Cambridge A Level Physics 9702 — 2018 Feb/March Paper 5 · Variant 2

1 A student is investigating how the extension e of an elastic cord depends on the diameter d of the cord when a force is applied. The student has a number of elastic cords of the same material with different diameters. The elastic cords have circular cross-sections. Each cord has an unstretched length of approximately 50 cm. It is suggested that the extension e and the cross-sectional area A of the cord are related by the expression FL = E Ae where E is the Young modulus of the material of the cord, F is the force applied and L is the unstretched length of the cord. Design a laboratory experiment to test the relationship between e and d. Explain how your results could be used to determine a value for E. 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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Mark scheme: 1 Defining the problem d is the independent variable and e is the dependent variable, or vary d and measure e. 1 Keep F constant. 1 Methods of data collection Labelled workable diagram including elastic cord fixed at one end to a support and load attached to the other end Cord and weight must be labelled 1 Use of ruler to measure unstretched length and stretched length. or labelled ruler drawn parallel to cord and original length L and either e or stretched length indicated. 1 Use of a micrometer / (calipers) to determine d. 1 Weigh load on a balance or use of balance to measure mass of load and multiply by g 1 Method of Analysis Plots a graph of e against 1 / d2 or equivalent 1 Relationship valid if a straight line passing through the origin is produced 1 4FL E gradient π = × 1 Additional detail including safety considerations Max 6 Use safety goggles / safety screen to prevent injury to eyes from (moving) elastic cord / load or use cushion / sand box in case load falls. D1 Keep L constant D2 Method to keep L constant, e.g. check length of each cord / adjust through cork D3 Question Answer Marks 1 Additional detail on measuring e, e.g. record initial position, record final position and subtract D4 Repeat measurement of d along cord / different diameters and average D5 Method to ensure Hooke’s law is obeyed, e.g. check that the length is constant after removing load or do not exceed elastic limit D6 Wait for cord to extend to its maximum value / stop oscillating D7 Use of 2 4 d A π = D8 Use of set square to check that ruler is vertical to bench, or use of set square as a fiducial mark to read measurements D9

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2 A student is investigating monochromatic light passing through a double slit. Bright and dark fringes are produced on a screen as shown in Fig. 2.1. w Fig. 2.1 The distance w between 10 bright fringes is measured. The fringe spacing P between neighbouring bright fringes is then determined. The experiment is repeated for light of different wavelengths m. It is suggested that the fringe spacing P and the wavelength m are related by the equation P m = D s where D is the distance from the double slit to the screen and s is the slit separation. (a) A graph is plotted of P on the y-axis against m on the x-axis. Determine an expression for the gradient. = gradient ...........................................................[1] (b) Values of m and w are given in Fig. 2.2. m / 10–7 m w / mm P / mm 4.3 39.5 ± 0.5 4.8 43.5 ± 0.5 5.3 48.0 ± 0.5 5.8 52.0 ± 0.5 6.2 55.5 ± 0.5 6.6 59.0 ± 0.5 Fig. 2.2 Calculate and record values of P / mm in Fig. 2.2. Include the absolute uncertainties in P. [2] (c) (i) Plot a graph of P / mm against m / 10–7 m. Include the error bars for P. [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 = D S 1 2(b) 3.95 4.35 4.80 5.20 5.55 5.90 1 Absolute uncertainties in P ± 0.05 1 Question Answer Marks 2(c)(i) Six points plotted correctly. Must be within half a small square. Diameter of points must be less than half a small square. 1 Error bars in P 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 2(c)(ii) Line of best fit drawn. Points must be balanced. Line must pass between (6.45, 5.8) and (6.55, 5.8) and between (4.55, 4.2) and (4.65, 4.2) 1 Worst acceptable line drawn. Steepest or shallowest possible line that passes through all the error bars. Mark scored only if all error bars are plotted. 1 2(c)(iii) Gradient determined with clear substitution of data points into ∆y/∆x; distance between data points must be at least half the length of the drawn line. 1 Gradient determined of WAL 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) Substitution of gradient to determine s 2.20 gradient D s = = (c)(iii) 1 s determined using gradient, given to 2 or 3 significant figures 1 s determined using gradient and correct unit and correct power of ten 1 Question Answer Marks 2(d)(ii) Percentage uncertainty in s %uncertainty in D + %uncertainty in gradient 0.02 gradient % 100 2.20 gradient s   ∆ = + ×     gradient % 0.91 100 gradient s ∆ = + × Or correct maximum/minimum method max max mingradient D s = or min min max gradient D s = 1 Question Answer Marks 2(e) Correct substitution of numbers must be seen, λ in the range 7 4.05 10 m − × to 7 4.24 10 m − × 3 3 3.5 10 3.5 10 gradient λ − − × × = = (c)(iii) OR 3 3 s 3.5 10 3.5 10 2.2 D λ − − = × × = × × (d)(i) 1 Correct substitution of numbers must be seen, Determines absolute uncertainty in λ. Using (c)(iii) 0.05 gradient uncertainty 3.50 gradient λ   ∆ = + ×     or 3 3.55 10 max mingradient λ − × = , or 3 3.45 10 min max gradient λ − × = , OR Using (d) 0.05 0.02 uncertainty 3.50 2.20 100 λ   = + + ×     (d)(i) uncertainty 0.0234 100 λ   = + ×     (d)(i) or 3 3.55 10 max max 2.18 s λ − × × = , or 3 3.45 10 min min 2.22 s λ − × × = . 1

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