2025 NYJC H2 Phy 9749 P3 Answers

NYJC 2025 9749/03/J2Prelim/25 [Turn over NANYANG JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME Solution CLASS TUTOR’S NAME CENTRE NUMBER S INDEX NUMBER PHYSICS 9749/03 Paper 3 Longer Structured Questions 19 September 2025 2 hours Candidates answer on the Question Paper. No Additional Materials are required. READ THESE INSTRUCTIONS FIRST Write your name, class, Centre number and index number 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 or rough working. 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. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. For Examiner’s Use Section A 1 / 6 2 / 8 3 / 12 4 / 10 5 / 9 6 / 8 7 / 7 Section B 8 / 20 9 / 20 Total / 80 This document consists of 24 printed pages. H

2 NYJC 2025 9749/03/J2Prelim/25 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 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  = 0 2 IB d magnetic flux density due to a flat circular coil = 0 2 NIB r magnetic flux density due to a long solenoid = 0B nI radioactive decay 0 exp( )x x t =− decay constant 1 2 ln2 t =

3 NYJC 2025 9749/03/J2Prelim/25 [Turn over Section A Answer all the questions in the spaces provided. 1 A projectile is fired from ground level with initial velocity u at an angle θ to the horizontal as shown in Fig. 1.1. The projectile strikes a target which is at a horizontal displacement x from the point of projection and a vertical height y above ground level. (a) Neglecting the effect of air resistance , show that the vertical height y is given by the expression 2 tan 4.91 cos xyx u  =−  [3] (b) Given that the angle  is 60°, the horizontal displacement x is 115 m and the vertical height y is 23 m, calculate the speed u. u = m s−1 [1] (c) Fig. 1.2 shows the variation with time t of the vertical velocity vy of the projectile when air resistance is negligible. On the same axes, sketch a graph to show the variation with time t of the vertical velocity vy of the projectile when air resistance is not negligible. [2] [Total: 6]  y u x Fig. 1.1 2 [M1] [C1]2 c ) ( cos ) os 1 2 1( sin ) ( 9.812 xx y y y s u t x u t xt u s u t a t y u t t    = = = =+ = + − target 2 1 11523 115 tan60 4.91 cos60 38 m s u u − =−  = vy t The curve must decrease at a decreasing rate throughout, with following details: 1. Positive area larger than negative area 2. a larger initial gradient 3. a smaller t-intercept 4. at vy = 0, gradient parallel 0 [M1] [A0] 2 2 sin 4.91 ( )cos cos tan 4.91 co s xxyu uu xyx u     =  − =−  Fig. 1.2

4 NYJC 2025 9749/03/J2Prelim/25 2 (a) State the two conditions necessary for a body to be in equilibrium. 1. 2. [2] (b) Fig. 2.1 shows a uniform beam AB of length 6.0 m and weight 2700 N suspended by two ropes AC and BC, each of length 6.0 m. The tensions in ropes AC and BC are T1 and T2 respectively. A worker of weight 900 N is holding onto the beam at point D, where AD = 4.0 m and DB = 2.0 m. The beam makes an angle  to the horizontal. The point M is the mid-point of the beam and the point G on the beam is the position of the centre of gravity of the beam and the worker. (i) Explain in terms of forces acting on the beam, why the point G must lie directly below C. [2] C A B M D G  Fig. 2.1 The combined weight of the beam and the worker can be considered to be acting at G. Considering moments about C, moments due to T1 and T2 about C is zero as their lines of action pass through C . Hence for the resultant moments about C to be zero, the line of action of the combined weight of the beam and worker must pass through C as well so that the moment of the combined weight about C is zero. Hence the vertical line through G must pass through C. ground The net force is zero. [B1] The net torque is zero./ The net moment about any point is zero. [B1]

5 NYJC 2025 9749/03/J2Prelim/25 [Turn over (ii) Calculate the distances MG and DG. distance MG = m distance DG = m [2] (iii) If the angle  is 2.8, determine the magnitude of the tension T2. tension T2 = N [2] [Total: 8] For horizontal equilibrium, T1 cos (60.0o – 2.8o) = T2 cos (60.0o + 2.8o) o 12 o 12 cos62.8 cos57.2 0.8438 (1) TT TT =   = For vertical equilibrium, T1 sin 57.2o + T2 sin 62.8o = 2700 + 900 (2) Substituting equation (1) into equation (2) and solving for T2, T2 = 2250 N (3 s.f.) Considering moments due to the weight of beam, Wb, and that of the worker, Ww , about C, Let distance MG be x and distance DG be y Wb . x = Ww . y (2700) x = (900) y y = 3 x MD = AD – AM = 4.0 – 3.0 = 1.0 m MD = MG + GD = x + 3 x = 1.0 m, 4x = 1.0 m, x = 0.25 m. MG = 0.25 m DG = 0.75 m

6 NYJC 2025 9749/03/J2Prelim/25 3 (a) Explain why gravitational potential is a negative value for an isolated mass. [3] (b) A satellite can orbit the Earth along an east-to-west direction (known as a retrograde orbit) as well as along the west-to-east direction (known as a prograde orbit). (i) A satellite is launched in the west -to-east direction from a launch pad on the Equator to the geostationary orbit. Explain why this launch direction is preferred. [2] (ii) The Earth may be considered to be a uniform sphere of radius 6400 km with its mass of 6.0 × 1024 kg concentrated at its centre. Show that the geostationary satellite is 3.59 × 107 m above the Earth’s surface.