Cambridge A Level Physics 9702 — 2014 Oct/Nov Paper 5 · Variant 3

1 A thin card is inserted between two separate iron cores. A coil is wound around one core as shown in Fig. 1.1. thin card iron cores Fig. 1.1 A current in the coil may induce an e.m.f. in another coil wound on the other core. The induced e.m.f. V depends on the thickness t of the card. A student suggests that V = V0e–σt where V0 is the induced e.m.f. without card between the cores and σ is a constant. Design a laboratory experiment to test the relationship between V and t and determine the value of σ. You should draw a diagram, on page 3, showing the arrangement of your equipment. In your account you should pay particular attention to (a) the procedure to be followed, (b) the measurements to be taken, (c) the control of variables, (d) the analysis of the data, (e) the safety precautions to be taken. 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Defining the Methods of Method of Safety Additional problem data collection analysis considerations detail

Mark scheme: 1 Planning (15 marks) Defining the problem (3 marks) P t is the independent variable or vary t. [1] P V is the dependent variable or measure V. [1] P Keep the current (in the primary coil) constant. [1] Methods of data collection (5 marks) M Diagram showing two independent labelled coils wound on iron cores. [1] M AC power supply / signal generator connected to one coil. [1] M Voltmeter / oscilloscope connected to other coil in a workable circuit. [1] M Measure thickness of card using micrometer / vernier calipers / digital calipers. [1] M Method to keep current constant – rheostat (or variable power supply) and ammeter correctly positioned in primary circuit and explained. Diagram and text required. [1] Method of analysis (2 marks) M Plot a graph of ln V against t (allow lg V against t) or ln V / V0 against t [1] M σ = – gradient [1] Safety considerations (1 mark) S Precaution linked to hot coil(s) e.g. switch off when not in use / do not touch / wear gloves. [1] Additional detail (4 marks) D Relevant points might include [4] 1 Use large current (in primary coil)/large number of turns on the secondary to achieve measurable V (allow more turns on secondary than primary). 2 Keep frequency of power supply constant or keep the number of turns on each coil constant. 3 Use laminated cores or use insulated wire for turns. 4 Repeat measurements of t and average. 5 Measurement of V0 stating that no card is present. 6 Logarithmic equation: ln V = ln V0 – σt 7 Relationship is valid if the graph is a straight line with y–intercept = ln V0 8 Discussion of compression of card / measure t when secured. Do not allow vague computer methods. [Total: 15]

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2 A student investigates how the final velocity v of a cylinder rolling down a board varies with the height h of the board as shown in Fig. 2.1. cylinder h board v Fig. 2.1 For different values of h, the velocity v is determined using a light sensor connected to a data logger. It is suggested that v and h are related by the equation 2gh = v 2Z where g is the acceleration of free fall and Z is a constant. (a) A graph is plotted of v 2 on the y-axis against h on the x-axis. Determine an expression for the gradient in terms of g and Z. gradient = ................................................. [1] (b) Values of h and v are given in Fig. 2.2. h / m v / m s–1 0.230 1.40 ± 0.05 0.280 1.55 ± 0.05 0.320 1.65 ± 0.05 0.360 1.75 ± 0.05 0.400 1.85 ± 0.05 0.450 1.95 ± 0.05 Fig. 2.2 Calculate and record values of v 2 / m2 s–2 in Fig. 2.2. Include the absolute uncertainties in v 2. [3] (c) (i) Plot a graph of v 2 / m2 s–2 against h / m. Include error bars for v 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 uncertainty in your answer. gradient = ................................................. [2] 4.0 3.8 v 2 / m2 s–2 3.6 3.4 3.2 3.0 2.8 2.6 2.4 2.2 2.0 1.8 0.22 0.26 0.30 0.34 0.38 0.42 0.46 h / m

Mark scheme: 2 Analysis, conclusions and evaluation (15 marks) Mark Expected Answer Additional Guidance (a) A1 2g gradient = Z (b) T1 v2 / m2 s–2 Allow v2 (m2 s–2) T2 Must be table values. 1.96 or 1.960 Allow a mixture of significant figures. 2.40 or 2.403 2.72 or 2.723 3.06 or 3.063 3.42 or 3.423 3.80 or 3.803 U1 From ± 0.1 to ± 0.2 with Allow more than one significant figure. uncertainties increasing (c) (i) G1 Six points plotted correctly Must be within half a small square. Penalise “blobs”. Ecf allowed from table. U2 Error bars in v2 plotted correctly All error bars to be plotted. Must be accurate to less than half a small square. (c) (ii) G2 Line of best fit If points are plotted correctly then lower end of line should pass between (0.24, 2.04) and (0.24, 2.10) and upper end of line should pass between (0.42, 3.54) and (0.42, 3.60). Line should not be from top point to bottom point. G3 Worst acceptable straight line. Line should be clearly labelled or dashed. Steepest or shallowest possible Examiner judgement on worst acceptable line. line that passes through all the Lines must cross. Mark scored only if all error error bars. bars are plotted. (c) (iii) C1 Gradient of best fit line The triangle used should be at least half the length of the drawn line. Check the read-offs. Work to half a small square. Do not penalise POT. (Should be about 8.4.) U3 Uncertainty in gradient Method of determining absolute uncertainty: difference in worst gradient and gradient. (d) (i) C2 Gradient must be used (no substitution v = gradient × h methods). Should be between 2.39 and 2.46. (ii) U4 1 ∆ gradient ∆ v Allow ecf from (d)(i). × 100 = × 100 2 gradient v (e) C3 K in the range 7.20 × 10–4 to 2  2g  7.80 × 10–4 and given to K = mr  − 1  gradient  2 or 3 s.f. C4 kg m2 U5 Absolute uncertainty in K Uses worst gradient. Allow ecf. [Total: 15] Uncertainties in Question 2 (c) (iii) Gradient [U3] Uncertainty = gradient of line of best fit – gradient of worst acceptable line Uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient) (d) (ii) [U4] max v = max gradient × h min v = min gradient × h (e) [U5] 2  2g  max K = mr  −1  min gradient  2  2g  min K = mr  −1  max gradient 

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