2024 TJC H2 Prelim Paper 1_Solutions

2024 H2 Physics Paper 1 Solutions 2024 H2 Prelim Paper 1 Solutions: 1 2 3 4 5 6 7 8 9 10 A B B C A A D A D B 11 12 13 14 15 16 17 18 19 20 C C B A A A C A A C 21 22 23 24 25 26 27 28 29 30 D C A C D B D C D B 1 A Units of ∆𝑃 = 𝑘𝑔𝑚−1𝑠−2, units of 𝜌 = 𝑘𝑔𝑚−3 Units of ∆𝑃 𝜌 = 𝑚2𝑠−2, hence n =0.5 2 B Accurate → How close they are to the true value Precise → How close the measured readings are from one another Readings d/ mm Ave Value Largest – smallest value 1.49 1.46 1.52 1.50 1.4925 0.06 1.48 1.58 1.51 1.40 1.4925 0.18 (not precise) 1.35 1.37 1.42 1.42 1.39 0.07 1.32 1.37 1.41 1.50 1.4 0.18 3 B Vrelative to ground = 22350 100− = 335 km/h 4 C 5 A Change in momentum = area under F-t graph Area under F-t graph from 0 to 10 seconds = 0.5 x (8+4) x 12 + [(-0.5) x (2) x (12)] = 60 N s Change in momentum = final momentum – initial momentum 60 = mv – 0 v = 200 m s−1 Vplane = 100 km/h Vplane = 350 km/h Vrelative to ground

2024 H2 Physics Paper 1 Solutions 6 A 2 221000 2 0 10 0 50 0 314 N net 2 2 2 Force by wall on water= F mass per unit time v m ρ(Vol)Δv Δv since m ρ(Vol) Δt Δt ρ r v(0 - v) = - ρ r v . ) ( . ) . −   =   = = = =  = − (  =− By Newton’s 3rd Law, Force by water on wall = Force by wall on water = 0.314 N 7 D On each cuboid ; Volume of water displaced is the same, therefore U is constant for all cuboids of equal volume. T + U = mg T +U =Vg (where  is density of material/cubpid) T = Vg – U Since cuboid W has largest density, T on string is the largest. 8 A Extra energy stored = area of trapezium = ( )( )1212 2 1 xx TT −+ 9 D Since P and Q have the same angular velocity, the linear speed (v = rω) of P is greater than Q as r1>r2. Also, centripetal force (F = mrω2) is greater for P than Q as r1>r2. 10 B At P: N + mg = mv2/r N = mv2/r – mg rgv = when N  0 Since GPE at P and Q is the same, KE at P = KE at Q. At Q:mg – N = mv2/2r N = mg - m(rg)/2r N = mg/2 F x X1 T2 X2 T1

2024 H2 Physics Paper 1 Solutions 11 D By conservation of energy, KEi + GPEi = Kef + GPEf    + − = + −       20½ 7 GMm GMmmvRR = 12 7 GMv R 12 C The antenna will have the same speed as the space -craft and is still bound in orbit. There is still gravitational force acting on it which causes it to move in circular motion. 13 B 21 3=  PV Nm c V and m remain unchanged  Nc P  2 = constant  ( ) N cN. PP  = 2 5 3 3 0 10 4 2  N = 2.0 105 m2 s2 14 A This is an SHM question and the Ep graph is A, lowest at the equilibrium and maximum at the amplitudes (opposite from the KE-r graph). 15 A ma = mg – R Object will remain in contact when R is greater than 0. Object will lose contact when R = 0, ma = mg For SHM, a = ω2xo = (2πf)2xo = g xo = g (2πf)2 = 9.81 (2π(2.0))2 = 9.81/(2𝜋(3.0))2 = 0.028 m 16 A I kA2= Resu