Cambridge A Level Physics 9702 — 2025 May/June Paper 5 · Variant 4

1 A ball is dropped on to an inclined thin metal sheet, as shown in Fig. 1.1. ball sheet bench z θ d Fig. 1.1 (not to scale) The angle between the sheet and the horizontal bench is θ. The height of the point of contact of the ball and the sheet is z. The horizontal distance travelled by the ball between its points of contact with the sheet and the bench is d, as shown in Fig. 1.1. It is suggested that d is related to θ by the relationship Pv 2 sin 4θ d = + Q z g where v is the speed of the ball as it makes contact with the sheet, g is the acceleration of free fall, and P and Q are constants. Plan a laboratory experiment to test the relationship between d and θ. Draw a diagram showing the arrangement of your equipment. Explain how the results could be used to determine values for P and Q. 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. 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Mark scheme: Question Answer Marks 1 Defining the problem vary and measure d or is the independent variable and d is the dependent variable 1 keep z constant 1 Methods of data collection labelled diagram of workable experiment including: 1 • sheet supported with stand or block(s) • path of ball (from sheet to bench) or distance d indicated • at least two labels from ball, bench, sheet, block/clamp/stand/support description of method to determine position of ball as it makes contact with the bench, e.g. paint ball, sand, video and 1 playback in slow motion / frame by frame (with rule in view) measure d with a rule 1 and measure z with a rule or calipers measure with a protractor 1 or determine using (a rule for) appropriate distances with correct trigonometric relationship Method of analysis plot a graph of d against sin (4) or equivalent 1 Do not accept logarithms. 1 1 d against sin (4) sin (4) against d g  gradient g P = P = v 2 v 2  gradient 1 d against sin (4) sin (4) against d y -intercept Pv 2  y -intercept Q = Q = − z g z or y -intercept Q = − gradient  z Additional detail including safety considerations 6 D1 precaution linked to prevent ball rolling (off surface), e.g. use of cushion / sand box / barrier to stop ball (rolling onto the floor) or precaution linked to (bouncing) ball or sand (spray) e.g. use of goggles to protect eyes D2 keep v constant D3 method to keep v constant, e.g. keep the distance the ball falls constant to keep v constant D4 method to determine v or v2, e.g. use of v = 2g  ( height ball falls ) or v 2 = 2g  ( height ball falls ) 1 D5 set square correctly positioned between rule and bench to ensure that rule is vertical to measure z or determine initial height of ball D6 method to ensure z is constant, e.g. as is changed, (re-)mark point of contact or change release point of ball D7 description of method to ensure ball is released above point of contact on sheet, e.g. use a plumb line from the ball through point of contact with sheet or (vertical) rule (and set square) D8 drop ball from large height to obtain greater values of d D9 repeat experiment for each and average d D10 relationship valid if a straight line produced (with y-intercept of e.g. Q Z ). ( ) Do not accept line passing through the origin.

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2 A student investigates the relationship between the luminosity of a star and its mass. The student obtains data of relative luminosity λ and relative mass µ for six stars, where luminosity of star λ = luminosity of Sun and mass of star µ = mass of Sun. It is suggested that λ and µ are related by the equation λ = kµn where k and n are constants. (a) A graph is plotted of lg λ on the y-axis against lg µ on the x-axis. Determine expressions for the gradient and y-intercept. gradient = ............................................................... y-intercept = ............................................................... [1] (b) Values of µ and λ are given in Table 2.1. Table 2.1 µ λ lg µ lg λ 4.6 ± 0.4 500 5.4 ± 0.4 800 8.4 ± 0.4 3200 11 ± 1 7000 16 ± 1 25 000 18 ± 1 38 000 Calculate and record values of lg µ and lg λ in Table 2.1. Include the absolute uncertainties in lg µ. [2] (c) (i) Plot a graph of lg λ against lg µ. Include error bars for lg µ. [2] (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]

Mark scheme: 2(a) gradient = n 1 y-intercept = lg k 2(b) 1 lg  lg  0.66 or 0.663  0.04 2.70 or 2.699 0.73 or 0.732  0.03 2.90 or 2.903 0.92 or 0.924  0.02 3.51 or 3.505 1.04 or 1.041  0.04 3.85 or 3.845 1.20 or 1.204  0.03 4.40 or 4.398 1.26 or 1.255  0.02 or  0.03 4.58 or 4.580 Values of lg and lg correct as shown above. Uncertainties in lg 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 lg 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) Straight line of best fit drawn. 1 Thickness of the line must be less than half a small square. Do not accept line from top point to bottom point. Line must pass between (0.82, 3.20) and (0.84, 3.20) and between (1.13, 4.20) and (1.16, 4.20). Worst acceptable straight line drawn (steepest or shallowest possible line that passes through all the error bars). 1 Thickness of the line must be less than half a small square. All error bars must be plotted. 2(c)(iii) Gradient determined with clear substitution of data points into y / x. 1 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 y / x. 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 x 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 methods using a false origin. 2(d) Value of k determined using y-intercept. 1 k = 10y −intercept n = gradient and n and k given to 2 or 3 significant figures. 1 absolute uncertainty in n = absolute uncertainty in gradient 1 and absolute uncertainty in k = (10y-intercept – 10y-intercept of WAL) Correct method must be seen. 2(e) M determined to a minimum of 2 significant figures from (d) or (c)(iii) and (c)(iv) with correct substitution and correct 1 power(s) of ten. Do not accept incorrect POT for n or k. Correct substitution must be seen.  0.46 lg0.46 − y -intercept lg0.46 − lg k = n = gradient or lg = or lg = k (d) gradient gradient and M =  2.0  1030

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