid marked on glass / place drop on waterproof graph paper
Use smaller (or 1 mm) grid / use large drops to reduce uncertainty
Use coloured water
Use masses / small load / other defined workable method of producing consistent force

PHYSICS 9702/42
Paper 4 A Level Structured Questions March 2018
MARK SCHEME
Maximum Mark: 100
Published
Teachers.
Cambridge IGCSE®
components.

PUBLISHED
March 2018
marking principles.
descriptors.

PUBLISHED
March 2018

PUBLISHED
March 2018
1(a)(i) either direction of force on a (small test) mass
or direction of acceleration of a (small test) mass
B1
1(a)(ii) Any three from:
• the lines are radial
• near the surface the lines are (approximately) parallel
• parallel lines so constant field strength
• constant field strength hence constant acceleration of free fall
B3
1(b)(i) g = GM / R2
g = (6.67 × 10–11
× 7.35 × 1022
) / (1.74 × 103
× 103
)2
C1
g = 1.62 N kg–1
A1
1(b)(ii) either xω2
= GM / x2
and ω = 2π / T
or v2
/ x = GM / x2
and v = 2πr / T
C1
(1.74 × 106
+ 320 × 103
)3
× 4π2
/ T2
= (6.67 × 10–11
× 7.35 × 1022
) C1
T2
= 7.04 × 107
T = 8400 s (8390)
A1

PUBLISHED
March 2018
2(a) pV = nRT
T = (5.60 × 105
× 3.80 × 10–2
) / (5.12 × 8.31)
C1
T = 500 K A1
2(b)(i) V / T is constant
V = (3.80 × 104
) × (500 + 125) / 500
C1
V = 4.75 × 104
cm3
A1
2(b)(ii) (for ideal gas,) change in internal energy is change in (total) kinetic energy (of molecules) B1
∆U = 3 / 2 × 1.38 × 10–23
× 125 × 5.12 × 6.02 × 1023
C1
= 7980 J A1
2(c)(i) w = p∆V
= 5.60 × 105
× (4.75 – 3.80) × 10–2
C1
= 5320 J A1
2(c)(ii) total = 7980 + 5320
= 13300 J
A1

PUBLISHED
March 2018
3(a) reasonably shaped circle or oval surrounding the origin B1
closed loop passing through (0,±v0) and (±x0,0) B1
3(b) line from (0,0) to (90, F0) B1
curve with decreasing positive gradient, zero gradient at θ = 90 B1
3(c) reasonable sinusoidal wave, one cycle, period 4.0 ms B1
amplitude at 4.0 V B1
3(d) U near right-hand end of line with Ba between U and peak of graph B1
Ba on right hand side of peak and Kr between Ba and peak of graph B1
4(a) frequency at which body will vibrate when there is no (resultant external) resistive force acting on it
OR
frequency at which body will vibrate when there is no driving force / external force acting on it
B1
4(b)(i) resonance B1
4(b)(ii) peak is not sharp / peak not infinite height M1
so damped A1
4(c) e.g. (quartz crystal) to produce ultrasound
(quartz crystal) in watch to keep timing
NMR / MRI
microwave ovens
tuning circuits
B1

PUBLISHED
March 2018
5(a) pulses of ultrasound B1
reflected at boundaries (between media) B1
reflected pulses detected by (ultrasound) generator B1
Any three from:
(reflected signal) processed and displayed (B1)
time delay (between transmission and receipt) gives information about depth (of boundary) (B1)
intensity of reflected pulse gives information about (nature of) boundary (B1)
gel used to minimise reflection at skin / maximise transmission into skin (B1)
degree of reflection depends upon impedances of two media (at boundary) (B1)
B3
5(b)(i) product of density and speed M1
of sound in the medium A1
5(b)(ii) (Z1 about equal to Z2,) coefficient very small / nearly 0 B1
(Z1 very different to Z2,) coefficient nearly 1 B1

PUBLISHED
March 2018
6(a)(i) 0101 A1
6(a)(ii) 1000 A1
6(b) sketch: series of steps B1
changes every 0.25 ms B1
correct heights 0, 5, 10, 12, 15, 8 at correct times
Two marks for all levels correct
One mark if one mistake
B2
7(a) work done per unit charge B1
(work done) moving positive charge from infinity (to the point) B1
7(b)(i) potential always same sign / potential is always positive so same sign of charge B1

PUBLISHED
March 2018
7(b)(ii) 1 from x = 12 cm to x = 25 cm: speed increases and
from x = 27 cm to x = 31 cm: speed decreases
B1
(from x = 12 cm to x = 25 cm: speed increases) at decreasing rate or
(from x = 27 cm to x = 31 cm: speed decreases) at increasing rate
B1
at x = 26 cm: speed maximum B1
at 32 cm: speed still decreasing B1
2 q ∆V = ½mv2
3.2 × 10–19
× (2.14 – 1.43) × 104
= ½ × 6.6 × 10–27
× v2
v2
= 6.88 × 1011
C1
v = 8.3 × 105
m s–1
(8.30) A1
8(a)(i) all frequencies have the same gain B1
8(a)(ii) output changes at the same time as input changes B1
8(b)(i) RT / 800 = 1.8 / 1.2
RT = 1200 Ω
A1
8(b)(ii) stepped from –9 V to +9 V or v.v. B1
Vout negative < RT and Vout positive > RT B1
8(b)(iii) correct LED symbol with connection between VOUT and earth B1
diode pointing upwards B1

PUBLISHED
March 2018
9(a)(i) PSYV and QRXW B1
9(a)(ii) electron moving in magnetic field deflected towards face QRXW M1
so face PSYV is more positive A1
9(b)(i) PV or SY or RX or QW B1
9(b)(ii) number of charge carriers per unit volume B1
9(b)(iii) negative and positive charge (carriers) would deflect in opposite directions M1
so no change in polarity A1
10(a)(i) either product of flux density and area M1
direction of flux normal to area A1
or flux density × area × sinθ (M1)
where θ is angle between direction of flux and area (A1)
10(a)(ii) (induced) e.m.f. proportional to rate M1
of change of (magnetic) flux linkage A1
10(b) e.m.f. = ∆( φN) / ∆t
= (6.8 × 10–6
× 2 × 3.5 × 96) / (2.4 × 10–3
)
C1
= 1.9 V A1

PUBLISHED
March 2018
10(c) alternating C1
with same frequency as supply A1
11(a) no forbidden band / valence and conduction bands overlap B1
no change in number of charge carriers (as temperature rises) B1
increased lattice vibrations so resistance increases B1
11(b) photons captured / absorbed by electrons in valence band B1
electrons promoted to conduction band B1
leaving holes in the valence band B1
more holes and / or electrons so resistance decreases B1
12(a) Any 2 from:
scattering of X-ray beam / no lead grid
lack of collimation of beam / aperture large
anode area large
beam p.d. low / photon energy low / X-ray soft
B2
12(b)(i) 0.81 = (e–1.5
× 0.32
) / (e–1.5
× x
) C1
x = 1.8 mm A1

PUBLISHED
March 2018
12(b)(ii) ratio/dB = 10 lg(0.81) C1
= (–) 0.92 dB A1
13(a)(i) probability of decay (of a nucleus) M1
per unit time A1
13(a)(ii) A = A0 e–λt
after one half-life, ½A0 = A0 e–λt1/2
M1
½ = exp(–λt½) and hence taking logs, ln2 = λt½ A1
13(b) activity = 3.8 × 104
exp(–ln2 × 36 / 15) C1
= 7200 Bq C1
or activity = 3.8 × 104
/ 22.4
(C1)
= 7200 Bq (C1)
volume = (7200 / 1.2) × 5.0 C1
= 3.0 × 104
cm3
A1
OR activity of 5.0 cm3
= 1.2 × 22.4
(C1)
= 6.3336 Bq (C1)
volume = (3.8 × 104
/ 6.3336) × 5.0 (C1)
= 3.0 × 104
cm3
(A1)

PHYSICS 9702/52
Paper 5 Planning, Analysis and Evaluation March 2018
MARK SCHEME
Maximum Mark: 30
Published
Teachers.
Cambridge IGCSE®
components.

PUBLISHED
March 2018
marking principles.
descriptors.

PUBLISHED
March 2018

PUBLISHED
March 2018
1 Defining the problem
d is the independent variable and e is the dependent variable, or vary d and measure e. 1
Keep F constant. 1
Methods of data collection
Labelled workable diagram including elastic cord fixed at one end to a support and load attached to the other end
Cord and weight must be labelled
1
Use of ruler to measure unstretched length and stretched length.
or labelled ruler drawn parallel to cord and original length L and either e or stretched length indicated.
1
Use of a micrometer / (calipers) to determine d. 1
Weigh load on a balance or use of balance to measure mass of load and multiply by g 1
Method of Analysis
Plots a graph of e against 1 / d2
or equivalent 1
Relationship valid if a straight line passing through the origin is produced 1
4FL
E
gradient π
=
×
1
Additional detail including safety considerations Max 6
Use safety goggles / safety screen to prevent injury to eyes from (moving) elastic cord / load or
use cushion / sand box in case load falls.
D1
Keep L constant D2
Method to keep L constant, e.g. check length of each cord / adjust through cork D3

PUBLISHED
March 2018
1 Additional detail on measuring e, e.g. record initial position, record final position and subtract D4
Repeat measurement of d along cord / different diameters and average D5
Method to ensure Hooke’s law is obeyed, e.g. check that the length is constant after removing load or do not exceed elastic
limit
D6
Wait for cord to extend to its maximum value / stop oscillating D7
Use of
2
4
d
A
π
=
D8
Use of set square to check that ruler is vertical to bench, or
use of set square as a fiducial mark to read measurements
D9
2(a)
Gradient =
D
S
1
2(b) 3.95
4.35
4.80
5.20
5.55
5.90
1
Absolute uncertainties in P ± 0.05 1

PUBLISHED
March 2018
2(c)(i) Six points plotted correctly.
Must be within half a small square. Diameter of points must be less than half a small square.
1
Error bars in P plotted correctly.
All error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.
1
2(c)(ii) Line of best fit drawn.
Points must be balanced.
Line must pass between (6.45, 5.8) and (6.55, 5.8) and between (4.55, 4.2) and (4.65, 4.2)
1
Worst acceptable line drawn.
Steepest or shallowest possible line that passes through all the error bars.
Mark scored only if all error bars are plotted.
1
2(c)(iii) Gradient determined with clear substitution of data points into ∆y/∆x; distance between data points must be at least half the
length of the drawn line.
1
Gradient determined of WAL
uncertainty = (gradient of line of best fit – gradient of worst acceptable line)
or
uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)
1
2(d)(i) Substitution of gradient to determine s
2.20
gradient
D
s = =
(c)(iii)
1
s determined using gradient, given to 2 or 3 significant figures 1
s determined using gradient and correct unit and correct power of ten 1

PUBLISHED
March 2018
2(d)(ii) Percentage uncertainty in s
%uncertainty in D + %uncertainty in gradient
0.02 gradient
% 100
2.20 gradient
s
 ∆
= + × 
 
gradient
% 0.91 100
gradient
s
∆
= + ×
Or correct maximum/minimum method
max
max
mingradient
D
s = or
min
min
max gradient
D
s =
1

PUBLISHED
March 2018
2(e) Correct substitution of numbers must be seen,
λ in the range 7
4.05 10 m−
× to 7
4.24 10 m−
×
3 3
3.5 10 3.5 10
gradient
λ
− −
× ×
= =
(c)(iii)
OR
3 3s
3.5 10 3.5 10
2.2D
λ − −
= × × = × ×
(d)(i)
1
Correct substitution of numbers must be seen,
Determines absolute uncertainty in λ.
Using (c)(iii)
0.05 gradient
uncertainty
3.50 gradient
λ
 ∆
= + × 
 
or
3
3.55 10
max
mingradient
λ
−
×
= , or
3
3.45 10
min
max gradient
λ
−
×
= ,
OR
Using (d)
0.05 0.02
uncertainty
3.50 2.20 100
λ
 
= + + × 
 
(d)(i)
uncertainty 0.0234
100
λ
 
= + × 
 
(d)(i)
or
3
3.55 10 max
max
2.18
s
λ
−
× ×
= , or
3
3.45 10 min
min
2.22
s
λ
−
× ×
= .
1

9702 m18 ms_all

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