RI 9749 2021

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

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 3 Initially, when middle filter is 45° to X and Y, the intensity of light emerging from Y is non-zero. When middle filter is 90° to X and Y, the intensity of light emerging from Y is zero (since cos 2 90° = 0) . 18 D Rayleigh criterion 9 6620 10 1.24 100.50b λθ − −×= = = × Using 6 16 10 where : separation of sources; : distance from source to observer 1.24 10 5.7 10 7.1 10 d dLL d d θ − ≈ ×= × = × 19 D 20 C 12 12 2 0 12 2 0 12 23 2 0 23 4 2.0 ..........(1)4 (2 ) .......(2)4 (2 ) 2(2)/(1) : 2 1 4 QQF r QQ r QQF r F πε πε πε = = = =×= 21 B R A R GA l l ρ ρ = = = For 4 wires arranged in parallel, the total cross-sectional area = 4 A 4 4 LR A LG ρ= = 22 A Both components are in series, hence have same current. The total p.d. across them has to be 3.0 V. From graph, when current is 0.10 A, the p.d. across both components = 1.0 + 2.0 = 3.0 V. 23 C For lamp to glow more brightly, p.d. across LDR has to be small and p.d. across lamp and thermistor has to be large. LDR has low resistance in bright light. Thermistor has high resistance in low temperature.

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 4 24 A Using right-hand grip rule for the circular coil, the B-field is pointing into paper. Using FLHR, the force on short wire is upwards. 25 D Magnetic flux = BA 26 A 2 () (3000)(1.8)( 0.01 ) 00.06 28 V NBAE tt π ∆Φ ∆< >= = = − = 27 B 16 8.51.88 s p V V = = ps ps V V I I= Hence Is = 0.32 / 8.5 = 0.038 A = 38 mA 28 B Longer wavelength corresponds to transition between smaller ∆E. Hence 590 nm corresponds to transition from n = 3 to 2. ∆E3→1 = ∆E3→2 + ∆E2→1 590 440 252 hc hc hc λ λ = + = 29 D intensity P Nhf A tA= = 30 B Energy released = (mreactants – massproducts) c 2 = (136.90709 – 136.90583 – 5.49 ×10–4)(1.66×10–27)(3.00 ×108)2 =1.1×10–13

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 1 2021 A-level H2 Physics Suggested Solutions Paper 2 1 (a) The block will accelerate uniformly along a straight path down the slope at a rate less than 29.81 m sg −= . [1] (b) (i) Using points (0.120, 0.030) and (0.320, 0.280) on the tangent line, 10.280 0.030speed gradient 1. 25 m s0.320 0.120 −−= = =− [1] [1] (ii) 2 1.25 0 0.20 6.25 m s v u at a a − = + =+× = [1] [1]

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 2 2 (a) From graph, the momentum of the ball as it makes contact with the surface (at 0.53 s) is 3.20 N s. 22 3.20E 2 2 0.62 8.26 J k p m= = × = [1] [1] [1] (b) From graph, the ball is in contact with the surface from t = 0.53 s to 0.68 s, during which its momentum changes from 3.20 N s to −1.80 N s. The magnitude of the average force on the ball is equal to the magnitude of the total change in momentum over the total time interval. (This is basically from Newton’s 2 nd Law.) 1.80 3.20 33.3 N0.68 0.53 pF t ∆−−= = =∆− But note that this average force represents only the net force on the ball – not the force the surface acts on the ball! That would be the normal contact force, which is what we are tasked to find. When in contact with the surface, two forces act on the ball – the normal contact force N upwards, and the weight of the ball mg downwards. (Note that the magnitude of N can vary with time.) However, since F on the ball is upwards, the average normal contact force N must be larger than the weight of the ball. ( )33.3 0.62 9.81 39.4 N F N mg N F mg = − =+= + = Note: The final answer has to be positive, since it is the magnitude. [1] [1] [1] [1] (c) 21.80KE after the first rebound 2. 61 J2 0.62= =× 2.61fraction of KE retained after every rebound 0.3168.26= = After n-th rebound, the fraction of the initial KE retained is 0.316n 0.316n = 5.0% = 0.050 (solved by taking logarithm on both sides) n = 2.60 = 3 (round up, integer number of times of rebounding) Note: Since the fractional loss of KE is constant for every rebound, the fraction of KE retained is also constant, because (fractional loss) + (fraction retained) = 1. [1] [1] [1]

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 3 3 (a) Let the length of PQ be L. clockwise moment due to F = anticlockwise moment due to weight 2 sin43 weight cos5832 2.3 9.81 3 cos58 4 sin 43 13.1 N LLF F × × = ×× × ××= × =    [1] [1] [1] [1] (b) Force F has a rightward component and an upward component. Since weight points only vertically down, force at Q must have a leftward horizontal component to counter the rightward horizontal component of F, so that PQ can be in equilibrium. Hence, the force at Q cannot be vertical. [1] [1]