EJC 2023 J2 H2 PRELIM P2 MS

©EJC 2023 9749/J2H2PRELIM/2023 EUNOIA JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS 2023 9749 PHYSICS MARK SCHEME Paper 2 – Structured Qns Answer Marks 1(a)(i) Precision is determined by the range in the values. Range of Set A = 1.55 – 1.43 = 0.12 cm Range of Set B = 1.58 – 1.42 = 0.16 cm Range of Set C = 1.56 – 1.45 = 0.11 cm Therefore, Set C is more precise as the range of values are smaller than the other two sets. M1 for all correct values A1 1(a)(ii) Accuracy is determined by the closeness of the measured value to the true value. Average values for Set A = 1.52 +1.49 +1.43 +1.52 +1.55 =1.5025 = 1.50 Average values for Set B = 1.42 +1.45 +1.43 +1.51+1.58 =1.4785 = 1.48 Average values for Set C = 1.45 +1.56 +1.47 +1.53 +1.46 =1.4945 = 1.49 Therefore, Set A is more accurate as the average values are closer to the true value of the diameter. M1 for all correct values A1

2 ©EJC 2023 9749/J2HPRELIM/2023 Qns Answer Marks 1(b)(i) Alternative s-t graph For s-t graph: 1m – for correct sketch of graph for car B and correct displacement value Also, gradient of the line for first 4.0s must be greater than the gradient of the line for the last 9.0s For a-t graph: 1m – for correct sketch of graph for car B and correct acceleration value B1 B1

3 ©EJC 2023 9749/J2PRELIM/2023 Qns Answer Marks 1(b)(ii) Minimum distance occurs when both cars have the same speed (v = 60 m s−1). ( ) ( ) ( ) m 1 Distance covered by Car A during the 6.0 s = × 25 + 60 × 6.0 = 255 m 2 1 Distance covered by Ca 4 r B during the 6.0 s : 80 × 4.0 + × 80 + 60 × 2.0 u = 460 m 2 Minim m distance = (500 + 255) - 60 = 295     B1 C1 for both correct answer A1

4 ©EJC 2023 9749/J2HPRELIM/2023 Qns Answer Marks 2(a) No resultant force (in any direction) No resultant moment (about any point) B1 B1 2(b) Let the length of the rod be L. Take moment about the wall hinge. Sum of clockwise moment = Sum of anticlockwise moment ( )180 sin60 sin502 ooL T L = T = 45.2 N Let the vertical force at hinge be Ry. Let the horizontal force at hinge be Rx. No resultant force in vertical direction: Ry + T cos (180o – 60o – 50o) = W Ry + (45.2) cos 70o = 80 Ry = 64.5 N No resultant force in horizotal direction: Rx = T sin (180o – 60o – 50o) Rx = (45.2) sin 70o Rx = 42.5 N Force by hinge on rod 22 xyR R R=+ 2242.5 64.5R=+ = 77.2 N Let θ be the angle of R from the wall tan x y R R = 42.5tan 64.5 = θ = 33.4o from wall M1 A1 C1 A1 A1

5 ©EJC 2023 9749/J2PRELIM/2023 Qns Answer Marks 3a Let speed of X before hitting Y be vi Gain in KE = Loss in GPE 2 1 1 2 2 2 9.81 1.50sin40 4.35 m s ii ii i mv mgh v gh v − = = =    = B0 M0 A1 3b