2021 H2 Phy Prelim Paper 1 Solution (forJCs)

2021 YIJC JC2 H2 Preliminary Examination Physics Paper 1 Answer Key Question Answer Question Answer Question Answer 1 B 11 B 21 A 2 C 12 D 22 C 3 B 13 B 23 C 4 A 14 A 24 C 5 D 15 A 25 B 6 C 16 B 26 C 7 C 17 C 27 B 8 A 18 C 28 C 9 B 19 B 29 B 10 C 20 D 30 C 2021 YIJC JC2 H2 Preliminary Examination Paper 1 Suggested Solutions S/N Answe r Explanation 1 B [σ] 2 C A implies large random error. B implies non-ohmic conductor. C Graph should be a straight line passing through the origin if no systematic error. D implies a possible error in the experiment (one data point) 3 B Estimation of values required: Mass of car = 1000 kg Speed limit on expressway = 90 km h-1 p =1000 (90 ×1000 3600  0) = 25 000 = 2.5 × 104 kg m s1 4 A The gas bubbles initially accelerate upwards before travelling at a terminal velocity. Hence, spacing between bubbles increases before becoming constant. 5 D Take vertically upwards as positive direction, the acceleration ( g) of the ball before and after the collision with the ceiling is negative (downwards). During collision with the ceiling, the ball experiences an downward force from the ceiling and together with its weight leads to a larger negative (downwards) acceleration.

2 6 C For the bob, T cos 11 = mg and T sin 11 = ma  tan 11 = a / g  a = g tan 11 = 1.90687 m s2 For the block and bob, F = (m) a 250 – Ff = 100  1.90687 Ff = 59.3 N 7 C Area under F-t graph gives the change in momentum of the body. From t = 75 ms to t = 150 ms, p = mv – mu = ½ (60 + 80)(100 – 75)(103) + ½ (80 + 40)(150 – 100)(103) = 4.75  mv = 4.75 + mu = 4.75 + (0.200)(15) = 7.75 v = 7.75 / 0.200 = 38.8 m s−1 8 A The resultant force acting on both blocks could be shown below. Hence, F = 3 ma a = F / 3m For Block X, let resultant force = RX RX = ma = For Block Y, let resultant force = RY RY = 2 ma = 9749 / YIJC / 2021 / JC2 Preliminary Examination / Paper 1 X YF a X RX Y RY

3 9 B For an object to be in equilibrium with three coplanar forces acting on it, (a) The vector sum of the three forces must form a closed loop diagram. (Option C not possible) (b) Net moment about any point must be zero. (Option D not possible) (c) Line of action of the three forces must pass through a common point (Option A not possible) 10 C Assume the sphere is dropped from a height h. Considering its downward motion to the plate, Gain in KE = Loss in GPE, Upon rebound, considering its upward motion to its maximum height, Loss in KE = Gain in GPE, Hence, 11 B Using conservation of energy, KEX + GPEX + Energy input = KEY + GPEY + Work Done by dissipative forces Taking Y as the reference point, 0 + mghX + 0 = ½ mvY 2 + 0 + Energy by frictional forces Energy by frictional forces = 500(9.81)(30) – ½(500)(11)2 = 1.2  105 J 9749 / YIJC / 2021 / JC2 Preliminary Examination / Paper 1

4 12 D At