2024 JVS Physics Prelim P1&2 marking scheme

Jurongville Secondary School Science Department 2024 Marking Scheme Assessment: Physics Preliminary Examinations Level: Sec 4 Exp Paper 1 Qn Ans Qn Ans Qn Ans Qn Ans Qn Ans Qn Ans Qn Ans Qn Ans Qn Ans Qn Ans 1 D 2 D 3 D 4 B 5 A 6 A 7 B 8 C 9 B 10 B 11 C 12 A 13 D 14 C 15 B 16 C 17 C 18 C 19 B 20 D 21 D 22 B 23 D 24 A 25 B 26 A 27 D 28 A 29 C 30 C 31 A 32 C 33 A 34 A 35 D 36 D 37 B 38 A 39 C 40 B

Qn Marking Scheme Remarks Marks 1(a) 1(b) 1(c) Acceleration due to gravity = 10.0 (6.0 – 0) / t = 10.0 t = 6.0 / 10.0 = 0.6 s Height = Area under velocity-time graph = ½ (0.6)(6) = 1.8 m Correct working Correct answer and unit Each correct straight line Allow for ecf from (a) Correct working Correct answer and unit Allow for ecf from (a) C1 A1 B1, B1 C1 A1 0.2 0 0.6 6 time / s 0.8 1.0 1.2 0.4 4 2 -6 -2 -4 0.0 m / s velocity

Qn Marking Scheme Remarks Marks 2(a) 2(b) Scale 1 cm rep 5 N Magnitude of wind force = 42 N ± 0.5 N Direction of wind force = 26° clockwise from the horizontal Mass of kite = 5 / 10 = 0.5 kg Resultant force = m a 42 = 0.5 x a a = 42 / 0.5 = 84 m/s2 Correct diagram with arrows Appropriate scale Correct magnitude Correct angle Correct working Correct answer and unit Allow for ecf from (a) B1 B1 B1 C1 A1 3(a) 3(b) 3(c) The load is placed vertically above support B. The perpendicular distance between the load and support B is zero. Thus there is no moment about support B. Taking moments about support B, Force at support A x 1.4 = (24)(1.4 / 2) Force at support A = 16.8 / 1.4 = 12 N For maximum distance that load can move to the right, the weight of plank just balance the load at support B and no force at support A. Taking moments about support B 76 x Maximum distance that load can move to the right = (24)(1.4 / 2) Maximum distance that load can move to the right = 16.8 / 76 = 0.22 m OWTTE Correct working Correct answer and unit Correct working Correct answer and unit B1 C1 A1 C1 A1 4(a) 4(b) 4(c) Pressure is the force per unit area. Pressure = Force / Area 23000 = 29 x 10 / Area Area = 290 / 23000 = 0.126 m2 Pressure in gas cylinder + 13600 x 10 x (0.059 – 0.035) = 100000 Pressure in gas cylinder = 100000 – 3264 = 96736 = 96700 Pa (3 s.f.) OWTTE Correct working Correct answer and unit Correct working Correct answer and unit B1 C1 A1 C1 A1 weight of kite tension of string resultant force

Qn Marking Scheme Remarks Marks 5(a) 5(b) 5(c)(i) 5(c) (ii) There dust particles