Physics 9702/51 May/June 2021: question paper and mark scheme
Cambridge AS & A Level Physics (9702), Paper 5 Planning, Analysis and Evaluation, May/June 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A student investigates the current in a coil and a resistor connected in series, as shown in Fig. 1.1.
coil
R
Fig. 1.1
The student connects a high-voltage d.c. power supply and a switch across the series combination.
When the switch is closed, it takes time t for the current in the resistor of resistance R to reach a maximum value. The time t is a few milliseconds.
There are a number of different unmarked resistors available.
It is suggested that the relationship between t and R is
KN2A t=
LR where N is the number of turns of wire on the coil, A is the cross-sectional area of the coil, L is the length of the coil and K is a constant.
Design a laboratory experiment to test the relationship between t and R.
Explain how your results could be used to determine a value for K.
You should draw a diagram, on page 3, showing the arrangement of your equipment. In your account you should pay particular attention to:
● 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
R is the independent variable and t is the dependent variable or vary R and measure t 1
keep the number of turns on the coil/N constant 1
Methods of data collection
labelled diagram or correct symbols including: 1
• labelled (d.c.) power supply
• switch in series with power supply, resistor and coil
• complete workable circuit
circuit diagram to measure R, e.g. ammeter and voltmeter correctly positioned or R connected to ohmmeter with no other 1
connections (not ohmmeter in main circuit)
method to determine t (of a few milliseconds) e.g. use (storage) oscilloscope or current/voltage sensor connected to 1
datalogger/computer
method to determine A, e.g. micrometer/calipers to determine diameter of coil and A = πd2 / 4 1
Method of analysis
plot a graph of t against 1 / R 1
(allow log t against log R)
relationship valid if a straight line passing through the origin is produced 1
(allow gradient = –1 for graph of log t against log R)
gradient × L 1
K= .
AN 2
1 Additional detail including safety considerations 6
D1 open switch/switch off (high voltage) circuit before changing the resistor/touching components or
ensure no bare wires/use shrouded connectors
D2 wear (insulating) gloves to prevent electric shock/electrocution
D3 keep A and L constant
D4 use ruler/calipers to measure L
D5 repeat measurements of diameter in different directions/at points along the coil and average
D6 method to determine R e.g. R = V / I linked to correct circuit diagram for ammeter/voltmeter method or measure
resistance using ohmmeter
D7 repeat experiment for each value of R and average t
D8 method to determine t:
use of time-base from oscilloscope explained
or
use of time axis of output from data logger/computer explained
D9 use smaller values of R to increase I
D10 reduce L or increase N or increase A to increase t
A student investigates the collision of two gliders A and B on a linear air-track. A card is attached to glider B, as shown in Fig. 2.1.
card light gate connected to data logger glider B glider A
air-track bench
Fig. 2.1
Glider B has a mass M. A mass m is added to glider B.
Glider A travels at a constant velocity u towards the stationary glider B. The gliders then collide and move together towards the light gate.
The card passes through the light gate which is connected to a data logger. The student records the velocity v of the two gliders from the data logger.
The student changes the mass m and repeats the experiment.
It is suggested that v and m are related by the equation
Au = (M + m + A)v
where A is the mass of glider A.
1
(a) A graph is plotted of v on the y-axis against (M + m) on the x-axis.
Determine expressions for the gradient and y-intercept.
gradient = ……
y-intercept = ……
[1]
(b) Values of m and v are given in Table 2.1.
The value of M is 330 g ± 5%.
Each value of m has a percentage uncertainty of ± 5%.
Table 2.1
m/g (M + m) / g v / cm s–1 1 / s cm–1 v
50 4.42
150 3.92
250 3.40
350 3.02
500 2.58
600 2.33
1
Calculate and record values of (M + m) / g and v / s cm–1 in Table 2.1.
Include the absolute uncertainties in (M + m). [2]
1
(c) (i) Plot a graph of v / s cm–1 against (M + m) / g.
Include error bars for (M + m). [2]
(ii) Draw the straight line of best fit and a worst acceptable straight line on your graph. Both lines should be clearly labelled. [2]
(iii) Determine the gradient of the line of best fit. Include the absolute uncertainty in your answer.
gradient = …… [2]
0.44
0.42
1
/ scm –1 v
0.40
0.38
0.36
0.34
0.32
0.30
0.28
0.26
0.24
0.22
300 400 500 600 700 800 900 1000
(M + m) / g
(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 u and A.
Include appropriate units.
u = ……
A = ……
[2]
(ii) Determine the percentage uncertainty in A.
percentage uncertainty in A = …… % [1]
(e) The experiment is repeated. Determine the value of m that would give a velocity v of
2.0 cm s–1.
m = …… g [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 =
uA
1
y-intercept =
u
2(b) 1
1
(M + m) / g / s cm–1
v
380 0.226 or 0.2262
480 0.255 or 0.2551
580 0.294 or 0.2941
680 0.331 or 0.3311
830 0.388 or 0.3876
930 0.429 or 0.4292
1
Values of (M + m) and as shown above.
v
Absolute uncertainties in (M + m) from ± (19 or 20) to ± (46.5 or 47 or 50). 1
2(c)(i) Six points plotted correctly. 1
Must be accurate to the nearest half a small square. Diameter of points must be less than half a small square.
Error bars in (M + m) 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) Line of best fit drawn covers all points. 1
Points must be balanced. Do not allow line from top point to bottom point.
Line must pass between (425, 0.240) and (440, 0.240) and between (850, 0.400) and (865, 0.400).
Worst acceptable line drawn (steepest or shallowest possible line that passes through all error bars). 1
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 at least half the length of the drawn line.
Gradient of worst acceptable line determined. 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 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 ECF from false origin method.
2(d)(i) u determined using y-intercept and u and A given to two or three significant figures. 1
1
u=
y -intercept
A determined using gradient with correct substitution and units with correct power of ten for u and A. 1
y -intercept 1
A= or A =
gradient u × gradient
2(d)(ii) Percentage uncertainty in A determined, e.g. 1
Δgradient Δy -intercept
percentage uncertainty in A = +
gradient y -intercept
or
Δu clearly determined using the value of u and
Δgradient Δu
percentage uncertainty in A = + × 100
gradient u
or
correct substitution for max/min methods e.g.
1
max A =
minu × min gradient
1
min A =
max u × max gradient
2(e) Value of m determined from (d)(i) or (c)(iii) and (c)(iv), with correct number substitution and correct power of ten. 1
A×u
m= − ( 330 + A )
2
or
0.5 − y -intercept
m= − 330
gradient
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On this page, free: the mark scheme for AS & A Level Physics 9702/51, May/June 2021 (9702_s21_ms_51), and the question paper (9702_s21_qp_51), to read or download, with each question's marking points written out underneath.
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Physics 9702/51 for May/June 2021 is out of 30 marks, with 1 hour 15 minutes to answer it.
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