PHYSICS PAPER MARK SCHEME

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Cambridge International AS & A Level
PHYSICS 9702/12
Paper 1 Multiple Choice March 2020
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 2020 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 2020
© UCLES 2020 Page 2 of 3
Question Answer Marks
1 C 1
2 D 1
3 D 1
4 C 1
5 D 1
6 D 1
7 A 1
8 D 1
9 C 1
10 B 1
11 A 1
12 A 1
13 B 1
14 B 1
15 B 1
16 D 1
17 A 1
18 C 1
19 B 1
20 C 1
21 B 1
22 A 1
23 B 1
24 D 1
25 D 1
26 C 1
27 B 1
28 C 1

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March 2020
29 D 1
30 A 1
31 D 1
32 D 1
33 B 1
34 D 1
35 B 1
36 B 1
37 C 1
38 D 1
39 D 1
40 C 1

PHYSICS 9702/22
Paper 2 AS Level Structured Questions March 2020
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.

PUBLISHED
March 2020
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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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.

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March 2020
5 ‘List rule’ guidance (see examples below)
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.
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 2020
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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March 2020
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.

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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.
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.

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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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1(a) time
(electric) current
allow amount of substance
allow luminous intensity
any two of the above quantities, 1 mark each
B2
1(b)(i) g = (4π2 × 1.50) / (2.482)
= 9.63 m s–2
A1
1(b)(ii) percentage uncertainty = 2 + (3 × 2)
or fraction uncertainty = 0.02 + (0.03 × 2)
C1
percentage uncertainty = 8% A1
1(b)(iii) absolute uncertainty = 0.08 × 9.6
= 0.8 m s–2
A1
2(a) f0 = fS v / (v – vS)
9560 = f × 1510 / (1510 – 4.50)
C1
f = 9530 Hz A1
2(b)(i) v 2 = u 2 + 2as
height = 5.62 / (2 × 9.81)
C1
= 1.6 m A1
2(b)(ii) downward sloping straight line starting from a point on the speed axis and ending at point (T, 0) B1

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2(b)(iii) (Δ)E= mg(Δ)h
= 0.45 × 9.81 × 1.6
C1
= 7.1 J A1
2(b)(iv) air resistance increases (and weight constant) B1
(resultant force decreases so) acceleration decreases B1
3(a) force × displacement in the direction of the force B1
3(b)(i) displacement = 4.4 × 30 C1
work done = 140 cos 30° × 4.4 × 30 C1
= 1.6 × 104 J A1
3(b)(ii) p = F / A C1
F = 860 – 140 sin 30° (= 790) C1
A= 790 / 2400
= 0.33 m2
A1
3(b)(iii) σ = F / A or F / πr 2 or 4F / πd 2 C1
9.6 × 106 = 4 × 140 / πd 2
d = 4.3 × 10–3 m
A1

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3(c) E = ½Fx or ½kx2 or area under graph C1
(Δ)E = ½ × (140 + 210) × 0.20 × 10–3
or
(Δ)E = (½ × 210 × 0.60 × 10–3) – (½ × 140 × 0.40 × 10–3)
or
(Δ)E = (140 × 0.20 × 10–3) + (½ × 0.20 × 10–3 × 70)
or
(Δ)E = [½×3.5 × 105 × (0.60 × 10–3)2 ] – [½ × 3.5 × 105 × (0.40 × 10–3)2 ]
C1
ΔE = 0.035 J A1
4(a)(i) distance moved by wavefront / energy during one cycle / vibration / oscillation / period (of source)
or
minimum distance between two wavefronts
or
distance between two adjacent wavefronts
B1
4(a)(ii) maximum displacement (of particle / point on wave) B1
4(b)(i) 1 light / waves spread (at each slit) B1
2 constant phase difference (between light / waves) B1
4(b)(ii) nλ = d sinθ C1
d = 3 × 650 × 10–9 / sin34° C1
d = 3.5 × 10–6 m A1
4(b)(iii) wavelength of blue light is shorter (than 650 nm / red light) M1
so angle (between third order diffraction maxima) decreases A1

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5(a) volt / ampere B1
5(b) R = ρ L / A C1
L = (1.8 × 0.38 × 10–6) / 9.6 × 10–7 C1
= 0.71 m A1
5(c)(i) thermal energy is dissipated in resistor Y B1
5(c)(ii) V / 1.2 = 1.8 / (1.8 + 0.6) C1
V = 0.90 V A1
or
I = 1.2 / (1.8 + 0.6) (= 0.50) (C1)
V= 0.50 × 1.8
= 0.90 V
(A1)
5(d)(i) remain the same B1
5(d)(ii) decrease B1
5(e)(i) 1 / R = 1 / 1.8 + 1 / 3.6
R = 1.2 Ω
A1

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5(e)(ii) I = 1.2 / (1.2 + 0.60) C1
= 0.67 A A1
or
VY = 1.2 × 0.60 / (1.2 + 0.60) (= 0.40) (C1)
I = 0.40 / 0.60
= 0.67 A
(A1)
6(a) E= V / d
d = 350 / 1.4 × 104
C1
= 0.025 m A1
6(b)(i) E = F / Q C1
Q = 6.7 × 10–15 / 1.4 × 104 (= 4.8 × 10–19 C)
= (4.8 × 10–19 / 1.6 × 10–19) e
C1
= 3.0 e A1
6(b)(ii) mass = 8.3 × 10–27 / 1.66 × 10–27
= 5.0 u
A1
6(b)(iii) number = 5 – 3
= 2
A1

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7(a) made up of quarks (so) not a fundamental particle B1
7(b)(i) beta plus / β+ (particle) B1
(electron) neutrino / ν(e) B1
7(b)(ii) kinetic energy of nucleus B1
gamma / γ radiation B1

PHYSICS 9702/33
Paper 3 Advanced Practical Skills 1 March 2020
MARK SCHEME
Maximum Mark: 40
Published
Teachers.

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March 2020
marking principles.
descriptors.

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March 2020

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your working’.

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1(a) Value of x in range 10.0 to 20.0 cm 1
1(b) Value of T in range 0.50 to 1.50 s 1
Evidence of repeat readings: at least two values of at least 5T 1
1(c) Six sets of readings of x and T with correct trend and without help scores 4 marks, five sets scores 3 marks etc. 4
Range:
xmin ⩽ 12.0 cm and xmax ⩾ 25.0 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 / x (cm–1)
1
Consistency:
All values of raw x must be given to the nearest mm
1
Significant figures:
All values of 1 / x should be to the same s.f. as (or one more than) the s.f. in the corresponding x value
1
Calculation:
Values of 1 / x calculated correctly
1

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1(d)(i) Axes:
Sensible scales must be used, no awkward scales (e.g. 3:10)
Scales must be chosen so that the plotted points occupy at least half the graph grid in both x and y directions
Scales must be labelled with the quantity which is being plotted.
Scale markings should be no more than 3 large squares apart.
1
Plotting of points:
All observations must be plotted on the grid.
Diameter of plotted points must be ⩽ half a small square (no blobs).
Plots must be accurate to within half a small square in both x and y directions.
1
Quality:
All points in the table must be plotted (at least 5) for this mark to be awarded. Scatter of plots must be no more than ± 0.5 m–1
(± 0.005 cm–1) from a straight line in the x-direction.
1
1(d)(ii) Line of best fit:
Judged by balance of all points on the grid (at least 5) about the candidate’s line. There must be an even distribution of points
either side of the line along the full length
One anomalous point is allowed only if clearly indicated (i.e. circled or labelled) by the candidate.
Lines must not be kinked or thicker than half a square.
1
1(d)(iii) Gradient:
The hypotenuse of the triangle used must be greater than half the length of the drawn line.
Method of calculation must be correct.
Both read-offs must be accurate to half a small square in both the x and y directions.
1
y-intercept:
Either
Correct read-off from a point on the line substituted into y = mx + c or an equivalent expression, with read-off accurate to half a
small square in both x and y directions.
Or
Intercept read directly from the graph, with read-off at x = zero accurate to half a small square in y direction.
1

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1(e) a equal to candidate’s gradient, and b equal to candidate’s intercept.
a and b are not fractions.
a has two or more significant figures.
1
1(e) Units for a and b correct (e.g. s cm or s m for a, and s for b). 1
2(a) Value of N, with no unit, in range 10 to 14. 1
2(b) Value of L to nearest mm, with unit, in range 5.0 to 15.0 cm. 1
2(c) Value(s) of raw θ to nearest degree and < 90°. 1
Value of I to nearest mA or nearest 0.1 mA, with unit 1
2(d) Absolute uncertainty in θ value of 2 to 5° and correct method of calculation to obtain percentage uncertainty.
If several readings have been taken, then the absolute uncertainty can be half the range (but not zero if values are equal) if
the working is clearly shown.
1
2(e) Correct calculation of B 1
2(f) Second values of N and L 1
Second values of θ and I 1
Quality: second θ > first θ 1
2(g)(i) Two values of k calculated correctly. 1
2(g)(ii) Justification based on s.f. in θ, I and L. 1
2(g)(iii) Sensible comment relating to the calculated values of k, testing against a criterion specified by the candidate. 1

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2(h)(i) Two readings are not enough to draw a valid conclusion 4 max
Difficult to determine I as reading not steady
Parallax error when measuring θ
Large % uncertainty in θ
Difficult to re-wind and fix the wire / wires may touch / wire is kinked
Channel moves on bench
4
2(h)(ii) Take more readings and plot a graph / calculate more k values and compare 4 max
Method of cleaning crocodile clips / contacts
Method of overcoming parallax error,
e.g. measure change of angle by turning channel on bench / measure θ on a photo / angle markings on compass / mark
‘before’ and ‘after’ positions on compass
Use larger current / voltage
Use two ready-wound channels / use new wire
Method of fixing channel, e.g. tape channel to bench
4

PHYSICS 9702/42
Paper 4 A Level Structured Questions March 2020
MARK SCHEME
Maximum Mark: 100
Published
Teachers.

PUBLISHED
March 2020
marking principles.
descriptors.

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March 2020

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March 2020
your working’.

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March 2020
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
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)

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March 2020
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 2020
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 2020
1(a) work done per unit mass B1
work done moving mass from infinity (to the point) B1
1(b)(i) gravitational force provides centripetal force C1
mv2 / r = GMm / r2 and v = 2πr / T OR mrω2 = GMm / r2 and ω = 2π / T
OR r3 = GMT2 / 4π2
C1
r3 = 6.67 × 10-11 × 6.0 × 1024 × (13.7 × 24 × 3600)2 / 4 π2
so r = 2.4 × 108 m
A1
1(b)(ii) (EP = –) GMm / r
work done = GMm / r1 – GMm / r2
C1
= 6.67 × 10–11 × 360 × 6.0 × 1024 (1/6.4 × 106 – 1 / 2.4 × 108) C1
= 2.2 × 1010
J A1
1(b)(iii) g = GM / r2 C1
ratio = rTESS
2 / rearth
2
= (2.4 × 108 / 6.4 × 106)2
= 1400
A1

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March 2020
2(a) n = 110 / 0.032 or 110000 / 32 or 3440 C1
pV = nRT C1
T = (1.0 × 105 × 85) / (8.31 × (110 / 0.032)) = 300 K A1
2(b) E= mcΔθ
= 110 × 0.66 × 50
C1
= 3600 J A1
2(c) Any 3 from:
• molecule collides with wall
• momentum of molecule changes during collision (with wall)
• force on molecule so force on wall
• many forces act over surface area of container exerting a pressure
B3
2(d) KE ∝ T
v ∝ √T
C1
ratio = √(350 / 300)
= 1.1
A1

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March 2020
3(a)(i) 0.050 m A1
3(a)(ii) ω = vo / xo C1
T = 2π / ω
0.42 = (2π × 0.050) / T
C1
T = 0.75 s A1
3(a)(iii) one point labelled P where ellipse crosses displacement axis marked A1
3(b)(i) (induced) e.m.f. proportional to rate M1
of change of (magnetic) flux (linkage) A1
3(b)(ii) (there is) current in the circuit B1
either
current causes thermal energy (dissipated) in resistor B1
thermal energy comes from energy of magnet B1
or
current causes magnetic field around coil (B1)
two fields cause an opposing force on magnet (B1)

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4(a)(i) Any 2 from:
• allows the reflected signal to be distinguished from the emitted signal
• detection occurs in the time between emitted pulses
• (reflection of ultrasound) detected by same probe / transducer / crystal
• cannot emit and detect at same time (hence pulses)
B2
4(a)(ii) piezo-electric crystal B1
ultrasound makes crystal vibrate / resonate B1
vibration produces (alternating) e.m.f. / p.d. across crystal B1
4(b)(i) = (1.6 × 106 – 4.3 × 102)2 / (1.6 × 106 + 4.3 × 102)2
= 0.999
B1
4(b)(ii) without the gel most of the ultrasound is reflected B1
Z values more similar / α reduces
so less (ultrasound) is reflected / more (ultrasound) is transmitted
B1

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5(a) Any 2 from:
• noise can be filtered out / noise can be removed / signal can be regenerated
• can carry more information per unit time / greater rate of transmission of data
• can have extra bits of data to check for errors
• can be encrypted
B2
5(b)(i) v ∝ λ C1
ratio = vair / vfibre
= 3.00 × 108 / 2.07 × 108
= 1.45
A1
5(b)(ii) attenuation = 10 log (P2/P1) C1
0.40 × L = 10 log (1.5 / 0.06)
0.40 × L = 13.979
C1
L = 35 km A1

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March 2020
6(a) 2.0 cm B1
6(b) At 16 (cm) from A the electric fields are equal or EA = EB B1
E = Q / 4πεor2
QA / (4πεorA
2) = QB / (4πεorB
2)
3.6 × 10-9 / 0.162 = QB / 0.082
C1
QB = 9.0 × 10–10
C A1
6(c)(i) V = Q / 4πεorA
V = 3.6 × 10–9 / (4 × π × 8.85 × 10–12 × 0.020)
C1
V = 1600 V A1
6(c)(ii) C= Q / V
= 3.6 × 10–9 / 1600
C1
= 2.3 × 10–12
F A1

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March 2020
7(a) axes labelled with resistance and temperature M0
concave curve not touching temperature axis A1
line with negative gradient throughout A1
7(b) resistance of thermistor decreases B1
total circuit resistance decreases so voltmeter reading increases
or
current increases so voltmeter reading increases
or
greater proportion of resistance in fixed resistor so voltmeter reading increases
or
p.d. across thermistor decreases so voltmeter reading increases
B1
7(c) (0.020 strain means) ΔR / R = 0.090 C1
ΔR = 0.090 × 120 = 10.8 Ω C1
resistance = 120 + 10.8 = 130 Ω A1

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March 2020
8(a) a region where a magnet / magnetic material / moving charge / current carrying conductor experiences a force B1
8(b) B = F / Il
e.g. = 9 × 10–3 / (5.0 × 0.045)
C1
= 0.040 T A1
8(c)(i) force is (always) perpendicular to the velocity / direction of motion B1
magnetic force provides the centripetal force
or force perpendicular to motion causes circular motion
B1
magnitude of force (due to the magnetic field) is constant
or
no work done by force
or
the force does not change the speed
B1
8(c)(ii) Applying the list rule, any 2 from:
accelerating p.d.
radius of path / radius of semicircle
magnetic flux density
B2

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March 2020
9(a)(i) 9.0 / √2 =
6.4 V
A1
9(a)(ii) ω = 20
ω = 2π / T
T = 2π / 20
C1
T = 0.31 s A1
9(b) the r.m.s. voltages are different, so no B1
9(c)(i) P= Vr.m.s. × Ir.m.s. C1
= 120 × 0.64
= 76.8 W
C1
efficiency = (76.8 / 80) × 100
= 0.96 or 96 %
A1
9(c)(ii) Any one from:
• heat losses due to resistance of windings / coils
• heat losses in magnetising and demagnetising core / hysteresis losses in core
• heat losses due to eddy currents in (iron) core
• loss of flux linkage
B1

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March 2020
10(a) energy of a photon required to remove an electron B1
either: energy to remove electron from a surface
or: minimum energy to remove electron
or: energy to remove electron with zero kinetic energy
B1
10(b)(i) Correct read off from graph of f as 5.45 × 1014 Hz when EMAX = 0
5.45 × 1014 × 6.63 × 10–34
C1
= 3.6 × 10–19
J A1
10(b)(ii) 3.6 × 10–19 / 1.6 × 10–19 = 2.3 eV so potassium A1
10(c)(i) each photon has same energy so no change B1
10(c)(ii) more photons (per unit time) so (rate of emission) increases B1

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March 2020
11(a) eV = hf
f = 1.60 × 10–19 × 100 000 / 6.63 × 10–34
C1
= 2.41 × 1019
Hz A1
11(b) (aluminium filter) absorbs (most) low energy X-rays B1
Any 2 from
• X-ray beam contains many wavelengths
• so low energy X-rays are not absorbed in the body
• low energy X-rays can can cause harm but do not contribute to the image
B2
11(c)(i) I / Io = e– μx
e–0.23 × 0.80 = 0.83
C1
17% absorbed A1
11(c)(ii) bone is seen as lighter / muscle is seen as darker B1
either bone has a higher µ value so absorbs more
or muscle has a lower µ value so transmits more
B1

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March 2020
12(a) (minimum) energy required to separate the nucleons M1
to infinity A1
12(b)(i) 37
2
B1
12(b)(ii) fission B1
12(b)(iii) binding energy per nucleon smaller for U than for Cs B1
12(c) Current ratio 2 Y to 1 Zr, so initially 3 Y
2 = 3 e–λt
λ = 0.693 / 2.7
C1
ln(2 / 3) = – (ln 2 / 2.7)t C1
t = 1.6 days A1
or
(½)n = 2 / 3 (C1)
n = 0.585 (C1)
time = 0.585 × 2.7
= 1.6 days
(A1)

PHYSICS 9702/52
Paper 5 Planning, Analysis and Evaluation March 2020
MARK SCHEME
Maximum Mark: 30
Published
Teachers.

PUBLISHED
March 2020
marking principles.
descriptors.

PUBLISHED
March 2020

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March 2020
your working’.

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March 2020
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.
Link ‘additional objects’.
Annotations
Tick
Correct point
Use in question 1 to represent analysis marks
Cross Incorrect point
^ Omission mark
BOD Benefit of the doubt
NBOD No benefit of doubt given
ECF Error carried forward
P
Defining the problem mark in question 1
Also indicates POT in question 2
M0 Method of data collection mark in question 1

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March 2020
Tick 1 Additional Detail mark point 1
Highlighter Highlighting areas of text
On-page
comment
Allows comments to be entered in speech bubbles on the candidate response.
etc.

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March 2020
1 Defining the problem
x is the independent variable and E is the dependent variable, or vary x and measure E. 1
Keep B or m constant and keep k or N constant. 1
Methods of data collection
Labelled diagram of workable experiment including:
• labelled spring supported by stand and clamp
• labelled magnet
• labelled coil positioned so that magnet is vertically above the coil by eye in the correct orientation.
1
Circuit diagram showing voltmeter / multimeter set to p.d. range / oscilloscope connected to the ends of the coil. Do not accept
other electrical components.
1
Method to measure x, e.g.
labelled ruler drawn parallel to spring/magnet and equilibrium position and displaced position indicated and x indicated or
difference determined or
description of use of ruler to measure equilibrium position and displaced position and difference determined.
1
Method to measure mass of magnet e.g. use balance or use newton-meter to measure weight and divide by g. 1
Method of Analysis
Plots a graph of E against x or equivalent.
Allow lg E against lg x.
1
Relationship valid if a straight line passing through the origin is produced.
(for lg E against lg x: relationship valid if a straight line with gradient = 1).
1
α =
gradient m
BN k
(for lg E against lg x: α =
-intercept
10y
m
BN k
)
1

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March 2020
1 Additional detail including safety considerations Max 6 6
Use safety goggles / safety screen to prevent injury (to eyes) from (detached) spring/magnet; do not accept from D1
oscillating magnet or
use cushion / sand box in case magnet falls or
use g clamp / weights on stand to prevent toppling.
Keep distance between equilibrium position and coil constant. D2
Check that the unstretched length of the spring has not changed or is not permanently deformed (after removing D3
load / magnet).
Expression to determine k from relevant experiment, e.g. k = mg / extension or gradient of F – extension graph. D4
Weight / force must be defined.
Measure B using a (calibrated) Hall probe. D5
Additional detail on use of Hall probe, e.g. D6
adjust probe until maximum value
or
measure B using Hall probe first in one direction and then in the opposite direction and average.
Method to maximise E, e.g. position magnet so that equilibrium position is at the centre of the coil or use D7
a large number of turns.
Explanation to determine max E e.g. use of video and slow-motion playback. D8
Repeat experiment for each x and average E. D9
Method to ensure clamped rule to measure x is vertical, e.g. correctly positioned set square indicated at D10
right angles between the rule and the horizontal surface or plumb line supported on a surface shown in
appropriate position.

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March 2020
2(a)
Gradient =
−1
CR
y-intercept = ln
E
R
1
2(b)
3.83 or 3.829
3.69 or 3.689
3.53 or 3.526
3.33 or 3.332
3.18 or 3.178
3.00 or 2.996
1
Absolute uncertainties in ln I
from ± 0.04 to ± 0.1
1
2(c)(i) Six points plotted correctly.
Must be within half a small square. Diameter of points must be less than half a small square.
1
Error bars in ln I plotted correctly.
All error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.
1
2(c)(ii) Line of best fit drawn.
Points must be balanced.
Line must pass between (5.5, 3.75) and (8.0, 3.75) and between (56, 3.05) and (58, 3.05)
1
Worst acceptable line drawn.
Steepest or shallowest possible line that passes through all the error bars.
Mark scored only if all error bars are plotted.
1

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March 2020
2(c)(iii) Negative 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 determined of WAL
uncertainty = (gradient of line of best fit – gradient of worst acceptable line)
or
uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)
1
2(c)(iv) y-intercept read from y-axis to less than half a small square, or
y-intercept determined from substitution into
y = m x + c.
1
2(d)(i) C determined using gradient and C given to two or three significant figures
Correct substitution of numbers must be seen,
− −
= =
× × × ×
3 3
1 1
150 10 gradient 150 10 (c)(iii)
C
1
E determined using y-intercept
Correct substitution of numbers must be seen,
( )
−
= × = × × ×
-intercept 3 (c)(iv) 6
150 10 e 10
y
E R e
Or
= +
ln ln -intercept
E R y
1
C determined using gradient and E determined using y-intercept and dimensionally correct SI unit for C: F or s Ω–1 or C V–1
or A s V–1
and E: V or A Ω.
1

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March 2020
2(d)(ii) Absolute uncertainty in C.
 
Δ
Δ = + ×
 
 
gradient
0.05
gradient
C C
OR
Correct substitution for max/min methods
−
=
× ×
3
1
max
142.5 10 minnumericalgradient
C
−
=
× ×
3
1
min
157.5 10 max numericalgradient
C
1
2(e) I determined from (d)(i) OR (c)(iii) and (c)(iv) with correct substitution and correct power of ten(s).
Do not accept ecf for POT from (c)(iii), (iv) or (d).
−
= ×
120
CR
E
e
R
I
OR
( )
× −
= × ×
gradient 120
-intercept 6
10
y
e e
I
OR
= × +
ln 120 gradient -intercept
y
I
× + −
= ×
120 gradient -intercept 6
10
y
e
I
1

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