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Cambridge International AS & A Level
PHYSICS 9702/12
Paper 1 Multiple Choice March 2021
MARK SCHEME
Maximum Mark: 40
Published
This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the
examination.
Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for
Teachers.
Cambridge International will not enter into discussions about these mark schemes.
Cambridge International is publishing the mark schemes for the March 2021 series for most Cambridge
IGCSE™, Cambridge International A and AS Level components and some Cambridge O Level components.

9702/12 Cambridge International AS & A Level – Mark Scheme
PUBLISHED
March 2021
© UCLES 2021 Page 2 of 3
Question Answer Marks
1 D 1
2 C 1
3 C 1
4 B 1
5 D 1
6 D 1
7 C 1
8 D 1
9 A 1
10 A 1
11 A 1
12 D 1
13 C 1
14 C 1
15 D 1
16 A 1
17 B 1
18 B 1
19 B 1
20 C 1
21 B 1
22 B 1
23 C 1
24 C 1
25 D 1
26 B 1
27 D 1
28 C 1

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March 2021
29 D 1
30 B 1
31 A 1
32 D 1
33 C 1
34 B 1
35 C 1
36 C 1
37 D 1
38 A 1
39 B 1
40 B 1

PHYSICS 9702/COMP
Paper 2 AS Level Structured Questions March 2021
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.

9702/COMP Cambridge International AS & A Level – Mark Scheme
PUBLISHED
March 2021
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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March 2021
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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March 2021
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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March 2021
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
(4 responses) 2 Correct 
3 Correct
CON (of 2.)

(discount 2)

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March 2021
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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March 2021
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
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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March 2021
Question Answer Mark
1(a) acceleration: vector
work: scalar
power: scalar
Three correct scores 2 marks. Two correct scores 1 mark.
B2
1(b)(i) a = (v – u) / t or a = gradient or a = Δv / (Δ)t
e.g. a = (1.40 – 0.70) / 4.0
C1
= 0.18 m s–2 A1
1(b)(ii) distance =0.5 × (0.70 + 1.40) × 4.0
or
(0.70 × 4.0) + (0.5 × 0.70 × 4.0)
C1
= 4.2 m A1
1(c)(i) (force equal to) rate of change of momentum B1
1(c)(ii) horizontal line starting from t = 0 and ending at t = 4.0 s at a positive value of F B1
horizontal line starting from t = 4.0 s and ending
at t = 8.0 s at F = 0
B1
horizontal line starting from t = 8.0 s and ending at t = 12.0 s at a negative value of F and the magnitude of F is larger than
from t = 0 to 4.0 s
B1

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March 2021
2(a) force × displacement in the direction of the force B1
2(b)(i) E = ½mv 2 C1
(m =) 23 × 2 / 162 = 0.18 (kg) A1
2(b)(ii) (Δ)E = mg(Δ)h
60 = 0.18 × 9.81 × h
C1
h = 34 m A1
2(b)(iii) (work done =) 60 – 23
= 37 (J)
C1
average resistive force = 37 / 34
= 1.1 N
A1
2(c) air resistance (acting on ball) increases B1
resultant force (on ball) decreases
or weight constant and air resistance increases
B1
acceleration decreases B1

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March 2021
3(a) Hooke’s (law) B1
3(b)(i) k = F / x or k = gradient
e.g. k = 7.0 / 5.0 × 10–2
C1
= 140 N m–1 A1
3(b)(ii) E= ½ F x or E = ½ k x 2 or E = area under graph
= ½ × 5.6 × 4.0 × 10–2 or ½ × 140 × (4.0 × 10–2)2
C1
= 0.11 J A1
3(c)(i) (upthrust =) 6.20 – 5.60 = 0.60 (N) A1
3(c)(ii) Δp = ΔF / A
= 0.60 / 1.2 × 10–3
C1
= 500 Pa A1
3(c)(iii) (Δ)p = ρg(Δ)h
ρ = 500 / (9.81 × 5.8 × 10–2)
C1
= 880 kg m–3 A1
3(d)(i) (upthrust) increases B1
3(d)(ii) (extension) decreases B1

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4(a) (two or more) waves meet (at a point) B1
(resultant) displacement is the sum of the individual displacements B1
4(b)(i) it is a (wave) reflector / it reflects (the wave) B1
4(b)(ii) v = fλ or c = fλ C1
f = 3.0 × 108 / 0.040
= 7.5 × 109 (Hz)
= 7.5 × 109 / 109 (GHz)
C1
= 7.5 GHz A1
4(b)(iii) 1 distance = 0.020 m A1
2 number = 5 A1
5(a) fo = fs v / (v + vs)
fo = 951 × 330 / (330 + 12)
C1
= 918 Hz A1
5(b) t = d / 12
= (π × 2.4 ) / 12
C1
= 0.63 s A1

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March 2021
6(a) sum of current(s) into junction = sum of current(s) out of junction
or
(algebraic) sum of current(s) at a junction is zero
B1
6(b)(i) I = 3.6 – 2.1
= 1.5
C1
V= 4.4 C1
R= 4.4 / 1.5
= 2.9 Ω
A1
6(b)(ii) 12.0 = 4.4 + 3.6r or 12.0 = 3.6 (1.2 + r ) C1
r = 2.1 Ω A1
6(b)(iii) t = (470 × 103 – 240 × 103 ) / (12 × 3.6) C1
= 5300 s A1
6(b)(iv) I = Anvq
ratio= (360A / A) × (2.5n / n) or 360 × 2.5
C1
= 900 A1

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March 2021
7(a)(i) most of the atom is empty space
or
the nucleus (volume) is very small compared to the atom
B1
7(a)(ii) the nucleus is charged B1
the mass is concentrated in nucleus / small region / small volume / small core
or
the majority of the mass is in nucleus / small region / small volume / small core
B1
7(b)(i) proton number = 84 A1
nucleon number = 214 A1
7(b)(ii) up down down changes to up up down / udd → uud
or
down changes to up / d → u
B1

PHYSICS 9702/33
Paper 3 Advanced Practical Skills 1 March 2021
MARK SCHEME
Maximum Mark: 40
Published
Teachers.

PUBLISHED
March 2021
marking principles.
descriptors.

PUBLISHED
March 2021

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March 2021
your working’.

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March 2021
1(a)(i) Value of Co to the nearest mm and in range 1.5 cm to 3.0 cm 1
1(a)(ii) Value of C > Co 1
1(a)(iii) Value of x in range 14.0 to 33.0 cm 1
1(b) Six sets of readings of C and x with correct trend and without help scores 4 marks, five sets scores 3 marks etc. 4
Range:
xmax - xmin ⩾ 14.0 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 / E4 / cm–4
1
Consistency:
All values of raw x must be given to the nearest mm
1
Significant figures:
Values of 1 / E4 should be to the same s.f. (or one more than) the s.f. in the corresponding value of E.
1
Calculation:
Values of 1 / E4 calculated correctly
1

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March 2021
1(c)(i) Axes:
Sensible scales must be used, no awkward scales (e.g. 3:10)
Scales must be chosen so that the plotted points occupy at least half the graph grid in both x and y directions
Scales must be labelled with the quantity which is being plotted.
Scale markings should be no more than 3 large squares apart.
1
Plotting of points:
All observations must be plotted on the grid.
Diameter of plotted points must be ⩽ half a small square (no blobs).
Plots must be accurate to within half a small square in both x and y directions.
1
Quality:
All points in the table must be plotted (at least 5) for this mark to be awarded. Scatter of plots must be no more than ±
2.0 cm (to scale) from a straight line in the x direction.
1
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.
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.
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 x = zero accurate to half a small square in y direction.
1

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March 2021
1(d) a equal to candidate’s gradient, and b equal to candidate’s intercept. Values are not written as fractions. 1
Unit for a is dimensionally correct, e.g. cm–5
and unit for b is dimensionally correct, e.g. cm–4
1
2(a) Value of L in range 2.5 cm to 3.5 cm and given to nearest mm or better 1
Value of D in range 6.0 to 9.0 cm and given to nearest mm or better 1
2(b)(i) Value of raw d to nearest 0.001 cm and final value in range 0.040 cm to 0.090 cm 1
2(b)(ii) Absolute uncertainty of 0.001 cm 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.
1
2(c) Value for t in s, with unit 1
Evidence of repeat readings for t 1
2(d) Second values for d and t 1
Quality: second t shorter than first t 1
2(e)(i) Two values of k calculated correctly. 1
2(e)(ii) Justification based on sig. fig. in t and d 1
2(e)(iii) Sensible comment relating to the calculated values of k, testing against a criterion specified by the candidate. 1
2(f) Correct calculation of S 1

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March 2021
2(g)(i) Two k values are not enough to draw a valid conclusion
Hole may not be same size as pin
Difficult to measure D as bottle distorts
Difficulty with 1 cm depth with reason, e.g. holding bottle steady (when timing) / seeing what the depth is
Difficulty with measuring time with reason, e.g. lines are thick / difficult to judge when level passes line / difficult to see
water level
4 max
4
2(g)(ii) Take more readings and plot a graph / calculate more k values and compare
Measure hole itself, with detail of method
Stand bottle on graph grid / use a more rigid named material for bottle / measure with bottle between blocks
Place bottle on a support in the bowl / clamp bottle (after filling) / mark line 1 cm from bottom of bottle
Video with timer in view (or frame by frame) / use coloured water
4 max
4

PHYSICS 9702/42
Paper 4 A Level Structured Questions March 2021
MARK SCHEME
Maximum Mark: 100
Published
Teachers.

PUBLISHED
March 2021
marking principles.
descriptors.

PUBLISHED
March 2021

PUBLISHED
March 2021
your working’.

PUBLISHED
March 2021
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
(4 responses) 2. Correct 
3. Correct
CON (of 2.)

(discount 2)

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March 2021
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.

PUBLISHED
March 2021
AE Indicates an arithmetic error. Do not allow the mark where the error occurs. Then follow through the working/calculation giving full
POT Indicates a power of ten error. Do not allow the mark where the error occurs. Then follow through the working/calculation giving full
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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March 2021
1(a) (gravitational) force is (directly) proportional to product of masses B1
force (between point masses) is inversely proportional to the square of their separation B1
1(b) correct read offs from the graph with correct power of ten for R3 C1
( )
2 34
2
11
4 1.2 10
6.67 10 2.4 365 24 3600
M
π
−
× × ×
=
× × × × ×
C1
30
3.0 10 kg
= × A1
1(c)(i) potential energy is zero at infinity B1
(gravitational) forces are attractive B1
work must be done on the rock to move it to infinity B1
1(c)(ii) 2
2
2
GMm mv GM GM
OR v OR v
r r r
r
= = =
M1
use of ½ mv2 (e.g. multiplication by ½ m) leading to
2
GMm
r
A1
1(c)(iii)
Ep = φ m and φ =
GM
r
−
or p
GMm
E
r
−
=
Total energy = Ek + Ep
C1
Total energy
2
GMm GMm
r r
−
= +
2
GMm
r
−
=
A1

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March 2021
2(a)(i) pV NkT
= or pV nRT
= and A
N nN
=
5 3
23
2.3 10 3.5 10
1.38 10 294
N
−
−
× × ×
=
× ×
C1
= 2.0 × 1023 A1
2(a)(ii) 2
1
3
pV Nmc
=
5 3
2
23 27
3 2.3 10 3.5 10
2.0 10 40 1.66 10
c
−
−
× × × ×
=
× × × ×
= 182 000
r.m.s. speed = 430 m s–1
C1
or
2 3
1
2 2
mc kT
=
A1
23
2
27
3 1.38 10 294
40 1.66 10
c
−
−
× × ×
=
× ×
= 183 000
(C1)
r.m.s.speed = 430 m s–1 (A1)

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March 2021
2(b) ( )
23 23
2
23 27
3 2.0 10 1.38 10 294 84
2.0 10 40 1.66 10
c
−
−
× × × × × +
=
× × × ×
2
236000
c =
485
c =
C1
485
1.1
430
ratio
 
= =
 
 
A1
OR
v T
∝ or 2
v T
∝
(C1)
273 21 84
273 21
ratio
+ +
=
+
or
378
294
ratio = 1.1
(A1)
3(a) Any 2 from:
• particles / atoms / molecules / ions (very) close together / touching
• regular, repeating pattern
• vibrate about a fixed point
B2
3(b) (much) greater increase in spacing of molecules (for vaporisation compared with fusion) B1
3(c)(i) –100 °C B1

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March 2021
3(c)(ii) time = 8.5 – 3.0
= 5.5 min
C1
Pt = mL
energy = power × time = 150 × 5.5 × 60
= 49 500 J
E
L
m
=
49500
0.045
=
C1
1
1100kJ kg−
= A1
3(c)(iii) gas has a higher specific heat capacity (than liquid) B1
4(a) acceleration and displacement are in opposite directions B1
4(b)(i) F kx
=
( ) ( )
8.0 0.060 0.048 8.0 0.060 0.048
or
= × − × +
or 8.0 0.012 8.0 0.108
or
× ×
M1
( ) ( )
8.0 0.012 8.0 0.108 0.77
F N
Σ = × − × =
or
0.864 0.096 0.77
F N
Σ = − =
A1

PUBLISHED
March 2021
4(b)(ii) F
a
m
=
0.77
0.25
=
2
3.1ms−
=
A1
4(b)(iii) a = – ω2x
3.1
0.048
ω =
8.04
ω =
C1
T= 2 π / ω C1
T= 2π / 8.04
= 0.78 s
A1
4(b)(iv) (resultant) force halved and distance halved B1
same T B1
5(a)(i) amplitude of the carrier wave varies M1
in synchrony with the displacement of the (information) signal A1
5(a)(ii) Any 2 from:
• fewer transmitters needed / each transmitter can cover a greater distance
• more stations can share waveband
• transmitters and receivers are cheaper
B2

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March 2021
5(b)(i) v
f
λ =
8
6
3.0 10
200
1.5 10
m
×
= =
×
A1
5(b)(ii) 10 kHz B1
5(c) 1520 kHz B1
6(a) (both have) radial field lines B1
6(b)(i) 2.1 cm B1
6(b)(ii)
2
4 o
Q
E
r
πε
=
e.g. r = 2.1 cm, E = 1.30 × 105 V m–1
2
4 o
Q r E
πε
=
12 2 5
4 8.85 10 0.021 1.30 10
π −
= × × × × × ×
C1
9
6.4 10 C
−
= × A1

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March 2021
6(c) Q
C
V
=
either
4 o
Q
V
r
πε
= leading to 4 o
C r
πε
=
C1
12
4 8.85 10 0.021
C π −
= × × × × C1
( ) 12
2.3 10
C −
= × F A1
or
4 o
Q
V
r
πε
=
9
12
6.4 10
4 8.85 10 0.021
π
−
−
×
=
× × × ×
2740V
=
9
6.4 10
2740
C
−
×
=
(C1)
12
2.3 10 F
−
= × (A1)

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March 2021
7(a)(i) non-inverting (amplifier) B1
7(a)(ii)
gain 1
f
R
R
= +
3.6
gain 1 6.0
0.72
= + =
B1
7(a)(iii) straight line from (0,0) to (T / 2, 3) B1
line from origin to 3.0 V then horizontal line at 3.0 V to T B1
7(a)(iv) ldr / light dependent resistor replaces one of the two resistors B1
7(b)(i) relay coil B1
7(b)(ii) relay coil between op-amp and earth B1
diode with correct polarity (pointing away from output) connected between output and device and no other connections
or diode with correct polarity (pointing towards earth) between device and earth and no other connections
B1
switch connected to high voltage circuit B1
8(a)(i) at least one anticlockwise arrow and no clockwise arrows B1
8(a)(ii) (force is to the) left B1
8(a)(iii) force is the same B1
Newton’s third law (of motion)
or force depends on the product of the two currents
B1

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March 2021
8(b)(i) frequency of radio waves is equal to natural frequency of protons B1
resonance of protons occurs / protons absorb energy B1
8(b)(ii) in between pulses / when pulse stops B1
Any 1 from:
• protons de-excite
• protons emit r.f. pulses
• emitted (r.f.) pulse (from proton) detected
B1
9(a) (magnetic) flux density × area × number of turns M1
area is perpendicular to (magnetic) field A1
9(b) use of t = 1.2 s C1
BAN
t
ε
Δ
=
Δ
2
0.250 0.030 540
1.2
π
× × ×
=
C1
0.32V
= A1
9(c)(i) light damping B1

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March 2021
9(c)(ii) sheet cuts (magnetic) flux and causes induced emf B1
(induced) emf causes (eddy) currents (in sheet) B1
either currents (in sheet) cause resistive force
or currents (in sheet) dissipate energy
B1
smaller currents in Y or larger currents in X, so dashed line is X B1
10(a) 230 V A1
10(b) ω = 100π
2 2
100
T
π π
ω π
= =
C1
0.020s
= A1
10(c)(i) half-wave (rectification) B1
10(c)(ii) sinusoidal half waves in positive V only or negative V only, peak at 320 V B1
line at zero for second half of cycle B1
two time periods shown, each of 0.020 s B1
10(c)(iii) capacitor added in parallel with resistor B1

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March 2021
11(a)(i) electrons decelerate (on hitting target) so X-ray photons produced B1
range of decelerations B1
photon energy depends on (magnitude of) deceleration B1
11(a)(ii) hc
eV
λ
=
C1
34 8
19
6.63 10 3.0 10
1.6 10 15000
λ
−
−
× × ×
=
× ×
C1
11
8.3 10 m
−
= × A1
or
E = hf and c = fλ and electron energy = eV
or
E = hc / λ and electron energy = eV
electron energy = 1.6 × 10–19 × 15000
= 2.4 × 10–15
(C1)
34 8
15
6.63 10 3.0 10
2.4 10
λ
−
−
× × ×
=
×
(C1)
11
8.3 10 m
λ −
= × (A1)
11(b)(i) μ = – gradient or ln (I / Io) = x
μ
− C1
(e.g. 2.08 / 10.0) = 0.21 cm–1 A1

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March 2021
11(b)(ii) ( )
ln 0.05 x
μ
= − C1
ln0.05
x
μ
=
−
. . 14
e g x cm
=
A1
12(a) 1 not affected by external factors B1
2 cannot predict when a (particular) nucleus will decay
or cannot predict which nucleus will decay (next)
B1
12(b)(i)
Number of atoms =
9
27
1.0 10
90 1.66 10
−
−
×
× ×
or
9 23
3
1.0 10 6.02 10
90 10
−
−
× × ×
×
15
6.693 10
= ×
C1
A N
λ
=
λ =
6
15
5.2 10
6.693 10
×
×
C1
10
7.8 10
λ −
= × s–1 A1
12(b)(ii) daughter nucleus is unstable B1

PHYSICS 9702/52
Paper 5 Planning, Analysis and Evaluation March 2021
MARK SCHEME
Maximum Mark: 30
Published
Teachers.

PUBLISHED
March 2021
marking principles.
descriptors.

PUBLISHED
March 2021

PUBLISHED
March 2021
your working’.
General Marking Points
When marking at the computer:
• ensure your sitting position is comfortable
• take regular breaks
• don’t mark when very tired
• try to mark some scripts every day
• don’t leave it all to the last minute
• there may not be sufficient scripts in the pot if you are the last to finish!
Check Blank Pages e.g. pages 2 and 5 and Additional Objects:
Before marking each script check any blank pages at the end for student answers and add some annotation to show the page has been viewed. It
is useful to highlight any written notes.
Annotations
etc.

PUBLISHED
March 2021
1 Defining the problem
Mass of cylinder m is the independent variable and period T is the dependent variable, or vary mass of cylinder m and
measure period T.
1
Keep radius of cylinder constant. 1
Methods of data collection
Labelled diagram of workable experiment including:
• beaker with (cooking) oil on a bench or container supported by stand where stand is on a bench
• cylinder partially submerged in (cooking) oil
• cylinder and (cooking) oil labelled.
1
Method to determine mass m of cylinder, e.g. use a (top pan) balance. 1
Method to determine period or T, e.g. use a stopwatch / timer to time oscillations. 1
Method to determine diameter of cylinder, e.g. micrometer or calliper 1
Method of Analysis
Plots a graph of T2 against m.
(Allow other valid graphs, e.g. lg T against lg m)
1
Relationship valid if a straight line passing through the origin is produced.
(Allow gradient = 0.5 for log T against log m).
1
σ
=
× 2
4π
gradient
K
r
(
σ
×
=
×
2 y-intercept 2
4π
10
K
r
for lg T against lg m).
1

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March 2021
1 Additional detail including safety considerations
Max 6
6
Use gloves to prevent oil contacting skin / slippery hands OR
Perform experiment in a tray to prevent oil spillages.
D1
Keep density / temperature of the (cooking) oil constant or keep σ constant. D2
Mass of oil = mass of beaker and oil – mass of beaker and
use a measuring cylinder to determine the volume of the oil.
Do not accept (calibrated) beaker.
D3
Methods to measure volume of oil and determine mass of oil and use equation density σ = mass / volume for
measurements.
D4
Time n oscillations and divide nT by n
where n ⩾ 5.
D5
Description of method of counting oscillations with position of fiducial mark / mark on cylinder / beaker / fixed point shown in
diagram.
D6
Repeat experiment for each value of m and average T. D7
r = diameter / 2 provided diameter measured. D8
Repeat measurements of diameter in different directions and average. D9
Wait for oscillations to become even / steady. D10

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March 2021
2(a)
Gradient =
1
2uA
y-intercept =
1
2u
.
1
2(b)
0.046
0.052
0.062
0.072
0.080
0.088
First mark for values of
1
v
/ s cm–1; allow 3sf.
1
Second mark for absolute uncertainties from
± 0.003 to ± 0.004.
1
2(c)(i) Six points plotted correctly.
Must be accurate to the nearest half small square. Diameter of points must be less than half a small square.
1
Error bars in
1
v
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

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March 2021
2(c)(ii) Line of best fit drawn.
Points must be balanced.
Do not allow line from top plot to bottom plot.
Line must pass between
(320, 0.050) and (345, 0.050) and between (795, 0.085) and (815, 0.085).
1
Worst acceptable line drawn.
Steepest or shallowest possible line.
Mark scored only if all error bars are plotted.
1
2(c)(iii) Gradient determined with clear substitution of data points into Δy / Δx; distance between data points must be at least half
the length of the drawn line.
1
Gradient of WAL determined and
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
y-intercept of worst acceptable line determined by substitution into y = mx + c.
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)
Do not accept ecf from false origin method.
1

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March 2021
2(d)(i) u determined using y-intercept and
u and A given to 2 or 3 sf.
=
× −
1
2 intercept
u
y
1
A determined using gradient with correct substitution and
Units with correct power of ten for u and A.
−
= =
× ×
intercept 1
or
gradient 2 gradient
y
A A
u
1
2(d)(ii) Percentage uncertainty in A.
 
Δ Δ
= + ×
 
 
gradient -intercept
%uncert. 100
gradient -intercept
y
y
OR
Δu clearly determined and
 
Δ Δ
= + ×
 
 
gradient
%uncert. 100
gradient
u
u
OR
Correct substitution for max/min methods.
1
2(e) Value of m determined from (d)(i) OR (c)(iii) and (c)(iv) with correct number substitution into relevant equation and correct
power of ten.
e.g. = − = −
2 2
10
uAt uA
m A A
L
, or
 
= − ×
 
 
1
2
2
t
m uA
L u
or
−
=
-intercept
gradient
t
y
L
m .
1

9702 m21 ms_all

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