Physics 9702/53 May/June 2024: question paper and mark scheme
Cambridge AS & A Level Physics (9702), Paper 5 Planning, Analysis and Evaluation, May/June 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
Fig. 1.1 shows a small solid metal cylinder of mass m, length L and diameter d.
L
d
Fig. 1.1
The cylinder is heated to a uniform temperature. The cylinder is then removed from the heat source and the cylinder is wrapped in an insulating material.
The temperature of the room is TR. At time t after the cylinder starts to cool, the surface temperature of the cylinder is TC.
It is suggested that TC is related to t by the relationship
– UAt
(TC – TR) = Ze mc
where A is the total surface area of the cylinder, c is the specific heat capacity of the metal, and U and Z are constants.
Plan a laboratory experiment to test the relationship between TC and t.
Draw a diagram showing the arrangement of your equipment.
Explain how the results could be used to determine values for U and Z.
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
t is the independent variable and TC is the dependent variable or vary t and measure TC 1
keep TR constant 1
Methods of data collection
labelled diagram of workable experiment including: 1
solid cylinder cooling
insulation surrounding all of the cylinder
thermometer touching cylinder inside insulation
insulation and thermometer labelled
method to heat the cylinder uniformly, e.g. place in oven/immerse in hot water or diagram showing cylinder in oven or hot 1
water
method to determine time t, e.g. stopwatch or temperature sensor connected to a data logger 1
method to measure L e.g. use a ruler/calipers/micrometer 1
and
method to measure d e.g. use calipers/micrometer
Method of Analysis
plot a graph of ln (TC – TR) against t or equivalent 1
mc × gradient 1
U =−
A
Z = ey-intercept 1
1 Additional detail including safety considerations 6
D1 precaution to prevent burns or use of hot cylinder / oven / hot water e.g. use of gloves, use of tongs
D2 keep thickness of the insulating material constant (for each TC)
D3 method to measure m, e.g. use a (top-pan) balance
D4 for water bath/oven methods, wait for initial temperature of the cylinder to become uniform or constant throughout
the cylinder
πd 2 πd 2
D5 (surface) A = πdL + or πdL + 2
2 4
D6 repeat measurements of d along the length of the cylinder / in different directions and determine the average value
of d
D7 description of how c is determined from a separate experiment by heating the cylinder using electrical heater and
ΔE
c=
mΔθ
D8 method of determining energy supplied to electrical heater to determine c, e.g. use of joulemeter for ΔE or electrical
method using ammeter and voltmeter to determine IVt
D9 use several temperature sensors and determine the average TC
D10 relationship valid if a straight line is produced (with y-intercept = ln Z)
Do not accept line passing through the origin.
A student investigates the sound from a horn attached to a car, as shown in Fig. 2.1.
v horn
Frequency:
894.2 Hz
Fig. 2.1 (not to scale)
A microphone is placed at the side of the road and connected to a frequency meter. The car travels towards the microphone. The frequency f of the sound detected by the microphone is read from the frequency meter.
The speed of the car is measured by two speed detectors. The two measurements of speed are v1 and v2. The average speed v of the car is determined from v1 and v2.
The experiment is repeated for different speeds of the car.
It is suggested that f and v are related by the equation
fsk f= k –v where fs is the frequency of the sound emitted by the horn and k is a constant.
1
(a) A graph is plotted of on the y-axis against v on the x-axis. f
Determine expressions for the gradient and y-intercept.
gradient = ……
y-intercept = ……
[1]
(b) Values of v1, v2 and f are given in Table 2.1.
Table 2.1
1 v1 / m s–1 v2 / m s–1 v / m s–1 f / Hz –3 –1 f / 10 Hz
3.1 3.9 894.2
6.7 5.9 901.2
9.2 8.2 908.0
11.9 10.9 915.8
13.3 14.5 923.6
15.6 16.8 931.2
1
Calculate and record values of v / m s–1 and / 10–3 Hz–1 in Table 2.1. f
Include the absolute uncertainties in v. [2]
1
(c) (i) Plot a graph of / 10–3 Hz–1 against v / m s–1. Include error bars for v. [2]
f
(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]
1.125
1.120
1
/ 10–3 Hz–1 f
1.115
1.110
1.105
1.100
1.095
1.090
1.085
1.080
1.075
1.070
3.0 5.0 7.0 9.0 11.0 13.0 15.0 17.0 v / m s–1
(iv) Determine the y-intercept of the line of best fit. Include the absolute uncertainty in your answer.
y-intercept = …… [2]
(d) (i) Using your answers to (a), (c)(iii) and (c)(iv), determine the values of fs and k. Include appropriate units.
fs = ……
k = ……
[2]
(ii) Determine the percentage uncertainty in k.
percentage uncertainty in k = …… % [1]
(e) The experiment is repeated. Determine the speed v that gives a value of f of 987.8 Hz.
v = …… m s–1 [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) 1 1
gradient = −
kfs
1
y-intercept =
fs
2(b) 1
1
v / m s–1 / 10–3 Hz–1
f
3.5 ± 0.4 1.118 or 1.1183
6.3 ± 0.4 1.110 or 1.1096
8.7 ± 0.5 1.101 or 1.1013
11.4 ± 0.5 1.092 or 1.0919
13.9 ± 0.6 1.083 or 1.0827
16.2 ± 0.6 1.074 or 1.0739
1
Values of v and correct as shown above.
f
Uncertainties in v 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 v plotted correctly. 1
All error bars to 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
Do not accept line from top point to bottom point.
Line must pass between (14.5, 1.080) and (14.9, 1.080) and between (4.5, 1.115) and (4.8, 1.115).
Worst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1
All error bars must be plotted.
2(c)(iii) Gradient determined with clear substitution of data points into Δy / Δx. 1
Gradient must be negative.
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 y 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 accept ECF from false origin method.
2(d)(i) fs determined using y-intercept and fs given to 2, 3 or 4 significant figures and k given to 2 or 3 significant figures. 1
1
fs =
y -intercept
k determined using gradient with method shown and fs and k given with SI units with appropriate powers of ten. 1
y -intercept 1
k=− or k = −
gradient gradient × fs
Units of fs: Hz
Units of k: m s–1
2(d)(ii) Percentage uncertainty in k with method shown. 1
Δy -intercept Δgradient
percentage uncertainty = + × 100
y -intercept gradient
or
correct substitution for max/min methods.
2(e) v determined (non-zero) to a minimum of 2 significant figures from (c)(iii) and (c)(iv) or (d)(i) with correct substitution. 1
1
− y -intercept
v= f
gradient
or
kf
v =k− s
f
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Physics 9702/53 for May/June 2024 is out of 30 marks, with 1 hour 15 minutes to answer it.
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