NJC_2024_H2_Physics_Prelim_P2_Ans

NJC Preliminary Examination 2024 H2 Physics Paper 2 Solutions and Mark Scheme 1 (a) Use 𝑣𝑦 = 𝑢𝑦 + 𝑎𝑦𝑡 At max height, 𝑣𝑦 = 0 0 = 30 sin 60o – (9.81) t M1 t = 2.6 s A0 (b) initial velocity = 26 m s–1, final velocity = –26 m s–1 B1 straight line intersects x-axis at 2.6 s and ends at 5.2 s (or 5.3 s) B1 (c) (i) horizontal line at 𝑢𝑥 = 15 m s–1 from t = 0 to 5.2 s (or 5.3 s) B1 (ii) downward sloping curve from 15 m s–1 and its gradient decreases numerically B1 ending before t = 5.20 s B1 (d) Calculate displacements of first and second object to be 34.4 m and 5.84 m respectively using 2.6 s (or 34.5 m and 5.31 m using 2.65 s) C1 Displacement between objects = 29 m A1 2 (a) Force is the rate of change of momentum B1 (b) (i) Resultant force = change of momentum / time taken = [0.140 x (5.5 – ( – 4.0)) ]/0.04 M1 = 33 N A1 (ii) resultant force on ball = Δp/Δt = 33.25 N Taking forces on the ball, (N is force on ball by bar) 33.25 = N – W M1 N = 33.25 + 0.14*9.81 = 34.62 N By N3L, force on bar by ball is 35 N. A0 (c) (i) Taking pivot about support B, clockwise moments = FA x (45 – 25) C1 anti clockwise moments = (0.450 x 9.81) (25) + 35 (25 + 50) C1 Clockwise moments = anti clockwise moments FA = 136.768 N = 140 N (to 2 sf) A1 (ii) net force = 0 Upward force = downward force FB = 35 + 140 + 0.450 x 9.81 M1 = 180 N (to 2 sf) A1

3 (a) Pressure = ρgh = 900 x 9.81 x 6.0/100 M1 = 530 Pa A1 (b) weight and upthrust drawn in correct places (weight starts along the central blue dotted line and upthrust from centre of submerged part, ignore lengths) B1 the object turn anticlockwise (rotating ACW and CW with decreasing angle) until the weight and upthrust are on the same line of action (dotted line is vertical) B1 4 (a) The gravitational field strength at a point in a gravitational field is defined as the gravitational force exerted per unit mass acting on a small mass placed at that point. B1 (b) (i) point X is on the line and closer to Moon B1 (ii) Gravitational field strength at A = 𝐺(5.97×1024) (6.37×106)2 − 𝐺(7.34×1022) (3.84×108−6.37×106)2 = 9.81 m s-2 Gravitational field strength at B = 𝐺(7.34×1022) (1.74×106)2 − 𝐺(5.97×1024) (3.84×108−1.74×106)2 = 1.61 m s-2 Marks awarded as follow for both calculations: - correct equations that include contributions by both Earth and Moon M1 - correct substitutions (e.g. unit conversion, etc) M1 - correct final values A1 (iii) Shape of graph, one positive one negative, cut x-axis closer to B. B1 W U

5 (a) (i) Length of path 1 = 1.000 - 0.300 = 0.700 m Length of path 2 = 1.000 + 0.300 = 1.300 m Path difference = 1.300 – 0.700 = 0.600 m B1 (ii) Phase difference = ( 0.600 1.000 × 2𝜋) ± 𝜋 M1 = 2.2𝜋 or 6.91 r