NYJC 2026 J2 H2 Prelim P3 (Teacher) Final (with comments)

NYJC 2026 9478/03/J2Prelim/26 [Turn over NANYANG JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME CLASS TUTOR’S NAME CENTRE NUMBER S INDEX NUMBER PHYSICS 9478/03 Paper 3 Longer Structured Questions 21 September 2026 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. 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. S ection B Answer one question only. Y ou are advised to spend one and a half hours on Section A and half an hour on Section B. A t 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 / 7 2 / 8 3 / 7 4 / 8 5 / 9 6 / 6 7 / 10 Section B 8 / 20 9 / 20 Total / 75 This document consists of 28 printed pages. SOLUTION

2 NYJC 2026 9 478/03/J2Prelim/26 Data speed of light in free space c = 3.00 × 108 m s−1 permeability of free space 0µ = 4π × 10−7 H m−1 permittivity of free space 0ε = 8.85 × 10−12 F m−1 0 1(4πε = 8.99 × 109 m F−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 Avogadro constant NA = 6.02 × 1023 mol−1 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 NYJC 2026 9478/03/J2Prelim/26 [Turn over Formulae uniformly accelerated motion s = 21 2ut at+ v2 = u2 + 2as work done on / by gas W = pV∆ pressure p = F A gravitational potential φ = GM r− temperature T / K = T / °C + 273.15 pressure of an ideal gas p = 21 3 Nm cV <> mean translational kinetic energy of an ideal gas particle E = 3 2 kT displacement of particle in s.h.m. x = 0 sinxt ω velocity of particle in s.h.m. v = ( ) 22 00cosv t xx = ±−ωω electric current I = Anvq resistors in series R = R1 + R2 + … resistors in parallel 1/R = 1/R1 + 1/R2 + … capacitors in series 1/C = 1/C1 + 1/C2 + … capacitors in parallel C = C1 + C2 + … energy in a capacitor U = 2 21 11 2 22 QQV CVC= = charging a capacitor Q = 0 1[] t Qe − − τ discharging a capacitor Q = 0 t Qe − τ RC time constant τ = RC electric potential V = 04 Q rπε alternating current / voltage x = 0 sinxt ω magnetic flux density due to a long straight wire B = 0 2 d µ π I magnetic flux density due to a flat circular coil B = 0 2r µ NI magnetic flux density due to a long solenoid B = 0µ nI energy states for quantum particle in a box En = 2 2 28 h nmL radioactive decay x = 0e tx −λ radioactive decay constant λ = 1 2 ln2 t

4 NYJC 2026 9478/03/J2Prelim/26 L1 L2 A B Fig 1.1 4.0 m s−1 Section A Answer all the questions in the spaces provided. 1 Two spheres A and B of masses 3.7 kg and 2.7 kg respectively are attached to two light inextensible strings L1 and L2 each of length 0.50 m. The combination of masses is made to swing in a vertical circle with both strings always in a straight line, as shown in Fig. 1.1. At the top of its motion, mass B is travelling at 4.0 m s−1. (a) Determine the speed of mass A at this instant. speed of mass A = m s−1 [2] (b) Show that the tension in L1 at this instant is 200 N. [1] Both masses have the same angular velocity. AB AB vr vv rr = ω = ⇒ A 1 A 4.0 [M1]0.5 0.5 0.5 8.0 m s [A1] v v − =+ = 2 AA 2 AA 2(3.7)(8.0) (3.7)(9.81) [M1]0.50 0.50 200 N [A0] mvWT r mvTW r += = − = −+ =

5 NYJC 2026 9478/03/J2Prelim/26 [Turn over (c) Hence or otherwise, determine the tension in L2 at this instant. tension in L 2 = N [2] (d) At another instant in time, the combination of masses is at the bottom of its motion as shown in Fig. 1.2. A student suggests that because the tensions in strings L 1 and L2 change continuously during the moti on, the total mechanical energy (i.e. kinetic and potential energy) of the system is not conserved. With reference to tensions in the strings, s tate whether you agree with this statement and explain your reasoning. [2] [Total: 7] L1 L2 A B Fig 1.2 B A 2 BB BBA B 2 BB B BA B 2(2.7)(4.0) (2.7)(9.81) 201 [M1]0.50 261 N [A1] mvWTT r mvT WT r +−= = −+ =−+ = I disagree with the statement, as tension is always acting perpendicular to the direction of motion of both masses [M1] Hence there is no work done by tension and hence total mechanical energy of the system is conserved. [A1] or I disagree with the statement, as tension is always acting perpendicular to the direction of motion of both masses. [M1] Hence there is no work done by tension but there may be other resistive forces and hence mechanical energy is not conserved. [A1] Comments: Always sketch FB Ds to help visualise the forces. The addition of the forces will give rise to the ne t resultant force towards the center of the circular path. Comments: Badly done. The question is asking whether the changing tensions affect the conservation of energy between KE and GPE of the system. If you recall our treatment of non-uniform vertical circular motion, the normal contact forces always change but we still use COE to calculate the minimum speeds etc. Why? Because tension doesn’t do work on the system. When given new scenarios, always try to recall existing concepts /questions that have similarities and try to rationalise from there.

6 NYJC 2026 9478/03/J2Prelim/26 2 (a) Define gravitational potential at a point [2] (b) (i) The dwarf planet Ceres may be considered to be an isolated uniform sphere of mass 9.4 × 1020 kg and radius 4.7 × 105 m. A particle is launched vertically upwards from the surface of the planet. Assuming negligible resistance from the planet's atmosphere, determine the minimum speed of this particle that will result in it escaping from the gravitational pull of the planet Ceres. speed = m s−1 [3] (ii) Hydrogen may be assumed to be an ideal gas. The mass of a hydrogen molecule is 3.3 × 10–27 kg. Calculate the root-mean-square (r.m.s.) speed of a hydrogen molecule in hydrogen gas that is at a temperature of 240 K. r.m.s. speed = m s−1 [2] The gravitational potential at a point is defined as the work done per unit mass [B1] by an external force in bringing a small test mass from infinity to that point. [B1] Comments: It was entertaining to see gravitational potential about charges. That being said, it is a good sign that you save those 2 definitions in the same part of your brain. Just don’t make that mistake again. 2 2 11 20 2 5 ] By conservation of energy, Total GPE + KE at surface = Total GPE + KE at infinite distance away GPE KE GPE KE [ 1 002 1 2 (6.67 10 )(9.4 10 ) 1 24. C1 [C1]7 10 surface surface r r GMm mvr GM vr v v = ∞= ∞ − +=+ −+ = + = ×× =× 1520 m s 1 [ A ]−= Comments: Majority marks loss is from careless mistakes. More concerning are the students who used 2mv r . Nobody is circulating around anybody here. O yes, and 516.5 is rounded to 517