Physics 9702/52 October/November 2020: question paper and mark scheme

Mark scheme

1 Defining the problem m is the independent variable and v is the dependent variable or vary m and measure v 1 keep P constant 1 Methods of data collection labelled diagram of workable experiment including: 1 • P attached to string • string passing over a supported pulley • string horizontally attached to trolley • labels for pulley and P method to determine v e.g. 1 • light gate attached to a timer to measure v • motion sensor connected to data logger • ticker tape timer • measure d, and t with timer/stopwatch/light gate and timer (to calculate v) use a (top pan) balance to measure m 1 measure d with a ruler 1 Method of Analysis plot a graph of 1 / v2 against m or m against 1 / v2 1 1 gradient 1 Q = Pg − or Q = Pg – (consistent with graph) 2d × gradient 2d y -intercept 1 R = 2d × ( Pg − Q ) × y -intercept ( = ) or R = –y-intercept (consistent with graph) gradient 1 Additional detail including safety considerations 6 D1 safety precaution related to falling mass P or preventing trolley falling e.g. cushion/sandbox for P, barrier for trolley D2 keep d constant D3 use a large distance for d D4 method to keep d constant, e.g. mark starting and end positions D5 v = distance / time for appropriate small distance on ticker tape/card/distance between light gates (not d) or use of v = 2d / t if d and total time measured D6 method to ensure wooden surface is horizontal, e.g. spirit level D7 for same m repeat experiment to find average v or t 1 m R 2d ( Pg − Q ) D8 = + or m = − R (consistent with graph) v 2 2d × ( Pg − Q ) 2d × ( Pg − Q ) v2 D9 relationship valid if a straight line (not passing through the origin) D10 method of attaching mass m to trolley e.g. tape, adhesive putty

Mark scheme

2(a) 4σ 1 gradient = ρg 2(b) 1 (1 / d) / mm–1 0.91 or 0.909 0.77 or 0.769 0.67 or 0.667 0.59 or 0.588 0.50 or 0.500 0.43 or 0.435 1 1 Absolute uncertainties in from ± (0.08 or 0.09) to ± (0.01 or 0.02). d 2(c)(i) Six points plotted correctly. 1 Must be accurate to nearest half a small square. Diameter of points must be less than half a small square. 1 1 Error bars in plotted correctly. d All error bars must be plotted. Total length of bar must be accurate to less than half a small square and symmetrical. 2(c)(ii) Line of best fit drawn. 1 Points must be balanced. Do not accept line from top to bottom point. Line must pass between (0.52, 10.5) and (0.55, 10.5) and between (0.84, 17.0) and (0.87, 17.0). 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 into Δy / Δx. 1 Distance between data points must be at least half the length of the drawn line. Gradient of worst acceptable line determined. 1 uncertainty = (gradient of line of best fit – gradient of worst acceptable line) or uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient) 2(d)(i) 0.606 − 0.422 0.184 1 ρ = −6 = = 1260 kg m –3 146 × 10 146 × 10 −6 and given to three or four significant figures. 2(d)(ii) % uncertainty in ρ = % uncertainty in m + % uncertainty in V 1 2 2 = + × 100 = 1.087 + 1.37 = 2.5% 184 146 or using max ρ = 186 / 144 = 1292 and/or min ρ = 182 / 146 = 1230 2(e) σ determined using gradient with correct substitution shown. 1 ρ g × gradient (d)(i) × 9.81× (c)(iii) σ = = 4 4 σ determined using gradient and correct SI unit given (N m–1 or kg s–2). 1 Absolute uncertainty in σ determined with correct substitution shown. 1 (d)(ii) Δ gradient uncertainty = + ×σ 100 gradient or 2 2 Δgradient uncertainty = + + ×σ 184 146 gradient or max (d)(i) × 9.81× max(c)(iii) using max σ = 4 or min (d)(i) × 9.81 × min (c)(iii) using min σ = 4 2(f) d determined to a minimum of two significant figures. 1 gradient d= h or 4 × (e)(i) d = h × (d)(i) × 9.81