Cambridge A Level Physics 9702 — 2025 Oct/Nov Paper 5 · Variant 4

1 Fig. 1.1 shows a horizontal turntable. turntable P C r Fig. 1.1 Point C is at the centre of the turntable. Point P is a distance r from the centre. Fig. 1.2 shows a side view of a d.c. motor attached to the turntable with a belt. terminals turntable C belt motor turntable base Fig. 1.2 The motor is used to rotate the turntable at frequency f0. The motor is switched off and the turntable continues to rotate at frequency f0. A sphere of adhesive putty of mass m is dropped onto the turntable at point P. The frequency of the turntable is now f. It is suggested that f is related to m by the relationship Kf0 = βK + mr 2 f where β and K are constants. Plan a laboratory experiment to test the relationship between f and m. Draw a diagram showing the arrangement of your equipment. Explain how the results could be used to determine values for β and K. 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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[15]

Mark scheme: Question Answer Marks 1 Defining the problem vary m and measure f or m is the independent variable and f is the dependent variable 1 keep f0 constant 1 Methods of data collection labelled diagram of workable experiment including: 1 • turntable base and motor on bench • two labels from turntable, motor, belt, C, putty, bench, release mechanism workable circuit showing terminals of motor connected to (d.c.) supply 1 method to measure time to determine f and f0, e.g. use a stopwatch to measure (n)t 1 and method to measure m, e.g. use a (top-pan) balance measure the time t for n revolutions 1 and T = t / n and f = 1 / T or f = n / t Method of Analysis 1 1 plots a graph of against m or equivalent f Do not accept logarithms. 1 1 1 1 against m m against f f r 2 r 2 × gradient K = K = f0 × gradient f0 1 1 1 against m m against f f = f0  y -intercept f0  y -intercept  = − gradient or r 2×y -intercept = − K 1 Additional detail including safety considerations 6 D1 precaution linked to putty leaving turntable, e.g. screens around apparatus / cushions to prevent putty from leaving bench D2 keep r constant D3 use rule(r) to measure r D4 stand and fixed point located above P (to ensure r is constant) D5 method to ensure putty is dropped at (constant) distance r, e.g.; fixed location above the turntable to drop putty to keep r constant or draw a circle of radius r on the turntable D6 method to keep f0 constant, e.g. adjust the rheostat to keep the current in the motor constant / check ammeter D7 clamp motor and/or turntable (base) to bench D8 lubricate the turntable to reduce friction D9 repeat experiment for the same value of m and determine the average f D10 relationship valid if a straight line is produced (with y-intercept = ) f0 Do not accept line through the origin.

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2 A student places a slide with a double slit on a support clamped to the bench as shown in Fig. 2.1. slide with D double slit light screen supports bench Fig. 2.1 The distance between the slide and the screen is D. The separation s of the slits is determined. Light from a laser is incident normally on the double slit. An interference pattern is observed on the screen. The distance w across 10 fringes is measured. The distance y between the centres of adjacent fringes is calculated using the equation w y = 10. The experiment is repeated with slides of different slit separation s. It is suggested that y and s are related by the equation sy λ = D where λ is the wavelength of the incident light. 1 (a) A graph is plotted of y on the y-axis against on the x-axis. s Determine an expression for the gradient. gradient = ......................................................... [1] (b) Values of s and w are given in Table 2.1. Table 2.1 s / mm 1 / mm–1 w / mm y / mm s 0.18 ± 0.01 33.0 0.21 ± 0.01 28.9 0.24 ± 0.01 25.1 0.27 ± 0.01 22.6 0.31 ± 0.01 19.6 0.38 ± 0.01 15.9 1 Calculate and record values of / mm–1 and y / mm in Table 2.1. Include the absolute s 1 uncertainties in s. [2] 1 1(c) (i) Plot a graph of y / mm against / mm–1. Include error bars for [2] s s. (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 = D 1 2(b) 1 1 / mm–1 y / mm s 5.6 or 5.56 3.30 4.8 or 4.76 2.89 4.2 or 4.17 2.51 3.7 or 3.70 2.26 3.2 or 3.23 1.96 2.6 or 2.63 1.59 1 Values of and y correct as shown above. s 1 1 Uncertainties in from  0.3 to  0.07. s 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. 1 1 Error bars in plotted correctly. s 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 (2.90, 1.80) and (3.00, 1.80) and between (5.30, 3.20) and (5.40, 3.20). Worst acceptable 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(d) (0.921  0.008) (m) 1 2(e)(i) determined using gradient and given to 2 or 3 significant figures. 1 gradient = D determined using gradient and given with SI units with appropriate power of ten. 1 gradient = D Unit of : m or mm. 2(e)(ii) Percentage uncertainty in determined with clear method shown. 1  gradient D  %uncertainty =  +   100  gradient D  or correct substitution for max/min methods. 2(f) s determined to a minimum of 2 significant figures from (c)(iii) or (d) and (e)(i) with correct substitution and correct power of 1 ten. gradient s = 0.005 (m) or D   (d)  (e)(i) s = = 0.005 (m) 0.005 (m) Absolute uncertainty determined with correct substitution. 1 For use of gradient from (c)(iii):  gradient 0.005  =s  +   s  gradient 0.500  or correct substitution for max/min methods. For use of D and from (d) and (e)(i):  D  0.005  =s + +  s    D  0. 500  or correct substitution for max/min methods:  max D  max   min D  min  =s   − s or =s s −   min y   max y 

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