Cambridge A Level Physics 9702 — 2019 May/June Paper 5 · Variant 2

1 A student is investigating the stability of a wooden block resting on a bench. A strip is attached by a nail to the centre of the top of the block and is able to rotate, as shown in Fig. 1.1 and Fig. 1.2. nail strip P strip nail θ block block bench SIDE VIEW TOP VIEW w Fig. 1.1 Fig. 1.2 A load of mass m is attached to the free end of the strip at point P. The student is investigating the position of the strip indicated by angle θ, as shown in Fig. 1.2, at which the block just topples. It is suggested that the relationship between m and θ is αVw = 2mL cos θ – mw where αis a constant, V is the volume of the block, w is the width of the block and L is the distance between the centre of the nail and the centre of the load. Design a laboratory experiment to test the relationship between m and θ. Explain how your results could be used to determine a value for α. 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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[15]

Mark scheme: 1 Defining the problem m is the independent variable and θ is the dependent variable or vary m and measure θ or θ is the independent variable and m is the dependent variable or vary θ and measure m 1 keep position of load constant or L constant 1 Methods of data collection labelled diagram of workable experiment including: • load shown touching at P • load labelled and at least one other label 1 use a protractor to measure θ 1 if m is the independent variable: (slowly) change the angle until the block (just) topples or if θ is the independent variable: (slowly) change mass until block (just) topples 1 use a balance to measure m 1 Question Answer Marks 1 Method of analysis plot a graph of cos θ against 1 / m (or reverse axes 1 / m against cos θ) (not 1 / cos θ against m or m against 1 / cos θ) 1 relationship valid if a straight line (not ‘through the origin’ unless the choice of axes indicates the intercept is zero) 1 2 gradient L V w α × = × or gradient -intercept V y α = × for reverse axes: 2 1 gradient -intercept L V w V y α α = = − × × × or 1 Question Answer Marks 1 Additional detail including safety considerations Max. 6 D1 use cushion/foam/sandbox in case block/load falls D2 measure L with a rule D3 correctly positioned protractor to measure θ, e.g. protractor with its centre over the nail and its straight edge parallel to an edge of the block D4 method to fix m to strip D5 method to determine volume of block, e.g. V = w × h × l D6 measure w, h and l with calipers/micrometer/rule D7 (if m is the independent variable:) repeat experiment for θ and determine the average θ or (if θ is the independent variable:) repeat experiment for m and determine the average m D8 equation must be in the form of y = mx + c with cos θ on one side and m on the other side e.g. for a graph of cos θ against 1 / m cos 2 2 Vw w Lm L α θ = + or for a graph of 1 / m against cos θ 1 2 cos 1 L m Vw V θ α α = − D9 method to determine centre of load or centre of block e.g. measure diameter/width and halve or diagonals across the block D10 method to ensure that block or strip is horizontal, e.g. check with a spirit level that table/block is horizontal or use a rigid strip

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2 A student is investigating the oscillations of a mass attached to an arrangement of springs. Fig. 2.1 shows a mass attached to two springs connected in series. springs mass Fig. 2.1 The student determines the spring constant k for the arrangement of the springs. A stopwatch is used to measure the time t for 20 oscillations. The measurement of t is repeated and the average period T is determined. The experiment is repeated for different arrangements and different numbers of springs. It is suggested that T and k are related by the equation M T = 2 π k where M is the mass. 2 1 (a) A graph is plotted of T on the y-axis against on the x-axis. k Determine an expression for the gradient. gradient = ......................................................... [1] 1 (b) Values of k, and the measurements of t are given in Fig. 2.2. k 1 2k / N m–1 / m N–1 t / s t / s T / s T / s2 k 7.9 0.13 22.2 22.6 11 0.091 19.2 18.8 15 0.067 16.6 16.0 24 0.042 12.8 13.4 32 0.031 11.0 11.8 49 0.020 9.8 9.0 Fig. 2.2 Calculate and record values of T / s and T 2 / s2 in Fig. 2.2. Include the absolute uncertainties in T and T 2. [4] 2 1 (c) (i) Plot a graph of T / s2 against / m N–1. k Include error bars for T 2. [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) 1 2(b) T / s T2 / s2 1.12 or 1.120 1.25 or 1.254 0.950 or 0.9500 0.903 or 0.9025 0.815 or 0.8150 0.664 or 0.6642 0.655 or 0.6550 0.429 or 0.4290 0.570 or 0.5700 0.325 or 0.3249 0.47 or 0.470 0.22 or 0.221 Values of T as above. 1 Values of T2 as above. 1 Uncertainties in T increase from ±0.01 to ±0.02. 1 Uncertainties in T2 about ±0.02. 1 2(c)(i) Six points plotted correctly. Must be accurate to the nearest half a small square. Diameter of points must be less than half a small square. 1 Error bars in T2 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. If points are plotted correctly then lower end of line should pass between (0.048, 0.5) and (0.052, 0.5) and upper end of line should pass between (0.098, 1.0) and (0.104, 1.0). 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 points from the line of best fit into ∆y / ∆x. Distance between 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(d)(i) M determined from gradient and given to 2 or 3 significant figures and with correct unit. 2 gradient 39.478 4 M = = π (c)(iii) 1 2(d)(ii) % uncertainty in M = % uncertainty in gradient 1 Question Answer Marks 2(e) k calculated. Correct substitution of numbers required. 2 2 2 2 4 4 6.3165 2.5 M k T   π π = = ×     (d)(i) or (d)(i) or 2 2 gradient 6.25 2.5 k T   = =     (c)(iii) (c)(iii) or 1 Absolute uncertainty in k. Correct substitution of numbers required. Using M: uncertainty in 2 M T k k M T ∆ ∆   = + × ×     uncertainty in 0.008 100 k k   = + ×     (d)(ii) 2 2 2 2 4 max 4 min max min min max M M k k T T π × π × = = or Using gradient: gradient uncertainty in 0.008 gradient k k   ∆ = + ×     2 2 maxgradient mingradient max min min max k k T T = = or 1

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