Cambridge A Level Physics 9702 — 2015 Oct/Nov Paper 5 · Variant 3

1 A beaker contains water and some metal blocks as shown in Fig. 1.1. water heater metal block metal block Fig. 1.1 A student uses an electrical heater to produce a particular temperature increase in the water. It is suggested that the electrical energy E supplied to the heater is related to the mass m of metal blocks by the relationship E = am + b where a and b are constants. Design a laboratory experiment to test the relationship between E and m. Explain how your results could be used to determine values for a and b. 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 m is the independent variable and E is the dependent variable or vary m and measure E. Do not allow time. [1] P Keep the temperature change of water constant. Allow two specified temperatures. Do not allow “keep temperature constant”. [1] P Keep the mass or volume of water constant. [1] Methods of data collection (5 marks) M Labelled diagram including labelled thermometer with bulb in water and at least one other label. [1] M Workable circuit diagram to determine E: power supply, heater and ammeter and voltmeter, or joulemeter or wattmeter. [1] M Method to determine change in temperature: measure initial temperature, measure final temperature and subtract, or measure initial temperature and specific temperature change. [1] M Use balance/scales to measure mass of blocks. [1] M Stir water (so that metal is in thermal equilibrium). [1] Method of analysis (2 marks) A Plot a graph of E against m. Do not allow log–log graphs. [1] A a = gradient and b = y-intercept; must be consistent with suggested graph. [1] Safety considerations (1 mark) S Precaution linked to hot heater/water, e.g. use gloves or use tongs for hot blocks. Do not allow goggles. [1] Additional detail (4 marks) D Relevant points might include [4] 1 Method to ensure that e.m.f. of the power supply is constant/current in heater is constant, e.g. adjust variable power supply/variable resistor to ensure p.d./current is constant 2 Keep the starting temperature of water/metal constant 3 Wait for water and metal temperatures to equalise 4 Add insulation to sides of beaker/lid (to prevent energy losses) 5 Use of timer and equation, e.g. E = Pt = ItV for candidate’s method 6 Use large temperature change to reduce percentage uncertainty 7 Relationship is valid if the graph is a straight line that does not pass through the origin Do not allow vague computer methods.

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2 A student is investigating circular motion. A small mass m attached to a larger mass P is rotated at constant speed in a horizontal circle, as shown in Fig. 2.1. r m rigid tube string P Fig. 2.1 The student changes the radius r of the circle and measures the time t for ten revolutions. The student then determines the period T of a revolution and then the speed v. It is suggested that v and r are related by the equation mv 2 Pg = r where g is the acceleration of free fall. (a) A graph is plotted of v 2 on the y-axis against r on the x-axis. Determine an expression for the gradient. gradient = ..................................................[1] (b) The speed v is given by 2πr v = . T Values of r and t are given in Fig. 2.2. r / m t / s 0.160 3.4 ± 0.2 0.280 4.0 ± 0.2 0.400 4.8 ± 0.2 0.520 5.4 ± 0.2 0.640 6.0 ± 0.2 0.760 6.6 ± 0.2 Fig. 2.2 Calculate and record values of T / s, v / m s–1 and v 2 / m 2 s–2 in Fig. 2.2. Include the absolute uncertainties in v 2. [3] (c) (i) Plot a graph of v 2 / m 2 s–2 against r / 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 absolute uncertainty in your answer. gradient = ..................................................[2]

Mark scheme: 2 Analysis, conclusions and evaluation (15 marks) Mark Expected Answer Additional Guidance (a) A1 Pg gradient = m (b) T1 T / s, v / m s–1 and v2 / m2 s–2 Allow T (s), v (m s–1) and v2 (m2 s–2). T2 Must be values of v2 in table (if v not rounded). 8.7 or 8.74 All values of v2 must be 2 s.f. or 3 s.f. 19 or 19.3 Allow a mixture of significant figures. 27 or 27.4 37 or 36.6 45 or 44.9 52 or 52.3 U1 From ± 0.9 or ± 1 to ± 3 Allow more than one significant figure. (c) (i) G1 Six points plotted correctly Must be within half a small square. Do not allow “blobs”. ECF allowed from table. U2 Error bars in v2 plotted All error bars to be plotted. Must be accurate to correctly less than half a small square. Length of bar must be accurate to less than half a small square. (ii) G2 Line of best fit If points are plotted correctly then lower end of line should pass between (0.16, 10) and (0.18, 10) and upper end of line should pass between (0.70, 50) and (0.72, 50). Line should not go from top to bottom points. G3 Worst acceptable straight line. Line should be clearly labelled or dashed. Steepest or shallowest Examiner judgement on worst acceptable line. possible line that passes Lines must cross. Mark scored only if error bars through all the error bars. are plotted. (iii) C1 Gradient of line of best fit 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 72.) U3 Absolute uncertainty in Method of determining absolute uncertainty: gradient difference in worst gradient and gradient. (d) (i) C2 m Must use gradient. Should be about 0.19. P = × gradient g = 2.55 × 10–3 × gradient C3 kg (ii) U4 Percentage uncertainty in P Must be greater than 4%. (e) (i) C4 v in the range 4.70 to 4.90 and given to 2 or 3 s.f. (ii) U5 Percentage uncertainty in v Allow credit if absolute uncertainty in mass used correctly. 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]  ∆ gradient 0.001   ∆ gradient  percentage uncertainty =  +  × 100 =  + 0 . 04  × 100  gradient 0.025   gradient  0.026 max. P = × max. gradient 9.81 0.024 max. P = × min. gradient 9.81 (e) (ii) [U5] 1  ∆ P 0.005  percentage uncertainty = ×  +  × 100 2  P 0.5  max. P × 9.81 × 0.505 max. v = 0.040 min. P × 9.81 × 0.495 min. v = 0.040

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