CJC.2025.H2.Phy.PRELIM.P2.Ans

CANDIDATE NAME MARK SCHEME CLASS 2T PHYSICS 9749/02 Paper 2 Structured Questions August 2025 2 hours Candidates answer on the Question Paper. READ THESE INSTRUCTIONS FIRST Write your name and class 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 an HB pencil for any diagrams or 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. This document consists of 26 printed pages and 0 blank page. [Turn over FOR EXAMINER’S USE Q1 / 10 Q2 / 8 Q3 / 12 Q4 / 8 Q5 / 9 Q6 / 11 Q7 / 22 PAPER 2 / 80 Catholic Junior College JC2 Preliminary Examinations Higher 2

2 DATA speed of light in free space c = 3.00 x 108 m s-1 permeability of free space 0 = 4 x 10-7 H m-1 permittivity of free space 0 = 8.85 x 10-12 F m-1 (1/(36)) x 10-9 F m-1 elementary charge e = 1.60 x 10-19 C the Planck constant h = 6.63 x 10-34 J s unified atomic mass constant u = 1.66 x 10-27 kg rest mass of electron me = 9.11 x 10-31 kg rest mass of proton mP = 1.67 x 10-27 kg molar gas constant R = 8.31 J K-1 mol-1 the Avogadro constant NA = 6.02 x 1023 mol-1 the Boltzmann constant k = 1.38 x 10-23 mol-1 gravitational constant G = 6.67 x 10-11 N m2 kg-2 acceleration of free fall g = 9.81 m s-2

3 [Turn over FORMULAE uniformly accelerated motion s = u t + ½ a t2 v2 = u2 + 2as work done on / by a gas W = p V hydrostatic pressure p = gh gravitational potential  = - Gm r temperature T / K = T / ˚C + 273.15 pressure of an ideal gas p = 1 3 Nm V 〈c2〉 mean translational kinetic energy of an ideal gas molecule E = 3 2 kT displacement of particle in s.h.m. x = x0 sin t velocity of particle in s.h.m. v = v0 cos t = 22 0 xx − electric current I = Anvq resistors in series R = R1 + R2 + ... resistors in parallel 1/R = 1/R1 + 1/R2 + ... electric potential V = Q 4πεor alternating current / voltage x = x0 sin t magnetic flux density due to a long straight wire B = μoI 2πd magnetic flux density due to a flat circular coil B = μoNI 2r magnetic flux density due to a long solenoid B = μonI radioactive decay x = x0 exp(-t) decay constant λ = 1 2 ln2 t

4 Answer all questions in the spaces provided. 1 A toy car of mass 0.42 kg is released from rest and accelerates along a straight track towards a wall. It hits the wall and rebounds in the opposite direction. The variation with time t of the momentum p of the toy car when not in contact with the wall is shown in Fig. 1.1 Fig 1.1 (a) Using Fig 1.1, calculate the impulse acting on the toy car during the collision. impulse = ……………….. N s [2] Solution: Impulse = change in momentum fi= p -p = -1.80 - (3.20) = - 5.00 N s M1 A1

5 [Turn over (b) Calculate the magnitude of the average acceleration of the car during the collision and state the direction of this acceleration relative to the initial motion of the car. average acceleration = …………………………..m s-2 direction = ………………..………………………………… [3] Solution: -1 -2 -2 Δp -5.00Δv= = =-11.9 m sm 0.42 Δv -11.9a= = =-79.3 m sΔt 0.15 |a|=79.3 m s Direction: opposite to the initial motion of the car M1 A1 A1 (c) Explain why the collision was inelastic. ……………………………………………………………………………………………………... …………………………………………………………………………………………………….. …………………………………………………………………………………………………….. ……………………………………………………………………………………………… [2] Solution: The momentum and hence speed of the car before and after the collision is different. The wall is stationary and hence has no momentum and no KE. The total initial KE and the total final KE of the system of the car and wall is not the same. Therefore, collision is inelastic. B1 B1 A0

6 (d) Calculate the percentage change in the kinetic energy of the car as a result of the collision. percentage change = ……………………… % [3] Solution: 22 22 if K,i K,f K,f K,i K K,i pp 3.20 1.80E = = =12.2 J and E = = =3.86 J2m 2(0.42) 2m 2(0.42) E -E%ΔE = ×100%E 3.86 - 12.2= ×100%12.2 =-68.4% (including -ve sign) M1 C1 A1 [Total: 10] 2 (a) State what is meant by the centre of gravity of an object. ………………………………………………………………………………………...…………... ……………………………………………………….....………….…….…….….............. [1] Solution: The centre of gravity of an object is a point where the entire weight of the object is taken to act. B1

7 [Turn over (b) A hollow plastic sphere is attached at one end of a bar. The sphere is partially submerged in water and the bar is attached to a fixed vertical support by a pivot P, as shown in Fig. 2.1. Fig. 2.1 (not to scale) The sphere has weight 0.30 N. The distance from P to the centre of gravity of the sphere is 0.29 m. The weight of the bar is negligible. The system shown in Fig. 2.1 is part of a mechanism that controls the amount of water in a tank. Water enters the tank and causes the sphere to rise. This results in the bar becoming horizontal as shown in Fig. 2.2. Fig. 2.2 (not to scale) At the position shown in Fig. 2.2, the system is stationary and in equilibrium. The rod R exerts a force to compress a horizontal spring that controls the water supply to the tank. The spring has a spring constant of 2100 N m −1. R is positioned at a perpendicular distance of 0.017 m above P. (i) The radius of the sphere is 0.0480 m and 26.0% of the volume of the sphere is submerged. The density of water is 1.00  103 kg m−3. Show that the upthrust on the sphere is 1.18 N.

8 [2] Solution: Volume V of the sphere = 34 3 r = ( ) 34 0.04803 = 4.63  10−4 m3 Upthrust, U = Vg = 1000(0.26)(4.63  10−4)(9.81) = 1.18 N M1 M1 (ii) For the position shown in Fig 2.2, by taking moments about P, determine the force exerted on the spring by the rod R. force = ………………………..………. N [2] Solution: The force exerted on the spring by the rod is equal to the force exerted on the rod by the spring. Let the force be F. Taking moments about P, (U – 0.30)(0.29) = F (0.017) (1.18 – 0.30)(0.29) = F (0.017) F = 15 N M1 A1 (iii) Calculate the elastic potential energy EP of the compressed spring. EP = ………………………..………. J [2]

9 [Turn over Solution: Compression x of the spring = F / k = 15 / 2100 (ecf for F) = 0.00715 m EP = ½ kx2 = ½ (2100)(0.00715)2 = 0.054 J M1 A1 (c) When the sphere moves from the position shown in Fig. 2.1 to the position shown in Fig. 2.2, the upthrust on the sphere does work. Assume that resistive forces are negligible.