2024 CCHY P1 AND P2 PRELIM MS

Marking scheme for 2024 Physics Prelim Paper 1 1 C 11 D 21 C 31 C 2 D 12 A 22 A 32 D C 3 D 13 B 23 C 33 B 4 B 14 D 24 B 34 A 5 C 15 B 25 D 35 C 6 C 16 B 26 B 36 A 7 B 17 C 27 B 37 B 8 A 18 D 28 D 38 B 9 C 19 B 29 B D 39 C 10 C 20 B 30 C 40 B Paper 2 Section A Qn Mark Scheme Marks Sub- total [-1] per section for any errors in sig. fig. or unit. 1(a) a = F/m OR 28000 / 25000 1.1 m/s2 B1 A1 [2] (b) The forward force is equal to the air resistance / opposing forces acting in the opposite direction. Resultant force is zero, hence acceleration is zero. B1 B1 [2] 2(a) mass and weight of hammer is larger than feather weight equals resultant force which is the product of mass and acceleration (FR = ma, W = ma) acceleration is equal since Fhammer / mhammer = Ffeather / mfeather B1 B1 A1 [3] 2(b)(i) v = u + at OR 1.6 x 1.5 2.4 m/s B1 B1 [2] 2(b)(i) Straight line from origin with positive gradient with v = 2.4 m/s at t = 1.5s indicated B1 [1]

Qn Mark Scheme Marks Sub- total 2(b)(iii) ½ x 1.5 x 2.4 OR distance travelled = area under the graph 1.8 m C1 A1 [2] 3(a) The principle of moments states that the sum of clockwise moments about a pivot is equal to the sum of anticlockwise moments about the same pivot When object is in equilibrium B1 B1 [2] 3(b) The body is in equilibrium, Taking moments about point P, Anti clockwise moment = Clockwise moment F X 150 = 70 X 15 F = 7.0 N B1 A1 [2] 3(c) Weight of chair acting from centre of gravity passes through pivot No resultant moment produced OR line of action of weight acting slightly to the left of pivot, producing anticlockwise restoring moment C1 A1 [2] 4(a) Pressure is the force acting per unit area A1 [1] 4(b) Pgas + Pmercury = Patm Pmercury = hρg OR 0.16 x 1.4 × 104 x 10 1.0 × 105 – (0.16 x 1.4 × 104 x 10) = 77600 =78000 Pa B1 B1 A1 [3] 4(c)(i) 8.0cm A1 [1] 4(c)(ii) Gas molecules move faster and gain kinetic energy as the gas is heated Gas molecules collide against the mercury surface more frequently with greater force, increasing pressure exerted on the right column of mercury Mercury column on right falls and mercury column on left rises until gas pressure is equal to atmospheric pressure, i.e. equilibrium B1 C1 A1 [3] 5(a) sin i / sin r = 1.5 OR sin 60 / sin r = 1.5 35⁰ M1 A1 [2] 5(b) n = 1/sin c M1

Qn Mark Scheme Marks Sub- total critical angle = 41.8 = 42⁰ A1 [2] 5(c) The angle of incidence of the light ray at KL is more than the critical angle and undergoes total internal reflection. The angle of incidence of the light at LM is less than the critical angle and undergoes refraction, bending away from the normal. B1 B1 [2] 6(a) Brownian motion is due to the random continuous motion of invisible air molec