Cambridge A Level Physics 9702 — 2020 May/June Paper 5 · Variant 2

1 A student investigates stationary waves with an elastic cord of circular cross-section attached to a load, as shown in Fig. 1.1. L pulley vibrator load cord Fig. 1.1 When the frequency of the vibrator is f, the cord vibrates with the stationary wave pattern shown. The student investigates how f varies with the cross-sectional area A of the cord. It is suggested that the relationship between f and A is 1 M f = 2L kA where L is the distance between the two nodes, M is the mass of the load and k is a constant. Design a laboratory experiment to test the relationship between f and A. 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. 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Mark scheme: 1 Defining the problem A is the independent variable and f is the dependent variable or vary A and measure f 1 keep M constant 1 Methods of data collection labelled diagram of workable experiment including: • elastic cord fixed at one end to a support • other end passed over a pulley • labelled pulley • labelled load 1 vibrator connected to signal generator 1 increase/decrease the frequency of the signal generator until stationary wave pattern is observed 1 measure diameter of cord with micrometer/calipers 1 Method of analysis plot a graph of f 2 against 1 / A or equivalent (e.g. lg f against lg A) 1 relationship valid if a straight line passing through the origin is produced (for lg f against lg A, relationship valid if a straight line with gradient –½) 1 2 gradient 4 M k L = × × (for lg f against lg A, k = M / [10(2 × y-intercept) × 4 × L2] ) 1 Question Answer Marks 1 Additional detail including safety considerations 6 D1 use safety goggles/safety screen to prevent injury to eyes from (moving) elastic cord/load or use cushion/sand box in case load falls D2 keep L constant D3 use cords of the same material/density D4 use CRO to determine f (or T) D5 method to determine T from CRO, e.g. period = time base × length of one wave D6 f = 1 / T D7 repeat measurement of diameter along cord and average D8 use of 2 4 d A π = D9 measure mass of the load on top-pan balance D10 detail on determining frequency at the maximum amplitude, e.g. increase frequency until the amplitude starts to decrease, then decrease frequency

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2 A student investigates how the viscous force in a liquid varies with temperature. The student releases a ball from the surface of the liquid in a container. The ball falls as shown in Fig. 2.1. ball container liquid P Q Fig. 2.1 The student determines the speed of the ball between P and Q and measures the thermodynamic temperature T of the liquid. Viscosity is a term used to describe the viscous forces acting in a liquid. Viscosity has the unit pascal second (Pa s). The viscosity η of the liquid is calculated from the speed of the ball. The experiment is repeated for the same liquid at different temperatures. It is suggested that η and T are related by the equation c kTE m η = He where E and H are constants and k is the Boltzmann constant. 1 (a) A graph is plotted of ln η on the y-axis against on the x-axis. T Determine expressions for the gradient and y-intercept. gradient = ............................................................... y-intercept = ............................................................... [1] (b) Values of T and η are given in Table 2.1. Table 2.1 1 T / K η/ 10–4 Pa s / 10–3 K–1 ln (η/ 10–4 Pa s) T 292 12.3 ± 0.2 3.42 303 9.8 ± 0.2 3.30 311 8.4 ± 0.2 3.22 323 6.8 ± 0.2 3.10 335 5.6 ± 0.2 2.99 346 4.8 ± 0.2 2.89 Calculate and record values of ln (η/ 10–4 Pa s) in Table 2.1. Include the absolute uncertainties in ln (η/ 10–4 Pa s). [2] 1(c) (i) Plot a graph of ln (η/ 10–4 Pa s) against / 10–3 K–1. T Include error bars for ln η. [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]

Mark scheme: 2(a) gradient = E k y-intercept = lnH 2(b) ln (η / 10–4 Pa s) 2.510 or 2.5096 2.28 or 2.282 2.13 or 2.128 1.92 or 1.917 1.72 or 1.723 1.57or 1.569 1 Absolute uncertainties in ln η from ± 0.02 (or ± 0.016) to about ± 0.04 1 2(c)(i) Six points plotted correctly. Must be accurate to nearest half a small square. Diameter of points must be less than half a small square. 1 Error bars in ln η plotted correctly. All error bars to be plotted. Length of bar must be accurate to less than half a small square and symmetrical. 1 2(c)(ii) Line of best fit drawn. Points must be balanced. Do not accept top point to bottom point. Line must pass between (2.93, 1.65) and (2.96, 1.65) and between (3.38, 2.45) and (3.41, 2.45). 1 Worst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). All error bars must be plotted. 1 Question Answer Marks 2(c)(iii) Gradient determined with clear substitution of data points into Δy / Δx. Distance between data points must be at least half the length of the drawn line. 1 uncertainty = (gradient of line of best fit – gradient of worst acceptable line) or uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient) 1 2(c)(iv) y-intercept determined from substitution into y = mx + c. 1 2(d)(i) E determined using gradient and given to two or three significant figures. Correct substitution of numbers required. E = 1.38 × 10–23 × gradient = 1.38 × 10–23 × (c)(iii) 1 H determined using y-intercept. H = ey-intercept = e(c)(iv) (× 10–4) 1 E determined using gradient and H determined using y-intercept and dimensionally correct units for E (J) and H (Pa s). 1 2(d)(ii) Absolute uncertainty in E. 23 1.38 10 absolute uncertainty in gradient E − Δ = × × or gradient gradient E E Δ Δ = × 1 Question Answer Marks 2(e) η determined from (d)(i) or (c)(iii) and (c)(iv) with correct substitution shown and correct power of ten. 23 4 273 1.38 10 273 10 E k H e e e η − − × × × = × = × × (d)(i) (c)(iv) or gradient intercept 4 273 10 y e e η − − = × × or 4 273 10 e e η − = × × (c)(iii) (c)(iv) or 23 ln ln 1.38 10 273 E H kT η − = + = + × × (d)(i) (c)(iv) or gradient ln -intercept 273 273 y η = + = + (c)(iii) (c)(iv) 1

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