2024 CHS Sec 4 Physics Prelim Answer

2024 CHS SEC 4 PHYSICS PRELIMINARY EXAMINATION Paper 1 1 C 2 D 3 B 4 A 5 D 6 D 7 D 8 B 9 D 10 C 11 A 12 C 13 A 14 A 15 B 16 C 17 B 18 A 19 A 20 D 21 A 22 C 23 C 24 D 25 C 26 D 27 C 28 B 29 D 30 A 31 C 32 A 33 A 34 D 35 D 36 D 37 B 38 C 39 D 40 D Paper 2 Section A S/N Mark Scheme Remarks 1ai ● Ball’s speed is decreasing [M1] at a decreasing rate [A1] OR ● Deceleration [M1] is decreasing [A1] OR ● Acceleration is decreasing [M1] and negative [A1] 1 mark awarded for “ball is decelerating at decreasing rate”. 1aii ● (weight and) air resistance acts (on ball) downwards (or in opposite direction to motion/velocity) (resulting in decreasing speed) [B1] ● As speed (or velocity) decreases, air resistance decreases [B1] ● (downward) resultant force decreases (resulting in decreasing deceleration) [B1] No credit is awarded for 1st mark if the answer mentions that upward force is acting on the ball as it is moving up. 1b 1.750 s [B1] No penalty if answer in 2 d.p. 1c ● graph is below x-axis and gradient decreases smoothly without reaching 0 at 3.50 s. [B1] ● final velocity, v is below x-axis and 0 m / s < v < 25 m / s [B1] No penalty if draw past 3.50 s. 2a ● energy transferred mechanically to gravitational potential (and kinetic) store of load [B1] ● (some) energy is transferred o mechanically due to (work done against) friction (or air resistance) [B1] to the internal store of surrounding air (and load). [B1] OR o by heating (due to temperature difference) [B1] to the internal store of surrounding air (and load). [B1] OR o by propagation of electromagnetic waves / infra-red radiation [B1] to the internal store of surrounding air (and load). [B1] OR o by propagation of mechanical / sound waves [B1] to the kinetic store of surrounding air. [B1] If student writes “store” at least once, BOD all answers without “store”. If student does not write “store” at all, penalise 1 mark. No credit awarded for the method of energy transfer to surrounding air if the energy transfer is not from the motor. 2b Moment = F x d [formula] = 2.5 x 0.20 [B1] = 0.50 Nm (or 50 Ncm) [B1]

2 S/N Mark Scheme Remarks 2c As load is stationary, Upward force due to liquid (F) = Weight of Load M = 2.5 N [B1] F = pA = (pbottom – ptop) x A = [(hbottom – htop) ρ g] x A 2.5 = [0.030 x ρ x 10] x (0.12 x 0.02) [B1] ρ = 3 500 kg m-3 Accept if student calculated density of load M [B1] and concluded that density of load M and the liquid are the same [B1] 3a ● molecules are in constant random motion [B1] ● collide with (inner walls of) piston, exerting a force per unit area on it [B1] 3b ● (when gas is heated), molecules collide more frequently and forcefully with (inner walls of) piston, exerting a larger force / pressure (than that due to atmosphere) [B1] ● hence, a downward force (or pressure) is needed to balance this larger force (or pressure) / keep the piston stationary /