Cambridge A Level Physics 9702 — 2020 Oct/Nov Paper 5 · Variant 2

1 A student investigates the motion of a trolley on a wooden surface, as shown in Fig. 1.1. mass m trolley wooden surface Fig. 1.1 A mass m is placed on the trolley. A mass P is attached to the trolley by string which passes over a pulley. When this mass falls, it pulls the trolley along the surface. The trolley is initially at rest. The student investigates how the speed v of the trolley at a distance d from the initial position of the trolley varies with m. It is suggested that the relationship between v and m is 2d m + R = v2 Pg – Q where g is the acceleration of free fall and Q and R are constants. Design a laboratory experiment to test the relationship between m and v. Explain how your results could be used to determine values for Q and R. 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 m is the independent variable and v is the dependent variable or vary m and measure v 1 keep P constant 1 Methods of data collection labelled diagram of workable experiment including: • P attached to string • string passing over a supported pulley • string horizontally attached to trolley • labels for pulley and P 1 method to determine v e.g. • light gate attached to a timer to measure v • motion sensor connected to data logger • ticker tape timer • measure d, and t with timer/stopwatch/light gate and timer (to calculate v) 1 use a (top pan) balance to measure m 1 measure d with a ruler 1 Method of Analysis plot a graph of 1 / v2 against m or m against 1 / v2 1 1 2 gradient Q Pg d = − × or Q = Pg – gradient 2d (consistent with graph) 1 ( ) 2 -intercept R d Pg Q y = × − × ( -intercept gradient y = ) or R = –y-intercept (consistent with graph) 1 Question Answer Marks 1 Additional detail including safety considerations 6 D1 safety precaution related to falling mass P or preventing trolley falling e.g. cushion/sandbox for P, barrier for trolley D2 keep d constant D3 use a large distance for d D4 method to keep d constant, e.g. mark starting and end positions D5 v = distance / time for appropriate small distance on ticker tape/card/distance between light gates (not d) or use of v = 2d / t if d and total time measured D6 method to ensure wooden surface is horizontal, e.g. spirit level D7 for same m repeat experiment to find average v or t D8 ( ) ( ) 2 1 2 2 m R d Pg Q d Pg Q v = + × − × − or m ( ) 2 2 d Pg Q R v − = − (consistent with graph) D9 relationship valid if a straight line (not passing through the origin) D10 method of attaching mass m to trolley e.g. tape, adhesive putty

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2 A student investigates the behaviour of a liquid inside a narrow tube, as shown in Fig. 2.1. d h liquid Fig. 2.1 (not to scale) When the tube is placed in the liquid, the liquid rises in the tube. The student measures the internal diameter d of the tube and the maximum height h that the liquid rises in the tube. The experiment is repeated for tubes of different diameter. It is suggested that h and d are related by the equation 4σ h = dρg where ρ is the density of the liquid, g is the acceleration of free fall and σ is a constant. 1 (a) A graph is plotted of h on the y-axis against on the x-axis. d Determine an expression for the gradient. gradient = ......................................................... [1] (b) Values of d and h are given in Table 2.1. Table 2.1 1 d / mm / mm–1 h / mm d 1.1 ± 0.1 18.3 1.3 ± 0.1 15.1 1.5 ± 0.1 13.1 1.7 ± 0.1 11.6 2.0 ± 0.1 9.9 2.3 ± 0.1 8.6 1 Calculate and record values of / mm–1 in Table 2.1. d 1 Include the absolute uncertainties in . [2] d 1(c) (i) Plot a graph of h / mm against / mm–1. d 1 Include error bars for . [2] d (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 = 4 g σ ρ 1 2(b) (1 / d) / mm–1 0.91 or 0.909 0.77 or 0.769 0.67 or 0.667 0.59 or 0.588 0.50 or 0.500 0.43 or 0.435 1 Absolute uncertainties in 1 d from ± (0.08 or 0.09) to ± (0.01 or 0.02). 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 1 d plotted correctly. All error bars must be plotted. Total 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 line from top to bottom point. Line must pass between (0.52, 10.5) and (0.55, 10.5) and between (0.84, 17.0) and (0.87, 17.0). 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 Gradient of worst acceptable line determined. 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(d)(i) ρ − − −   = = =   × ×   6 6 0.606 0.422 0.184 1260 146 10 146 10 kg m–3 and given to three or four significant figures. 1 2(d)(ii) % uncertainty in ρ = % uncertainty in m + % uncertainty in V 2 2 100 1.087 1.37 2.5% 184 146     = + × = + =         or using max ρ = 186 / 144 = 1292 and/or min ρ = 182 / 146 = 1230 1 Question Answer Marks 2(e) σ determined using gradient with correct substitution shown. ρ σ × × × = = gradient 9.81 4 4 g (d)(i) (c)(iii) 1 σ determined using gradient and correct SI unit given (N m–1 or kg s–2). 1 Absolute uncertainty in σ determined with correct substitution shown. σ   Δ = + ×     gradient uncertainty 100 gradient (d)(ii) or σ   Δ = + + ×     2 2 gradient uncertainty 184 146 gradient or using σ × × = max 9.81 max max 4 (d)(i) (c)(iii) or using σ × × = min 9.81 min min 4 (d)(i) (c)(iii) 1 2(f) d determined to a minimum of two significant figures. gradient d h = or × = × × 4 9.81 d h (e)(i) (d)(i) 1

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