Physics 9702/52 May/June 2020: question paper and mark scheme

Mark scheme

1 Defining the problem A is the independent variable and f is the dependent variable or vary A and measure f 1 keep M constant 1 Methods of data collection labelled diagram of workable experiment including: 1 • elastic cord fixed at one end to a support • other end passed over a pulley • labelled pulley • labelled load vibrator connected to signal generator 1 increase/decrease the frequency of the signal generator until stationary wave pattern is observed 1 measure diameter of cord with micrometer/calipers 1 Method of analysis plot a graph of f 2 against 1 / A or equivalent (e.g. lg f against lg A) 1 relationship valid if a straight line passing through the origin is produced 1 (for lg f against lg A, relationship valid if a straight line with gradient –½) M 1 k= gradient × 4 × L2 (for lg f against lg A, k = M / [10(2 × y-intercept) × 4 × L2] ) 1 Additional detail including safety considerations 6 D1 use safety goggles/safety screen to prevent injury to eyes from (moving) elastic cord/load or use cushion/sand box in case load falls D2 keep L constant D3 use cords of the same material/density D4 use CRO to determine f (or T) D5 method to determine T from CRO, e.g. period = time base × length of one wave D6 f = 1/T D7 repeat measurement of diameter along cord and average πd 2 D8 use of A = 4 D9 measure mass of the load on top-pan balance D10 detail on determining frequency at the maximum amplitude, e.g. increase frequency until the amplitude starts to decrease, then decrease frequency

Mark scheme

2(a) E 1 gradient = k y-intercept = lnH 2(b) 1 ln (η / 10–4 Pa s) 2.510 or 2.5096 2.28 or 2.282 2.13 or 2.128 1.92 or 1.917 1.72 or 1.723 1.57or 1.569 Absolute uncertainties in ln η from ± 0.02 (or ± 0.016) to about ± 0.04 1 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. Error bars in ln η 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 (2.93, 1.65) and (2.96, 1.65) and between (3.38, 2.45) and (3.41, 2.45). 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. 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 from substitution into y = mx + c. 1 2(d)(i) E determined using gradient and given to two or three significant figures. 1 Correct substitution of numbers required. E = 1.38 × 10–23 × gradient = 1.38 × 10–23 × (c)(iii) H determined using y-intercept. 1 H = ey-intercept = e(c)(iv) (× 10–4) E determined using gradient 1 and H determined using y-intercept and dimensionally correct units for E (J) and H (Pa s). 2(d)(ii) Absolute uncertainty in E. 1 ΔE = 1.38 × 10 −23 × absolute uncertainty in gradient or Δgradient ΔE = ×E gradient 2(e) η determined from (d)(i) or (c)(iii) and (c)(iv) with correct substitution shown and correct power of ten. 1 E (d)(i) (c)(iv) −4 −23 η = H ×e k × 273 =e × 10 × e 1.38×10 ×273 or gradient η = e y −intercept × 10−4 × e 273 or (c)(iii) η = e(c)(iv) × 10−4 × e 273 or E (d)(i) lnη = + ln H = + (c)(iv) kT 1.38 × 10−23 × 273 or gradient (c)(iii) lnη = + y -intercept = + (c)(iv) 273 273