Physics 9702/31 May/June 2023: question paper and mark scheme

Mark scheme

1(a) Value of S1 with unit and 10.0 cm ⩾ S1 ⩾ 2.0 cm. 1 1(b)(i) Values of L1 and L2 to the nearest mm with unit. 1 1(b)(ii) e1 and e2 correctly calculated and e2 > e1. 1 1(c) Five sets of readings of x (different values), L1 and L2 with correct trend (as x increases L1 decreases and L2 increases) 5 and without help from Supervisor scores 5 marks, four sets scores 4 marks, etc. Range: 50.0 cm ⩽ xmin ⩽ 55.0 cm and xmax ⩾ 95.0 cm. 1 Column headings: 1 Each column heading must contain a quantity and a unit where appropriate. The presentation of quantity and unit conforms to accepted scientific convention e.g. x / cm, L1 / cm, no unit for e2 / e1. Calculation: Correct calculation of e2 / e1. 1 1(d)(i) Axes: 1 Axes must be labelled with the correct quantities. Scales must be chosen so that the plotted points occupy at least half the graph grid in both x and y directions. Scale markings are no more than 2 cm apart (one large square). Sensible scales must be used. Scale must not be awkward (e.g. 3:10 or fractions). Plotting of points: 1 All observations in the table must be plotted on the grid. Diameter of plotted points must be less than half a small square. Points must be plotted to an accuracy of half a small square in both x and y directions. Quality: 1 Trend of points must be positive. All points in the table must be plotted on the grid. It must be possible to draw a straight line that is within ± 0.1 (to scale) on the e2 / e1 axis of all plotted points. 1(d)(ii) Line of best fit: 1 ‘Best fit’ is judged by the balance of all points on the grid (at least 4 points) about the candidate’s line. There must be an even distribution of points either side of the line along the full length. Lines must not be kinked or thicker than half a small square. Some candidates may choose to identify an anomalous point. If they identify one point as anomalous (e.g. by circling or labelling) then this point is to be disregarded when judging the line of best fit. There must be at least 5 points left after the anomalous point is disregarded. 1(d)(iii) Gradient: 1 The hypotenuse of the triangle used should be greater than half the length of the drawn line. Both read-offs must be accurate to half a small square in both x and y directions. The method of calculation must be correct, not Δx / Δy. The gradient sign on the answer line must be consistent with the graph drawn. y-intercept: 1 Intercept read directly from the graph, with read-off at x = 0, accurate to half a small square in y direction. or Correct read-off from a point on the line and substituted correctly into y = mx + c or an equivalent expression. Read-off is accurate to half a small square in both x and y directions. 1(e) Value of P = candidate’s gradient and value of Q = – candidate’s intercept value. 1 Values must not be written as fractions or given to one significant figure. Units for P consistent with readings: m–1 or cm–1 or mm–1 1 and no unit given for Q. 1(f) W line shown with larger gradient (and larger negative value of y-intercept). 1

Mark scheme

2(a) Value of a to the nearest mm and (final) a in the range 23.0–25.0 cm with unit. 1 2(b) Value of final t in the range 10.50–12.50 s with unit. 1 Evidence of repeat readings for t. 1 2(c)(i) Value of l in the range 27.0−39.0 cm with unit and correct calculation of (l − a). 1 2(c)(ii) Absolute uncertainty in (l − a) in range 3–8 mm. 1 Correct method of calculation to find percentage uncertainty e.g. (absolute uncertainty / value from (c)(i)) × 100. If repeated readings have been taken, then the uncertainty can be half the range (but not zero) if the working is shown clearly. 2(d) Second values of a and t. 1 Second value of l. 1 Second value of (l ─ a) > first value of (l ─ a). 1 2(e)(i) Two values of C calculated correctly. 1 The final C values must not be written as fractions or given to one significant figure. 2(e)(ii) Justification for significant figures in C linked to significant figures in a and (l – a). 1 2(f) Calculation of percentage difference between candidate’s two C values. 1 Comparison of percentage difference with 5% leading to a consistent conclusion. 2(g) Correct calculation of g in the range 9.0−12.0 m s–2 with consistent unit. 1 2(h)(i) A Two readings are not enough to draw a (valid) conclusion (not “not enough for accurate results”, “few readings”). 4 B Difficult to ensure twooden strip = tpendulum. C Difficult to measure a with a reason e.g. finding centre of hole/strip moves on pivot. D Difficult to measure l with a reason e.g. parallax/bob moves when touched by rule/rule moves/rule not vertical/deciding position of centre of bob. E Difficult to measure t or time for 10 oscillations with a reason e.g. judging/determining the start of or end of or one complete oscillation/strip wobbling/strip oscillating in different planes. F Times for holes G and H are close/small difference. G Difficult to balance wooden strip. 1 mark for each point up to a maximum of 4. 2(h)(ii) A Take more readings and plot a graph or take more readings and compare C values (not “repeat readings” on its own). 4 B Improved method to ensure t values are the same e.g. use two stands allowing wood and pendulum to oscillate at the same time. C Improved method to measure a e.g. draw a scale on the strip/mark position of the pivot on strip/grooves on lower side/roughen strip lower surface. D Improved method for measuring l e.g. rule with pointers/clamp ruler/measure diameter of bob and add or subtract or find radius. E Improved method for measuring t e.g. video or record or film and timer or replay frame-by-frame/use a (fiducial) marker at the centre of the oscillation. F Holes further apart. 1 mark for each point up to a maximum of 4.