ASRJC Prelim H2P1 MS final

1 9749/01/ASRJC/2023Prelim [Turn Over Anderson Serangoon Junior College 2023 H2 Physics Prelim P1 Exam Solution Paper 1 (30 marks) E – Easy, A – Average, D – Difficult 1 2 3 4 5 6 7 8 9 10 D A B B B B C A C C 11 12 13 14 15 16 17 18 19 20 C B C C A D C D D A 21 22 23 24 25 26 27 28 29 30 D C C D A D A B C A Easy: 6, Average: 15, Difficult: 9 1 D II= = =W Q Fs masV tt Base units of V = -2 2 -3 -1kg×ms ×m = kgm s AA×s E 2 A . .I. 1 200 250 48= = = VR Reading in ammeter is recorded to 2 dp. For division, the s.f. of calculated data is recorded to the lowest s.f. of the raw data A 3 B Using s = ut + ½ t2 9 = u(2) + ½ a(2)2 --- (1) 21 = u(4) + ½ a(4)2 --- (2) Solving (1) & (2), a = 0.75 m s–2 D 4 B Initial vertical velocities for both stones, uy = 0 sy = uyt + ½ gt2 = ½ gt2 Both stones will take the same time to fall through the same height since they experience the same gravitational acceleration g. E 5 B Change in momentum = 2(5.0 – 30) = -50 Area under F-t graph = Change in momentum -F x (4 + 6) / 2 = -50 F = 10 N A

2 9749/01/ASRJC/2023Prelim 6 B Due to smaller weight, a smaller air resistance is needed by the foam ball to achieve zero net for to reach terminal velocity. Hence time required to reach terminal velocity is shorter. So option A and C are eliminated. In addition. since air resistance depends on speed, a smaller air resistance will require a smaller terminal velocity. H ence option D is eliminated. D 7 C Tension in string is 3.0W (2.0 W + 1.0W) Let distance between string and CG be d Taking moments abouts the CG (which is at 0.5x) (2.0 W) × (0.3x) = (3.0 W) × d d = 0.2x When load is moved, it will be 0.3x away from the CG (but on the other side) Hence, the string needs to be placed 0.2x from the CG (but on the other side) for rod to be in equilibrium. Distance moved by string = 0.2x + 0.2x = 0.4x A 8 A Option A: WD to change speed of object from 2 m s-1 to 3 m s-1 = ½ m (32 – 22) = ½ m (5) WD to change speed of object from 1 m s-1 to 2 m s-1 = ½ m (22 – 12) = ½ m (3) So option A is correct. Option B: When an object slows down, there has to be negative work done by a force (opposing the motion of the object). Hence option B is not correct. Option C: When lifting an object 1 m at constant velocity, WD = mgh = mg(1) When moving an object 1 m across a smooth horizontal surface at constant speed, there is no need for a force to keep it at constant velocity. Therefore, WD is zero. Hence option C is not correct. Option D: When an object is dropped, the force of gravity is in the direction of the motion of the object. There is WD by the force of gravity and is given by Wd. Hence option D is not correct. A 9 C Distance moved in 60 = 0.40 × ( )60 2360  = 0.4189 m Since both masses were displaced v