ACSI 2024 Promo HL Phys MS

Y5 IBDP EOY HL Physics Examination 2024 Marking Scheme PHYSICS P1a 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 C B C D A D D B A B D C C D B C B A C B 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 B A B B B D B D D C C B D B C D A D C B PHYSICS P1b Answer Marks 1(a)(i) uncertainty in period reduces; (you can show that % uncertainty in T reduces from 8.7% to 0.43%, a reduction of about 20 times) by 20 times; improves the precision of period; 2 1(a)(ii) Human reaction time in using stopwatch; Likely to be at least 1 10 th a second (0.1 s); value student recorded is too precise; 2 1(a)(iii) T = 0.5750 ± 0.0025 s; T2 = 0.3306 s2 % uncertainty in T = 0.435 % Hence % uncertainty in T2 = 0.870 %; T2 = 0.870% of 0.57502 = ± 0.0029 = ± 0.0003; 3 1(a)(iv) % uncertainty in L= % uncertainty in F + 2(% uncertainty in T) + % uncertainty in m = 5.88 + 0.8696 + 0.5; = 7.3 %; This question has an ecf full marks from 1(a)(iii). 2 1(b) i Plot T2 (y-axis) vs m (x-axis) or T (y-axis) vs √𝑚(x-axis); 1 1(b) (ii) <m> = 202 g (± 1 g) m = ± ½ (210 – 195) = ±7.5 g = ± 8 g 2 2(a)(i) The effective acceleration along the slope is equal to 𝑔 = 𝑎 sin 𝜃 or 𝑎 = 𝑔 𝑠𝑖𝑛 𝜃 1 2(a)(ii) t = 0.4 s, x = 0.2  1002 cm hence g = 2x / t2 = 250 cm s-2 (Choose at least greater than 0.35 s (half the graph) and easy and accurate to read the value of x.) Hence g = a ÷ sin 15 = 970 cm s-2 ; This question has an ecf full marks from 2(a)(i). 3 2(a)(iii) human reaction in the time taken; take several repeats of the same measurement; determine average; Alterative answer: direction of wind is random and so time taken is either lower or higher than true value; minimize this random error by taking several repeats of the same measurement and determine average; 2 2(b) (smaller acceleration) hence larger time; (% / fractional) uncertainty of time decreases; 2

PHYSICS P2 Answer Marks 1(a) Using s = ut + ½ at2 y =½ (9.8) t2 100 = ½ (9.8) t2 t = 4.5 s 1 1 1 1(b) 1 mark for each correct graph 1 1 1(c) Total horizontal displacement travelled by parcel = 72 × 4.5 = 324 m Speed of truck = 324 125 444.5 − = m s−1 (if student uses 4.52 s from part (a), then the answer could be 44.5 or 45 ms-1) 1 1 2(a) Correct labelling and directions of forces. Correct relative magnitude of forces (vertical components). Minus 1 if relative magnitude of vertical components is obviously not correct. 1 1 2 (b) o4.00 2.50 sin 28.0 5.2 m= + =r 1 2 (c) 1 x/ m t/ s 125 4.5 T B Tension Weight

2 2 T cos θ= mg mvT sin θ= r vtan θ = rg o(tan 28.0 )2v = r g o5.17(9.81)( tan 28.0 ) 5.2 m==v s−1 1 1 2 (d) 50 9.8 cos 28 560 N = = T cos θ= mg mgT= cos θ 1 1 3 (a) F = kx = 75 × 0.085 = 6.38 N 1 1 3 (b) mgh = ½ kx2