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 r and I=received EarthP Area of solar panel so  = 24 sun received PP Area of solar panel r =received by Earth received by MarsPP   =   11 2 11 2(1.0) ( ) 4 (1.5 10 ) 4 (2.3 10 ) sun sunPP Area of solar panel on Mars Area of solar panel = 2.4 m2 11 12417 Ans: D Resolving to vertical direction: A0cos15° 12 12418 Ans: D Refer to definition of Rayleigh criterion 14 12419 Ans: C I = nAvq Same current (I) and same material (so same n) Smaller diameter, smaller A, drift velocity is larger. (v 14 12420 Ans: B l=R A Biggest area and shortest length to give minimum resistance 15 12421 Ans: A Using potential divider, Pd across 2.0Ω,  == + 2.0 2.0 (0.60) 0.402.0 1.0V V Same direction of polarity as the driver cell of 0.60 V.

Topic PYYQQ Part Suggested solutions 15 12422 Ans: A T increases, R (thermistor) decreases R3, the effective of R and R2, will decrease Using potential divider between R3 and R1, p.d. across R3 and hence p.d. across R2 will decrease So A2 will decrease. Overall resistance of circuit will decrease. So A1 will increase. 16 12423 Ans: C Option A, Force by electric field on negative ion is out of the page Force by magnetic field on negative ion is into the page Both can cancel off for ion to go straight Option B, Force by electric field on negative ion is downward and forward Force by magnetic field is upwards. Vertical forces can balance and ion can accelerate forwards in a straight line Option C Force by electric field on negative ion is into the page Force by magnetic field on negative ion is downward Not possible Option D Force by electric field on negative ion is forward Force by magnetic field on negative ion is zero Ion can accelerate forward. 16 12424 Ans: B Magnetic force provides the centripetal force == 2 B mvF Bqv r ➔ = Bqrv m Period, = = =222 / rrmT v Bqr m Bq 17 12425 Ans: D Area swept = 80 × (150 × 60) = 7.2 × 105 m2 Flux cut = field density × area swept = (1.0 × 10–5) (7.2 × 105) = 7.2 Wb 18 12426 Ans: B T/2 V0 –V0 V T R R1 R2 R3

Topic PYYQQ Part Suggested solutions 18 12427 Ans: A Resistance of cables = (2 × 100)(0.030) = 6.0 Ω Current in cables = P/V = (100 × 106) / (400 × 103) = 250 A Power loss in cable = I2R = (250)2(6.0) = 3.75 × 105 W Power input to station = power output – power loss in cable = 100 × 106 – 3.75 × 105 = 99.63 MW 19 12428 Ans: D = hp and == /E hc pc 19 12429 Ans: B 20 12430 Ans: D Mass defect = total mass of nucleons + electrons – mass of atom = 26(1.00728+0.00055) u + (56-26) (1.00867 u) – 55.93493 u = 0.52875 u = 8.78 × 10–28 kg 2, 3, 5 22401 a Increase in GPE = 27 J 27mg h= (0.43)(9.81) 27h= 6.40 6.4h = = m b At maximum height, 0yv = Using 22 2y y yv u as=+ and take upwards as positive 20 2( 9.81)(6.40)yu= + − 11.2 11yu == m s–1 c i 11.2tan32 y xx u uu= = 17.9 18xu == m s–1 Apply conservation of momentum (0.43)(17.9) + (0.67)(0) = (0.43 + 0.67) v v = 6.997 = 7.0 m s–1 c ii Vertical height is the same, so time of fall (or rise) is the same Horizontal speed is smaller for the combined object (by a ratio of 7.0/18 = 0.39), so the horizontal distance travelled is smaller by 0.39 of the horizontal distance travelled by the original object. 8, 9 22402 a There are no inter-molecular forces between the molecules, except during a collision. b Internal energy of an ideal gas is the total kinetic energy Ek of the molecules due to the random motion of the molecules. (No potential energy Ep due to intermolecular / inter-atomic forces between molecules.)

Topic PYYQQ Part Suggested solutions c i Total KE = 3 2 NkT ( )( )( ) 23 233 2.50 10 1.38 10 21 273.152 −=   + 1522= = 1520 J c ii Increase in internal energy ΔU = ΔKE ( )( )( ) 23 233 2.50 10 1.38 10 15 77.62 −=   = J Thermal energy is supplied to the gas and the gas will expand Work done by the gas = PV = ( )( ) 541.0 10 6.3 10 − = 63 J Applying 1st law of thermodynamics to onU Q W = + 77.6 63 toQ=− ➔ 140.6 140toQ == J 12 22403 a i So that waves from the loudspeakers have the same frequency / wavelength to ensure coherence. a ii • Microphone connected to cathode ray oscilloscope (CRO) • Metre rule, clamped on retort stand, is placed between two speakers Move the microphone directly along a horizontal line from centre of loudspeaker X to the centre of loudspeaker Y. Record the reading L1