2013_MJC_H2_Physics_Solutions

2013 Meridian JC H2 Physics Preliminary Examination Solutions 1 Answers to 2013 Prelim Paper 1 (H2 Physics) 1 D 11 A 21 B 31 C 2 A 12 C 22 C 32 A 3 B 13 A 23 B 33 B 4 B 14 A 24 D 34 A 5 D 15 B 25 A 35 A 6 A 16 D 26 B 36 C 7 A 17 C 27 B 37 C 8 C 18 A 28 C 38 D 9 A 19 C 29 B 39 C 10 B 20 B 30 D 40 C MCQ 1: (D) Reasoning: Random errors cannot be eliminated but it can be reduced by averaging repeated measurements. By nature, random errors are of varying sign and magnitude. MCQ 2: (A) Reasoning: √ MCQ 3: (B) Reasoning: Since the area under the acceleration time graph is the change in speed and the initial speed of the object is zero, the biggest area under the graph will correspond to the largest speed. At B, the area under the graph is the largest, following which the area is negative which means that the speed decreases from B. MCQ 4: (B) Reasoning: Using: 152 = 202 + 2a (70) a = -1.25 m s-2 02 = 152 + 2a(x) x = 90 m MCQ 5: (D) Reasoning: The resultant of the two forces 3 N and 4 N will be 5 N pointing in this direction. Hence the resultant of all three forces will then be 1 N pointing in the similar direction.

2013 Meridian JC H2 Physics Preliminary Examination Solutions 2 MCQ 6: (A) Reasoning: The 3 forces acting on the rod is weight, F and the hinge force For equilibrium, all 3 forces must pass through a common point (concurrent) and form a closed triangle. MCQ 7: (A) Reasoning: By Hooke’s Law, 3 00 9 81 25 2 0 500 0 0208 mg sin f ke . ( . )(sin ) .e .m θ   MCQ 8: (C) Reasoning For elastic collision, Relative speed of approach = relative speed of separation v2 – v1 = u1 – u2 (where the sign conventions of u1, u2, v1, v2 are to the right) Hence ux – (-uy) = vy – (-vx) ux + uy = vx + vy MCQ 9: (A) Reasoning: Work done by friction = 60 x 10 = 600 J = Increase in internal energy ∆Ek + ∆Ep +∆Ee = Wsupplied – Wdissipated Kinetic energy = 150(10) -600 – 100(10 sin 30) = 400 J MCQ 10: (B) Reasoning: For each jet engine, Power output 0.80Power input  power outputPower input = 0.8 Power input = 0.8 Fv where F = half of total thrust Power input = 8.0 )250(000 200 x 2 1       31.3 MW

2013 Meridian JC H2 Physics Preliminary Examination Solutions 3 MCQ 11: (A) Reasoning: ( ) MCQ 12: (C) Reasoning: Net force acting on the satellite at that instant is now zero. Since the satellite is already moving with a speed tangent to the orbit, Newton’s 1st law says that it will continue to move along the tangent to the orbit. MCQ 13: (A) Reasoning: ( ) ( ) MCQ 14: (A) Reasoning: 3 2 2 2 4() 43 3 GRGM G V G Ra R R R aR          Since  and R is doubled, a’ is 4a. MCQ 15: (B) Reasoning: Frequency = 50/47 Hz KEmax =