9702 s17 ms_all

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Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level

PHYSICS 9702/11
Paper 1 Multiple Choice May/June 2017
MARK SCHEME
Maximum Mark: 40

Published

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the
examination.

Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for
Teachers.

Cambridge will not enter into discussions about these mark schemes.

Cambridge is publishing the mark schemes for the May/June 2017 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 2017

© UCLES 2017 Page 2 of 3

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

9702/11 Cambridge International AS/A Level – Mark Scheme
PUBLISHED May/June 2017

© UCLES 2017 Page 3 of 3

Question Answer Marks
29 A 1
30 A 1
31 B 1
32 A 1
33 C 1
34 A 1
35 C 1
36 B 1
37 C 1
38 A 1
39 B 1
40 C 1

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Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level

PHYSICS 9702/12
Paper 1 Multiple Choice May/June 2017
MARK SCHEME
Maximum Mark: 40

Published

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the
examination.

Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for
Teachers.

Cambridge will not enter into discussions about these mark schemes.

Cambridge is publishing the mark schemes for the May/June 2017 series for most Cambridge IGCSE
®
,
Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level
components.

9702/12 Cambridge International AS/A Level – Mark Scheme
PUBLISHED May/June 2017

© UCLES 2017 Page 2 of 3

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

9702/12 Cambridge International AS/A Level – Mark Scheme
PUBLISHED May/June 2017

© UCLES 2017 Page 3 of 3

Question Answer Marks
29 B 1
30 C 1
31 C 1
32 A 1
33 A 1
34 C 1
35 A 1
36 B 1
37 B 1
38 A 1
39 B 1
40 A 1

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Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level

PHYSICS 9702/13
Paper 1 Multiple Choice May/June 2017
MARK SCHEME
Maximum Mark: 40

Published

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the
examination.

Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for
Teachers.

Cambridge will not enter into discussions about these mark schemes.

Cambridge is publishing the mark schemes for the May/June 2017 series for most Cambridge IGCSE
®
,
Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level
components.

9702/13 Cambridge International AS/A Level – Mark Scheme
PUBLISHED May/June 2017

© UCLES 2017 Page 2 of 3

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

9702/13 Cambridge International AS/A Level – Mark Scheme
PUBLISHED May/June 2017

© UCLES 2017 Page 3 of 3

Question Answer Marks
29 D 1
30 D 1
31 B 1
32 B 1
33 C 1
34 B 1
35 C 1
36 A 1
37 A 1
38 D 1
39 A 1
40 A 1

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Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level

PHYSICS 9702/21
Paper 2 AS Level Structured Questions May/June 2017
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 will not enter into discussions about these mark schemes.

Cambridge is publishing the mark schemes for the May/June 2017 series for most Cambridge IGCSE
®
,
Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level
components.

9702/21
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 2 of 7

Question
Answer
Marks
1(a) (stress =) force / area or kg m s
–
2
/ m
2
B1
= kg m
–
1
s
–
2
A1
1(b)(i) 0.58 = 2π × [(4 × 0.500 × 0.600
3
) / (E × 0.0300 × 0.00500
3
)]
0.5
C1
E = [4π
2
× 4 × 0.500 × (0.600)
3
] / [(0.58)
2
× 0.0300 × (0.00500)
3
]

= 1.35 × 10
10
(Pa)
C1
= 14 (13.5) GPa A1
1(b)(ii)1. (accuracy determined by) the closeness of the value(s)/measurement(s) to the true value B1
(precision determined by) the range of the values/measurements B1
1(b)(ii)2. l is (cubed so) 3 × (percentage/fractional) uncertainty
and T is (squared so) 2 × (percentage / fractional) uncertainty
and (so) l contributes more
B1

9702/21
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 3 of 7

Question
Answer
Marks
2(a) resultant force (in any direction) is zero B1
resultant torque/moment (about any point) is zero B1
2(b)(i) a = (v − u) / t or gradient or ∆v / (∆)t C1
e.g. a = (8.8 − 4.6) / (7.0 – 4.0) = 1.4 m s
–
2
A1
2(b)(ii) s = 4.6 × 4 + [(8.8 + 4.6) / 2] × 3 C1
= 18.4 + 20.1

= 39 (38.5) m
A1
2(b)(iii) ∆E = ½ × 95 [(8.8)
2
− (4.6)
2
] C1
= 3678 – 1005

= 2700 (2673) J
A1
2(b)(iv)1. weight = 95 × 9.81 (= 932 N) C1
vertical tension force = 280 sin 25° or 280 cos 65° (=118.3 N) C1 F = 932 + 118

= 1100 (1050) N
A1
2(b)(iv)2. horizontal tension force = 280 cos 25° or 280 sin 65° (= 253.8 N) C1
resultant force = 95 × 1.4 (= 133 N) C1 133 = 253.8 – R

R = 120 (120.8) N
A1

9702/21
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 4 of 7

Question
Answer
Marks
3(a)
ρ
= m / V C1
V = πd
2
L / 4 or πr
2
L C1
weight = 2.7 × 10
3
× π (1.2 × 10
–
2
)
2
× 5.0 × 10
–
2
× 9.81 = 0.60 N A1
3(b)(i) the point from where (all) the weight (of a body) seems to act B1 3(b)(ii) W × 12 C1
(0.25 × 8) + (0.6 × 38) C1 W = (2 + 22.8) / 12

= 2.1 (2.07)
N
A1
3(c)(i) pressure changes with depth (in water)
or
pressure on bottom (of cylinder) different from pressure on top
B1
pressure on bottom of cylinder greater than pressure on top
or
force (up) on bottom of cylinder greater than force (down) on top
B1
3(c)(ii) anticlockwise moment reduced and reducing the weight of X reduces clockwise moment
or
anticlockwise moment reduced so clockwise moment now greater than (total) anticlockwise moment
B1

9702/21
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 5 of 7

Question
Answer
Marks
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)
λ
= 12 / 250 (= 0.048 m)

C1
distance = 1.5 × 0.048

= 0.072 m
A1
4(b)(ii) T = 1 / 250

= 0.004 (s) or 4 (ms)
C1
1. curve drawn is mirror image of that in Fig. 4.2 and labelled P A1 2. horizontal line drawn between A and B and labelled Q A1
Question
Answer
Marks
5(a) observed frequency is different to source frequency when source moves relative to observer B1 5(b) 360 = (400 × 340) / (340 ± v) C1
v = 38 (37.8) m s
–
1
A1
away (from the observer) B1

9702/21
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 6 of 7

Question
Answer
Marks
6(a) volt / ampere B1
6(b)(i) R
T
= [1
/
3.0 + 1
/
6.0]
–
1
+ 4.0 (= 6.0 Ω) C1

Ι
= 1.5 / 6.0 C1
= 0.25 A A1
6(b)(ii) V
B
= 0.5 V

I = 0.5 / 3.0

= 0.17 (0.167) A
A1
6(b)(iii) P = I
2
R or VI or V
2
/
R C1
ratio = (0.167
2
× 3.0) /
(0.25
2
× 4.0)

= 0.33
A1
6(c)(i) vary/change/different radius/diameter/cross-sectional area (of wire)
B1
6(c)(ii) v = I / Ane

(
)
()
=
B
C
/
ratio
/
A
A
B
C
I
I
or
×
C
B
A
A
B
C
I
I

C1

(R ∝ 1 / A so) ratio =
×
B
C
R
R
B
C
I
I
=
×
0.167 3.0
0.25 4.0

= 0.50
A1
6(d)(i) 0.25 A to 0.13 (0.125) A or halved A1 6(d)(ii) no change A1

9702/21
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 7 of 7

Question
Answer
Marks
7(a)(i) (proton is uud so) (2 / 3)e + (2 / 3)e – (1 / 3)e = e B1 7(a)(ii) (neutron is udd so) (2 / 3)e – (1 / 3)e –(1 / 3)e = 0 B1 7(b)(i)

β
–
β
+

nucleon number 90 64 proton number 39 28

all correct
B1
7(b)(ii) weak (nuclear force/interaction) B1 7(b)(iii) β
–
decay: electron and (electron) antineutrino
β
+
decay: positron and (electron) neutrino
all correct
B1

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Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level

PHYSICS 9702/22
Paper 2 AS Level Structured Questions May/June 2017
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 will not enter into discussions about these mark schemes.

Cambridge is publishing the mark schemes for the May/June 2017 series for most Cambridge IGCSE
®
,
Cambridge International A and AS Level and Cambridge Pre–U components, and some Cambridge O Level
components.

9702/22
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 2 of 7

Question
Answer
Marks
1(a) kelvin, mole, ampere, candela
any two
B1 1(b) use of resistivity = RA / l and V = IR (to give
ρ
= VA/ Il) C1
units of V: (work done / charge) kg m
2
s
–
2
(A s)
–
1
C1
units of resistivity: (kg m
2
s
–
3
A
–
1
A
–
1
m)
= kg m
3
s
–3
A
–2

A1
or use of R =
ρ
L / A and P = I
2
R (gives
ρ
= PA / I
2
L) (C1)
units of P: kg m
2
s
–
3
(C1)
units of resistivity: (kg m
2
s
–
3
× m
2
) / (A
2
× m)
= kg m
3
s
–3
A
–2

(A1)
1(c)(i)
ρ
= (RA/l) C1
= (0.03 × 1.5 × 10
–
6
) / 2.5 (= 1.8 × 10
–
8
) C1
= 18 nΩ m A1
1(c)(ii) 1. precision is determined by the range in the measurements/values/readings/data/results B1
2. metre rule measures to ± 1 mm and micrometer to ± 0.01 mm (so there is less (percentage) uncertainty/random error) B1

9702/22
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 3 of 7

Question
Answer
Marks
2(a) rate of change of displacement or change in displacement/time taken B1
2(b)(i) s = ut + ½at
2
C1
t = [(2 × 1.25) / 9.81]
1/2
(= 0.5048
s) C1
or v
2
= u
2
+ 2as

v
vert
= (2 × 9.81 × 1.25)
1/2
(= 4.95)
(C1)
t = [2s / (u + v)] = 2 × 1.25 / 4.95 (= 0.5048
s) (C1)
v = d / t = 1.5 / 0.50(48)

= 3.0 (2.97)
m
s
–1

A1
2(b)(ii) vertical velocity = at

= 9.81 × 0.5048 (= 4.95) [using t = 0.50 gives 4.9]
C1
velocity = [(v
h
)
2
+ (v
v
)
2
]
1/2
C1
= [(2.97)
2
+ (4.95)
2
]
1/2

= 5.8 (5.79) [using t = 0.50 leads to 5.7]
A1
direction (= tan
–
1
4.95/2.97) = 59° A1
2(b)(iii) kinetic energy = ½mv
2
C1
= ½ × 0.45 × (5.8)
2

= 7.6 (7.57) J [using t = 0.50 leads to 7.3 J]
A1

9702/22
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 4 of 7

Question
Answer
Marks
2(b)(iv) potential energy = mgh C1
= (0.45 × 9.81 × 1.25)

= 5.5 (5.52) J
A1
2(c) there is KE of the ball at the start/leaving table
or
the ball has an initial/constant horizontal velocity
or
the ball has velocity at start/leaving table
B1

Question
Answer
Marks
3(a) E = stress / strain or (F / A) / (e / l) C1
= [gradient × 3.5] / [π × (0.19 × 10
–
3
)
2
]

e.g. E = [{(40 – 5) / ([11.6
– 3.2] × 10
–3
)} × 3.5] / [π × (0.19 × 10
–3
)
2
]
or
[4170 × 3.5] / [π × (0.19 × 10
–3
)
2
]
C1
E (= 1.3 × 10
11
) = 0.13 TPa (allow answers in range 0.120–0.136
TPa) A1
3(b) a larger range of F required or range greater than 35 N
B1

9702/22
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
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Question
Answer
Marks
4(a) a body/mass/object continues (at rest or) at constant/uniform velocity unless acted on by a resultant force B1
4(b)(i) initial momentum = final momentum

m
1
u
1
+ m
2
u
2
= m
1
v
1
+ m
2
v
2

C1
0.60 × 100 − 0.80 × 200 = −0.40 × 100 + v × 200

v = (−) 0.3(0) m s
–1

A1
4(b)(ii) kinetic energy is not conserved/is lost (but) total energy is conserved/constant
or
some of the (initial) kinetic energy is transformed into other forms of energy
B1
Question
Answer
Marks
5(a) frequency is the number of vibrations/oscillations per unit time or the number of wavefronts passing a point per unit time B1 5(b) vibrations/oscillation of the air particles are parallel to the direction of it (the direction of travel of the sound wave ) B1
5(c)(i) T = 2(.0) (ms) C1
f = 500 Hz A1
5(c)(ii) 1. amplitude increases
(time) period decreases

2. amplitude decreases
(time) period increases

any 3 points
B3

9702/22
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 6 of 7

Question
Answer
Marks
6(a)(i) waves at (each) slit/aperture spread
B1
(into the geometric shadow) wave(s) overlap/superpose/sum/meet/intersect
B1
6(a)(ii) there is not a constant phase difference/coherence (for two separate light source(s))
or
waves/light from the double slit are coherent/have a constant phase difference
B1
6(b) x =
λ
D / a C1

λ
= (36 × 10
–
3
× 0.48 × 10
–
3
) / (16 × 2.4) C1
= 4.5 × 10
–
7
m A1
6(c)(i) no movement of the water/water is flat/no ripples/disturbance
B1
the path difference is 2.5
λ
or the phase difference is 900° or 5π rad B1
6(c)(ii) 1. surface/water/P vibrates/ripples
and
as (waves from the two dippers) arrive in phase
B1
2. surface/water/P vibrates/ripples
and
as amplitudes/displacements are no longer equal/do not cancel
B1

9702/22
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 7 of 7

Question
Answer
Marks
7(a) energy transformed from chemical to electrical /
unit charge (driven around a complete circuit)
B1
7(b)(i) the current decreases (as resistance of Y increases) M1
lost volts go down (as resistance of Y increases) M1 p.d. AB increases (as resistance of Y increases) A1
7(b)(ii)1. 1.50 = 0.180 × (6.00 + 0.200 + R
X
) C1
R
X
= 2.1(3) Ω A1
7(b)(ii)2. p.d. AB = 1.5 − (0.180 × 0.200) or 0.18 × (2.13 + 6.00) C1
= 1.46(4) V A1
7(b)(ii)3. efficiency = (useful) power output / (total) power input or IV / IE C1
( = 1.46 / 1.5) = 0.97 [0.98 if full figures used] A1
Question
Answer
Marks
8(a) β
–
emission: neutron changes to proton (+ beta
–
/electron)
and
β
+
emission: proton changes to neutron (+ beta
+
/positron)
B1
β
–
emission: (electron) antineutrino also emitted
and
β
+
emission: (electron) neutrino also emitted
B1
8(b) proton: up up down (and zero strange)

neutron: up down down (and zero strange)
B1

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Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level

PHYSICS 9702/31
Paper 3 Advanced Practical Skills 1 May/June 2017
MARK SCHEME
Maximum Mark: 40

Published

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the
examination. 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 will not enter into discussions about these mark schemes.

Cambridge is publishing the mark schemes for the May/June 2017 series for most Cambridge IGCSE
®
,
Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level
components.

9702/31
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 2 of 5

Question
Answer
Mark
1(b)(iii) Value(s) of I
1
with unit in the range 30–200 mA. 1
1(c)(iii) I
2
> I
1
(ECF unit from 1(b)(iii)). 1
1(d) Six sets of readings of x (different values), I
1
, I
2
with correct trend and without help from Supervisor scores 5 marks, five sets
scores 4 marks etc.
5
Range: x
ma
x
⩾ 90 cm. 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. x / m, I
2
/ mA, I
2
/ I
1
no unit.
1
Consistency:
All values of x must be given to the nearest mm.
1 Significant figures:
All values of I
2
/ I
1
must be given to the same number of s.f. as (or one more than) the s.f. in raw I
1
and I
2
(using the s.f. of the
I value with the lowest number of s.f.).
1
Calculation:
Values of I
2
/ I
1
are correct.
1
1(e)(i) Axes:
Sensible scales must be used, no awkward scales (e.g. 3:10 or fractions).
Scales must be chosen so that the plotted points occupy at least half the graph grid in both x and y directions.
Scales must be labelled with the quantity that is being plotted.
Scale markings should be no more than three large squares apart.
1
Plotting of points:
All observations must be plotted on the grid.
Diameter of plotted points must be ⩽ half a small square (no “blobs”).
Points must be plotted to an accuracy of half a small square.
1

9702/31
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
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Question
Answer
Mark
Quality:
All values of I
2
/ I
1
must
be greater than 1.
All points in the table must be plotted for this mark to be awarded.
It must be possible to draw a straight line that is within ± 0.05 on the I
2
/ I
1
axis (normally y 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 li ne (at least 5). There must be an even distribution of poin ts
either side of the line along the full length.
Allow one anomalous point only if clearly indicated (i.e. circled or labelled) by the candidate. There must be at least five
points left after the anomalous point is disregarded.
Lines must not be kinked or thicker than half a small square.
1 1(e)(iii) Gradient:
Gradient sign on answer line matches graph drawn.
The hypotenuse of the triangle used must be greater than half the length of the drawn line.
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.
1
y-intercept:
Correct read-off from a point on the line substituted correctly into y = mx + c or an equivalent expression.
Read-off accurate to half a small square in both x and y directions.
or
Intercept read directly from the graph, with read-off at x = 0, accurate to half a small square in y direction.
1
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 is correct (e.g. m
–
1
) and Q stated without a unit. 1

9702/31
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 4 of 5

Question
Answer
Mark
2(b)(ii) Values of time in the range 3–7
s and within 0.4
s of each other.
All raw readings stated to same precision and to at least 0.1
s.
1
2(c)(ii) x on the answer line in the range 35.0–40.0 cm and all value(s) of raw x to nearest mm with unit. 1 2(c)(iii) Absolute uncertainty in x in range 2–5 mm.
If repeated readings have been taken, then the uncertainty can be half the range (but not zero) if the working is clearly shown .
Correct method of calculation to obtain percentage uncertainty.
1 2(e)(i) Value of n on answer line in the range 10 to 20. No unit. 1
Evidence of repeat readings. 1
2(e)(ii) Correct calculation of (n + 1)
2
/
n
2
. 1
2(f) Second value of x. 1
Second value of n. 1 Quality: second value of n < first value of n (if x
2
< x
1
allow n
2
> n
1
). 1
2(g)(i) Two values of k calculated correctly. 1 2(g)(ii) Valid comment consistent with calculated values of k, testing against a criterion specified by the candidate. 1
2(h) Correct calculation of L written to 3 s.f. and correctly rounded to 3 s.f. 1

9702/31
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 5 of 5

Question
Answer
Mark
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 judge when the balls are back in phase.

C Balls hit each other/blown by moving air/balls move sideways/irregular movement.

D Difficult to ensure both lengths 35 cm or difficult to measure x with a reason e.g. string kinked/parallax error.

E Damping/oscillations die out quickly.

F Difficulty linked to fixing block/wood with reason e.g. longer block requires more tape (therefore an unfair test)/block
comes unstuck/blocks attached at an angle (x not constant).

G For two/both balls the release point is different or release instant is different.

1 mark for each point up to a maximum of 4.

4
2(i)(ii) A Take more readings and plot a graph/take more readings and compare k values (not “repeat readings” on its own).

B Video/film/record and perpendicular to plane of oscillation/from the side.

C Turn off air conditioning/close windows/use windshield/use a longer rod.

D Improved method of measuring 35 cm or x e.g. use clamps to fix loop and ball.

F Improved method of attachment e.g. Blu-Tack/glue/Velcro/stickier tape/stronger adhesive on the tape.

G Improved method of release e.g. bar to hold both balls then drop away like a gate.

1 mark for each point up to a maximum of 4.
4

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Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level

PHYSICS 9702/32
Paper 3 Advanced Practical Skills 2 May/June 2017
MARK SCHEME
Maximum Mark: 40

Published

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the
examination. 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 will not enter into discussions about these mark schemes.

Cambridge is publishing the mark schemes for the May/June 2017 series for most Cambridge IGCSE
®
,
Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level
components.

9702/32
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 2 of 5

Question
Answer
Marks
1(a)(v) Value of p in the range 20.0–30.0 cm, with unit. 1
Value of q less than p. 1
1(b) Six sets of readings of p, q and V (with correct trend and without help from Supervisor) scores 5 marks, five sets scores 4
marks etc.
5
Range:
p
ma
x
⩾ 40.0 cm and p
min
⩽ 10.0 cm.
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
/
p
/
cm
–1
.
1
Consistency:
All values of p and q must be given to the nearest mm.
1 Significant figures:
Significant figures for every value of 1
/
q must be same as, or one greater than, the s.f. of q as recorded in table.
1
Calculation:
Values of 1
/
q calculated correctly.
1
1(c)(i) Axes:
Sensible scales must be used, no awkward scales (e.g. 3:10 or fractions).
Scales must be chosen so that the plotted points occupy at least half the graph grid in both x and y directions
Scales must be labelled with the quantity that is being plotted.
Scale markings should be no more than three large squares apart.
1
Plotting of points:
All observations must be plotted on the grid.
Diameter of plotted points must be ⩽ half a small square (no “blobs”).
Plots must be accurate to within half a small square in both x and y directions.
1

9702/32
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 3 of 5

Question
Answer
Marks
Quality:
All points in the table must be plotted (at least 5) for this mark to be awarded.
It must be possible to draw a straight line that is within ± 1.0 m
–1
(± 0.01 cm
–1
) of all the plotted points in the 1
/
q direction.
1
1(c)(ii) Line of best fit:
Judge by balance of all points on the grid (at least 5) about the candidate’s line. There must be an even distribution of poin ts
either side of the line along the full length.
Allow one anomalous point only if clearly indicated (i.e. circled or labelled) by the candidate.
Lines must not be kinked or thicker than half a small square.
1
1(c)(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. Do not allow ∆x / ∆y.
Both read-offs must be accurate to half a small square in both the x and y directions.
1
y-intercept:
Correct read-off from a point on the line substituted into y = mx + c or an equivalent expression.
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 x = 0 accurate to half a small square in y direction.
1

1(d) Value of a = candidate’s gradient and value of b = candidate’s intercept.
Values must not be fractions.
1
There must be no unit for a.
Unit for b correct e.g. m
–1
or cm
–1
.
1

9702/32
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 4 of 5

Question
Answer
Marks
2(a)(ii) Value of h
1
to nearest 0.1 cm. 1
h
2
< h
1
. 1
2(a)(iv) Absolute uncertainty in h
2
of 2–5 mm 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(b)(i) Correct calculation of k. 1 2(b)(ii) Justification for s.f. in k linked to s.f. in m, g and (h
1
–h
2
) or m, g, h
1
and h
2
. 1
2(c)(iii) Value of T in range 1.00–3.00 s. 1
Evidence of repeat readings of nT with n ⩾ 5. 1
2(d) Second values of h
1
and h
2
.1
Second value of T. 1 Quality: T for three springs > T for two springs. 1
2(e)(i) Two values of C calculated correctly. 1 2(e)(ii) Valid comment consistent with the calculated values of C, testing against a criterion specified by the candidate. 1

9702/32
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 5 of 5

Question
Answer
Marks
2(f)(i) A Two readings are not enough to draw a (valid) conclusion (not “not enough for accurate results”, “few readings”).

B (Difficulty) measuring h because of parallax/metre rule not vertical.

C Oscillation dies away quickly/oscillation damped.

D Difficult to judge/tell end of oscillation/decide when to operate stopwatch.

E Other modes of oscillation occur.

F Some movement at joints still occurs.

1 mark for each point up to a maximum of 4.
4 2(f)(ii) A Take more readings and plot a graph/take more readings and compare C values (not “repeat readings” on its own).

B Use set square against rule (with detail)/use pointer attached to bottom of mass/use set square from rule to bottom of
mass.

C Twist through larger angle to get more rotations.

D View (video) recording/film of motion with timer in view or put mark on mass to make it easier to see motion.

E Workable method of restricting vertical/swinging movement (e.g. enclose in transparent tube).

F Better method of fixing joints/use two complete springs.

1 mark for each point up to a maximum of 4.
4

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r

Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level

PHYSICS 9702/33
Paper 3 Advanced Practical Skills 1 May/June 2017
MARK SCHEME
Maximum Mark: 40

Published

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the
examination. 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 will not enter into discussions about these mark schemes.

Cambridge is publishing the mark schemes for the May/June 2017 series for most Cambridge IGCSE
®
,
Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level
components.

9702/33
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 2 of 5

Question
Answer
Marks
1(a) Value of L with unit to the nearest mm in the range 2.5–3.5 cm. 1 1(c) Value of T with unit in the range 0.7
s to 1.5
s. 1
Evidence of repeated timings. Must see nT repeated where n ⩾ 5. 1
1(d) Six sets of readings of n (different values) and time with correct trend and without help from Supervisor scores 4 marks, five
sets scores 3 marks etc.
4
Range of n ⩾ 9. 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. No unit for n or √n.
1
Consistency:
All values of raw time must be given to either 0.1 s or 0.01 s.
1 Significant figures:
All values of √n must be given to 3 significant figures.
1
1(e)(i) Axes:
Sensible scales must be used, no awkward scales (e.g. 3:10 or fractions).
Scales must be chosen so that the plotted points occupy at least half the graph grid in both x and y directions.
Scales must be labelled with the quantity that is being plotted.
Scale markings should be no more than three large squares apart.
1
Plotting of points:
All observations must be plotted on the grid.
Diameter of plotted points must be ⩽ half a small square (no “blobs”).
Points must be plotted to an accuracy of half a small square.
1 Quality:
All points in the table must be plotted for this mark to be awarded.
It must be possible to draw a straight line that is within ± 0.10 on the √n axis of all plotted points.
1

9702/33
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 3 of 5

Question
Answer
Marks
1(e)(ii) Line of best fit:
Judge by balance of all points on the grid about the candidate’s li ne (at least 5). There must be an even distribution of poin ts
either side of the line along the full length.
Allow one anomalous point only if clearly indicated (i.e. circled or labelled) by the candidate. There must be at least five
points left after the anomalous point is disregarded.
Lines must not be kinked or thicker than half a small square.
1
1(e)(iii) Gradient:
The hypotenuse of the triangle used must be greater than half the length of the drawn line.
Method of calculation must be correct. Do not allow ∆x / ∆y.
Both read-offs must be accurate to half a small square in both x and y directions.
1
y-intercept:
Correct read-off from a point on the line substituted correctly into y = mx + c or an equivalent expression.
Read-off accurate to half a small square in both x and y directions.
or
Intercept read directly from the graph, with read-off at x = 0, accurate to half a small square in y direction.
1
1(f) Value of P = candidate’s gradient and value of Q = candidate’s intercept.
The values must not be fractions.
1
Units for P and Q both s. 1
1(g) Correct calculation of g = Lπ
2
/ P
2
with consistent unit. 1

9702/33
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 4 of 5

Question
Answer
Marks
2(a) Value(s) of A with unit in the range 97.5–99.5 cm. 1
2(b)(ii) Values of all raw x to nearest mm. 1 2(c)(iii) Value of y ˃ x. 1 2(c)(iv) Correct calculation of (y – x). 1 2(c)(v) Percentage uncertainty in (y – x) based on absolute uncertainty of 2–5 mm.
If repeated readings have been taken, then the uncertainty can be half the range (but not zero) if the working is clearly shown .
Correct method of calculation to obtain percentage uncertainty.

1
2(d)(i) Correct calculation of m(A – 2y). 1 2(d)(ii) Justification for s.f. in m(A – 2y) linked to s.f. in A, y and m or (A – 2y) and m. 1 2(e)(i) Second value of y. 1
Quality: second value of y > first value of y. 1
2(f)(i) Two values of k calculated correctly. 1 2(f)(ii) Valid comment consistent with calculated values of k, testing against a criterion specified by the candidate. 1
2(g) Correct calculation of M = 1
/
k – B. 1

9702/33
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 5 of 5

Question
Answer
Marks
2(h)(i) A Two readings are not enough to draw a (valid) conclusion (not “not enough for accurate results”, “few readings”).

B Difficult to measure x or y with reason e.g. markings on rule obscured because of thickness of string/twist in string/ruler
oscillating/ruler swinging/rule slides in loop/string not vertical.

C Large (%) uncertainty in (y – x) or (y – x) is small.

D Little difference in y values.

E Difficult to balance rule.

1 mark for each point up to a maximum of 4.
4
2(h)(ii) A Take more readings and plot a graph/take more readings and compare k values (not “repeat readings” on its own).

B Use thinner string/tie a knot in loop/use two loops/use a hook/use longer loop/string.

C Heavier (added/slotted) masses (not heavier hangers).

D Use wider range of (added) masses.

E Suspend rule from its top edge/balance on fulcrum/use thicker string with reason such as to make easier to
balance/prevent rule from slipping.

1 mark for each point up to a maximum of 4.
4

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This document consists of 5 printed pages.

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r

Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level

PHYSICS 9702/34
Paper 3 Advanced Practical Skills 2 May/June 2017
MARK SCHEME
Maximum Mark: 40

Published

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the
examination. 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 will not enter into discussions about these mark schemes.

Cambridge is publishing the mark schemes for the May/June 2017 series for most Cambridge IGCSE
®
,
Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level
components.

9702/34
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 2 of 5

Question
Answer
Marks
1(a)(v) Value of T in the range 0.10–0.90 s. 1
Evidence of repeated readings. Must see nT repeated where n ⩾ 5. 1
1(b) Six sets of readings of M, h and T showing the correct trend and without help from the Supervisor scores 5 marks, five sets
scores 4 marks etc.
5
Range: M
max
⩾ 450
g and M
min
⩽ 200
g. 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
3
/ s
3
.
1
Consistency:
All values of h must be given to the nearest mm.
1 Significant figures:
Significant figures for every value of T
3
must be the same as (or one greater than) the s.f. of raw times as recorded in table.
1
Calculation:
Values of T
3
calculated correctly.
1
1(c)(i)

Axes:
Sensible scales must be used, no awkward scales (e.g. 3:10 or fractions).
Scales must be chosen so that the plotted points occupy at least half the graph grid in both x and y directions.
Scales must be labelled with the quantity that is being plotted.
Scale markings should be no more than three large squares apart.
1
Plotting of points:
All observations must be plotted on the grid.
Diameter of plotted points must be ⩽ half a small square (no “blobs”).
Points must be plotted to an accuracy of half a small square.
1
Quality:
All points in the table must be plotted (at least 5) for this mark to be awarded.
It must be possible to draw a straight line that is within ± 1.0 cm (to scale) of all the plotted points in the h direction.
1

9702/34
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 3 of 5

Question
Answer
Marks
1(c)(ii) Line of best fit:
Judge by balance of all points on the grid (at least 5) about the candidate’s line. There must be an even distribution of poin ts
either side of the line along the full length.
Allow one anomalous point only if clearly indicated (i.e. circled or labelled) by the candidate. There must be at least five
points left after the anomalous point is disregarded.
Lines must not be kinked or thicker than half a square.
1
1(c)(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. Do not allow ∆x / ∆y.
Both read-offs must be accurate to half a small square in both the x and y directions.
1
y-intercept:
Correct read-off from a point on the line substituted into y = mx + c or an equivalent expression.
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 h = 0, accurate to half a small square in y direction.
1

1(d)

Value of a = candidate’s gradient and value of b = candidate’s intercept.
The values must not be fractions.
1
Unit for a correct (e.g. s
3
cm
–
1
).
Unit for b is s
3
.
1

9702/34
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 4 of 5

Question
Answer
Marks
2(a) Value for x
0
to nearest 0.1
cm. 1
2(b)(ii) x greater than x
0
. 1
2(b)(iii) Raw
θ
in range 91–110° and recorded to nearest degree. 1
2(b)(v) Absolute uncertainty in (
φ
– 90°) of 2–5° and correct method of calculation to obtain percentage uncertainty.
If repeated readings of
φ
have been taken, then the absolute uncertainty can be half the range (but not zero) only if working
shown clearly.
1
2(c)(ii) Second value of x. 1
Second value of
φ
. 1
Quality: second value of
φ
⩾ first value of
φ
. 1
2(d)(i) Two values of k calculated correctly. 1 2(d)(ii) Justification for s.f. in k linked to s.f. in
θ
, x and x
0
. 1
2(d)(iii) Valid comment consistent with calculated values of k, testing against a criterion specified by the candidate. 1 2(e)(ii) Raw value(s) of D and d recorded to the nearest 0.1 cm. 1 2(e)(iv) Value of
ρ
calculated correctly. 1

9702/34
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 5 of 5

Question
Answer
Marks
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 get/adjust wooden strip so that it is horizontal/parallel to bench.

C Difficulty when measuring x (or x
0
), with reason e.g. parallax error/thickness of string/difficult to judge centre of nail.

D Difficult to measure angle with reason, e.g. parallax error/hard to hold protractor steady/hard not to touch wooden
strip/difficult to align protractor correctly.
(Do not credit parallax error twice for both C and D. )

E Large (percentage) uncertainty in (
φ
– 90°) or (
φ
– 90°) is small.

F Difficult to measure D (or d) with reason linked to use of ruler.

1 mark for each point up to a maximum of 4.
4
2(f)(ii) A Take more readings and plot a graph/take more readings and compare k values (not “repeat readings” on its own).

B Use thinner nail/fulcrum/prism.

C Mark a scale on the wooden strip/use a thinner string.

D Hold protractor in a clamp/
take photograph and measure angle on photo/
trigonometric method with detail e.g. measure height(s) of end(s) of wooden strip

E Method to increase (
φ
– 90°) e.g. more paper clips/larger values of x/move pipe and pivot closer.

F Use vernier/digital calipers/travelling microscope.

1 mark for each point up to a maximum of 4.
4

® IGCSE is a registered trademark.

This document consists of 5 printed pages.

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r

Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level

PHYSICS 9702/35
Paper 3 Advanced Practical Skills 1 May/June 2017
MARK SCHEME
Maximum Mark: 40

Published

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the
examination. 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 will not enter into discussions about these mark schemes.

Cambridge is publishing the mark schemes for the May/June 2017 series for most Cambridge IGCSE
®
,
Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level
components.

9702/35
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 2 of 5

Question
Answer
Marks
1(a)(iii) Value of x with unit to the nearest mm in the range 10.0–15.0 cm. 1 1(b)(ii) Evidence of repeated timings. Must see nT repeated where n ⩾ 5. 1
1(c) Six sets of readings of x (different values) and time with correct trend and without help from Supervisor scores 5 marks, five
sets scores 4 marks etc.
5
Range: x
min
⩽ 5.0
cm and x
max
⩾ 20.0
cm. 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. x
2
/ m
2
.
1
Consistency:
All values of raw x must be given to the nearest mm.
1 Significant figures:
All values of x
2

must be given to the same number of s.f. as (or one more than) the s.f. in raw 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 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 for this mark to be awarded.
It must be possible to draw a straight line that is within ± 0.025
s on the T axis of all plotted points.
1

9702/35
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 3 of 5

Question
Answer
Marks
1(d)(ii) Line of best fit:
Judge by balance of all points on the grid about the candidate’s li ne (at least 5). There must be an even distribution of poin ts
either side of the line along the full length.
Allow one anomalous point only if clearly indicated (i.e. circled or labelled) by the candidate. There must be at least five
points left after the anomalous point is disregarded.
Lines must not be kinked or thicker than half a small square.
1
1(d)(iii) Gradient:
The hypotenuse of the triangle used must be greater than half the length of the drawn line.
Gradient sign must match graph.
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.
1
y-intercept:
Correct read-off from a point on the line substituted correctly into y = mx + c or an equivalent expression.
Read-off accurate to half a small square in both x and y directions.
or
Intercept read directly from the graph, with read-off at x
2
= 0, accurate to half a small square in the y direction.
1

1(e) Value of P = candidate’s gradient and value of Q = candidate’s intercept.
The values must not be fractions.
1
Unit for P dimensionally correct (e.g. s
m
–
2
or s
cm
–
2
or s
mm
–
2
).
Unit for Q correct (s).
1
1(f)(iii) Correct calculation of x. 1

9702/35
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 4 of 5

Question
Answer
Marks
2(a) Value of raw L with unit in range 1.5–2.5
cm. 1
2(c)(ii) Value of y ⩾ L and to the nearest mm. 1 2(c)(iii) Value of raw
θ
to the nearest degree. 1
2(d) Percentage uncertainty in
θ
based on absolute uncertainty of 2–10°.
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(e)(i) Correct calculation of (y – L).1 2(e)(ii) Correct calculation of cos
(θ
/ 2). 1
2(e)(iii) Justification for s.f. in cos
(θ
/ 2) linked to s.f. in
θ
. 1
2(f) Second value of y. 1
Second value of
θ
. 1
Quality: second value of
θ
> first value of
θ
. 1
2(g)(i) Two values of k calculated correctly. 1 2(g)(ii) Valid comment consistent with calculated values of k, testing against a criterion specified by the candidate. 1

9702/35
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 5 of 5

Question
Answer
Marks
2(h)(i) A Two readings are not enough to draw a (valid) conclusion (not “not enough for accurate results”, “few readings”).

B Moving hands affecting keeping 15
cm the same or measuring y or
θ
.

C Parallax error affecting the measurement of y or
θ
.

D Reason for d not remaining constant e.g. loops slip on stand.

E Difficult to know where to measure from for the angle or y.

1 mark for each point up to a maximum of 4.
4
2(h)(ii) A Take more readings and plot a graph/take more readings and compare k values (not “repeat readings” on its own).

B Use another stand for valid purpose (e.g. instead of hand to hold spring).

C Photo or still from video to measure
θ
or including scale in frame to measure y.

D Improved method of fixing springs onto stand e.g. use Blu-Tack/tape/vertically clamped hacksaw blade/bulldog clips or
use sandpaper to make stand rough.

E Use a grid behind the springs/markers on spring.

1 mark for each point up to a maximum of 4.
4

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Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level

PHYSICS 9702/41
Paper 4 A Level Structured Questions May/June 2017
MARK SCHEME
Maximum Mark: 100

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 will not enter into discussions about these mark schemes.

Cambridge is publishing the mark schemes for the May/June 2017 series for most Cambridge IGCSE
®
,
Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level
components.

9702/41
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 2 of 12

Question
Answer
Marks
1(a) gravitational force (of attraction between satellite and planet) B1
provides / is centripetal force (on satellite about the planet)
B1
1(b) M = (4/3) × πR
3
ρ

B1

ω
= 2π / T or v = 2πnR / T B1
GM / (nR)
2
= nR
ω
2
or v
2
/ nR M1
substitution clear to give
ρ
= 3πn
3
/ GT
2
A1
1(c) n = (3.84 × 10
5
) / (6.38 × 10
3
) = 60.19 or 60.2C1

ρ
= 3π × 60.19
3
/ [(6.67 × 10
–
11
) × (27.3 × 24 × 3600)
2
] C1

ρ
= 5.54 × 10
3
kg m
–
3
A1

9702/41
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 3 of 12

Question
Answer
Marks
2(a) e.g. period = 3 / 2.5 C1
frequency = 0.83 Hz A1
2(b) light (damping) B1 2(c) at 2.7 s, A
0
= 1.5 (cm) B1
energy = ½ m × 4π
2
f
2
A
0
2
B1
= ½ × 0.18 × 4π
2
× 0.83
2
× (1.5 × 10
–
2
)
2

= 5.51 × 10
–4
(J)
C1
at 7.5 s, A
0
= 0.75 (cm) B1
energy = ¼ × 5.51 × 10
–
4

or
energy = ½ × 0.18 × 4π
2
× 0.83
2
× (0.75 × 10
–2
)
2

C1
energy = 1.38 × 10
–
4
(J)

change = (5.51 × 10 –4
– 1.38 × 10
–4
) = 4.13
J
A1

9702/41
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 4 of 12

Question
Answer
Marks
3(a)(i) signal consists of (a series of) 1s and 0s or offs and ons or highs and lows B1 3(a)(ii) component X: parallel-to-serial converter B1
component Y: DAC/digital-to-analogue converter B1
3(a)(iii) sample the (analogue) signal M1
at regular intervals and converts the analogue number to a digital number A1
3(b)(i) attenuation in fibre = 84 × 0.19 (= 16 dB) C1
ratio = 16 + 28

= 44 dB
A1
3(b)(ii) ratio / dB = 10 lg (P
2
/ P
1
) C1
44 = 10 lg ({9.7 × 10
–
3
} / P)
or
–44 = 10 lg (P / {9.7 × 10
–3
})
C1
power = 3.9 × 10
–
7
W A1

9702/41
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 5 of 12

Question
Answer
Marks
4(a) random/haphazard B1
constant velocity or speed in a straight line between collisions
or
distribution of speeds/different directions
B1
4(b) (small) specks of light/bright specks/pollen grains/dust particles/smoke particles M1
moving haphazardly/randomly/jerky/in a zigzag fashion A1
4(c)(i) pV = ⅓ Nm〈c
2
〉

1.05 × 10
5
× 0.0240 = ⅓ × 4.00 × 10
–3
× 〈c
2
〉
C1
〈c
2
〉 = 1.89 × 10
6
C1
or ½ m〈c
2
〉 = (3 / 2) kT

0.5 × (4.00 × 10
–3
/ 6.02 × 10
23
) × 〈c
2
〉 = 1.5 × 1.38 × 10
–23
× 300
(C1)
〈c
2
〉 = 1.87 × 10
6
(C1)
or nRT = ⅓ Nm〈c
2
〉

1.00 × 8.31 × 300 = ⅓ × 4.00 × 10
–3
× 〈c
2
〉
(C1)
〈c
2
〉 = 1.87 × 10
6
(C1)
c
r.m.s.
= 1.37 × 10
3
m s
–
1
A1

9702/41
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
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© UCLES 2017 Page 6 of 12

Question
Answer
Marks
4(c)(ii) 〈c
2
〉 ∝ T C1
〈c
2
〉 at 177 °C = 1.89 × 10
6
× (450 / 300) C1
c
r.m.s.
at 177 °C = 1.68 × 10
3
m s
–
1
A1
Question
Answer
Marks
5(a) (loss in) kinetic energy of α-particle = Qq / 4π
ε
0
r
or
7.7 × 10
–13
= Qq / 4π
ε
0
r
C1
7.7 × 10
–
13
= 8.99 × 10
9
× 79 × 2 × (1.60 × 10
–
19
)
2
/ r M1
r = 4.7 × 10
–
14
m

r is closest distance of approach so radius less than this
A1
5(b) force = Qq / 4π
ε
0
r
2
= 4u × a C1
8.99 × 10
9
× 79 × 2 × (1.60 × 10
–
19
)
2
/ (4.7 × 10
–
14
)
2
= 4 × 1.66 × 10
–
27
× a C1
a = 2.5 × 10
27
m s
–
2
A1
5(c) so that single interactions between nucleus and α-particle can be studied
or
so that multiple deflections with nucleus do not occur
B1

9702/41
Cambridge International AS/A Level – Mark Scheme
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Question
Answer
Marks
6(a)(i) lamp needs ‘high’ power/‘large’ current/‘large’ voltage B1
op-amp can deliver only a small current/small voltage B1
6(a)(ii) correct symbol for relay coil connected between output and earth B1
switch between mains supply and lamp B1
6(b)(i) vary light intensity at which lamp is switched on/off B1 6(b)(ii) so that relay operates for only one current/voltage direction
or
so that relay/lamp operates for either dark or light conditions
B1
6(c) when light level increases, LDR resistance decreases B1
(R
LDR
low,) so V
–
> V
+
, so V
OUT
negative/–5 V (must be consistent with B1 mark) M1
or when light level decreases, LDR resistance increases (B1) (R
LDR
high,) so V
–
< V
+
, so V
OUT
is positive/+5 V (must be consistent with B1 mark) (M1)
lamp comes on as light level decreases
or
lamp goes off as light level increases
A1

9702/41
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
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© UCLES 2017 Page 8 of 12

Question
Answer
Marks
7(a) (magnetic) force (always) normal to velocity/direction of motion M1
(magnitude of magnetic) force constant
or
speed is constant/kinetic energy is constant
M1 so provides the centripetal force A1
7(b) increase in KE = loss in PE or ½ mv
2
= qV M1
p = mv with algebra leading to p = √(2mqV) A1
7(c) Bqv = mv
2
/ r

mv = Bqr or p = Bqr
C1
(2 × 9.11 × 10
–
31
× 1.60 × 10
–
19
× 120)
1/2
= B × 1.60 × 10
–
19
× 0.074 C1
B = 5.0 × 10
–
4
T A1
7(d) greater momentum M1
(p = Bqr and) so r increased A1

9702/41
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
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Question
Answer
Marks
8 strong (uniform) magnetic field B1
* nuclei precess/rotate about field (direction)

radio frequency pulse/RF pulse (applied) B1 * RF or pulse is at Larmor frequency / frequency of precession causes resonance / excitation (of nuclei)/nuclei to absorb energy B1 on relaxation/de-excitation, nuclei emit RF/pulse B1 * (emitted) RF/pulse detected and processed non-uniform field (superposed on uniform field) B1 allows positions of (resonating) nuclei to be determined
B1
* allows for position of detection to be changed/different slices to be studied max. 2 of additional detail points marked * B2

9702/41
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
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© UCLES 2017 Page 10 of 12

Question
Answer
Marks
9(a)(i) core reduces loss of (magnetic) flux linkage/improves flux linkage B1 9(a)(ii) reduces (size of eddy) currents in core B1
(so that) heating of core is reduced B1
9(b) alternating voltage gives rise to changing magnetic flux in core M1
(changing) flux links the secondary coil A1 induced e.m.f. (in secondary) only when flux is changing/cut B1
Question
Answer
Marks
10(a)(i) penetration of beam M1
greater hardness means greater penetration/shorter wavelength/higher frequency/higher photon energy A1
10(a)(ii) greater accelerating potential difference
or
greater p.d. between anode and cathode
B1
10(b) I = I
0
exp(–
µ
x)

ratio = (exp {–1.5 × 2.9}) / (exp {–4.0 × 0.95}) (= exp {–0.55})
C1
= 0.58 A1

9702/41
Cambridge International AS/A Level – Mark Scheme
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Question
Answer
Marks
11(a) electrons (in gas atoms/molecules) interact with photons B1
photon energy causes electron to move to higher energy level/to be excited B1 photon energy = difference in energy of (electron) energy levels B1 when electrons de-excite, photons emitted in all directions (so dark line) B1
11(b)(i) photon energy ∝ 1 /
λ
C1
energy = 1.68 eV A1 or E = hc /
λ

E = 6.63 × 10
–34
× 3.0 × 10
8
/ (740 × 10
–9
)

= 2.688 × 10
–19
J
(C1)
energy = 1.68 eV (A1)
11(b)(ii) 3.4 eV → 1.5 eV
3.4 eV → 0.85 eV
3.4 eV → 0.54 eV
all correct and none incorrect 2/2
2 correct and 1 incorrect or only 2 correctly drawn 1/2
B2

9702/41
Cambridge International AS/A Level – Mark Scheme
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Question
Answer
Marks
12(a) x = 7 A1
12(b)(i) E = mc
2
C1
= 1.66 × 10
–
27
× (3.0 × 10
8
)
2

= 1.494 × 10
–10
J
C1
division by 1.6 × 10
–
13
clear to give 934 MeV A1
12(b)(ii) ∆m = (235.123 + 1.00863) – (94.945 + 138.955 + 2 × 1.00863 + 7 × 5.49 × 10
–
4
)
or
∆m = 235.123 – (94.945 + 138.955 + 1 × 1.00863 + 7 × 5.49 × 10
–4
)
C1
= 0.21053 u C1 energy = 0.21053 × 934

= 197 MeV
A1
12(c) kinetic energy of nuclei/particles/products/fragments B1
γ–ray photon energy B1

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This document consists of 13 printed pages.

© UCLES 2017 [Turn ove
r

Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level

PHYSICS 9702/42
Paper 4 A Level Structured Questions May/June 2017
MARK SCHEME
Maximum Mark: 100

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 will not enter into discussions about these mark schemes.

Cambridge is publishing the mark schemes for the May/June 2017 series for most Cambridge IGCSE
®
,
Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level
components.

9702/42
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 2 of 13

Question
Answer
Marks
1(a)
force per unit mass
B1
1(b)(i)
g = GM / r
2
= (6.67 ×10
–11
× 1.0 × 10
13
) / (3.6 × 10
3
)
2
C1

= 5.1 × 10
–
5
N kg
–
1

A1
1(b)(ii)
mass = (960 / 9.81) kg weight on comet = (960 / 9.81) × 5.1 ×10
–5
C1

= 5.0 × 10
–
3
N
A1
1(c)
similarity: e.g. both attractive/pointed towards the comet e.g. same order of magnitude
B1

difference: e.g. radial/non-radial e.g. same (over surface)/varies (over surface)
B1

9702/42
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Question
Answer
Marks
2(a)(i)
mean/average square speed/velocity
B1
2(a)(ii)
pV = NkT or pV = nRT
B1

ρ
= Nm /
V
or ρ
= nN
A
m / V and k = nR / N
B1

E
K
= ½ m〈c
2
〉 with algebra to (3 / 2)kT
B1
2(b)(i)
no (external) work done or ∆U = q or w = 0
B1

q = N
A
× (3 / 2)k × 1.0
M1

N
A
k = R so q = (3 / 2)R
A1
2(b)(ii)
specific heat capacity = {(3 / 2) × R} / 0.028
C1

= 450 J kg
–
1
K
–
1

A1

9702/42
Cambridge International AS/A Level – Mark Scheme
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Question
Answer
Marks
3(a)(i)
e.g. period = 6 / 2.5
C1

frequency = 0.42 Hz
A1
3(a)(ii)
energy = ½ m × 4π
2
f
2
y
0
2

C1

= ½ × 0.25 × 4π
2
× 0.42
2
× (1.5 × 10
–
2
)
2
C1

= 2.0 × 10
–
4
J
A1
3(b)(i)
(induced) e.m.f. proportional to rate of
M1

change of magnetic flux (linkage) or cutting of magnetic flux
A1
3(b)(ii)
coil cuts flux/field (of moving magnet) inducing e.m.f. in coil
B1

(induced) current in resistor causes heating (effect)
M1

thermal energy/heat derived from energy of oscillations (of magnet)
A1

9702/42
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Question
Answer
Marks
4(a)
pulse (of ultrasound)
B1

* produced by quartz crystal/piezo-electric crystal

* gel/coupling medium (on skin) used to reduce reflection at skin

reflected from boundaries (between media)
B1

reflected pulse/wave detectedby (ultrasound) transmitter
B1

reflected wave processed and displayed
B1

* intensity of reflected pulse/wave gives information about boundary

* time delay gives information about depth of boundary

max. 2 of additional detail points marked *
B2
4(b)
I
T
= I
0
exp (–
µ
x)
C1

2.9 = exp (4.6
µ
)
C1

µ
= 0.23 cm
–
1

A1

9702/42
Cambridge International AS/A Level – Mark Scheme
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Question
Answer
Marks
5(a)
any two reasonable suggestions e.g. • signal can be regenerated/noise removed (not “no noise”) • circuits more reliable • circuits cheaper to produce • multiplexing (is possible) • error correction/checking • easier encryption/better security
B2
5(b)(i)
samples the analogue signal
M1

at regular intervals and converts it (to a digital number)
A1
5(b)(ii)
1. smaller step depth
B1

2. smaller step height
B1

9702/42
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Question
Answer
Marks
6(a)
force proportional to product of charges and inversely proportional to the square of the separation
M1

reference to point charges
A1
6(b)(i)
(near to each sphere,) fields are in opposite directions or point (between spheres) where fields are equal and opposite or point (between spheres) where field strength is zero
M1

so same (sign of charge)
A1
6(b)(ii)
(at x = 5.0 cm,) E = 3.0 × 10
3
V m
–
1
and a = qE / m
C1

E = (1.60 × 10
–
19
× 3.0 × 10
3
) / (1.67 × 10
–
27
)
C1

= 2.9 × 10
11
m
s
–
2

A1
6(c)
field strength or E is potential gradient or field strength is rate of change of (electric) potential
M1

(field strength) maximum at x = 6 cm
A1

9702/42
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Question
Answer
Marks
7(a)
equal and opposite charges on the plates so no resultant charge
B1

+ve and –ve charges separated so energy stored
B1
7(b)
charge / potential difference
M1

reference to charge on one plate and p.d. between plates
A1
7(c)
energy = ½ CV
2

or energy = ½ QV and C = Q / V
C1

(1 / 16) × ½ CV
0
2
= ½ CV
2
V = ¼ V
0

A1

9702/42
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Question
Answer
Marks
8(a)(i)
circle around both diodes
B1
8(a)(ii)
indicates (whether) temperature
M1

(is) above or below a set value
A1
8(b)(i)
(when resistance of C > R
V
,) V
–
> V
+
or V
+
< 3
V
or p.d. across R
V
< p.d. across R/Y/3
V
or p.d. across C > p.d. across R/ X/3
V
M1

op-amp output is negative
M1

(only) green
A1
8(b)(ii)
resistance of C becomes less than R
V

or V
–
< V
+
B1

green (LED) goes out
A1

blue (LED) comes on
A1
8(c)
changes/determines temperature at which LEDs switch
B1

9702/42
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Question
Answer
Marks
9(a)(i)
Hall voltage depends on thickness of slice
C1

thinner slice, larger Hall voltage
A1
9(a)(ii)
Hall voltage depends on current in slice
B1
9(b)
sinusoidal wave, one cycle
B1

at
θ
= 0 and at
θ
= 360°, V
H
= V
MAX
B1

at
θ
= 180°, V
H
= –V
MAX

B1

9702/42
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Question
Answer
Marks
10(a)
two from: • frequency below which electrons not ejected • maximum energy of electron depends on frequency • maximum energy of electrons does not depend on intensity • instantaneous emission of electrons
B2
10(b)(i)
(λ
0
is the) threshold wavelength
or wavelength corresponding to threshold frequency or maximum wavelength for emission of electrons
B1
10(b)(ii)1.
intercept = 1
/
λ
0
= 2.2 × 10
6
m
–
1

λ
0
= 4.5 × 10
–7
m or 450 nm
A1
10(b)(ii)2.
gradient = hc
C1

gradient = 2.0 × 10
–
25
or correct substitution into gradient formula
C1

h = (2.0 × 10
–
25
) / (3.0 × 10
8
) = 6.7 × 10
–
34
J s
A1
10(c)
line: same gradient
B1

straight line, positive gradient, intercept at greater than 2.2 × 10
6
when candidate’s line extrapolated
B1

9702/42
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Question
Answer
Marks
11(a)
loss of (electric) potential energy = gain in kinetic energy or qV = ½ mv
2

or E
K
= p
2
/ 2m = qV
B1

p = mv with algebra leading to p = √(2mqV)
B1
11(b)(i)
particle/electron has a wavelength (associated with it)
B1

dependent on its momentum or when/because particle is moving
B1
11(b)(ii)
p = (2 × 9.11 × 10
–
31
× 1.60 × 10
–
19
× 120)
1/2
C1

λ
= (6.63 × 10
–
34
) / (5.91 × 10
–
24
)
C1

= 1.12 × 10
–
10
m
A1
11(c)
wavelength is similar to separation of atoms
M1

so diffraction observed
A1

9702/42
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Question
Answer
Marks
12(a)
7
e
0
1−

A1
12(b)(i)
E = mc
2
C1

= 1.66 × 10
–
27
× (3.00 × 10
8
)
2
M1

= 1.494 × 10
–
10
J
division by 1.60 × 10
–13
clear to give 934 MeV
A1
12(b)(ii)
∆m = (82 × 1.00863u) + (57 × 1.00728u) – 138.955u = (–) 1.16762 (u)
C1

energy = 1.16762 × 934
C1

energy per nucleon = (1.16762 × 934) / 139 = 7.85 MeV
A1
12(c)
above A = 56, binding energy per nucleon decreases as A increases
B1

U-235 has larger nucleon number
M1

so less (binding energy per nucleon)
A1

or

fission takes place with uranium
(B1)

fission reaction releases energy
(M1)

binding energy per nucleon less (for uranium than for products)
(A1)

® IGCSE is a registered trademark.

This document consists of 12 printed pages.

© UCLES 2017 [Turn ove
r

Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level

PHYSICS 9702/43
Paper 4 A Level Structured Questions May/June 2017
MARK SCHEME
Maximum Mark: 100

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 will not enter into discussions about these mark schemes.

Cambridge is publishing the mark schemes for the May/June 2017 series for most Cambridge IGCSE
®
,
Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level
components.

9702/43
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 2 of 12

Question
Answer
Marks
1(a) gravitational force (of attraction between satellite and planet) B1
provides / is centripetal force (on satellite about the planet)
B1
1(b) M = (4/3) × πR
3
ρ

B1

ω
= 2π / T or v = 2πnR / T B1
GM / (nR)
2
= nR
ω
2
or v
2
/ nR M1
substitution clear to give
ρ
= 3πn
3
/ GT
2
A1
1(c) n = (3.84 × 10
5
) / (6.38 × 10
3
) = 60.19 or 60.2C1

ρ
= 3π × 60.19
3
/ [(6.67 × 10
–
11
) × (27.3 × 24 × 3600)
2
] C1

ρ
= 5.54 × 10
3
kg m
–
3
A1

9702/43
Cambridge International AS/A Level – Mark Scheme
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May/June 2017
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Question
Answer
Marks
2(a) e.g. period = 3 / 2.5 C1
frequency = 0.83 Hz A1
2(b) light (damping) B1 2(c) at 2.7 s, A
0
= 1.5 (cm) B1
energy = ½ m × 4π
2
f
2
A
0
2
B1
= ½ × 0.18 × 4π
2
× 0.83
2
× (1.5 × 10
–
2
)
2

= 5.51 × 10
–4
(J)
C1
at 7.5 s, A
0
= 0.75 (cm) B1
energy = ¼ × 5.51 × 10
–
4

or
energy = ½ × 0.18 × 4π
2
× 0.83
2
× (0.75 × 10
–2
)
2

C1
energy = 1.38 × 10
–
4
(J)

change = (5.51 × 10 –4
– 1.38 × 10
–4
) = 4.13
J
A1

9702/43
Cambridge International AS/A Level – Mark Scheme
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© UCLES 2017 Page 4 of 12

Question
Answer
Marks
3(a)(i) signal consists of (a series of) 1s and 0s or offs and ons or highs and lows B1 3(a)(ii) component X: parallel-to-serial converter B1
component Y: DAC/digital-to-analogue converter B1
3(a)(iii) sample the (analogue) signal M1
at regular intervals and converts the analogue number to a digital number A1
3(b)(i) attenuation in fibre = 84 × 0.19 (= 16 dB) C1
ratio = 16 + 28

= 44 dB
A1
3(b)(ii) ratio / dB = 10 lg (P
2
/ P
1
) C1
44 = 10 lg ({9.7 × 10
–
3
} / P)
or
–44 = 10 lg (P / {9.7 × 10
–3
})
C1
power = 3.9 × 10
–
7
W A1

9702/43
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© UCLES 2017 Page 5 of 12

Question
Answer
Marks
4(a) random/haphazard B1
constant velocity or speed in a straight line between collisions
or
distribution of speeds/different directions
B1
4(b) (small) specks of light/bright specks/pollen grains/dust particles/smoke particles M1
moving haphazardly/randomly/jerky/in a zigzag fashion A1
4(c)(i) pV = ⅓ Nm〈c
2
〉

1.05 × 10
5
× 0.0240 = ⅓ × 4.00 × 10
–3
× 〈c
2
〉
C1
〈c
2
〉 = 1.89 × 10
6
C1
or ½ m〈c
2
〉 = (3 / 2) kT

0.5 × (4.00 × 10
–3
/ 6.02 × 10
23
) × 〈c
2
〉 = 1.5 × 1.38 × 10
–23
× 300
(C1)
〈c
2
〉 = 1.87 × 10
6
(C1)
or nRT = ⅓ Nm〈c
2
〉

1.00 × 8.31 × 300 = ⅓ × 4.00 × 10
–3
× 〈c
2
〉
(C1)
〈c
2
〉 = 1.87 × 10
6
(C1)
c
r.m.s.
= 1.37 × 10
3
m s
–
1
A1

9702/43
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Question
Answer
Marks
4(c)(ii) 〈c
2
〉 ∝ T C1
〈c
2
〉 at 177 °C = 1.89 × 10
6
× (450 / 300) C1
c
r.m.s.
at 177 °C = 1.68 × 10
3
m s
–
1
A1
Question
Answer
Marks
5(a) (loss in) kinetic energy of α-particle = Qq / 4π
ε
0
r
or
7.7 × 10
–13
= Qq / 4π
ε
0
r
C1
7.7 × 10
–
13
= 8.99 × 10
9
× 79 × 2 × (1.60 × 10
–
19
)
2
/ r M1
r = 4.7 × 10
–
14
m

r is closest distance of approach so radius less than this
A1
5(b) force = Qq / 4π
ε
0
r
2
= 4u × a C1
8.99 × 10
9
× 79 × 2 × (1.60 × 10
–
19
)
2
/ (4.7 × 10
–
14
)
2
= 4 × 1.66 × 10
–
27
× a C1
a = 2.5 × 10
27
m s
–
2
A1
5(c) so that single interactions between nucleus and α-particle can be studied
or
so that multiple deflections with nucleus do not occur
B1

9702/43
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Question
Answer
Marks
6(a)(i) lamp needs ‘high’ power/‘large’ current/‘large’ voltage B1
op-amp can deliver only a small current/small voltage B1
6(a)(ii) correct symbol for relay coil connected between output and earth B1
switch between mains supply and lamp B1
6(b)(i) vary light intensity at which lamp is switched on/off B1 6(b)(ii) so that relay operates for only one current/voltage direction
or
so that relay/lamp operates for either dark or light conditions
B1
6(c) when light level increases, LDR resistance decreases B1
(R
LDR
low,) so V
–
> V
+
, so V
OUT
negative/–5 V (must be consistent with B1 mark) M1
or when light level decreases, LDR resistance increases (B1) (R
LDR
high,) so V
–
< V
+
, so V
OUT
is positive/+5 V (must be consistent with B1 mark) (M1)
lamp comes on as light level decreases
or
lamp goes off as light level increases
A1

9702/43
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Question
Answer
Marks
7(a) (magnetic) force (always) normal to velocity/direction of motion M1
(magnitude of magnetic) force constant
or
speed is constant/kinetic energy is constant
M1 so provides the centripetal force A1
7(b) increase in KE = loss in PE or ½ mv
2
= qV M1
p = mv with algebra leading to p = √(2mqV) A1
7(c) Bqv = mv
2
/ r

mv = Bqr or p = Bqr
C1
(2 × 9.11 × 10
–
31
× 1.60 × 10
–
19
× 120)
1/2
= B × 1.60 × 10
–
19
× 0.074 C1
B = 5.0 × 10
–
4
T A1
7(d) greater momentum M1
(p = Bqr and) so r increased A1

9702/43
Cambridge International AS/A Level – Mark Scheme
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Question
Answer
Marks
8 strong (uniform) magnetic field B1
* nuclei precess/rotate about field (direction)

radio frequency pulse/RF pulse (applied) B1 * RF or pulse is at Larmor frequency / frequency of precession causes resonance / excitation (of nuclei)/nuclei to absorb energy B1 on relaxation/de-excitation, nuclei emit RF/pulse B1 * (emitted) RF/pulse detected and processed non-uniform field (superposed on uniform field) B1 allows positions of (resonating) nuclei to be determined
B1
* allows for position of detection to be changed/different slices to be studied max. 2 of additional detail points marked * B2

9702/43
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Question
Answer
Marks
9(a)(i) core reduces loss of (magnetic) flux linkage/improves flux linkage B1 9(a)(ii) reduces (size of eddy) currents in core B1
(so that) heating of core is reduced B1
9(b) alternating voltage gives rise to changing magnetic flux in core M1
(changing) flux links the secondary coil A1 induced e.m.f. (in secondary) only when flux is changing/cut B1
Question
Answer
Marks
10(a)(i) penetration of beam M1
greater hardness means greater penetration/shorter wavelength/higher frequency/higher photon energy A1
10(a)(ii) greater accelerating potential difference
or
greater p.d. between anode and cathode
B1
10(b) I = I
0
exp(–
µ
x)

ratio = (exp {–1.5 × 2.9}) / (exp {–4.0 × 0.95}) (= exp {–0.55})
C1
= 0.58 A1

9702/43
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Question
Answer
Marks
11(a) electrons (in gas atoms/molecules) interact with photons B1
photon energy causes electron to move to higher energy level/to be excited B1 photon energy = difference in energy of (electron) energy levels B1 when electrons de-excite, photons emitted in all directions (so dark line) B1
11(b)(i) photon energy ∝ 1 /
λ
C1
energy = 1.68 eV A1 or E = hc /
λ

E = 6.63 × 10
–34
× 3.0 × 10
8
/ (740 × 10
–9
)

= 2.688 × 10
–19
J
(C1)
energy = 1.68 eV (A1)
11(b)(ii) 3.4 eV → 1.5 eV
3.4 eV → 0.85 eV
3.4 eV → 0.54 eV
all correct and none incorrect 2/2
2 correct and 1 incorrect or only 2 correctly drawn 1/2
B2

9702/43
Cambridge International AS/A Level – Mark Scheme
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Question
Answer
Marks
12(a) x = 7 A1
12(b)(i) E = mc
2
C1
= 1.66 × 10
–
27
× (3.0 × 10
8
)
2

= 1.494 × 10
–10
J
C1
division by 1.6 × 10
–
13
clear to give 934 MeV A1
12(b)(ii) ∆m = (235.123 + 1.00863) – (94.945 + 138.955 + 2 × 1.00863 + 7 × 5.49 × 10
–
4
)
or
∆m = 235.123 – (94.945 + 138.955 + 1 × 1.00863 + 7 × 5.49 × 10
–4
)
C1
= 0.21053 u C1 energy = 0.21053 × 934

= 197 MeV
A1
12(c) kinetic energy of nuclei/particles/products/fragments B1
γ–ray photon energy B1

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Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level

PHYSICS 9702/51
Paper 5 Planning, Analysis and Evaluation May/June 2017
MARK SCHEME
Maximum Mark: 30

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 will not enter into discussions about these mark schemes.

Cambridge is publishing the mark schemes for the May/June 2017 series for most Cambridge IGCSE
®
,
Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level
components.

9702/51 Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017 © UCLES 2017 Page 2 of 6

Question
Answer
Marks
1 Defining the problem
(sin)
θ
is the independent variable and v is the dependent variable or vary (sin)
θ
and measure v 1
keep s (PQ) constant
1
Methods of data collection labelled diagram showing inclined plane with labelled support and P and Q marked
1
method to measure angle e.g. use a protractor to measure
θ
or use a ruler to measure marked distances from which sin
θ
or
θ

may be determined
1
method of timing for an appropriate distance to determine v (at Q) e.g. use a stopwatch/timer or correctly positioned light
gate(s) connected to a timer/data-logger or correctly positioned motion sensor connected to data-logger
1 measurement of an appropriate distance to determine v (at Q) e.g. rule to measure an appropriate length or length of a card
to interrupt light beam or distance from motion sensor to Q
1 Method of analysis plot a graph of v
2
against sin
θ
1
relationship valid if a straight line produced (not passing through the origin) 1
+
=−× gradient
2
Bm
g
ms

or
=
g

+
× -intercept
2
Bm
y
Bs

1

9702/51 Cambridge International AS/A Level – Mark Scheme
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Question
Answer
Marks
Additional detail including safety considerations Max. 6 D1 use cushion/foam/sandbox for falling body (B)

D2 (sin)
θ
determined using trigonometry relationship using marked lengths

D3 appropriate equation to determine v (at Q) e.g.
=
2s
v
t

D4 repeat experiment for each
θ
and average v or t

D5 use of balance to measure mass of wooden block m and falling body B and rule to measure s

D6
=
+
2
-intercept .
Bsg
y
Bm

D7 clean surfaces of blocks/inclined plane/ensure surface of the plane is smooth D8 keep B and m constant or keep mass of block and mass of falling body constant

D9 method to ensure that wooden block starts at the same position P, e.g. put a mark on the block or align front or back of
block
D10 method to prevent plane slipping so that angle being measured remains the same, e.g. a mass as a stop

9702/51 Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017 © UCLES 2017 Page 4 of 6

Question
Answer
Marks
2(a)
gradient =
−
2
v

y-intercept =
2L
v

1
2(b)
0.80 ± 0.01
0.77 ± 0.01 0.73 ± 0.01 0.70 ± 0.01 0.66 ± 0.01 0.62 ± 0.01
First mark for all values of t correct.
Second mark for uncertainties correct.
2
2(c)(i) Six points plotted correctly.
Must be accurate to less than half a small square. No “blobs”. Diameter of points must be less than half a small square.
1
Error bars in t plotted correctly.
All error bars to be plotted. Length of bar must be accurate to less than half a small square and symmetrical.
1

9702/51 Cambridge International AS/A Level – Mark Scheme
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Question
Answer
Marks
2(c)(ii) Line of best fit drawn.

If points are plotted correctly then upper end of line should pass between (4.8, 0.76) and (5.6, 0.76) and lower end of line
should pass between (17.6, 0.64) and (18.8, 0.64). Line should not be from first to last plot.
1
Worst acceptable line drawn (steepest or shallowest possible line).
All error bars must be plotted.
1
2(c)(iii) Gradient determined with a triangle that is at least half the length of the drawn line.
Gradient must be negative.
1
uncertainty = gradient of line of best fit – gradient of worst acceptable line
or
uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)
1
2(c)(iv) y-intercept read-off y-axis to less than half small square or determined by substitution into y = mx + c. 1
uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line
or
uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)
1
2(d)(i) v determined from gradient and units for v and L correct with correct power of ten.
=−=−
22
gradient 2(c)(iii)
v

1
L determined from y-intercept and v and L given to 2 or 3 significant figures.
Correct substitution of numbers must be seen.

=× =× =−=−
-intercept (c)(iv)
-intercept (c)(iv)
2 2 gradient (c)(iii)
vvy
Ly

1

9702/51 Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017 © UCLES 2017 Page 6 of 6

Question
Answer
Marks
2(d)(ii) % uncertainty in v = % uncertainty in gradient 1
% uncertainty in L = % uncertainty in y-intercept + % uncertainty in gradient
or
% uncertainty in L = % uncertainty in y-intercept + % uncertainty in v

Correct substitution of numbers must be seen.

Maximum/minimum methods:
=×
max -intercept
Max max -intercept max or
mingradient
y
Ly v

=×
min -intercept
Min min -intercept min or
max gradient
y
Ly v

1

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r

Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level

PHYSICS 9702/52
Paper 5 Planning, Analysis and Evaluation May/June 2017
MARK SCHEME
Maximum Mark: 30

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 will not enter into discussions about these mark schemes.

Cambridge is publishing the mark schemes for the May/June 2017 series for most Cambridge IGCSE
®
,
Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level
components.

9702/52
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 2 of 6

Question
Answer
Marks
1 Defining the problem
r is the independent variable and f (frequency of turntable) is the dependent variable or vary r and measure f (frequency of
turntable)
1 keep m constant
1
Methods of data collection labelled diagram showing power supply connected to motor (two leads) within turntable; circuits must be workable
1
method to change frequency of rotation of the turntable, e.g. adjust output of (variable) power supply or adjust variable
resistor
1 increase frequency until the cube moves (relative to the turntable) 1 method to determine period of rotation of the turntable, e.g. stopwatch, light gate attached to a timer/data-logger or
stroboscope
1 Method of analysis plots a graph of f against 1 / r (allow log f against log r) 1 relationship valid if a straight line produced passing through the origin
(for lg f vs. lg r straight line of gradient of –1)
1 K = gradient × 4π
2
m
(for lg f vs. lg r, K = 10
y-intercept
× 4π
2
m)
1

9702/52
Cambridge International AS/A Level – Mark Scheme
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Question
Answer
Marks
Additional detail including safety considerations Max. 6 D1 use safety screen D2 time at least 10 rotations of turntable or detailed use of stroboscope D3 f = 1 / T for correct determination of period of rotation of turntable D4 repeat experiment for each r and average f D5 use balance to measure mass of cube D6 wait for turntable to rotate steadily before increasing frequency
or
gradual/incremental/slowly increase in frequency
D7 use a spirit level to check that turntable is horizontal or clean cube/surface D8 use a rule to measure r D9 method to ensure r is measured to the centre of the cube, e.g. put a mark on the cube or align front or back of cube
by a set distance
D10 method to determine centre of the turntable e.g. measure two or more diameters/maximum distance ideas

9702/52
Cambridge International AS/A Level – Mark Scheme
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© UCLES 2017 Page 4 of 6

Question
Answer
Marks
2(a)
gradient =
1
E

y-intercept =
Q
E

1
2(b)
P / Ω

1

I
/ A
–1

± 9 29 or 29.4 ± 11 36 or 35.7 ± 16.5 53 or 52.6 ± 23.5 71 or 71.4 ± 28 83 or 83.3 ± 34 100

First mark for uncertainties correct. Allow 1 s.f. e.g. 10, 10, 20, 20, 30, 30.
Second mark for all second column correct. Allow a mixture of significant figures.
2
2(c)(i) Six points plotted correctly.
Must be accurate to less than half a small square. No “blobs”. Diameter of points must be less than half a small square.
1
Error bars in P plotted correctly.
All error bars to be plotted. Length of bar must be accurate to less than half a small square and symmetrical.
1

9702/52
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 5 of 6

Question
Answer
Marks
2(c)(ii) Line of best fit drawn.

If points are plotted correctly then lower end of line should pass between (200, 32) and (200, 34) and upper end of line should
pass between (600, 88) and (600, 91).
1
Worst acceptable line drawn (steepest or shallowest possible line).
All error bars must be plotted.
1
2(c)(iii) Gradient determined with a triangle that is at least half the length of the drawn line. 1
uncertainty = gradient of line of best fit – gradient of worst acceptable line
or
uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)
1
2(c)(iv) y-intercept determined by substitution of correct point into y = mx + c. 1
uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line
or
uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)
1

9702/52
Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017
© UCLES 2017 Page 6 of 6

Question
Answer
Marks
2(d)(i) E determined using gradient and units for E and Q with correct power of ten.
==
11
gradient 2(c)(iii)
E

1
Q determined using y-intercept and E and Q given to 2 or 3 significant figures.
Correct substitution of numbers must be seen.
=× =× = =
-intercept 2(c)(iv)
-intercept 2(c)(iv)
gradient 2(c)(iii)
y
QEy E

1
2(d)(ii) % uncertainty in E = % uncertainty in gradient 1
% uncertainty in Q = % uncertainty in E + % uncertainty in y-intercept
or
% uncertainty in Q = % uncertainty in gradient + % uncertainty in y-intercept.

Correct substitution of numbers must be seen.

Maximum/minimum methods:
=×
max -intercept
Max max -intercept max or
mingradient
y
Qy E

=×
min -intercept
Min min -intercept min or
max gradient
y
Qy E

1

® IGCSE is a registered trademark.

This document consists of 6 printed pages.

© UCLES 2017 [Turn ove
r

Cambridge International Examinations
Cambridge International Advanced Subsidiary and Advanced Level

PHYSICS 9702/53
Paper 5 Planning, Analysis and Evaluation May/June 2017
MARK SCHEME
Maximum Mark: 30

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 will not enter into discussions about these mark schemes.

Cambridge is publishing the mark schemes for the May/June 2017 series for most Cambridge IGCSE
®
,
Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level
components.

9702/53 Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017 © UCLES 2017 Page 2 of 6

Question
Answer
Marks
1 Defining the problem
(sin)
θ
is the independent variable and v is the dependent variable or vary (sin)
θ
and measure v 1
keep s (PQ) constant
1
Methods of data collection labelled diagram showing inclined plane with labelled support and P and Q marked
1
method to measure angle e.g. use a protractor to measure
θ
or use a ruler to measure marked distances from which sin
θ
or
θ

may be determined
1
method of timing for an appropriate distance to determine v (at Q) e.g. use a stopwatch/timer or correctly positioned light
gate(s) connected to a timer/data-logger or correctly positioned motion sensor connected to data-logger
1 measurement of an appropriate distance to determine v (at Q) e.g. rule to measure an appropriate length or length of a card
to interrupt light beam or distance from motion sensor to Q
1 Method of analysis plot a graph of v
2
against sin
θ
1
relationship valid if a straight line produced (not passing through the origin) 1
+
=−× gradient
2
Bm
g
ms

or
=
g

+
× -intercept
2
Bm
y
Bs

1

9702/53 Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017 © UCLES 2017 Page 3 of 6

Question
Answer
Marks
Additional detail including safety considerations Max. 6 D1 use cushion/foam/sandbox for falling body (B)

D2 (sin)
θ
determined using trigonometry relationship using marked lengths

D3 appropriate equation to determine v (at Q) e.g.
=
2s
v
t

D4 repeat experiment for each
θ
and average v or t

D5 use of balance to measure mass of wooden block m and falling body B and rule to measure s

D6
=
+
2
-intercept .
Bsg
y
Bm

D7 clean surfaces of blocks/inclined plane/ensure surface of the plane is smooth D8 keep B and m constant or keep mass of block and mass of falling body constant

D9 method to ensure that wooden block starts at the same position P, e.g. put a mark on the block or align front or back of
block
D10 method to prevent plane slipping so that angle being measured remains the same, e.g. a mass as a stop

9702/53 Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017 © UCLES 2017 Page 4 of 6

Question
Answer
Marks
2(a)
gradient =
−
2
v

y-intercept =
2L
v

1
2(b)
0.80 ± 0.01
0.77 ± 0.01 0.73 ± 0.01 0.70 ± 0.01 0.66 ± 0.01 0.62 ± 0.01
First mark for all values of t correct.
Second mark for uncertainties correct.
2
2(c)(i) Six points plotted correctly.
Must be accurate to less than half a small square. No “blobs”. Diameter of points must be less than half a small square.
1
Error bars in t plotted correctly.
All error bars to be plotted. Length of bar must be accurate to less than half a small square and symmetrical.
1

9702/53 Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017 © UCLES 2017 Page 5 of 6

Question
Answer
Marks
2(c)(ii) Line of best fit drawn.

If points are plotted correctly then upper end of line should pass between (4.8, 0.76) and (5.6, 0.76) and lower end of line
should pass between (17.6, 0.64) and (18.8, 0.64). Line should not be from first to last plot.
1
Worst acceptable line drawn (steepest or shallowest possible line).
All error bars must be plotted.
1
2(c)(iii) Gradient determined with a triangle that is at least half the length of the drawn line.
Gradient must be negative.
1
uncertainty = gradient of line of best fit – gradient of worst acceptable line
or
uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)
1
2(c)(iv) y-intercept read-off y-axis to less than half small square or determined by substitution into y = mx + c. 1
uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line
or
uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)
1
2(d)(i) v determined from gradient and units for v and L correct with correct power of ten.
=−=−
22
gradient 2(c)(iii)
v

1
L determined from y-intercept and v and L given to 2 or 3 significant figures.
Correct substitution of numbers must be seen.

=× =× =−=−
-intercept (c)(iv)
-intercept (c)(iv)
2 2 gradient (c)(iii)
vvy
Ly

1

9702/53 Cambridge International AS/A Level – Mark Scheme
PUBLISHED
May/June 2017 © UCLES 2017 Page 6 of 6

Question
Answer
Marks
2(d)(ii) % uncertainty in v = % uncertainty in gradient 1
% uncertainty in L = % uncertainty in y-intercept + % uncertainty in gradient
or
% uncertainty in L = % uncertainty in y-intercept + % uncertainty in v

Correct substitution of numbers must be seen.

Maximum/minimum methods:
=×
max -intercept
Max max -intercept max or
mingradient
y
Ly v

=×
min -intercept
Min min -intercept min or
max gradient
y
Ly v

1

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