Physics 9702/54 May/June 2025: question paper and mark scheme
Cambridge AS & A Level Physics (9702), Paper 5 Planning, Analysis and Evaluation, May/June 2025. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A ball is dropped on to an inclined thin metal sheet, as shown in Fig. 1.1.
ball
sheet
bench z
θ
d
Fig. 1.1 (not to scale)
The angle between the sheet and the horizontal bench is θ. The height of the point of contact of the ball and the sheet is z. The horizontal distance travelled by the ball between its points of contact with the sheet and the bench is d, as shown in Fig. 1.1.
It is suggested that d is related to θ by the relationship
Pv 2 sin 4θ d= g +Q z
where v is the speed of the ball as it makes contact with the sheet, g is the acceleration of free fall, and P and Q are constants.
Plan a laboratory experiment to test the relationship between d and θ.
Draw a diagram showing the arrangement of your equipment.
Explain how the results could be used to determine values for P and Q.
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.
Diagram
…… [15]
Mark scheme
1 Defining the problem
vary θ and measure d or θ is the independent variable and d is the dependent variable 1
keep z constant 1
Methods of data collection
labelled diagram of workable experiment including: 1
• sheet supported with stand or block(s)
• path of ball (from sheet to bench) or distance d indicated
• at least two labels from ball, bench, sheet, block/clamp/stand/support
description of method to determine position of ball as it makes contact with the bench, e.g. paint ball, sand, video and 1
playback in slow motion / frame by frame (with rule in view)
measure d with a rule 1
and
measure z with a rule or calipers
measure θ with a protractor 1
or
determine θ using (a rule for) appropriate distances with correct trigonometric relationship
Method of analysis
plot a graph of d against sin (4θ) or equivalent 1
Do not accept logarithms.
1 1
d against sin (4θ) sin (4θ) against d
g × gradient g
P= P= 2
v2 v × gradient
1
d against sin (4θ) sin (4θ) against d
y -intercept Pv 2 × y -intercept
Q= Q=−
z g z
or
y -intercept
Q=−
gradient × z
Additional detail including safety considerations 6
D1 precaution linked to prevent ball rolling (off surface), e.g. use of cushion / sand box / barrier to stop ball (rolling onto the
floor)
or
precaution linked to (bouncing) ball or sand (spray) e.g. use of goggles to protect eyes
D2 keep v constant
D3 method to keep v constant, e.g. keep the distance the ball falls constant to keep v constant
D4 method to determine v or v2, e.g. use of
v = 2g × (height ball falls )
or
v 2 = 2g × ( height ball falls )
1 D5 set square correctly positioned between rule and bench to ensure that rule is vertical to measure z or determine initial
height of ball
D6 method to ensure z is constant, e.g. as θ is changed, (re-)mark point of contact or change release point of ball
D7 description of method to ensure ball is released above point of contact on sheet, e.g. use a plumb line from the ball
through point of contact with sheet or (vertical) rule (and set square)
D8 drop ball from large height to obtain greater values of d
D9 repeat experiment for each θ and average d
( )
D10 relationship valid if a straight line produced (with y-intercept of e.g. Q Z ).
Do not accept line passing through the origin.
A student investigates the relationship between the luminosity of a star and its mass.
The student obtains data of relative luminosity λ and relative mass µ for six stars, where
luminosity of star λ= luminosity of Sun and mass of star µ= . mass of Sun
It is suggested that λ and µ are related by the equation
λ = kµn
where k and n are constants.
(a) A graph is plotted of lg λ on the y-axis against lg µ on the x-axis.
Determine expressions for the gradient and y-intercept.
gradient = ……
y-intercept = ……
[1]
(b) Values of µ and λ are given in Table 2.1.
Table 2.1
µ λ lg µ lg λ
4.6 ± 0.4 500
5.4 ± 0.4 800
8.4 ± 0.4 3200
11 ± 1 7000
16 ± 1 25 000
18 ± 1 38 000
Calculate and record values of lg µ and lg λ in Table 2.1.
Include the absolute uncertainties in lg µ. [2]
(c) (i) Plot a graph of lg λ against lg µ.
Include error bars for lg µ. [2]
(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]
4.8
4.6
lg λ
4.4
4.2
4.0
3.8
3.6
3.4
3.2
3.0
2.8
2.6
0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 lg μ
(iv) Determine the y-intercept of the line of best fit. Include the absolute uncertainty in your answer.
y-intercept = …… [2]
(d) Using your answers to (a), (c)(iii) and (c)(iv), determine the values of k and n. Include the absolute uncertainties in your values. You need not be concerned with units.
k = ……
n = ……
[3]
(e) The mass of the Sun is 2.0 × 1030 kg. The star Alpha Centauri B has a value of λ of 0.46.
Determine the mass M of Alpha Centauri B.
M = …… kg [1]
[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 = n 1
y-intercept = lg k
2(b) 1
lg μ lg λ
0.66 or 0.663 ± 0.04 2.70 or 2.699
0.73 or 0.732 ± 0.03 2.90 or 2.903
0.92 or 0.924 ± 0.02 3.51 or 3.505
1.04 or 1.041 ± 0.04 3.85 or 3.845
1.20 or 1.204 ± 0.03 4.40 or 4.398
1.26 or 1.255 ± 0.02 or ± 0.03 4.58 or 4.580
Values of lg μ and lg λ correct as shown above.
Uncertainties in lg μ correct as shown above. 1
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.
Error bars in lg μ plotted correctly. 1
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 (0.82, 3.20) and (0.84, 3.20) and between (1.13, 4.20) and (1.16, 4.20).
Worst acceptable straight 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(c)(iv) y-intercept determined by substitution of correct point with consistent power of ten in m and x into y = mx + c. 1
y-intercept of worst acceptable line determined by substitution into y = mx + c. 1
uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line
or
uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)
Do not allow methods using a false origin.
2(d) Value of k determined using y-intercept. 1
k = 10y −intercept
n = gradient and n and k given to 2 or 3 significant figures. 1
absolute uncertainty in n = absolute uncertainty in gradient 1
and
absolute uncertainty in k = (10y-intercept – 10y-intercept of WAL)
Correct method must be seen.
2(e) M determined to a minimum of 2 significant figures from (d) or (c)(iii) and (c)(iv) with correct substitution and correct 1
power(s) of ten.
Do not accept incorrect POT for n or k.
Correct substitution must be seen.
λ 0.46 lg0.46 − y -intercept lg0.46 − lg k
μ=n = gradient or lg μ = or lg μ =
k (d) gradient gradient
and
M = μ × 2.0 × 1030
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Physics 9702/54 for May/June 2025 is out of 30 marks, with 1 hour 15 minutes to answer it.
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