PHS 2024 4Exp PP Prelim P2 answer (Shared)

PRESBYTERIAN HIGH SCHOOL SCIENCE DEPARTMENT MARKING SCHEME Subject: Physics Paper 2 Exam: Preli Setter: Year: 2024 Level: Sec 4 Express Section A (70 marks) Q/N Suggested Answer Sub- total Total 1 a speed is a scalar quantity and velocity is a vector quantity. OR Speed is defined as distance travelled per unit time. Velocity is defined as the rate of change of displacement. B1 1 b Change in velocity = -22 – 25 = - 47 m/s OR Change in velocity = 22 – (-25) = 47 m/s acceleration = (v – u) / t = 47/0.013 = 3600 m/s2 M1 A1 2 c Resultant force, F = ma = (0.16)(3600) = 576 N or 580 N allow for e.c.f A1 1 2 a mass of concrete = (1750)(4 x 3 x 0.1) = 2100 kg W = mg = (2100)(10) = 21 000 N A1 A1 2 b Total CM = Total ACM F x 0.5 = 21 000 x 1 F = 42 000 N M1 A1 2 c total upward force = 42 000 + 21 000 = 63 000 N allow for e.c.f A1 1 3 a(i) (G.P.E)T + (K.E)T = (K.E)B (85)(10)h + ½ (85)(2)2 = ½ (85)(30)2 h = 44.8 m or 45 m M1 A1 2 a(ii) All energy in the gravitational potential store of the skier is transferred to the kinetic store. OR No energy is transferred to the internal store of skier and the surroundings due to work done against friction. B1 1 b WD against opposing force = K.E F x 33 = ½ (85)(30)2, F = 1159 N = 1160 N (3 s.f) or 1200 N M1 A1 2

2 4 a(i) Force = pressure × area = (3.8 × 105 ) ( 6.1 ÷ 1002) = 232 N or 230 N M1 A1 2 a(ii) force on other side of piston due to air / atmospheric pressure OR pressure of trapped air decreases as valve opens (and air enters the tyre) B1 1 b The temperature rise of the air will increases the average speed of the air molecules or K.E of the molecules inside the pump. This means the speed and frequency of collision by air molecules on the walls of piston and pump will also increase, causing an increase in the average force exerted by air molecules on walls of piston, increasing the air pressure inside the pump. B1 B1 2 5 A amplitude = 1.5 cm B1 1 b(i) Distance between A and B about = λ/4 = 38 cm λ = 38 x 4 = 152 cm V = f λ = (1/0.8) (152) = 190 cm/s ≈ 200 cm/s M1 M1 2 b(ii) Distance between A and B could be 5λ/4, or 9λ/4 etc… B1 1 6 a Some reflect (some) pass into new material or (some) absorbed by material B1 B1 2 b Time interval = 0.03 ms = 0.03 × 10-3 s distance between emitter and the child = 1500 x (0.03 × 10-3) /2 = 0.023 m M1 A1 2 7 a The light ray is not refracted when it enters the glass block at right angle such that the angle of incidence is zero.