Physics AS Level Multiple Choice and Structured Questions Mark Schemes

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© UCLES 2022 [Turn over
Cambridge International AS & A Level
PHYSICS 9702/12
Paper 1 Multiple Choice February/March 2022
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 February/March 2022 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
February/March 2022
© UCLES 2022 Page 2 of 3
Question Answer Marks
1 B 1
2 B 1
3 D 1
4 B 1
5 D 1
6 D 1
7 C 1
8 A 1
9 C 1
10 D 1
11 A 1
12 C 1
13 D 1
14 D 1
15 A 1
16 B 1
17 C 1
18 D 1
19 C 1
20 D 1
21 B 1
22 C 1
23 A 1
24 D 1
25 C 1
26 C 1
27 D 1
28 B 1

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February/March 2022
29 D 1
30 B 1
31 A 1
32 C 1
33 C 1
34 A 1
35 A 1
36 B 1
37 C 1
38 C 1
39 B 1
40 D 1

PHYSICS 9702/22
Paper 2 AS Level Structured Questions February/March 2022
MARK SCHEME
Maximum Mark: 60
Published
examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the
details of the discussions that took place at an Examiners’ meeting before marking began, which would have
considered the acceptability of alternative answers.
Teachers.
components.

PUBLISHED
February/March 2022
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.

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February/March 2022
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.
Science-Specific Marking Principles
1 Examiners should consider the context and scientific use of any keywords when awarding marks. Although keywords may be present, marks
should not be awarded if the keywords are used incorrectly.
2 The examiner should not choose between contradictory statements given in the same question part, and credit should not be awarded for any
correct statement that is contradicted within the same question part. Wrong science that is irrelevant to the question should be ignored.
3 Although spellings do not have to be correct, spellings of syllabus terms must allow for clear and unambiguous separation from other syllabus
terms with which they may be confused (e.g. ethane / ethene, glucagon / glycogen, refraction / reflection).
4 The error carried forward (ecf) principle should be applied, where appropriate. If an incorrect answer is subsequently used in a scientifically
correct way, the candidate should be awarded these subsequent marking points. Further guidance will be included in the mark scheme where
necessary and any exceptions to this general principle will be noted.
5 ‘List rule’ guidance
For questions that require n responses (e.g. State two reasons …):
• The response should be read as continuous prose, even when numbered answer spaces are provided.
• Any response marked ignore in the mark scheme should not count towards n.
• Incorrect responses should not be awarded credit but will still count towards n.
• Read the entire response to check for any responses that contradict those that would otherwise be credited. Credit should not be
awarded for any responses that are contradicted within the rest of the response. Where two responses contradict one another, this should
be treated as a single incorrect response.
• Non-contradictory responses after the first n responses may be ignored even if they include incorrect science.

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February/March 2022
6 Calculation specific guidance
Correct answers to calculations should be given full credit even if there is no working or incorrect working, unless the question states ‘show
your working’.
For questions in which the number of significant figures required is not stated, credit should be awarded for correct answers when rounded by
the examiner to the number of significant figures given in the mark scheme. This may not apply to measured values.
For answers given in standard form (e.g. a × 10n) in which the convention of restricting the value of the coefficient (a) to a value between 1
and 10 is not followed, credit may still be awarded if the answer can be converted to the answer given in the mark scheme.
Unless a separate mark is given for a unit, a missing or incorrect unit will normally mean that the final calculation mark is not awarded.
Exceptions to this general principle will be noted in the mark scheme.
7 Guidance for chemical equations
Multiples / fractions of coefficients used in chemical equations are acceptable unless stated otherwise in the mark scheme.
State symbols given in an equation should be ignored unless asked for in the question or stated otherwise in the mark scheme.

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February/March 2022
Examples of how to apply the list rule
State three reasons…. [3]
A 1 Correct 
2
2 Correct 
3 Wrong 
B 1 Correct, Correct , 
3
(4 responses) 2 Correct 
3 Wrong ignore
C 1 Correct 
2
(4 responses) 2 Correct, Wrong , 
3 Correct ignore
D 1 Correct 
2
(4 responses) 2 Correct, CON
(of 2.)
, (discount 2)
3 Correct 
E 1 Correct 
3
3 Correct, Wrong 
F 1 Correct 
2
3 Correct
CON (of 3.)

(discount 3)
G 1 Correct 
3
3 Correct
Correct
CON (of 4.)

ignore
ignore
H 1 Correct 
2
(4 responses) 2 Correct 
3 CON (of 2.)
Correct
(discount 2)

I 1 Correct 
2
3 Correct
CON (of 2.)

(discount 2)

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February/March 2022
Abbreviations
/ Alternative and acceptable answers for the same marking point.
( ) Bracketed content indicates words which do not need to be explicitly seen to gain credit but which indicate the context for an
answer. The context does not need to be seen but if a context is given that is incorrect then the mark should not be awarded.
___ Underlined content must be present in answer to award the mark. This means either the exact word or another word that has the
same technical meaning.
Mark categories
B marks These are independent marks, which do not depend on other marks. For a B mark to be awarded, the point to which it refers must
be seen specifically in the candidate’s answer.
M marks These are method marks upon which A marks later depend. For an M mark to be awarded, the point to which it refers must be seen
specifically in the candidate’s answer. If a candidate is not awarded an M mark, then the later A mark cannot be awarded either.
C marks These are compensatory marks which can be awarded even if the points to which they refer are not written down by the candidate,
providing subsequent working gives evidence that they must have known them. For example, if an equation carries a C mark and
the candidate does not write down the actual equation but does correct working which shows the candidate knew the equation, then
the C mark is awarded.
If a correct answer is given to a numerical question, all of the preceding C marks are awarded automatically. It is only necessary to
consider each of the C marks in turn when the numerical answer is not correct.
A marks These are answer marks. They may depend on an M mark or allow a C mark to be awarded by implication.
Annotations
 Indicates the point at which a mark has been awarded.
X Indicates an incorrect answer or a point at which a decision is made not to award a mark.
XP Indicates a physically incorrect equation (‘incorrect physics’). No credit is given for substitution, or subsequent arithmetic, in a
physically incorrect equation.
ECF Indicates ‘error carried forward’. Answers to later numerical questions can always be awarded up to full credit provided they are
consistent with earlier incorrect answers. Within a section of a numerical question, ECF can be given after AE, TE and POT errors,
but not after XP.

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February/March 2022
AE Indicates an arithmetic error. Do not allow the mark where the error occurs. Then follow through the working/calculation giving full
subsequent ECF if there are no further errors.
POT Indicates a power of ten error. Do not allow the mark where the error occurs. Then follow through the working/calculation giving full
subsequent ECF if there are no further errors.
TE Indicates incorrect transcription of the correct data from the question, a graph, data sheet or a previous answer. For example, the
value of 1.6 × 10–19 has been written down as 6.1 × 10–19 or 1.6 × 1019.
Do not allow the mark where the error occurs. Then follow through the working/calculation giving full subsequent ECF if there are
no further errors.
SF Indicates that the correct answer is seen in the working but the final answer is incorrect as it is expressed to too few significant
figures.
BOD Indicates that a mark is awarded where the candidate provides an answer that is not totally satisfactory, but the examiner feels that
sufficient work has been done (‘benefit of doubt’).
CON Indicates that a response is contradictory.
I Indicates parts of a response that have been seen but disregarded as irrelevant.
M0 Indicates where an A category mark has not been awarded due to the M category mark upon which it depends not having
previously been awarded.
^ Indicates where more is needed for a mark to be awarded (what is written is not wrong, but not enough). May also be used to
annotate a response space that has been left completely blank.
SEEN Indicates that a page has been seen.

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February/March 2022
1(a) Fv: kg m s–2
k: kg m s–2 / m × m s–1
C1
= kg m–1 s–1 A1
1(b) F = ρgV
V = 4 / 3 × π × (2.1 × 10–3)3 (=3.88 × 10–8 m3)
C1
ρ = 4.8 × 10–4 / 9.81 × V C1
= 1300 kg m–3 A1
1(c)(i) W downwards, U upwards, Fv upwards B1
1(c)(ii) FV = 7.2 × 10–4 – 4.8 × 10–4
= 2.4 × 10–4 (N)
C1
velocity = 2.4 × 10–4 / (17 × 2.1 × 10–3)
= 6.7 × 10–3 m s–1
A1
2(a) force (on droplet of water) in horizontal direction is zero. B1
2(b) (time taken =) 3.5 / 6.6 = 0.53 (s) A1
2(c) s = ut + ½at 2
s = ½ × 9.81 × 0.532
C1
h = 1.4 m A1

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February/March 2022
2(d) displacement is straight-line distance (from P to Q) so less (than distance along path)
or
displacement is the shortest distance (from P to Q).
B1
2(e) (displacement)2 = 3.52 + 1.42 C1
displacement = 3.8 m A1
3(a) (m =) ρV C1
= 1.0 × 103 × 1.5 × 10–4 × 5.0 × 1.6 = 1.2 (kg) A1
3(b)(i) (∆)p = 1.2 × 5.0
= 6.0 N s
A1
3(b)(ii) F = 6.0 / 1.6 or 1.2 × 5.0/1.6
= 3.8 N
A1
3(c) Newton’s third law applies (so) 3.8 N. B1
3(d) p = F / A
= 3.8 / 1.5 × 10–4
C1
= 2.5 × 104 Pa A1

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February/March 2022
4(a) ratio = 300 / 3200
= 0.094
A1
4(b) E = ½mv 2 or E ∝ v 2 C1
ratio = (0.094)0.5
= 0.31
A1
4(c) work (done against frictional force) = 3200 – 300 (=2900) C1
length = 2900 / 76
= 38 m
A1
4(d)(i) E = ½kx 2 or E = ½Fx and F = kx
140 = ½ × 63 × x 2 or 140 = ½Fx and F = 63x
C1
x = 2.1 m A1
4(d)(ii) percentage efficiency = (140 / 300) × 100
= 47%
A1
4(d)(iii) curved line from the origin M1
gradient of line increases A1
5(a)(i) (two) waves travelling (at same speed) in opposite directions overlap B1
waves (of the same type) have same frequency/wavelength B1
5(a)(ii) phase difference = 0 A1

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February/March 2022
5(b)(i) fO = fS v / v–vS
543 = f × 334 / (334 – 13)
C1
f = 522 Hz A1
5(b)(ii) (the speed is) decreasing B1
5(c)(i) I ∝ A2 B1
IT / I0 = cos2 20° or AT / A0 = cos 20° C1
ratio = 0.94 A1
5(c)(ii) angle = 140° A1
6(a)(i) P = VI C1
I= 36 / 8.0
= 4.5 A
A1
6(a)(ii) charge = 4.5 × 50
= 225
C1
number = 225 / 1.6 × 10–19
= 1.4 × 1021
A1
6(a)(iii) R = V 2 / P or R = V / I or R = P / I 2
= 8.02 / 36 or = 8.0 / 4.5 or = 36 / 4.52
C1
= 1.8 Ω A1

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February/March 2022
6(b) R = ρL / A C1
L = (1.8 × 0.25 × 10–6) / 1.4 × 10–6
= 0.32 m
A1
6(c) (larger cross-sectional area, same length, same resistivity and so) less resistance M1
(same p.d. and more current so) more power (dissipated) A1
6(d) current (in wire) is the same M1
(same p.d. across wire so) power stays the same A1
7(a)(i) (electron) neutrino B1
7(a)(ii) nucleon number = 22 A1
proton number = 10 A1
7(a)(iii) up up down changes to up down down
or
up changes to down
B1
7(b)(i) charge = – ⅔ e A1
7(b)(ii) antiup / anticharm / antitop B1

PHYSICS 9702/33
Paper 3 Advanced Practical Skills 1 February/March 2022
MARK SCHEME
Maximum Mark: 40
Published
Teachers.
components.

PUBLISHED
February/March 2022
marking principles.
descriptors.

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February/March 2022

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February/March 2022
your working’.

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February/March 2022
1(a)(i) Value of F to nearest mm and in range 6.5 to 8.5 cm 1
1(a)(ii) a and b, with unit 1
Correct calculation of y 1
1(b) Six sets of readings of h and y with correct trend and without help scores 4 marks, five sets scores 3 marks etc. 4
Range:
hmax ⩾ 18.0 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. h / cm.
1
Consistency:
All values of h and a and b must be given to the nearest mm
1
1(c)(i) Axes:
Sensible scales are used, no awkward scales (e.g. 3:10)
Scales are chosen so that the plotted points occupy at least half the graph grid in both x and y directions
Axes are labelled with the quantity which is being plotted.
Scale markings are no more than 2cm (one large square) apart.
1
Plotting of points:
All observations in the table are plotted on the grid.
Diameters of plotted points are less than half a small square (no blobs).
Points are plotted to an accuracy of half a small square in both x and y directions.
1
Quality:
All points in the table must be plotted (at least 5) for this mark to be awarded, and trend of points must have a negative
gradient.
It must be possible to draw a straight line that is within ± 0.5 cm on the y axis of all plotted points.
1

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1(c)(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. There must be at least 5
points left after the anomalous point is discarded.
Lines must not be kinked or thicker than half a square.
1
1(c)(iii) Gradient:
Sign of gradient matches graph.
The hypotenuse of the triangle used is greater than half the length of the drawn line.
Method of calculation is correct, not Δx / Δy.
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 h = zero accurate to half a small square in y direction.
1
1(d) P equal to candidate’s gradient, and Q equal to candidate’s intercept. Values must not be written as fractions. 1
Units for P and Q correct and consistent with value (e.g. no unit for P, cm for Q) 1
1(e) Correct calculation of ρ, with correct unit 1
Value of ρ on answer line given to 2 or 3 s.f. 1

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February/March 2022
2(a)(i) All raw values of d to nearest mm. 1
Evidence of repeat readings for d. 1
2(a)(ii) Absolute uncertainty of 2 to 5 mm and correct method of calculation to obtain percentage uncertainty in d.
If several readings have been taken, then the absolute uncertainty can be half the range (but not zero if values are equal)
provided the working is clearly shown.
1
2(b) n = 1 1
Value for T in range 0.35 to 0.55 s, with unit. 1
At least two measurements of at least 5T. 1
2(c) (second d) ÷ (first d) in range 0.85 to 0.95 1
Second value for T. 1
Quality: second T longer than first T. 1
2(d)(i) Two values of k calculated correctly. 1
2(d)(ii) Justification for sig. fig. in k linked to sig. fig. in d and T 1
2(e) Calculation of percentage difference between candidate’s two k values.
Comparison of percentage difference with 20% leading to a consistent conclusion
1
2(f)(i) Two k values are not enough to draw a valid conclusion
Difficult to roll the clay into a sphere / diameter of sphere varies
Parallax error when measuring d
Clay covers some coils as well as the joint
Difficult to judge the start and/or end of an oscillation
Time period is short so it is difficult to count the oscillations
Some horizontal oscillation as well as vertical
4 max
4

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February/March 2022
2(f)(ii) Take more readings and plot a graph / calculate more k values and compare
Improved method to make a perfect sphere
Improved method of measuring d, e.g. use calipers / measure between blocks
Method to prevent touching coils, e.g. denser material / improved shape / improved mounting method
Put a fiducial mark at the centre of the oscillation
Use video with timer in view / video and review frame by frame
4 max
4

PHYSICS 9702/42
Paper 4 A Level Structured Questions February/March 2022
MARK SCHEME
Maximum Mark: 100
Published
Teachers.
components.

PUBLISHED
February/March 2022
marking principles.
descriptors.

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February/March 2022

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February/March 2022
your working’.

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February/March 2022
A 1. Correct 
2
2. Correct 
3. Wrong 
B 1. Correct, Correct , 
3
(4 responses) 2. Correct 
3. Wrong ignore
C 1. Correct 
2
(4 responses) 2. Correct, Wrong , 
3. Correct ignore
D 1. Correct 
2
(4 responses) 2. Correct, CON (of 2.) , (discount 2)
3. Correct 
E 1. Correct 
3
3. Correct, Wrong 
F 1. Correct 
2
3. Correct
CON (of 3.)

(discount 3)
G 1. Correct 
3
3. Correct
Correct
CON (of 4.)

ignore
ignore
H 1. Correct 
2
(4 responses) 2. Correct 
3. CON (of 2.)
Correct
(discount 2)

I 1. Correct 
2
(4 responses) 2. Correct 
3. Correct
CON (of 2.)

(discount 2)

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February/March 2022
1(a) at least 4 straight radial lines to P B1
all arrows pointing along the lines towards P B1
1(b) Any 2 from:
gravitational force provides the centripetal force
(centripetal or gravitational) force has constant magnitude
(centripetal or gravitational) force is perpendicular to velocity (of moon) / direction of motion (of moon)
B2
1(c)(i) 2
2
GMm
= mr
r
ω
M1
3 2
r
M=
G
ω
and gradient = 3 2
r ω hence
gradient
M
G
=
or
r3 = GM × 1/ω2 so gradient = GM hence
gradient
M
G
=
A1
1(c)(ii) M = 4.1 × 1023 / (6.0 × 107 × 6.67 × 10–11) = 1.0 × 1026
kg B1

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February/March 2022
1(c)(iii) 2
2
GMm mv
=
r
r
2
GM
=v
r
C1
11 26
2
8
6.67 10 1.0 10
v =
1.2 10
−
× × ×
×
2 7 1
v 5.6 10 m s−
= ×
C1
1
v =7500 ms− A1
2(a) 0 B1
2(b) pV = nRT
(n =) 1.5 × 105 × 4.2 × 10–3 / 8.31 × 540
C1
= 0.14 mol A1
2(c) missing pressure 1.5 (× 105) B1
both missing volumes 1.8 (× 10–3) B1
2(d)(i) (ΔU:) increase in internal energy (of the system) B1
(q:) thermal energy supplied to the system B1
(W:) work done on system B1

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February/March 2022
2(d)(ii) volume increases and work is done by the gas B1
temperature decreases and internal energy decreases B1
3(a) upthrust, weight B1
3(b) upthrust greater than weight so (resultant force is) upwards B1
3(c)(i) A, g and ρ all constant so F ∝ x B1
minus sign means F and x are in opposite directions B1
3(c)(ii) F Agρx
(a = so) a = ( )
m m
−
M1
2 Ag Ag
so = hence =
m m
ρ ρ
ω ω
A1
3(d)(i) damping due to viscous forces B1
3(d)(ii) ( ) 2 2
0
1
E = m x
2
ω C1
ω2 = (–) gradient C1
( ) 2 2 2
1 2
1
E = m (x x )
2
ω −
2 2
2.3
1 0.57 ( )(0.020 0.016 )
2 0.020
= × × −
3
= 4.7 10 J
−
×
A1

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February/March 2022
4(a) direction of force B1
force on a positive charge B1
4(b)(i)
o
Q
V =
4 r
πε
9 9
o o
4.0 10 7.2 10
+ = 0
4 x 4 (0.120 x)
− −
× − ×
πε πε −
( )
4 0.120 x = 7.2 x
−
C1
x = 0.043 m A1
4(b)(ii) fields are in the same direction so no B1
4(b)(iii) straight arrow drawn leftwards from X in direction between extended line joining Q and X and the horizontal B1

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February/March 2022
5(a) (energy stored =) area under line or ½ QV
= ½ × 8.0 × 1.2 × 10-4
C1
= 4.8 × 10–4
J A1
5(b)(i) (τ=) RC C1
(τ=) 220 × 103 × (1.2 × 10-4/8.0) = 3.3 s A1
5(b)(ii) E ∝ V2 C1
(so time to) Vo / 3
t
RC
o
V =V e
−
C1
t
o 3.3
o
V
=V e
3
−
t
3.3
1
=e
3
−
C1
t = 3.6 s A1
5(c) (total) capacitance is doubled M1
time constant is doubled A1

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February/March 2022
6(a) less in smaller solenoid B1
6(b) greater in smaller solenoid B1
6(c)(i) direction of (induced) e.m.f. M1
such as to (produce effects that) oppose the change that caused it A1
6(c)(ii) change of flux (linkage) in smaller solenoid induces e.m.f. in smaller solenoid B1
(induced) current in smaller solenoid causes field around it B1
the two fields (interact to) create an attractive force B1
7(a)(i) two diodes added in correct directions (Both diodes pointing inwards and upwards), correct symbols only B1
7(a)(ii) ‘+’ anywhere on upper output wire B1
7(b)(i) ω = 2π / T
= 2π / 2.5
= 0.80 π or 4π / 5 or 2.5
C1
(V =) 3.5 sin (0.8π t) or 3.5 sin (4π t / 5) or 3.5 sin (2.5 t) A1

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February/March 2022
7(b)(ii) 2
V
(P=)
2R
or
2
. . .
(P=)
R
r m s
V
2
3.5
=
2 12
×
or
2
2.47
12
C1
= 0.51 W A1
8(a) h h
= or =
p mv
λ λ
M1
where h is the Planck constant and
p is the momentum (of particle) / mv is the momentum (of particle) / m is the mass (of particle) and v is the velocity (of
particle)
A1
8(b)(i) (electron) diffraction B1
8(b)(ii) moving electrons behave like waves B1
8(b)(iii) spacing between atoms ≈ wavelength of electron
or
diameter of atom ≈ wavelength of electron
B1
8(b)(iv) Any one of:
• wavelength has decreased
• electron had greater momentum
M1
so (accelerating) p.d. was increased A1

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February/March 2022
9(a) 207, 82 for lead B1
4, 2 for alpha B1
9(b)(i) (half-life found as) 0.52 s or correctly read points substituted into
0
t
N N e λ
−
=
1
2
0.693
=
t
λ
0.693
0.52
λ =
C1
λ = 1.3 s–1 A1
9(b)(ii) A= N
λ
12
= 1.3 24 10
× ×
13
= 3.1 10
× Bq
A1
9(b)(iii) upwards curve of decreasing gradient starting from (0,0) B1
passes through (0.52, 12) and (1.2, 18.8) B1
9(c)(i) 16 × 1012 and 7.2 × 1012 C1
6900 × 103 × 1.6 × 10-19
(16 × 1012 – 7.2 × 1012) × 6900 × 103 × 1.6 × 10-19
C1
= 9.7 J A1

PUBLISHED
February/March 2022
9(c)(ii) lead nuclei have kinetic energy
or
gamma photons are also emitted
B1
10(a) energy = mc T
Δ C1
energy = ItV
0.40 0.020 75 000 0.95
( T =)
0.015 130
× × ×
Δ
×
C1
=290 K A1
10(b) t
oe μ
−
=
I I
0.22t
0.20 = e−
C1
t = 7.3 cm A1

PUBLISHED
February/March 2022
10(c) either
(linear) attenuation coefficients / μ very different for bone and muscle
M1
(very) different amounts (of X-rays) absorbed so good contrast
or (very) different intensities transmitted so good contrast
A1
or
(linear) attenuation coefficients / μ similar for blood and muscle
(M1)
similar amounts (of X-rays) absorbed so poor contrast
or similar intensities transmitted so poor contrast
(A1)
11(a) substance containing radioactive nuclei that is introduced into the body or
substance containing radioactive nuclei that is absorbed by the tissue being studied
B1
11(b)(i) a particle interacting with its antiparticle so that mass is converted into energy B1
11(b)(ii) electron(s) and positron(s) B1
11(c)(i) 2
E = 2mc
31 82
= 2 9.11 10 3.00 10
− −
× × × ×
13
= 1.64 10 J
−
×
A1

PUBLISHED
February/March 2022
11(c)(ii) 2hc
=
E
λ
34 8
13
2 6.63 10 3.00 10
=
1.64 10
−
−
× × × ×
×
C1
12
= 2.43 10 m
−
× A1
11(d) Any 3 from:
• the two gamma photons travel in opposite directions
• gamma photons detected (outside body / by detectors)
• gamma photons arrive (at detector) at different times
• determine location of production (of gamma)
• image of tracer concentration in tissue produced
B3
12(a) total power of radiation emitted (by the star) B1
12(b)
2
L
F =
4 d
π
26
112
3.83 10
=
4 1.51 10
×
× π × ×
C1
2
= 1340 W m− A1

PUBLISHED
February/March 2022
12(c)
2
E
m
c
=
26
82
3.83 10
=
3.00 10
×
×
9
= 4.26 10 kg
×
A1
12(d) 2 4
L = 4 T
r
πσ
26 8 82 4
3.83 10 = 4 5.67 10 6.96 10 T
−
× × π× × × × × leading to T = 5770 K
B1
12(e)
(max)
1
T
∝
λ
7
5.00 10 9940
5770
λ
−
×
=
C1
7
2.90 10 m
−
λ = × A1

PHYSICS 9702/52
Paper 5 Planning, Analysis and Evaluation February/March 2022
MARK SCHEME
Maximum Mark: 30
Published
Teachers.
components.

PUBLISHED
February/March 2022
marking principles.
descriptors.

PUBLISHED
February/March 2022

PUBLISHED
February/March 2022
your working’.

PUBLISHED
February/March 2022
A 1 Correct 
2
2 Correct 
3 Wrong 
B 1 Correct, Correct , 
3
3 Wrong ignore
C 1 Correct 
2
(4 responses) 2 Correct, Wrong , 
3 Correct ignore
D 1 Correct 
2
(4 responses) 2 Correct, CON
(of 2.)
, (discount 2)
3 Correct 
E 1 Correct 
3
3 Correct, Wrong 
F 1 Correct 
2
3 Correct
CON (of 3.)

(discount 3)
G 1 Correct 
3
3 Correct
Correct
CON (of 4.)

ignore
ignore
H 1 Correct 
2
3 CON (of 2.)
Correct
(discount 2)

I 1 Correct 
2
3 Correct
CON (of 2.)

(discount 2)

PUBLISHED
February/March 2022
1 Defining the problem
θ is the independent variable and v is the dependent variable, or vary θ and measure v.
1
Keep d constant 1
Methods of data collection
Labelled diagram of workable experiment including:
• sheet supported by stand / jack
• light gate positioned at X
• support, light gate and X labelled.
1
Light gate connected to timer / datalogger. 1
Measure length (L) (of card) interrupted by beam for single light gate. 1
Method to measure θ, e.g. use protractor
or
Method to determine θ , e.g. use a rule(r) to measure two appropriate distances to use in a trigonometrical ratio
1
Method of Analysis
Plots a graph of v2 on y-axis and sin θ on x-axis.
Allow other valid graphs, e.g. sin θ against v2 Do not accept log graphs.
1
gradient
2
p
d
= for v2 against sin θ
or
1
2 gradient
p
d
=
×
for sin θ against v2
1

PUBLISHED
February/March 2022
1 intercept
2
m y
q
Bd
× −
= − for v2 against sin θ
or
intercept intercept
2 gradient
mp y m y
q
B dB
× − × −
= =
×
for sin θ against v2
1
Additional detail including safety considerations
Any six from:
6
D1 Method to stop the trolley once the trolley passes X, e.g. place a block / stop on the bench near the end of the sheet
Ignore trolley falls
D2 Keep B and m constant
D3 Use a rule(r) to measure d
D4 Method to keep d constant, e.g. mark distance d on the sheet or the starting position of the trolley on the sheet
D5 Method to measure mass of trolley (and magnet), e.g. use balance or use newton meter to measure weight and divide
by g
and
Measure B using a (calibrated) Hall probe
D6 Additional detail on use of Hall probe, e.g.
adjust probe until maximum value or
measure B using Hall probe first in one direction, then in the opposite direction and average
D7 Determine v (the velocity at X) from L / t (for a single light gate)
D8 Additional detail on measuring θ, e.g. protractor drawn in correct position on diagram, or
additional detail on determining θ , e.g. relationship between measured lengths and θ

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February/March 2022
1 D9 Relationship valid if a straight line is produced (not passing through the origin)
D10Repeat experiment for each θ and average v.
2(a)
Gradient =
1
2 fC
π
1
2(b)
1
R
/ 10–3 Ω–1
tan θ
83 or 83.3 6.17 or 6.174
63 or 62.5 4.51 or 4.511
45 or 45.5 3.27 or 3.271
30 or 30.3 2.16 or 2.164
26 or 25.6 1.86 or 1.857
23 or 23.3 1.68 or 1.684
1
Absolute uncertainties in
1
R
from ± 4 to ± 1
1

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February/March 2022
2(c)(i) Six points from (b) plotted correctly.
Must be within half a small square. Diameter of points must be less than half a small square.
1
Error bars in
1
R
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) Straight line of best fit drawn.
Points must be balanced.
Do not accept line from top plot to bottom plot.
Line must pass between
(33.5, 2.5) and (35.0, 2.5) and
(74.0, 5.5) and (76.0, 5.5)
1
Worst acceptable line drawn.
Steepest or shallowest possible line that passes through all the error bars.
All error bars must be plotted.
1
2(c)(iii) Gradient determined with clear substitution of data points into Δy/Δx; distance between data points must be greater than half
the length of the drawn line.
1
Gradient determined of WAL with clear substitution of data points into Δy/Δx;
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) 99 ± 2 (Hz) 1
2(e)(i) C determined using gradient and C given to two or three significant figures.
1 1
2 gradient 2
C
f
= =
π × π× ×
(d) (c)(iii)
1
C determined using gradient with correct SI unit and power of ten for C: F or s Ω–1 1

PUBLISHED
February/March 2022
2(e)(ii) Percentage uncertainty in C determined with method shown.
gradient
%uncertainty 100
gradient
f
f
 
Δ Δ
= + ×
 
 
OR
Correct substitution for max/min methods
1
max
2 min mingradient
C
f
=
π × ×
1
min
2 max maxgradient
C
f
=
π× ×
1
2(f) R determined to at least two significant figures with appropriate power of ten from (c)(iii) OR (d) and (e)(i) with correct
substitution seen.
gradient
tan 0.839
R
θ
= =
(c)(iii)
OR
1 1
2 tan 2 0.839
R
fC θ
= =
π π× × ×
(d) (e)(i)
1
Absolute uncertainty in R determined.
Method must be consistent with determination of R and correct substitution must be seen.
For R determined by using the gradient:
gradient
gradient
R R
Δ
Δ = ×
OR
For R determined by using (d) and (e)(i):
f C
R R
f C
Δ Δ
 
Δ = + ×
 
 
OR
ΔR determined by max / min methods.
1

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