2025 VJC H2 Phy Prelim P2 Solution

VJC 2025 9749/02/J2Prelim [Turn over VICTORIA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME SOLUTION CLASS TUTOR NAME PHYSICS 9749/02 Paper 2 Structured Questions 16 September 2025 2 hours Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your name, class and tutor name in the spaces at the top of this page. Write in dark blue or black pen on both sides of the paper. You may use a HB pencil for any diagrams, graphs. Do not use staples, paper clips, glue or correction fluid. The use of an approved scientific calculator is expected, where appropriate. Answer all questions. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 / 7 2 / 8 3 / 8 4 / 9 5 / 8 6 / 12 7 / 10 8 / 19 Total / 80 This document consists of 24 printed pages.

2 VJC 2025 9749/02/J2Prelim Data speed of light in free space c = 3.00 × 108 m s−1 permeability of free space  = 4 × 10−7 H m−1 permittivity of free space  = 8.85 × 10−12 F m−1 (1 / (36)) × 10−9 F m−1 elementary charge e = 1.60 × 10−19 C the Planck constant h = 6.63 × 10−34 J s unified atomic mass constant u = 1.66 × 10−27 kg rest mass of electron me = 9.11 × 10−31 kg rest mass of proton mp = 1.67 × 10−27 kg molar gas constant R = 8.31 J K−1 mol−1 the Avogadro constant NA = 6.02 × 1023 mol−1 the Boltzmann constant k = 1.38 × 10−23 J K−1 gravitational constant G = 6.67 × 10−11 N m2 kg−2 acceleration of free fall g = 9.81 m s−2

3 VJC 2025 9749/02/J2Prelim [Turn over Formulae uniformly accelerated motion 21 2s ut at=+ 22 2v u as=+ work done on / by a gas W p V= hydrostatic pressure p gh= gravitational potential /Gm r =− temperature / K / C 273.15TT =  + pressure of an ideal gas 21 3 Nmpc V=   mean translational kinetic energy of an ideal molecule 3 2E kT= displacement of particle in s.h.m. 0 sinx x t = velocity of particle in s.h.m. 0 cosv v t = 22 0xx=  − electric current =I Anvq resistors in series 12 . . .R R R= + + resistors in parallel 121/ 1/ 1/ . . .R R R= + + electric potential 04 QV r= alternating current/voltage 0 sinx x t = magnetic flux density due to a long straight wire  = IB d 0 2 magnetic flux density due to a flat circular coil = NIB r 0 2 magnetic flux density due to a long solenoid =B nI 0 radioactive decay 0 exp( )x x t =− decay constant 1 2 ln2 t =

4 VJC 2025 9749/02/J2Prelim 1 A golfer is practising his tee shot from a platform 7.0 m off the ground as shown in Fig. 1.1. The golf ball was launched at a speed of 50 m s−1, 40 above the horizontal. Assume air resistance is negligible. (a) Determine the maximum height above the ground attained by the ball. maximum height = m [3] (b) Calculate the time of flight of the ball. time of flight = s [2] 50 m s−1 7.0 m ground Fig 1.1 Initial vertical velocity [1] At the top of flight the vertical velocity Using , taking up as positive, we get [1] Max height = 52.647+7.0 = 59.647 = 60 m [1] Can also solve via energy considerations When ball touches the ground 7 0 mys. =− Using 21 2 yys u t at=+ , ( ) 217 32 139 9 812. t . t− = + − [1] 0 211 st.=− (reject) or 6 7639 6 8 st . .== [1]

5 VJC 2025 9749/02/J2Prelim [Turn over (c) A golf ball typically bounces a few times after a tee shot as shown in Fig. 1.2. The first time the ball touches the ground is indicated by A and the fourth time it touches the ground is indicated by B. Take the upward direction as positive. Fig. 1.2 Sketch, on Fig. 1.3, a graph to show the variation with time of the vertical velocity of the ball between from the instant it leaves A to the instant it reaches B. Fig. 1.3 [2] [Total: 7] A B vertical velocity time 3 parallel lines, same gradient, in between bounces; with vertical lines (or very slight slant) when ball is in contact with ground. [1] Speed after rebound is lower than speed before rebound and area above and below graph are equal [1]

6 VJC 2025 9749/02/J2Prelim 2 (a) Define Newton’s second law. [1] (b) A light rope is attached to a 120 kg box on the ground. The other end of the rope runs over a light frictionless pulley. A 80 kg man climbs up the free-hanging rope. As the man climbs up the rope, he pulls on the rope hard enough to cause himself to accelerate upwards. The only point of contact between the rope and the man occurs at his hands. Fig. 2.1 (i) Draw, on the outline of the man in Fig. 2.2, the forces acting on the man as he climbs. Fig. 2.1 [1] tension weight The rate of change of momentum of a body is proportional to the resultant external force acting on it and is in the direction of the resultant force.

7 VJC 2025 9749/02/J2Prelim [Turn over (ii) If the man climbs the rope with an acceleration of 8.0 m s −2, determine the acceleration of the box. acceleration = m s−2 [2] (c) The man releases the rope and the box falls. The box hits the ground with a speed of 2.0 m s −1 and sinks into the ground over a vertical distance of 10 cm before coming to a stop. Calculate the force exerted by the ground on the box during the deceleration. force = N [3] [Total: 8] Let m = mass of man. Let M = mass of box. Apply F = ma to man: T – mg = m aman T = m (g + aman) ---(1) Apply F = ma to box: T – Mg = M abox ---(2) Sub. T from (1) into (2): m (g + aman) – Mg = M abox 80 (9.81 + 8.0) − 120 × 9.81 = 120 abox abox = 2.1 m s−2. =+22 2v u as =+ 20 2.0 2 (0.10)a a = −20 m s−2 Apply F = ma to box: F – Mg = Ma F – 120 × 9.81 = 120 × 20 F = 3600 N F Mg a = 20 m s−2 v s = 10 cm

8 VJC 2025 9749/02/J2Prelim 3 A peg is fixed to the rim of a vertical turntable of radius r rotating with a constant angular speed , as shown in Fig. 3.1. Fig. 3.1 A parallel beam of light is incident on the turntable such that the shadow of the peg is observed on the screen. Initially, the peg is at position S  and its shadow is at S. After time t, the peg moves through an angle of  and it is positioned at T while its shadow is at T. The displacement x of the shadow from O is shown in Fig. 3.1 where the upward direction is taken to be positive. (a) (i) Express the angular displacement  of the peg in terms of  and t. [1] (ii) Write down an expression for the displacement x of the shadow on the screen in terms of , t and r . [1] (iii) Hence, prove that the shadow of the peg is moving in simple harmonic motion. Explain your working. [2]  r S S T T peg shadow of peg screen displacement x of shadow O t= cosx r t=−  ( ) 2 2 2 cos sin cos cos x r t dxv r t dt dva r t r t xdt =− == = = = − − = −        which is the defining equation of a simple harmonic motion.

9 VJC 2025 9749/02/J2Prelim