IJC_H1_PHY_P1_Solution

Innova JC 2016 Prelim H1 Physics Paper 1 Q Ans Q Ans Q Ans 1 A 11 D 21 B 2 B 12 D 22 C 3 B 13 C 23 A 4 A 14 B 24 C 5 A 15 B 25 D 6 D 16 B 26 C 7 B 17 B 27 D 8 D 18 A 28 C 9 D 19 A 29 C 10 B 20 D 30 D Solution to 2016 H1 Physics Paper 1 1 The dimensions of a typical household washing machine are 450mm×450 mm×700mm. So its volume is 1.4×108 mm3 ; its mass is around 70 – 80 kg ; power consumption 500 W The spin rate is around 20 cycles per sec. [Switched options in original question] A 2 21 2W kx ∆𝑊 𝑊 = ∆𝑘 𝑘 + 2 ∆𝑥 𝑥 ∆𝑊 𝑊 = 2 100 + 2 0.002 0.050 = 0.10 (10%) B 3 𝑅 = 𝑉 𝐼 = (𝑃 𝐼 ) (1 𝐼) = 𝑃 𝐼2 Power = Energy  time = Force × displacement  time SI base unit for power = (kg m s-2)(m) / s = kg m2 s-3 SI base unit for electrical resistance = kg m2 s-3 / A2 = kg m2 s-3 A-2 B 4 At the maximum height, the acceleration is 9.81 m s-2 not zero. A 5 The slope of the s-t graph gives the value of the instantaneous velocity. A 6 The force of P on Q and the force of Q on P are an action-reaction pair. Hence the two forces are equal in magnitude but opposite in direction. The F-t graph of Q on P is the mirror image of P on Q. Alternatively, D

  initial,P initial,Q after,P after,Q after,Q initial,Q after,P initial,P QP Total momentum before Total momentum afte r Area under F-t graph for Q Area under F-t graph for P P P P P P P P P PP          7 Newton’s 2nd law defines the net force acting on an object as being directly proportional to the rate of change of momentum of the object, where the constant of proportionality is 1. Hence, from this definition, options A and C (rate of change of energy), and option D (Rate of change of force) can be eliminated. water exerted by wall on water jet exerted by water jet on wall exerted by w all on water jet change in momentum of the water jet per unit time. The magnitude of the force exerted by the water je pF t FF    t on the wall is equal to the force exerted by the wall on the water jet. Thus the magnitude of the force exerted by the water jet on the wall is numerically equal to the change in momentum of the water jet per unit time. B 8 For a car travelling at constant speed, the net force acting on the car is zero. Consider the forces acting parallel to the slope, Sum of forces upward and parallel to slope = sum of forces downward and parallel to slope engine force = component of weight parallel to slope + resistive force engine force = mg sin  + F D 9 Increase in force on arrow = Increased in total horizontal forces on arrow 2 (120 cos 55) – 2 (100 cos 65) = 53 N D 10 The change in velocity, - - (- ) R Final velocity Initial Velocity QP QP    B

11 Weight is a vertically downward force. Hence Option A and C are not correct. The forces should form a closed triangle as there is equilibrium. Hence, only D is correct. D