Physics 9702/51 May/June 2020: question paper and mark scheme

Mark scheme

1 Defining the problem A is the independent variable and k is the dependent variable or vary A and measure k 1 keep N 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 of the spring • labelled load method to measure mass or weight of load: use top-pan balance to measure mass or newton meter to measure weight 1 use of a micrometer/calipers to determine t and rule/calipers to measure the diameter of the spring 1 method to measure extension, e.g. labelled ruler drawn parallel to spring, equilibrium position and displaced position 1 indicated and x indicated or description of use of ruler to measure equilibrium position and displaced position and difference determined Method of analysis plot a graph of k against 1 / A3/2 (or A–3/2) or equivalent e.g. lg k against lg A 1 relationship valid if a straight line passing through the origin is produced 1 (for lg k against lg A, relationship valid if a straight line with gradient –3/2) gradient × N 1 β= ρt 4 [for lg k against lg A, β = 10y-intercept × N / (ρt4) ] 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 t constant mg F D3 k= or x x D4 use of set square when taking measurements to determine extension of spring D5 repeat measurement of t along wire/spring and average D6 repeat measurement of diameter D of spring (to determine A) in different directions and average π D2 D7 use of A = 4 D8 method to ensure clamped rule to measure extension 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 D9 method to determine the density of the wire or additional detail on construction of coil D10 method to determine the mean diameter of the spring, e.g. subtract t from external diameter of spring

Mark scheme

2(a) −1 1 gradient = CR Q0 y-intercept = ln C 2(b) ln (V / V) 1 1.82 or 1.825 1.53 or 1.526 1.22 or 1.224 0.96 or 0.956 0.69 or 0.693 0.34 or 0.336 Absolute uncertainties in ln V from ± 0.03 or ± 0.04 to ± 0.13 or ± 0.14 or ± 0.15. 1 2(c)(i) Six points plotted correctly. 1 Must be within half a small square. Diameter of points must be less than half a small square. Error bars in ln V plotted correctly. 1 All error bars to 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 Points must be balanced. Do not accept top point to bottom point. Line must pass between (4.0, 1.6) and (5.0, 1.6) and between (26.5, 0.5) and (28.0, 0.5) 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 must be negative. 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 read from y-axis to less than half a small square 1 or y-intercept determined from substitution into y = mx + c. 2(d)(i) C determined using gradient and C and Q0 given to two or three significant figures. 1 Correct substitution of numbers required. −1 −1 C= 3 = 3 39 × 10 × gradient 39 × 10 × (c)(iii) Q0 determined using y-intercept. 1 Q0 = C × e y -intercept = C × e (c)(iv) C determined using gradient 1 and Q0 determined using y-intercept and dimensionally correct units for C (F or s Ω–1) and Q0 (C or V s Ω–1 or A s). 2(d)(ii) Absolute uncertainty in C. 1 Δgradient ΔC = 0.05 + ×C gradient 2(e) V determined from (d)(i) (or (c)(iii) and (c)(iv)) with correct substitution shown and correct power of ten. 1 −60 Q0 × e CR = e y -intercept × e( gradient×60 ) V= C or ln V = – (t / RC) + ln (Q0 / C) = – (60 / 39 000) × (d)(i) + ln (Q0 / C) ln V = 60 × gradient + y-intercept ln V = 60 × (c)(iii) + (c)(iv)