Physics 9702/52 October/November 2021: question paper and mark scheme
Cambridge AS & A Level Physics (9702), Paper 5 Planning, Analysis and Evaluation, October/November 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A student investigates the extension of a spring supporting a wooden strip, as shown in Fig. 1.1.
spring
wire
β d
wooden strip bench θ
P
L
Fig. 1.1
One end of the strip is at point P. The strip has length L and is at an angle θ to the bench.
The spring is attached to the strip by a wire at a distance d from point P. The wire is at an angle β to the strip. The spring has extension x.
It is suggested that the relationship between x and θ is
WL cos θ = kxd sin β
2 where k is the spring constant of the spring and W is a constant.
Design a laboratory experiment to test the relationship between x and θ.
Explain how your results could be used to determine a value for W.
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
θ is the independent variable and x is the dependent variable or vary θ and measure x 1
keep (angle) β constant 1
Methods of data collection
labelled diagram of workable experiment including: 1
• spring attached at both ends e.g. one end connected to a clamp and stand
• strip free to move
• at least two labels from: clamp, stand, wire, strip, spring, bench
(Do not accept extra masses added to strip.)
use a rule to measure L and d 1
use a protractor to measure θ 1
or
use a rule to measure appropriate distances to determine θ by trigonometry methods
measure original length of spring and new length of spring using rule/calipers 1
Method of analysis
plot a graph of x against cos θ or cos θ against x 1
(Allow log x against log (cos θ).)
relationship is valid if a straight line passing through the origin is produced 1
(Allow straight line with gradient = 1 for log-log graph.)
for x against cos θ for cos θ against x 1
gradient × 2kd sin β or 2kd sin β
W = W =
L gradient × L
1 Additional detail including safety considerations 6
D1 wear goggles to prevent spring/wire/strip entering into eyes
or
(retort) stand used to support spring is clamped to bench
D2 keep distance d constant
D3 description of (separate) experiment to determine k, e.g. weigh mass and measure extension
D4 k = weight / extension or mg / extension or gradient of weight–extension graph for candidate’s workable (separate)
experiment
D5 method to prevent strip at point P sliding, e.g. use adhesive putty/hinge
(Do not accept methods that prevent rotation at point P.)
D6 use fiducial markers on spring at both ends or measure length of spring on both sides and average
D7 method to attach wire to strip, e.g. wire wrapped around the strip/(strong) tape/drill hole and tie wire
D8 determine x by subtracting original length of spring from new length
D9 adjust support of spring to keep β constant
D10 protractor correctly positioned on diagram to measure θ
or
correct trigonometric relationship given for θ
A Geiger–Müller (G–M) tube is a device that can detect beta-radiation. A student places paper between a radioactive source emitting beta-radiation and a G–M tube, as shown in Fig. 2.1.
G–M tube paper
radioactive source to rate-meter bench
Fig. 2.1
The G–M tube is connected to a rate-meter which records the count rate R.
The thickness t of the paper is measured in two different places using a micrometer.
The student repeats the experiment for different thicknesses of paper.
It is suggested that R and t are related by the equation
R = R0e−μt
where R0 is the count rate without any paper and μ is a constant.
(a) A graph is plotted of ln R on the y-axis against t on the x-axis.
Determine expressions for the gradient and y-intercept.
gradient = ……
y-intercept = ……
[1]
(b) The two measurements of thickness are t1 and t2. Values of t1, t2 and R are given in Table 2.1.
Table 2.1
t1 / mm t2 / mm average t / mm R / s−1 ln (R / s−1)
0.19 0.13 47.7
0.22 0.28 44.0
0.39 0.45 38.2
0.58 0.54 34.3
0.64 0.68 31.7
0.78 0.74 29.7
Calculate and record values of average t / mm and ln (R / s−1) in Table 2.1.
Include the absolute uncertainties in average t. [2]
(c) (i) Plot a graph of ln (R / s−1) against average t / mm.
Include error bars for average t. [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]
3.90
3.85
3.80
In (R / s–1)
3.75
3.70
3.65
3.60
3.55
3.50
3.45
3.40
3.35
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 t / mm
(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 μ and R0. Include the absolute uncertainties in your values and include appropriate units.
μ = ……
R0 = ……
[3]
(e) The experiment is repeated using a different thickness of paper.
Determine the value of t that would give a value of R of 20.0 s−1.
t = …… mm [1]
[Total: 15]
Mark scheme
2(a) gradient = –μ 1
y-intercept = ln R0
2(b) 1
average t / mm ln (R / s–1)
0.16 ± 0.03 3.865 or 3.8649
0.25 ± 0.03 3.784 or 3.7842
0.42 ± 0.03 3.643 or 3.6428
0.56 ± 0.02 3.535 or 3.5351
0.66 ± 0.02 3.456 or 3.4563
0.76 ± 0.02 3.391 or 3.3911
Values of average t and ln R correct as shown above.
Absolute uncertainties in average t correct as shown above. 1
2(c)(i) Six points plotted correctly. 1
Must be accurate to nearest half a small square. Diameter of points must be less than half a small square.
Error bars in average t 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. 1
Points must be balanced. Do not accept line from top point to bottom point.
Line must pass between (0.22, 3.80) and (0.24, 3.80) and between (0.60, 3.50) and (0.62, 3.50).
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) Negative 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 point on line into y = mx + c. 1
y-intercept of worst acceptable line determined by substitution of point on line 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) μ = – gradient value 1
Do not accept negative values (from a negative gradient).
R0 determined using y-intercept and μ and R0 both given with valid SI unit. 1
R0 = e y -intercept
unit of μ: mm–1
unit of R0: s–1
absolute uncertainty in μ = absolute uncertainty in gradient 1
and
absolute uncertainty in R0 = e y −intercept of WAL − R0
Correct substitution of numbers must be seen.
2(e) Value of t determined to two or three significant figures from (d) or (c)(iii) and (c)(iv) with correct substitution and correct 1
power of ten(s).
Do not accept ECF for POT from (c)(iii), (c)(iv) or (d).
ln R − ln R0 ln 20 − ln R0
t= =
−μ −μ
or
ln 20 − y -intercept 2.996 − (c)(iv)
t= =
gradient (c)(iii)
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On this page, free: the mark scheme for AS & A Level Physics 9702/52, October/November 2021 (9702_w21_ms_52), and the question paper (9702_w21_qp_52), to read or download, with each question's marking points written out underneath.
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Physics 9702/52 for October/November 2021 is out of 30 marks, with 1 hour 15 minutes to answer it.
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