Cambridge A Level Physics 9702 — 2021 May/June Paper 5 · Variant 3

1 A student investigates the current in a coil and a resistor connected in series, as shown in Fig. 1.1. coil R Fig. 1.1 The student connects a high-voltage d.c. power supply and a switch across the series combination. When the switch is closed, it takes time t for the current in the resistor of resistance R to reach a maximum value. The time t is a few milliseconds. There are a number of different unmarked resistors available. It is suggested that the relationship between t and R is KN2A t = LR where N is the number of turns of wire on the coil, A is the cross-sectional area of the coil, L is the length of the coil and K is a constant. Design a laboratory experiment to test the relationship between t and R. 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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[15]

Mark scheme: 1 Defining the problem R is the independent variable and t is the dependent variable or vary R and measure t 1 keep the number of turns on the coil/N constant 1 Methods of data collection labelled diagram or correct symbols including: • labelled (d.c.) power supply • switch in series with power supply, resistor and coil • complete workable circuit 1 circuit diagram to measure R, e.g. ammeter and voltmeter correctly positioned or R connected to ohmmeter with no other connections (not ohmmeter in main circuit) 1 method to determine t (of a few milliseconds) e.g. use (storage) oscilloscope or current/voltage sensor connected to datalogger/computer 1 method to determine A, e.g. micrometer/calipers to determine diameter of coil and A = πd2 / 4 1 Method of analysis plot a graph of t against 1 / R (allow log t against log R) 1 relationship valid if a straight line passing through the origin is produced (allow gradient = –1 for graph of log t against log R) 1 2 gradient . L K AN × = 1 Question Answer Marks 1 Additional detail including safety considerations 6 D1 open switch/switch off (high voltage) circuit before changing the resistor/touching components or ensure no bare wires/use shrouded connectors D2 wear (insulating) gloves to prevent electric shock/electrocution D3 keep A and L constant D4 use ruler/calipers to measure L D5 repeat measurements of diameter in different directions/at points along the coil and average D6 method to determine R e.g. R = V / I linked to correct circuit diagram for ammeter/voltmeter method or measure resistance using ohmmeter D7 repeat experiment for each value of R and average t D8 method to determine t: use of time-base from oscilloscope explained or use of time axis of output from data logger/computer explained D9 use smaller values of R to increase I D10 reduce L or increase N or increase A to increase t

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2 A student investigates the collision of two gliders A and B on a linear air-track. A card is attached to glider B, as shown in Fig. 2.1. card light gate connected to data logger glider B glider A air-track bench Fig. 2.1 Glider B has a mass M. A mass m is added to glider B. Glider A travels at a constant velocity u towards the stationary glider B. The gliders then collide and move together towards the light gate. The card passes through the light gate which is connected to a data logger. The student records the velocity v of the two gliders from the data logger. The student changes the mass m and repeats the experiment. It is suggested that v and m are related by the equation Au = (M + m + A)v where A is the mass of glider A. 1 (a) A graph is plotted of on the y-axis against (M + m) on the x-axis. v Determine expressions for the gradient and y-intercept. gradient = ............................................................... y-intercept = ............................................................... [1] (b) Values of m and v are given in Table 2.1. The value of M is 330 g ± 5%. Each value of m has a percentage uncertainty of ± 5%. Table 2.1 1 m / g (M + m) / g v / cm s–1 / s cm–1 v 50 4.42 150 3.92 250 3.40 350 3.02 500 2.58 600 2.33 1 Calculate and record values of (M + m) / g and / s cm–1 in Table 2.1. v Include the absolute uncertainties in (M + m). [2] 1 (c) (i) Plot a graph of / s cm–1 against (M + m) / g. v Include error bars for (M + m). [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 = 1 uA y-intercept = 1 u 1 2(b) (M + m) / g 1 v / s cm–1 380 0.226 or 0.2262 480 0.255 or 0.2551 580 0.294 or 0.2941 680 0.331 or 0.3311 830 0.388 or 0.3876 930 0.429 or 0.4292 Values of (M + m) and 1 v as shown above. 1 Absolute uncertainties in (M + m) from ± (19 or 20) to ± (46.5 or 47 or 50). 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 (M + m) 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 Question Answer Marks 2(c)(ii) Line of best fit drawn covers all points. Points must be balanced. Do not allow line from top point to bottom point. Line must pass between (425, 0.240) and (440, 0.240) and between (850, 0.400) and (865, 0.400). 1 Worst acceptable line drawn (steepest or shallowest possible line that passes through all error bars). All error bars must be plotted. 1 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(c)(iv) y-intercept determined by substitution of correct point into y = mx + c. 1 y-intercept of worst acceptable line determined by substitution into y = mx + c. 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 ECF from false origin method. 1 2(d)(i) u determined using y-intercept and u and A given to two or three significant figures. 1 -intercept u y = 1 A determined using gradient with correct substitution and units with correct power of ten for u and A. -intercept 1 gradient gradient y A A u = = × or 1 Question Answer Marks 2(d)(ii) Percentage uncertainty in A determined, e.g. gradient -intercept percentage uncertainty in gradient -intercept y A y   Δ Δ = +     or Δu clearly determined using the value of u and gradient percentage uncertainty in 100 gradient u A u   Δ Δ = + ×     or correct substitution for max/min methods e.g. 1 max min min gradient A u = × 1 min max max gradient A u = × 1 2(e) Value of m determined from (d)(i) or (c)(iii) and (c)(iv), with correct number substitution and correct power of ten. ( ) 330 2 A u m A × = − + or 0.5 -intercept 330 gradient y m − = − 1

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