Physics 9702/54 October/November 2025: question paper and mark scheme
Cambridge AS & A Level Physics (9702), Paper 5 Planning, Analysis and Evaluation, October/November 2025. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
Fig. 1.1 shows a horizontal turntable.
turntable P
C r
Fig. 1.1
Point C is at the centre of the turntable. Point P is a distance r from the centre.
Fig. 1.2 shows a side view of a d.c. motor attached to the turntable with a belt.
terminals turntable C belt
motor turntable base
Fig. 1.2
The motor is used to rotate the turntable at frequency f0. The motor is switched off and the turntable continues to rotate at frequency f0.
A sphere of adhesive putty of mass m is dropped onto the turntable at point P. The frequency of the turntable is now f.
It is suggested that f is related to m by the relationship
Kf0
= βK + mr 2 f where β and K are constants.
Plan a laboratory experiment to test the relationship between f and m.
Draw a diagram showing the arrangement of your equipment.
Explain how the results could be used to determine values for β and K.
In your plan you should include:
•
the procedure to be followed
• the measurements to be taken
• the control of variables
• the analysis of the data
• any safety precautions to be taken.
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Diagram
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…… [15]
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Mark scheme
1 Defining the problem
vary m and measure f or m is the independent variable and f is the dependent variable 1
keep f0 constant 1
Methods of data collection
labelled diagram of workable experiment including: 1
• turntable base and motor on bench
• two labels from turntable, motor, belt, C, putty, bench, release mechanism
workable circuit showing terminals of motor connected to (d.c.) supply 1
method to measure time to determine f and f0, e.g. use a stopwatch to measure (n)t 1
and
method to measure m, e.g. use a (top-pan) balance
measure the time t for n revolutions 1
and
T = t / n and f = 1 / T
or
f = n/t
Method of Analysis
1 1
plots a graph of against m or equivalent
f
Do not accept logarithms.
1 1
1 1
against m m against
f f
r2 r 2 × gradient
K= K=
f0 × gradient f0
1
1 1
against m m against
f f
β = f0 × y -intercept f × y -intercept
β =− 0
gradient
or
r 2 ×y -intercept
β =−
K
1 Additional detail including safety considerations 6
D1 precaution linked to putty leaving turntable, e.g. screens around apparatus / cushions to prevent putty from leaving
bench
D2 keep r constant
D3 use rule(r) to measure r
D4 stand and fixed point located above P (to ensure r is constant)
D5 method to ensure putty is dropped at (constant) distance r, e.g.;
fixed location above the turntable to drop putty to keep r constant
or
draw a circle of radius r on the turntable
D6 method to keep f0 constant, e.g. adjust the rheostat to keep the current in the motor constant / check ammeter
D7 clamp motor and/or turntable (base) to bench
D8 lubricate the turntable to reduce friction
D9 repeat experiment for the same value of m and determine the average f
β
D10 relationship valid if a straight line is produced (with y-intercept = )
f0
Do not accept line through the origin.
A student places a slide with a double slit on a support clamped to the bench as shown in Fig. 2.1.
slide with D double slit
light screen
supports
bench
Fig. 2.1
The distance between the slide and the screen is D.
The separation s of the slits is determined.
Light from a laser is incident normally on the double slit. An interference pattern is observed on the screen. The distance w across 10 fringes is measured. The distance y between the centres of adjacent fringes is calculated using the equation
w y= .
10
The experiment is repeated with slides of different slit separation s.
It is suggested that y and s are related by the equation sy λ=
D where λ is the wavelength of the incident light.
(a) A graph is plotted of y on the y-axis against 1 on the x-axis. s
Determine an expression for the gradient.
gradient = …… [1]
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6
(b) Values of s and w are given in Table 2.1.
Table 2.1
s / mm 1 / mm–1 w / mm y / mm s
0.18 ± 0.01 33.0
0.21 ± 0.01 28.9
0.24 ± 0.01 25.1
0.27 ± 0.01 22.6
0.31 ± 0.01 19.6
0.38 ± 0.01 15.9
Calculate and record values of 1 / mm–1 and y / mm in Table 2.1. Include the absolute s uncertainties in 1 . [2]
s
(c) (i) Plot a graph of y / mm against 1 / mm–1. Include error bars for 1 . [2]
s s
(ii) Draw the straight line of best fit and a worst acceptable straight line on your graph. Label both lines. [2]
(iii) Determine the gradient of the line of best fit. Include the absolute uncertainty in your answer.
gradient = …… [2]
* 0000800000007 * DFD
7
3.6
3.4
3.2
3.0
y / mm
2.8
2.6
2.4
2.2
2.0
1.8
1.6
1.4
2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0
1 / mm–1 s
* 0000800000008 * DFD
8
(d) The distance between the slide and the screen is measured several times:
0.929 m 0.918 m 0.913 m 0.918 m 0.927 m.
Determine the mean distance D. Include the absolute uncertainty.
D = …… m [1]
(e) (i) Using your answers to (a), (c)(iii) and (d), determine the value of λ. Include an appropriate unit.
λ = …… [2]
(ii) Determine the percentage uncertainty in λ.
percentage uncertainty in λ = ……% [1]
(f) The experiment is repeated. Determine the slit separation s that gives a value of y of
(0.500 ± 0.005) cm. Include the absolute uncertainty.
s = …… m [2]
[Total: 15]
To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge
Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cambridgeinternational.org after the live examination series.
Mark scheme
2(a) gradient = Dλ 1
2(b) 1
1
/ mm–1 y / mm
s
5.6 or 5.56 3.30
4.8 or 4.76 2.89
4.2 or 4.17 2.51
3.7 or 3.70 2.26
3.2 or 3.23 1.96
2.6 or 2.63 1.59
1
Values of and y correct as shown above.
s
1 1
Uncertainties in from ± 0.3 to ± 0.07.
s
2(c)(i) Six points from (b) plotted correctly. 1
Must be within half a small square. Diameter of points must be less than half a small square.
1 1
Error bars in plotted correctly.
s
All error bars must be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.
2(c)(ii) Straight line of best fit drawn. 1
Thickness of the line must be less than half a small square.
Do not accept line from top point to bottom point.
Line must pass between (2.90, 1.80) and (3.00, 1.80) and between (5.30, 3.20) and (5.40, 3.20).
Worst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1
Thickness of the line must be less than half a small square.
All error bars must be plotted.
2(c)(iii) Gradient determined with clear substitution of data points into Δy / Δx. 1
Distance between data points must be greater than half the length of the drawn line.
Gradient determined of worst acceptable line with clear substitution of data points into Δy / Δx. 1
uncertainty = (gradient of line of best fit – gradient of worst acceptable line)
or
uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)
2(d) (0.921 ± 0.008) (m) 1
2(e)(i) λ determined using gradient and λ given to 2 or 3 significant figures. 1
gradient
λ=
D
λ determined using gradient and λ given with SI units with appropriate power of ten. 1
gradient
λ=
D
Unit of λ: m or mm.
2(e)(ii) Percentage uncertainty in λ determined with clear method shown. 1
Δgradient ΔD
%uncertainty = + × 100
gradient D
or
correct substitution for max/min methods.
2(f) s determined to a minimum of 2 significant figures from (c)(iii) or (d) and (e)(i) with correct substitution and correct power of 1
ten.
gradient
s=
0.005 (m)
or
D×λ (d) × (e)(i)
s= =
0.005 (m) 0.005 (m)
Absolute uncertainty determined with correct substitution. 1
For use of gradient from (c)(iii):
Δgradient 0.005
Δs = + ×s
gradient 0.500
or
correct substitution for max/min methods.
For use of D and λ from (d) and (e)(i):
ΔD Δλ 0.005
Δs = + + ×s
D λ 0. 500
or
correct substitution for max/min methods:
max D × max λ min D × min λ
Δs = − s or Δs = s −
min y max y
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Physics 9702/54 for October/November 2025 is out of 30 marks, with 1 hour 15 minutes to answer it.
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