Cambridge A Level Physics 9702 — 2024 May/June Paper 5 · Variant 2

1 A spring is attached to a strong cylindrical magnet of length L and cross-sectional area A. The magnet is placed on thin card on top of a magnetic sheet on the bench, as shown in Fig. 1.1. spring cylindrical magnet magnetic sheet card bench t Fig. 1.1 The thickness of the card is t. The magnetic flux density at one of the poles of the magnet is B. A force is applied upwards to the spring. The extension of the spring when the magnet just leaves the card is s. It is suggested that s is related to t by the relationship ALBZ ks = t where k is the spring constant of the spring and Z is a constant. Plan a laboratory experiment to test the relationship between s and t. Draw a diagram showing the arrangement of your equipment. Explain how the results could be used to determine a value for Z. 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: 1 Defining the problem t is the independent variable and s is the dependent variable or vary t and measure s 1 keep k constant 1 Methods of data collection labelled diagram of workable experiment including:  spring connected to magnet  vertical rule parallel to spring to determine s  rule held in position by a stand  stand resting on the bench  rule labelled and at least one other label from stand, clamp, card, (magnetic) sheet, (cylindrical) magnet, spring 1 s = (new) length/position of spring – original length/position of spring 1 use a micrometer to measure t 1 measure B using a (calibrated) Hall probe and rotate probe until maximum value or measure B using Hall probe first in one direction, then in the opposite direction and average 1 Question Answer Marks 1 Method of Analysis plot a graph of s against 1 t or equivalent (allow lg s against lg t) 1 relationship valid if a straight line that passes through the origin is produced (for lg s against lg t: relationship valid if a straight line with gradient 1) 1 gradient k Z ALB   (for lg s against lg t: -intercept 10y k Z ALB   ) 1 Question Answer Marks 1 Additional detail including safety considerations 6 D1 precaution related to spring and/or magnet hitting eyes, e.g. use of goggles/use of safety screen around experiment D2 keep A, L and B constant D3 use a rule to measure L D4 micrometer/calipers to measure diameter d of the magnet and A = d2 / 4 D5 description of method to determine k, e.g. add mass to spring and k = mg / extension or use newton meter to measure force applied to spring and k = force / extension or take several readings of force and extension, plot a force–extension graph and k = gradient D6 (magnetic) sheet clamped to bench D7 use pointer(s)/marker(s) on the spring to read off values from the rule D8 method to use video recorder and replay to determine maximum length of the spring or increase s or force gradually/slowly until magnet (just) leaves the card D9 repeat measurements of t in different positions on the card and average t or repeat measurements of s for each value of t and average s D10 method to check that the spring has not exceeded the elastic limit D11 use of non-magnetic stand or named non-magnetic material for stand, e.g. wood

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2 A student investigates the resonant frequency of a metal rod. The metal rod of length L is suspended from two rubber loops. A sensitive microphone with a cone is positioned at one end of the rod. The microphone is attached to an oscilloscope, as shown in Fig. 2.1. cone rubber loop microphone to oscilloscope hammer rod stand bench Fig. 2.1 (not to scale) The rod is hit gently with a hammer. The period T of the trace produced on the oscilloscope is determined. The experiment is repeated for different values of L. It is suggested that T and L are related by the equation 2Ln T = C where C and n are constants. (a) A graph is plotted of lg T on the y-axis against lg L on the x-axis. Determine expressions for the gradient and y-intercept. gradient = ............................................................... y-intercept = ............................................................... [1] (b) Values of L and T are given in Table 2.1. Table 2.1 L / cm T / 10–5 s lg (L / cm) lg (T / 10–5 s) 54 24 ± 1 70 32 ± 1 86 39 ± 1 108 49 ± 2 140 64 ± 2 167 74 ± 2 Calculate and record values of lg (L / cm) and lg (T / 10–5 s) in Table 2.1. Include the absolute uncertainties in lg T. [2] (c) (i) Plot a graph of lg (T / 10–5 s) against lg (L / cm). Include error bars for lg T. [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 y-intercept = lg 2 C 1 2(b) lg (L / cm) lg (T / 10–5 s) 1.73 or 1.732 1.38 or 1.380  0.02 1.85 or 1.845 1.51 or 1.505  0.01 1.93 or 1.934 1.59 or 1.591  0.01 2.033 or 2.0334 1.69 or 1.690  0.02 2.146 or 2.1461 1.81 or 1.806  0.01 2.223 or 2.2227 1.87 or 1.869  0.01 Values of lg (L / cm) and lg (T/ 10–5 s) correct as shown above. 1 Uncertainties in lg (T/ 10–5 s) correct as shown above. 1 2(c)(i) Six points from (b) plotted correctly. Must be within half a small square. Diameter of points must be less than half a small square. 1 Error bars in lg T plotted correctly. All error bars to 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) Straight line of best fit drawn. Do not accept line from top point to bottom point. Line must pass between (1.780, 1.45) and (1.800, 1.45) and between (2.085, 1.75) and (2.100, 1.75) 1 Worst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). All error bars must be plotted. 1 2(c)(iii) Gradient determined with clear substitution of data points into y / x. Distance between data points must be greater than half the length of the drawn line. 1 Gradient determined of worst acceptable line with clear substitution of data points into y / x. 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 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. 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 accept ECF from false origin method. 1 Question Answer Marks 2(d) Value of n determined using gradient (n = gradient) and C given to 2 or 3 significant figures. 1 Value of C determined using y-intercept with method shown. -intercept 2 10y C  1 Absolute uncertainties in n and C. uncertainty in n = uncertainty in gradient and   worst y-intercept 2 = 10 C C or   min worst -intercept max worst -intercept 2 2 10 10 = 2 y y C Clear method must be shown with C correctly evaluated. 1 2(e) Value of L determined (non-zero) to a minimum of 2 significant figures from (c)(iii) and (c)(iv) or (d) with correct substitution and correct power of ten. Units of T and either C or y-intercept must be consistent. 2 log log log10 -intercept log T y C L n n     log10 -intercept 10 y n L   or 2 n TC L  1

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