2025 VJC H2 Phy Prelim P3 Solution

VJC 2025 9749/03/J2Prelim [Turn over VICTORIA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME SOLUTION CLASS TUTOR NAME PHYSICS 9749/03 Paper 3 Longer Structured Questions 18 September 2025 2 hour 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. Section A Answer all questions. Section B Answer one question only. You are advised to spend one and a half hours on Section A and half an hour on Section B. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use 1 / 7 2 / 10 3 / 8 4 / 9 5 / 7 6 / 9 7 / 10 8 / 20 9 / 20 Total / 83 This document consists of 23 printed pages.

2 VJC 2025 9749/03/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 9749/03/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 9749/03/J2Prelim 1 Fig. 1.1 shows a 1000 N uniform thin rod being towed by a force T and moving at constant horizontal velocity. Fig. 1.1 (a) State the conditions required for a body to be in equilibrium. [2] (b) On Fig. 1.1, draw and label the two other forces acting on the rod. [2] (c) Given angle  is 70, determine force T. force T = N [3] A 30o floor T Weight of rod Reaction force by floor on rod A body is in equilibrium if the net force in any direction on the body is zero and the net torque about any axis on the body is zero. Let be the length of the rod. Taking moment about A, clockwise moment due to weight = 1000( co s30 ) [1]2 anticlockwise moment due to force = ( si n 40 ) [1] At equilibrium net moment is zero, 10 L L T T L   00( cos30 ) ( sin40 ) 2 670 N [1] L TL T  =  = 70 40

5 VJC 2025 9749/03/J2Prelim [Turn over [Total: 7] 2 The International Space Station (ISS) orbits the Earth at a height of 4.1105 m above the Earth’s surface. The radius of the Earth is 6.37106 m. (a) Both the ISS and the astronauts inside it are in free fall . Explain why this makes the astronauts feel weightless [1] (b) (i) Calculate the value of the gravitational field strength g at the height of the ISS above the Earth. 2 00 2 2 62 10 2 2 6 5 2 , [1] 9.81 (6.37 10 ), 8.66 N kg [1]( ) ( ) (6.37 10 4.1 10 ) − == = = = =+ +  +  GMg g R GMR gRGMgg R h R h g = N kg−1 [2] (ii) State the value of the centripetal acceleration of ISS at this height. aC = m s−2 [1] (iii) The speed of the ISS in its orbit is 7.7 km s−1. Show that the period of the ISS in its orbit is 92 minutes. 65 2 ( ) [1] 7700 2 (6.37 10 4.1 10 ) 92.2 = 92 min [1]   =+ =  +  = vT R h T T [2] (iv) The ISS is in a low Earth orbit. Suggest an advantage of this orbit as compared to higher orbits. [1] There is no contact force between the astronaut and the floor of the space station. It requires less fuel to launch, and hence it is less expensive. It is easier to access for maintenance and repair. Higher resolution of images captured by ISS of features on the surface of the Earth. 8.66

6 VJC 2025 9749/03/J2Prelim (c) The ISS has arrays of solar cells on its wings. These solar cells charge batteries which power the ISS. The wings always face the Sun. 7% of the energy of the sunlight incident on the cells is stored in the batteries. The total area of the cells facing the solar radiation is 2500 m2. The intensity of solar radiation at the orbit of the ISS is 1.4 kW m −2 outside of the Earth’s shadow and zero inside it. The ISS passes through the Earth’s shadow for 35 minutes during each orbit. By reference to (b)(iii), calculate the average power delivered to the batteries during one orbit. Power stored in batteries = 0.07IntensityArea = 0.071.41032500 = 2.45105 W [1] Cells are in the Sun for 92 −35 = 57 min [1] Average power = 5557 2.45 10 1.5 10 W92   =  [1] average power = W [3] [Total: 10] 3 (a) Define the term angular velocity. [1] (b) A 10 kg baggage is left on a rotating baggage carousel at an airport as shown in Fig. 3.1. Fig. 3.1 The baggage stays at a fixed position on the slope of the carousel and rotates about in a circle of radius 10 m. The angle θ that the slanted surface makes with the horizontal is 37. The frictional force acting on the baggage is 60 N. The baggage is moving in uniform circular motion. r  = 37º Side View Direction of rotation Angular velocity is the rate of change of angular displacement.

7 VJC 2025 9749/03/J2Prelim [Turn over (i) Explain, using Newton’s law(s) of motion, why the baggage will experience a net force towards the centre of the circle. [2] (ii) Considering the forces acting on the baggage, show that the normal contact force is 78 N. Consider forces perpendicular to slant surface, cos [1] (10 9.81)cos37 78.346 [1] 78 N (to 2 s.f.) NW = =   = = [2] (iii) Calculate the time required for the baggage to complete one full rotation. 2 2 Net force on baggage = cos sin [1] cos sin 2 cos sin [1 ] 2 cos sin (10)(10) 2 (60)cos37 (78)sin37 64 s fN mr f N mr f NT mrT fN          − =−  =− = − =  −  = [1] time = s [3] [Total: 8] wei