NYJC 2026 J2 H2 Prelim P2 (Teacher) Final (with comments)

NYJC 2026 9478/02/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 9478/02 Paper 2 Structured Questions 16 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. Answer all questions. 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 1 / 10 2 / 8 3 / 10 4 / 14 5 / 11 6 / 22 Total / 75 This document consists of 23 printed pages.

2 NYJC 2026 9 478/02/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 9478/02/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 9478/02/J2Prelim/26 Answer all the questions in the spaces provided. 1 (a) State the conditions required for a body to be in equilibrium. [2] (b) Block A of mass 1.7 kg is connected by a light inextensible string to a counterweight B of mass M via a system of two massless and frictionless pulleys P and Q. Block A is balanced on a smooth plane inclined at 45°. Pulley Q is a double pulley made up of two fixed wheels of different radii attached to a common axle, with the radii of the smaller and larger wheels r and 3r respectively as shown in Fig. 1.1. Fig.1.1 (i) Block A is in equilibrium. In Fig. 1.2, draw and label all forces acting on block A. Fig.1.2 [2] A B 45° pulley Q pulley P 3r r smooth slope string string axle block A normal contact force force of pulley P on A weight [B1] 3 correct forces [B1] correct labels The sum of forces acting on the body in any direction must be zero. (The resultant force on the body is zero / No resultant force) The sum of the moments of the forces about any point must be zero. (The resultant moment (or torque) on the body about any axis is zero. / No resultant torque) ER note: a small number did not sum the clockwise or anticlockwise moments or refer to the resultant or net force. Many left out the terms “net” or “resultant”, just simply just stated that the sum of forces is zero which is insufficient. Many mistakenly wrote momentum instead of moments! For rotational equilibrium, it doesn’t matter which point or axes you take (it holds for any point inside or outside of the body!). The resultant torque will still be zero. If you stated “ about a point” or “a point on the body”, it’s not general enough and no credit given. Force of pulley on A must start from the edge of the block and not the CG. That is not the tension from the string! Since object is in equilibrium, all forces must meet at the CG! “Normal” alone is not accepted. Normal just means perpendicular and not a force. Many did not label the forces in full. Lengths of the vectors should all balance out. Some lines were too short/long.

5 NYJC 2026 9478/02/J2Prelim/26 [Turn over (ii) Calculate the tension in the string connected to pulley P. tension = N [2] (iii) Determine the mass M of counterweight B. M = kg [2] (iv) Suggest and explain a change that can be made to the pulley such that block A will accelerate up the slope. [2] [Total: 10] // slopeBlock A is in equilibrium: 0 F =∑ A A sin45 2 sin45 (1.7)(9.81)sin45 [M1]22 5.90 N [A1] mg T mgT °= ° °= = = 5.90 N Taking pivot about axel, 0=∑τ B B CW moments ACW moments (3 ) 5.90 [M1](3 ) 3 (3)(9.81) 0.200 N [A1] m g r Tr Tr Tm gr g = = = = = 0.200 kg The larger wheel of the double pulley can be made larger with the smaller wheel remaining constant / more than 3 times larger than the smaller wheel. [M1] This is so that the clockwise moment will be larger than the anticlockwise moment / net clockwise torque causing the block A to slide up the slope. [A1] You need to read the question carefully and answer it. The question explicitly asked for changes that can be made to the pulley, so any answer not related to the pulley is not accepted. Since this is moments question, the main aim is to make the system have a net clockwise moment, the easiest is to increase the 3r radius to some value larger. Many did not state clearly which part of the pully’s radius is to be changed. (can also change the radius r to be smaller). Many jumped straight to the conclusion that tension on A till be greater, which wasn’t awarded credit as the main physics of having a net clockwise moment is missing. Badly done. The two strings connecting A and B are not the same!!! This is a moments question, and you need to take time to identify what is happening within the system. Most student simply equated the tension to the weight of B which does not make any sense. Badly done. Majority did not realise that the force from A is distributed between 2 strings. Most did not divide the force by 2.

6 NYJC 2026 9478/02/J2Prelim/26 2 A plane is flying with a velocity of 240 m s−1 at an angle of 30° with respect to the horizontal, as shown in Fig. 2.1. At an altitude of 2.0 km, a projectile is released from the plane. The projectile hits the target on the ground. Assume air resistance is negligible. Fig. 2.1 (a) Determine the horizontal displacement of the projectile from the point of release to the target. horizontal displacement =