Physics 9702/53 October/November 2019: question paper and mark scheme

Mark scheme

1 Defining the problem r is the independent variable and h is the dependent variable or vary r and measure/determine h 1 keep P constant 1 Methods of data collection labelled diagram of workable experiment including: 1 • labelled air blower • labelled ball vertically (by eye) above the blower • vertical (by eye) rule at least from top of blower to ball circuit diagram to determine P, e.g. voltmeter and ammeter connected to motor and power supply or wattmeter connected to motor 1 and power supply use of micrometer/calipers/rule to measure diameter of ball 1 measure the distance between the top/middle/bottom of the ball and the top of the blower 1 Method of Analysis plot a graph of h against 1 / r3 1 (or lg h against lg r) relationship valid if a straight line through (0,0) 1 (for lg h against lg r straight line with gradient = –3) 4πg × gradient 1 K= 3P 4πg × 10y −intercept (for lg h against lg r, K = ) 3P 1 Additional detail including safety considerations D1 use large box/tray to collect ball (to prevent ball rolling on floor/bouncing) 6 or reasoned method to avoid draughts, e.g. switch off fans, close windows, use a screen D2 method to determine P from correct circuit, e.g. P = I × V or use wattmeter to measure P D3 stand on bench with clamped rule vertically to measure vertical distance D4 method to ensure clamped rule to measure h is vertical, e.g. correctly positioned set square indicated at right angles between the rule and the horizontal surface or plumb line shown in appropriate position D5 r = d / 2 when diameter measured D6 repeat diameter measurement in different directions and find average D7 repeat experiment for each value of r and determine average h D8 method to determine h, e.g. h = reading top of ball – r – top of blower or h = reading bottom of ball + r – top of blower or h = (distance from top of ball to top of blower + distance from bottom of ball to top of blower) / 2 D9 wait for the ball to become stationary (vertically) D10 video (camera) shown level with elevated ball and description of playback frame-by-frame or slow motion

Mark scheme

2(a) gradient = q 1 and  2π  1 y-intercept = lg   = lg 2π − 2 lg k  k 2(b) T/s lg (T / s) 1.56 0.193 or 0.1931 1.79 0.253 or 0.2529 1.97 0.294 or 0.2945 2.14 0.330 or 0.3304 2.31 0.364 or 0.3636 2.45 0.389 or 0.3892 Values of T as above. 1 Values of lg T as above. 1 Uncertainties in T all ±0.04. 1 Uncertainties in lg (T / s) consistent with uncertainties in T e.g. from ±0.011 to ±0.007. 1 2(c)(i) Six points plotted correctly. 1 Must be accurate to the nearest half a small square. Diameter of points must be less than half a small square. Error bars in lg T plotted correctly. 1 All error bars must be plotted. Length of bar must be accurate to less than half a small square and symmetrical. 2(c)(ii) Line of best fit drawn. 1 Lower end of line should pass between (2.22, 0.22) and (2.25, 0.22) and upper end of line should pass between (2.49, 0.34) and (2.51, 0.34). Do not accept line from first to last plot. Worst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1 All error bars must be plotted. 2(c)(iii) Gradient determined with clear substitution of data points from the line of best fit into ∆y / ∆x. 1 Distance between data points must be greater than half the length of the drawn line. uncertainty = gradient of line of best fit – gradient of worst acceptable line 1 or uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient) 2(c)(iv) y-intercept determined by substitution of correct point from the line of best fit into y = mx + c. 1 2(d) k determined from y-intercept. 1   2π    2π  2 2 k  =  y −intercept   =  (c)(iv)    10    10   q = answer to (c)(iii) and given to 2 or 3 significant figures. 1 2(e) M determined from (d) or (c)(iii) and (c)(iv) with correct substitution shown. 1 T k q k 2q k M=q = = 2π 2π 4 π2 or  2π  (lg 1) − lg   lg M =  k q y -intercept (c)(iv) lg M = − =− gradient (c)(iii)  (c)(iv)  −   (c)(iii)  M = 10