2023 CHS Phy Prelims Ans

2023 SEC 4 PRELIMINARY EXAMINATION Paper 1 1 B 2 B 3 C 4 D 5 C 6 A 7 D 8 D 9 D 10 A 11 B 12 A 13 C 14 A 15 D 16 B 17 A 18 C 19 C 20 A 21 B 22 A 23 A 24 D 25 D 26 C 27 B 28 B 29 B 30 C 31 C 32 D 33 A 34 D 35 A 36 C 37 A 38 C 39 A 40 C Paper 2 Section A S/N Mark Scheme Remarks 1ai displacement = area under graph [formula] = ½ (2.50) (4.0) [B1] = 5.0 m [B1] 1aii Acceleration due to gravity = gradient of graph = (0.0 – 4.0) / (2.5 – 0.0) [B1] = -1.6 m / s2 Deceleration = 1.6 m / s2 [B1] 1bi Decreasing + -ve gradient; v = 0.0 m / s before t = 2.5 s [B1] 1bii Air resistance = W = mg [formula] = 7.5 (1.6) = 12 N [B1] 2a Fnet = ma [formula] F – W = ma F – mg = ma F – (2.8 × 106) (10) = (2.8 × 106) (1.4) [B2] F = 3.2 × 107 N or 3.19 × 107 N [B1] W = mg & F = ma calculations are one mark each 2b Mass of rocket decreases (as fuel is used up) [B1] Value of g on rocket decreases (as rocket rises) [B1] 3ai Point where (all) the weight (of the object) appears to act [B1] (regardless of the orientation of the object). 3aii No, because centre of gravity of XY is not at the (geometric) centre (or 2.0 m mark) of rod. [B1] 3b Taking moments about (pivot) P, Sum of clockwise moments = sum of anticlockwise moments M = F x d [formula – all above] (3.0 – 2.5) (2.90) + 44 (0.60) = F (1.10) [B1] F = 25.3 N [B1] 3c Resultant moment clockwise (when sphere is not in water) OR Larger anticlockwise moment due to F (needed to balance larger clockwise moment). [B1] Shifting P to the right increases perpendicular distance from (line of action of) F to P/pivot. [B1] Accept if discuss about smaller clockwise moment due to weight of sphere and / or rod

2 4a Same volume for two different temperatures. [B1] Volume does not vary continuously with temperature. [B1] 4bi Kinetic energy of the gas particles increases (with temperature). [B1] Force and frequency of collisions of gas particles on walls of container increases. [B1] Force per unit area (and hence pressure) of gas increases. [B1] 4bii Conduction [B1] Flask particles vibrate and collide against gas particles (causing them to vibrate and collide more often). [B1] 5a Change in mass = 120 – 80 = 40 g [B1] Lv = IVt2 = m (lv) [formula] 2.0 (12) (3750) [B1] = 40 (lv) [B1] lv = 2250 J / g (or 2.25 x 106 J / kg) [B1] 5b Large amount of energy needed to overcome (large) inte