Cambridge A Level Physics 9702 — 2011 Oct/Nov Paper 5 · Variant 3

1 A changing e.m.f. in a coil can induce an e.m.f. in another coil. For Examiner’s Fig. 1.1 shows a coil (coil X), which is wound on a cardboard tube. Use cardboard tube coil X Fig. 1.1 Coil X has cross-sectional area A. A student winds another coil (coil Y) tightly around coil X. The student wishes to investigate how the e.m.f. V in coil Y depends on A. It is suggested that V is directly proportional to A. Design a laboratory experiment to investigate the suggested relationship. You should draw, on page 3, a diagram 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 For problem data collection analysis considerations detail Examiner’s Use

Mark scheme: 1 Planning (15 marks) Defining the problem (3 marks) P A is the independent variable and V is the dependent variable or vary A and measure V. [1] P Keep the number of turns on coil Y or coil X constant. [1] P Keep the current in coil X constant. [1] Methods of data collection (5 marks) M1 Two independent coils labelled X and Y; coil Y wound over coil X. [1] M2 Alternating power supply/signal generator connected to coil X. [1] M3 Coil Y connected to voltmeter/c.r.o. in a workable circuit. [1] M4 Measure diameter/radius/lengths with a ruler/vernier callipers. [1] M5 Method to determine area. [1] Method of analysis (2 marks) A Plot a graph of V against A. [1] A Relationship valid if straight line through origin. [1] Safety considerations (1 mark) S Precaution linked to (large) current in coil/heating, e.g. switch off when not in use to avoid overheating coil; do not touch coil because it is hot. [1] Additional detail (4 marks) D Relevant points might include [4] 1 Use large current in coil X/large number of turns/high frequency a.c. to produce measurable e.m.f. 2 Detail on measuring e.m.f., e.g. height × y-gain on CRO. 3 Keep frequency of power supply constant. 4 Use of rheostat to keep current constant in coil X. 5 Monitor with a.c. ammeter. 6 Avoid other alternating magnetic fields. 7 Repeat measurement for r or d or lengths and average. Do not allow vague computer methods. [Total: 15] GCE AS/A LEVEL – October/November 2011 9702 53

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2 A student is investigating how a mass attached to a trolley affects the motion of the trolley. For Examiner’s The trolley is attached to a mass m by a string, passing over a pulley, as shown in Fig. 2.1. Use 0.200 m trolley card mass M pulley m h Fig. 2.1 A piece of card of length 0.200 m is fixed to the trolley. The mass M of the trolley and card is 0.800 kg. The mass m is released and falls through a fixed height h, accelerating the trolley. When the mass m hits the ground, the trolley continues to move with constant velocity v. This velocity v is determined by measuring the time t for the card to pass fully through a light gate connected to a timer. Question 2 continues on the next page. It is suggested that v and m are related by the equation For v 2 Examiner’sUse mg = (m + M ) 2h where g is the acceleration of free fall. 2 m(a) A graph is plotted of v on the y-axis against on the x-axis. Express the gradient m + M in terms of g. gradient = ................................................. [1] (b) Values of m and t are given in Fig. 2.2. m m / kg t / 10–3 s v / m s–1 v 2 / m2 s–2 m + M 0.100 174 ± 2 0.200 132 ± 2 0.300 112 ± 2 0.400 102 ± 2 0.500 95 ± 2 0.600 90 ± 2 Fig. 2.2 m Calculate and record values of , v and v 2 in Fig. 2.2. Include the absolute m + M uncertainties in v 2. [3] m (c) (i) Plot a graph of v 2 / m2 s–2 against . Include error bars for v 2. [2] m + M (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] 5.4 Examiner’sFor Use 5.0 v 2 / m2 s–2 4.6 4.2 3.8 3.4 3.0 2.6 2.2 1.8 1.4 1.0 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 m m + M

Mark scheme: 2 Analysis, conclusions and evaluation (15 marks) Part Mark Expected Answer Additional Guidance (a) A1 2gh Allow 2hg. (b) T1 T1 for ratio values: Ignore sf in 2nd row. 0.111 or 0.1111 1.320 – 1.321 T2 T2 for v 2. 0.200 or 0.2000 2.295 – 2.310 Rows 1–4 to 3 s.f. or 4 s.f. 0.273 or 0.2727 3.189 – 3.204 Rows 5–6 to 2 s.f. or 3 s.f. 0.333 or 0.3333 3.842 – 3.846 0.385 or 0.3846 4.41 – 4.45 0.429 or 0.4286 4.84 – 4.94 U1 From ± 0.02 or ± 0.03, to ± 0.2 Allow more than one significant figure. (c) (i) G1 Six points plotted correctly Must be within half a small square. Do not allow ‘blobs’ (more than half a small square). Ecf allowed from table. U2 Error bars in v2 plotted correctly All error bars to be plotted. Check third and fourth plot. 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 (0.10, 1.16) and (0.10, 1.24) and upper end of line should pass between (0.45, 5.12) and (0.45, 5.20). Allow ecf from points plotted incorrectly – examiner judgement. G3 Worst acceptable straight line. Line should be clearly labelled or dashed. Steepest or shallowest possible line Should pass from top of top error bar to that passes through all the error bars. bottom of bottom error bar or bottom of top error bar to top of bottom error bar. Mark scored only if error 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. U3 Uncertainty in gradient Method of determining absolute uncertainty. Difference in worst gradient and gradient. (d) C2 g = gradient / 2h = gradient / 1.2 Gradient must be used. Allow ecf from (c)(iii). U4 Absolute uncertainty in g Uses worst gradient. Do not check calculation. (e) C3 Ratio = 0.6/(0.6 + 1.8) = 0.25 Expect to see 1.00 added and largest m. C4 Between 1.66 and 1.70 given to 2 or v = 2 × 0.25 × g × 0.6 = 0 .3 × g 3 s.f. or v = gradient × 0.25 or v = v 2 read from graph for ratio 0.25. Must be in range. Allow 1.7. U5 Determines absolute uncertainty Allow ecf. Expect to see difference between best and worst values. [Total: 15] GCE AS/A LEVEL – October/November 2011 9702 53 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) [U4] Uncertainty = best g – worst g Uncertainty = uncertainty in gradient / 1.2 ∆m Uncertainty = g m (e) [U5] Uncertainty = best v – worst v 1 ∆m Uncertainty = × v 2 m 1 ∆ g Uncertainty = × v 2 g

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