2021 AHS Physics Prelim Paper 3 Mark Scheme

Mark scheme for 2021 Physics Prelim Practical Exam Question 1 Minus maximum 1 m for each type of unit. For wrong s.f., minus max 1 m for Q1. Allow ECF for calculations. (a) d = 0.38 mm (2 d.p. in mm, accept 0.36 to 0.40 mm) [1] (b) A =  (0.38)2 / 4 = 0.11 mm2 = 1.1 x 10-7 m2 (2 s.f.) (correct calculation with units) [1] OR: A =  (0.38 x 10-3)2 / 4 = 1.1 x 10-7 m2 (2 s.f.) (correct calculation with units) [1] (c) I = 0.30 A (2 d.p. with correct units, accept 0.20 to 0.40 A) [1] V = 2.40 V (2 d.p. with correct units, accept 1.60 V to 3.20 V) [1] (d)(i) R = 2.40 / 0.30 = 8.0  (2 s.f.) (correct calculation with units) [1] (ii)  = RA / 𝑙 = 8.0 x 1.1 x 10-7 / 0.800 (correct use of formula) [1] = 1.1 x 10-6 m (2 s.f.) (correct calculation and units) [1] (do not accept ohm per metre) (iii) P = 0.30 x 2.40 = 0.72 W (2 s.f.) (correct calculation and units) [1] (e) P = 0.40 x 1.60 = 0.64 W (2 s.f.) (correct calculation and units) [1] (f) P is not directly proportional to 𝑙 as the value of P did not decrease / increase by half when 𝑙 decrease / increase by half. (accept ECF according to their data)

Question 2 Minus maximum 1 m for each type of unit. No need to penalise s.f. for Q2. (a) Correct d.p.and units [1] h = 6.6 cm j = 30.4 cm k = 12.0 cm (b) Show evidence of taking average value of at least two readings. [1] t = 5.39 s (2 d.p. with units) [1] (accept 1 d.p due to human reaction time) (c) P1 = 5 x (10 x 10-3) / 5.39 (correct substitution) [1] = 9.3 x 10-3 W (2 s.f.) (correct calculation and units) [1] Accept non SI units: For example P1 = 5 N x (10 mm) / 5.39 s = 9.3 Nmm / s P1 = 5 N x (1.0 cm) / 5.39 s = 0.93 Ncm / s The value of W is a constant as it is not measured. So its s.f. does not affect P. The increase in height h is measured using a ruler. Hence its s.f. is taken into account in the calculation of P. (d) P2 = 2 x (10 x 10-3) / 4.84 = 4.1 x 10-3 W (2 s.f.) (W = 2 and value less than P1 with units) [1] (e) At least two constant variables: h and j &/or k, [1] (or implied in the procedure). 1. Set up the apparatus as shown in Fig. 2.1. 2. Rotate the weight clockwise till h = 10 mm. 3. Release the weight and record t, the time taken for the string to fully unwind. 4. Repeat step 3 to get an average value of t. [1] 5. Repeat step 2 to 4 for 4 more set of readings by changing the value of W by removing 1 slotted mass each time (i.e. W = 4N, 3N, 2N ,1N). [1] 6. Calculate power using the formula P = 𝑊ℎ 𝑡 7. Tabulate values of W, t and P. 8. Plot a graph of P against W. P / W W / N 0 Sketch of straight line or curve with positive gradient passing through the origin. [1] Should indicate the origin in the sketch. P / W W / N P / W W / N OR OR 0 0

Question 3 Note: There are 2 different kinds of slotted masses and 2 different kinds of springs used. (a) lo = 5.1 cm (1 d.p. in cm) (accept 5.0 to 5.4 cm) (BOD - Accept 2 d.p. in cm if student use vernier calipers) (b) L = 50.0 cm (1 d.p. in cm) [1] (accept 49.5 to 50.5 cm) (c) d = 3.18 cm or 3.78 cm (2 d.p. in cm) [1] (accept 3.15 to 3.81 cm) (d)(i) l = 8.4 cm (1 d.p. in cm) [1] OR: l = 6.6 cm (1 d.p. in cm) [1] (different spring in Chem lab 2) (no need to consider acceptable range of values as the rulers are not uniform) (ii) e = 3.3 cm (correct calculation and unit) [1] (correct d.p. penalty under (f)(i)) (e) l = 11.1 cm (1 d.p. in cm) [1] (no need to consider acceptable range of values as the rulers are not uniform) (ii) e = 6.0 cm (correct calculation and unit) [1] (correct d.p. penalty under (f)(i)) (f)(i) 1. Correct headings of n, l and e with correct units for l and e. [1] 2. At least 5 sets of readings with correct trend (l increases as n increases). (Must include n = 1 & n = 2. No need to look for acceptable range of values). [1] 3. Correct precision for l. [1] 4. Correct calculation of e and correct d.p. [1] n l / cm e / cm 1 8.4 3.3 2 11.1 6.0 3 13.8 8.7 4 16.7 11.6 5 19.2 14.5 6 22.8 17.7 7 25.1 20.0 (ii) 1. axes labelled with units and correct orientation [1] 2. suitable scale, not based on 3, 6, 7 etc. with plotted data occupying at least half the page in both directions (allow graph to start at the origin) [1] 3. all points plotted correctly (points must be at most half a small square from the correct position) [1] 4. best fit line and fine crosses [1] (no penalty if graph does not start from the origin) (iii) 1. use of a triangle that uses more than half the drawn line [1] 2. correct calculation of gradient [1] (G = 2.84 cm (not in mark scheme)) (g) k = mg (dn + L) / 2GL = 0.100 x 10 x (3.82 x 3 + 50.0) / 2 x 2.84 x 50.0 (correct substitution of values) [1] = 0.216 N / cm (if g is treated as a constant) or 0.22 N / cm (if consider g as 2 sf value) (correct calculation [1] and correct units) [1] (no need to penalize sf)