ASRJC H2 Physics 2024 A Level suggested solutions for P1-3 ASRJC

ANDERSON SERANGOON JUNIOR COLLEGE PHYSICS 9478 Suggested Solutions to 2024 A Level P1 Topic PYYQQ Part Suggested solutions 1 12401 Ans: A Area enclosed by the orbit = 2Ar ( )=    =  22 6 174 10 10 5 10 m2 = 5 × 10–1 Gm2 2 12402 Ans: B Acceleration = gradient of v – t graph 0 – t1: gradient is constant and positive 0 – t2: gradient is constant and negative 2 12403 Ans: C Consider the vertical direction and taking downwards as positive: uy = –15 sin40° m s–1, vy = 15 sin40° m s–1, g = 1.62 m s–2 =+v u at 15 sin40° = –15 sin40° + 1.62t t = 11.9 s 4 12404 Ans: B Equilibrium: Closed loop vector diagram 4 12405 Ans: D Considering forces on the beam which is in equilibrium, so no net force vertically. Initial: upward forces by 3 springs = W, so 3kx = W Final: upward forces by 2 springs = 2W, so 2kx’ = 2W = 2(3kx) x' = 3x 5 12406 Ans: B Net WD = area 1 – area 2 Fx Fy W 30° Area 1 = WD on specimen to give extension T Area 2 = Energy recovered when extension reduced back to 0

Topic PYYQQ Part Suggested solutions 5 12407 Ans: C Driving force by car FD − −  =600 900 sin(10 ) 0DFg = 2133DF N Power by engine = FDv = (2133)(16) = 34 kW 9 12408 Ans: D Work done on gas are thermal processes where volume decreases 6 12409 Ans: A =sr ➔ ( )l =r ➔ l= r l ==t rt 6 12410 Ans: A =  =24 3600 86400T s and  = 2 86400 rad s–1    = = =   27 2 1.3 10 2 0.0342 86400ar m s–2 Since the object is undergoing circular motion, the acceleration is directed towards the centre of the circle, which is the centre of the Earth. 7 12411 Ans: A Potential = r Gm− Magnitude of potential = Gm r Thus, as r increases, magnitudes of potentials decrease, and the magnitude of potential difference between them also decreases as magnitude of potential is a 1/r relationship. 8 12412 Ans: A Duration between collision of shaded side, = 2xt v Change in momentum of particle, = 2p mv Average Force by box on particle, == 2p mvF tx By N3L, magnitude of average force by particle on box = magnitude of force by box on particle. 9 12413 Ans: D Heat loss by V volume of water = heat gain by V/4 volume of water ( ) ( ) ( ) − = − 60 20 4 VV c T c T ➔ T = 52 °C 900g FD  r P Q P Q 0

Topic PYYQQ Part Suggested solutions 10 12414 Ans: C At t = 0, mass at max displacement with zero speed. In one oscillation, mass crosses the equilibrium position twice, hence there are two peaks per period. 10 12415 Ans: C For critical damping, there is no oscillation, and the object returns to its equilibrium in the minimum amount of time. 11 12416 Ans: C I  = 24 sun Earth P