2023 A Level H2 Physics P1 Soln

Suggested solutions to 2023 A-Level H2 Physics Paper 1 1 C 11 B 21 B 2 D 12 C 22 D 3 A 13 B 23 C 4 C 14 D 24 A 5 B 15 B 25 D 6 B 16 D 26 C 7 C 17 A 27 D 8 B 18 A 28 A 9 C 19 C 29 A 10 A 20 B 30 C Q Ans Working 1 C Person’s reaction time using a stopwatch- person could either be fast or slow in pressing the stopwatch (at the start or at the end of operating stopwatch). This could lead to random error instead of systematic error. 2 D Displacement of car can be found from the area under v-t graph. Hence, the trapezium area = 1 ()2 p q r+ 3 A By Newton’s 2nd law; ( 0) av mv m vF tt −==  ----- (1) Kinetic energy E = ½ mv2 ----(2) Combining (1) and (2) 2; but 2 2 Since m is constant EFt mv v m EFt m m Ft Em Ft E == = =  Please note that impulse (FΔt) is not equal to work done. 4 C For an elastic collision relative speed of approach = relative speed of separation 1 2 2 1 1 2 2 1 ()u u v v u u v v − − = − + = −

5 B Consider the sheet of metal is broken to 2 parts: Part 1: mass = 2m; c.g. is located 6.0 cm from point P Part 2: mass = m; c.g. is located 2.0 cm from point P Hence, taking moments about point P; Net moments due to part 1 and part 2 = 2m (6.0) + m (2.0) -------- (1) Assume the net c.g. is x away from point P Net moments due to entire shape (at the net c.g. position) = 3m (x) ----- (2) equating (1) and (2) 2m (6.0) + m (2.0) = 3m (x) x = 4.7 cm 6 B Energy lost by wind per second = ( ) 221(9.7) 4.0 1.5 66.7 J2 −= Power generated 0.60 66.7 40.0 W=  = 7 C Angular displacement in one year is 2 . Therefore, in half a year the angular displacement is ( )1 22 = 8 B Definition of gravitational field strength. 9 C For path C, there must be a force exerted on the rocket to balance the gravitational force acting on the rocket in order for it to continue travelling as shown. For path D, it is possible for the shuttle to travel as shown without firing its rockets – it would just be slowing down.

10 A Time interval between the molecule hitting the shaded wall 2p v= Force exerted by N molecules on the shaded wall ( ) 2 () 2 2 2 p t N mv m v p v Nmv p v Nmv p =  −−= = = 11 B U  T (note: T must be in K not °C) initially when T = 273 K (0 °C), internal energy is U later when T = 273+273 K, i.e. doubled, internal energy is also doubled hence answer 2U 12 C steeper gradient  smaller specific heat capacity (since temperature rises more for the same heat), so P has smaller specific heat capacity longer horizontal line  larger specific latent heat (since requires more heat to completely melt the solid), so P has larger specific latent heat 13 B max KE = ½ m ω2 x02 m = 2 23 2 9 10 6 02 10 . . −  ω = 3 2 3 10  − x0 = 2 × 10−6 substituting gives max KE = 4.2 × 10−31 J 14 D compare the peak times for wave X, the peak happens at t = 5 units for wave Y, the peak happens at t = 15 units period = 30 units so phase difference = 10360 36