Cambridge A Level Physics 9702 — 2017 May/June Paper 5 · Variant 1

1 A student is investigating the motion of a wooden block on an inclined plane, as shown in Fig. 1.1. A falling body causes the wooden block to accelerate. pulley string Q falling body wooden block plane P θ Fig. 1.1 The wooden block is initially at rest at point P and has velocity v at point Q. It is suggested that the relationship between v and the angle θ of the plane to the horizontal is (B + m)v 2 = Bg – mg sin θ 2s where B is the mass of the falling body, m is the mass of the wooden block, s is the distance between P and Q and g is the acceleration of free fall. Design a laboratory experiment to test the relationship between v and θ. Explain how your results could be used to determine a value for g. 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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[Total: 15]

Mark scheme: 1 Defining the problem (sin) θ is the independent variable and v is the dependent variable or vary (sin) θ and measure v 1 keep s (PQ) constant 1 Methods of data collection labelled diagram showing inclined plane with labelled support and P and Q marked 1 method to measure angle e.g. use a protractor to measure θ or use a ruler to measure marked distances from which sin θ or θ may be determined 1 method of timing for an appropriate distance to determine v (at Q) e.g. use a stopwatch/timer or correctly positioned light gate(s) connected to a timer/data-logger or correctly positioned motion sensor connected to data-logger 1 measurement of an appropriate distance to determine v (at Q) e.g. rule to measure an appropriate length or length of a card to interrupt light beam or distance from motion sensor to Q 1 Method of analysis plot a graph of v2 against sin θ 1 relationship valid if a straight line produced (not passing through the origin) 1 + = − × gradient 2 B m g ms or = g + × -intercept 2 B m y Bs 1 Question Answer Marks Additional detail including safety considerations Max. 6 D1 use cushion/foam/sandbox for falling body (B) D2 (sin) θ determined using trigonometry relationship using marked lengths D3 appropriate equation to determine v (at Q) e.g. = 2s v t D4 repeat experiment for each θ and average v or t D5 use of balance to measure mass of wooden block m and falling body B and rule to measure s D6 = + 2 -intercept . Bsg y B m D7 clean surfaces of blocks/inclined plane/ensure surface of the plane is smooth D8 keep B and m constant or keep mass of block and mass of falling body constant D9 method to ensure that wooden block starts at the same position P, e.g. put a mark on the block or align front or back of block D10 method to prevent plane slipping so that angle being measured remains the same, e.g. a mass as a stop

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2 A student is investigating how the time for an electrical pulse to travel in a coaxial cable varies with the length of the cable. The pulse is reflected at one end of the cable. An oscilloscope is used to display the initial pulse and the reflected pulse. The trace on the oscilloscope is shown in Fig. 2.1. d Fig. 2.1 The time t for the pulse to travel to the end of the cable and back is determined by measuring the distance d between the pulses on the screen, and then using the time-base and the relationship t = d × time-base. The initial length of the cable is L. A total length Z is removed from the cable and the experiment is repeated. It is suggested that t and Z are related by the equation 2 (L – Z ) v = t where v is the speed of the pulse. (a) A graph is plotted of t on the y-axis against Z on the x-axis. Determine expressions for the gradient and the y-intercept. gradient = ............................................................... y-intercept = ............................................................... [1] (b) Values of Z and d are given in Fig. 2.2. The time-base is 0.1 µs cm–1. Z / m d / cm t / µs 0.0 8.0 ± 0.1 4.0 7.7 ± 0.1 8.0 7.3 ± 0.1 12.0 7.0 ± 0.1 16.0 6.6 ± 0.1 20.0 6.2 ± 0.1 Fig. 2.2 Calculate and record values of t / µs in Fig. 2.2. Include the absolute uncertainties in t. [2] (c) (i) Plot a graph of t / µs against Z / m. Include error bars for t. [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 = −2 v y-intercept = 2L v 1 2(b) 0.80 ± 0.01 0.77 ± 0.01 0.73 ± 0.01 0.70 ± 0.01 0.66 ± 0.01 0.62 ± 0.01 First mark for all values of t correct. Second mark for uncertainties correct. 2 2(c)(i) Six points plotted correctly. Must be accurate to less than half a small square. No “blobs”. Diameter of points must be less than half a small square. 1 Error bars in t plotted correctly. All error bars to be plotted. Length of bar must be accurate to less than half a small square and symmetrical. 1 Question Answer Marks 2(c)(ii) Line of best fit drawn. If points are plotted correctly then upper end of line should pass between (4.8, 0.76) and (5.6, 0.76) and lower end of line should pass between (17.6, 0.64) and (18.8, 0.64). Line should not be from first to last plot. 1 Worst acceptable line drawn (steepest or shallowest possible line). All error bars must be plotted. 1 2(c)(iii) Gradient determined with a triangle that is at least half the length of the drawn line. Gradient must be negative. 1 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 read-off y-axis to less than half small square or determined by substitution into y = mx + c. 1 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) 1 2(d)(i) v determined from gradient and units for v and L correct with correct power of ten. = − = − 2 2 gradient 2(c)(iii) v 1 L determined from y-intercept and v and L given to 2 or 3 significant figures. Correct substitution of numbers must be seen. = × = × = − = − -intercept (c)(iv) -intercept (c)(iv) 2 2 gradient (c)(iii) v v y L y 1 Question Answer Marks 2(d)(ii) % uncertainty in v = % uncertainty in gradient 1 % uncertainty in L = % uncertainty in y-intercept + % uncertainty in gradient or % uncertainty in L = % uncertainty in y-intercept + % uncertainty in v Correct substitution of numbers must be seen. Maximum/minimum methods: = × max -intercept Max max -intercept max or mingradient y L y v = × min -intercept Min min -intercept min or max gradient y L y v 1

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