Physics 9702/51 October/November 2020: question paper and mark scheme

Mark scheme

1 Defining the problem A is the independent variable and x is the dependent variable or vary A and measure x 1 keep m constant 1 Methods of data collection labelled diagram of workable experiment including: 1 • spring fixed at one end (to a support) • load attached to the other end • labelled load/mass clamped labelled ruler perpendicular by eye and ruler positioned close to the spring 1 method to determine x: 1 record initial position, record final position and subtraction or record length of stretched spring, original length of spring and subtraction use of a micrometer/calipers to determine diameter d of wire 1 Method of Analysis plots a graph of lg x against lg A or equivalent 1 n = gradient and relationship valid if a straight line 1 mgw 3 N 1 γ = ρ × 10 y -intercept 1 Additional detail including safety considerations 6 D1 use safety goggles/safety screen to prevent injury to eyes from (moving) spring/load or use cushion/sand box in case load falls D2 keep constant all of: • w • N • ρ (or use same material for wire) D3 measure mass of load/m with a balance or use newton meter to measure weight of load D4 use ruler/calipers to measure the width of the spring/w D5 use the equation ρ = mass of metal / volume to calculate ρ D6 method to measure w, e.g. w measured to the outside and subtract d or method to measure ρ, e.g. mass (balance) and volume (measuring cylinder) D7 repeat measurement of x (for the same A) and average D8 repeat measurements of either d or w made in different places/directions and average D9 use of A = πd2 / 4 to determine A of wire D10 use of set square to check that clamped ruler is vertical to bench or use of set square/pin as a fiducial mark to read positions (on the ruler)

Mark scheme

2(a) 1 1 gradient = f t y-intercept = −1 2f 2(b) 1 hi ho 0.71 or 0.708 0.79 or 0.792 0.92 or 0.917 1.1 or 1.08 1.3 or 1.29 1.5 or 1.50 hi 1 Absolute uncertainties in from ± 0.07 to ± 0.1 (or ± 0.10 or ± 0.11). ho 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. hi 1 Error bars in plotted correctly. ho 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 (55, 0.75) and (56, 0.75) and between (77, 1.40) and (78, 1.40). 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(c)(iv) y-intercept determined from substitution into y = mx + c. 1 y-intercept determined using gradient from worst acceptable line. 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) No ECF from false origin method. 2(d)(i) f determined using gradient with correct substitution shown. 1 1 1 f = = gradient (c)(iii) f determined using gradient and given to two or three significant figures and correct SI unit shown with correct power of 1 ten e.g. 33 cm or 0.33 m or 33.1 cm or 0.331 m Absolute uncertainty in f determined. 1 Δgradient absolute uncertainty in f = ×f gradient 2(d)(ii) t determined using y-intercept and given to two or three significant figures. Correct substitution of numbers required. 1 t = 2f × ( y -intercept + 1) or 2 × ( y -intercept + 1) t= gradient