RI 2021 A-Level H2 Phys Solution

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 1 2021 A-Level H2 Physics Suggested Solutions Paper 1 1 C Relate to how heavy a 50 g slotted mass feels (during practical sessions). A handphone is approximately 150 g (around three pieces of 50 g). W = mg = (0.150)(9.81) = 1.45 N ≈ 1.5 N 150 cN = 150×10−2 N = 1.5 N 2 A Using F = B I L unit on left : kg m s−2 (using F = ma) unit on right : (unit of B)(A)(m) hence unit of B = kg A−1 s−2 3 B At the top of the projectile, ball has horizontal velocity to the right. Hence air resistance is to the left. It has a weight, vertically downwards. Therefore, resultant force is in the direction indicated by arrow B. 4 B Taking right as positive: relative speed of approach = relative speed of separation v − 0 = 0.67v − vx vx = − 0.33 v 5 A Common mistake is to choose C. Option C gives the N3L pair of floor and brick. But question is asking for N3L pair of W and that of S. W is force between brick and Earth. S is force between brick and floor. 6 B When ball is in water, weight of ball = upthrust + tension (0.100)(9.81) = (0.5v) ρw g + (0.75)(0.100)(9.81) where ρw : density of water (1000 kg m−3) hence v = 5×10−5 m3 = 50 cm3 7 D Extra elastic potential energy = area under force-extension graph (hence it is the area the line makes with the vertical axis) = ½ (W1)(x1) – ½ (W0)(x0) = ½ (W1x1 – W0 x0) 8 A Work done against resistive forces in moving 1 km = f × d = (400)(1000) = 400 000 J 16 % of fuel is converted to work done against resistive forces. Hence fuel needed is 400 000 / 0.16 = 2500 000 J Since 1 kg provides 48 MJ, the amount of fuel needed = 2500 000/ (48 ×106) = 0.052 kg = 52 g 9 A v Be rmω = = hence angular velocity is proportional to B (for constant e and m).

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 2 10 B Using proportion, 10 m : 6.0 J kg−1 2.5 m : 1.5 J kg−1 Work done = gain in gPE = m (∆φ) = (2)(1.5) = 3.0 J 11 D 1 2(36000000 6400000)( ) 24 3600 3083 ms vr ω π − = = + × = 12 C Assuming gas is ideal, using 21 3 Nmpc V= <> where Nm = total mass of gas = (5000)(0.029) hence 52 2 2 1 (5000)(0.029)10 3 (10)(3)(4) 248275 498 c c c = <> < >= < >= 13 D Q = mc ∆θ + ml =(5)(4190)(70) + (5)(2260 000) = 1.28 × 107 J 14 B No change in internal energy since there is no heat transferred or work done on gas. 15 C For SHM, when velocity is maximum, KE is maximum and PE is minimum. When velocity is zero, KE is zero and PE is maximum. Since PE is maximum at 0 and 2 s, only option A and C has velocity zero at 0 and 2 s. When t = 1s, PE = 0, so KE should be max. Hence answer is C. 16 D c = f λ λ = (3.00 × 108) / (5.0 × 1014) = 6.0 × 10−7 6 7 (2 ) 1.5 10 (2 )6.0 10 5 xφπ λ π π − − ∆∆= ×= × = Phase difference of 5π is equivalent to π radian. 17 B Malus’ law : intensity ∝ cos 2 θ

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