2022 NCHS Phy Prelims Ans

1 NAN CHIAU HIGH SCHOOL Sec 4 Express Physics Papers 1, 2 and 3 Solutions 2022 Preliminary Examination Paper 1 (40 marks) 1 2 3 4 5 6 7 8 9 10 C C B C A A C B C C 11 12 13 14 15 16 17 18 19 20 A B B C A B C C D C 21 22 23 24 25 26 27 28 29 30 A B B B C C D A C B 31 32 33 34 35 36 37 38 39 40 B C C D B A B B D D Paper 2 Section A (50 marks) Question Answer Remarks 1 a > The ball is at equilibrium as > it is at rest, no resultant moment/ no resultant forces / stationary / not moving. B1 B1 b W = m.g = (2.0).(10) = 20 N (2sf) c The force must be from the centre of the ball and pointing downwards. d Deduct 1 mark each for: -No, wrong arrows; -Out of range; -Diagram too small,(minimum 1cm:2N -Wrong orientation *2 or 3sf is accepted 2 a D should be between 12 to 18 secs W 1 cm = 1.0 N T1 = 16.4 N T2 = 11.5 N 35 º 20 N 55 º

2 b (23, 8000), (31, 4000) acceleration = change of speed / time taken = (4000 – 8000) / (31- 23) = - 500 mm s-2 = -0.50 m s-2. Deceleration = 0.50 m s-2 allowance for error 0.48 to 0.52 m s-2 M1 A1 c > The rock is likely to have hit the bottom of the pond as > its speed changes from 2.0 m s-2 to 0 in an extremely short time, speed is zero, not moving (etc of the same meaning) B1 (position) B1 (motion) 3 a When an object is at equilibrium, the sum of clockwise moments about any point equal to the sum of anti-clockwise moments about the same point. B1 B1 b Sum of anti-clockwise moments = sum of clockwise moments 90.RB = (40).(25) + (90-80).(600) RB = 77.77 N = 78 N (2 sf) M1 A1 c The position of the centre of gravity of the man-ladder system will rise and the system to be less stable. *do not accept “CG increase” , as a position cannot increase. reasoning of system to be less stable must be correct also. M1 A1 4 a The force per unit mass acting on an object under gravity. bi and bii 1m for both shapes correct 1m for both labels P and K correct 1m for total energy is constant c GPE at 40 m = m.g.h = (5.0).(10).(40) = 2000 J GPE at 40 km = m.g.h = (5.0).(9.7).(40 000) = 1,940,000 J Difference in GPE = 1,940,000 – 2000 = 1,938,000 = 1,900,000 J (2 sf) (both gpe must be correct to get 1m) 1m 5 a (h.ρ.g)air + (h.ρ.g)H at position P= (h.ρ.g)Hg at foot of mountain hair.(1.23).(10) = (0.760 – 0.700).(13600).(10) hair = 663 m (No marks will be awarded if students only state pressure difference as 6.0 cm Hg) M1 M1 A1 b As point P has a lower atmospheric pressure than at sea level, the molecules do not need to overcome so high downward force exerted by the atmospheric pressure than when it is at sea level, so less energy is needed to boil water a point P, so water boils at a lower temperature. 1m 1m Speed / m s-1 D time t/s Energy/ J h /m0 P K

3 Note: energy required to overcome intermolecular forces of attractions at P and at the foot of the mountain is constant thus the assumption by most students that less energy is required to overcome the interm