2025 EJC H2 Prelim P1 MS

©EJC 2025 9749/J2H2PRELIM/2025 EUNOIA JUNIOR COLLEGE JC2 Preliminary Examination 2025 9749 PHYSICS MARK SCHEME Paper 1 – Multiple Choice Questions 1 Answer: C The average value is 47.78 mm which is not close to the true value so not accurate. There is a variation in measurements. So it is not precise. 2 Answer: D distance 90time taken to travel first 90 km = 1.12 5 hour<speed> 80== time taken for remaining 110 km = 0.875 hour 1total distance 110 km 126 km htotal time 0.875 hourspeed − = = = 3 Answer: A 3 (1) 2 ' (2) (2) 2 ':1(1) 3 3' 2 3' 0.5 0.5(0.40) 0.20 cm2 kx mg kx mg x x xx x x x x x = −−− = −−− = = − = − = = =

2 ©EJC 2025 9749/J2H2PRELIM/2025 4 Answer: A 2 (0.50)(9.81) (0.50) (1) ' 3.5 (0.20) (2) Solving (1) and (2): 2.0 m s− −= − = −−− −= − = −−− = mg T ma Ta T friction m a Ta a 5 Answer: B Change in velocity = final velocity – initial velocity 2 2 10.40 0.30 0.50 m s −= + = Change in momentum = m v = (0.20)(0.50) = 0.10 kg m s–1 6 Answer: C ( ) 1 4 0.4004 0 1.20(9.81)1.00 7.36 m s Thrust weight m v mgt v v − = =  −= = 7 Answer: B (3m)(4)= mvsinθ (2m)(6) = mv cosθ tanθ = 1 θ = 45°

3 ©EJC 2025 9749/J2H2PRELIM/2025 8 Answer: B Constant speed, so no change in Ek. hence constant gradient p p E mgh dE hmg mgvdt t = == 9 Answer: A sin 60 sin100 60 80000 (1000)(9.81)(sin45 )100 6.5 450 N drive resistive resistive o resistive resistive F F W P F mgv F F   =+ =+  = + = 10 Answer: A Vertical equilibrium: coskx mg = --- (1) Vertical component of tension = weight Horizontal component of tension provides centripetal force 2 2sinkx mr T  =  --- (2) 2 2 (2) 4 (0.108sin27 ): tan27(1) (9.81) o o T = T = 0.62 s 11 Answer: A For satellite in orbit, TE = ‒ KE, so option A is correct. Since 2 GMmKE r= , larger m with same KE requires larger r, so option B is incorrect. Since 2 23 4Tr GM = , larger r will result in larger T, so option C is incorrect. Since 2 T  = , larger T means smaller ω, so option D is incorrect.

4 ©EJC 2025 9749/J2H2PRELIM/2025 12 Answer: C “It has a property that varies linearly with temperature” is not strictly essential but rather an assumption used in constructing empirical centigrade scales. 13 Answer: D 11 0.67 s1.5T f= = =   0 cos 0.12cos (2 )(1.5)(0.75) 0.0848 m 85 mm x x t x x   = = == Distance = 120 × 4 + (120 – 85) = 515 mm 14 Answer: C Particles are displaced away from the equilibrium positions towards the centre of the compression. Right is taken to be positive (distance from P) Particles displaced rightwards have positive displacements and leftwards have negative displacements. 15 Answer: C ( )( ) 22 0 0 cos (30 ) cos (60 ) 0.19 = = ooII II

5 ©EJC 2025 9749/J2H2PRELIM/2025 16 Answer: C 22 0 22 0 2 2 2 (2 ) (1) 23 ' (3 ) (2) (2) ' (3 ): ' 2.3(1) (2 ) net P Q net