Multiple Choice Questions on Newton’s Laws of Motion

You will have to apply Newton’s Laws of Motion in different branches of Physics, but you will find questions specifically meant for checking your understanding of these laws in AP Physics Examination, Graduate Record Examination (GRE) and Medical and Engineering Entrance Examinations.

Consider the following MCQ based on impulse:
The force ‘F’ acting on a particle of mass ‘m’ is indicated by a force-time graph (Fig.). The momentum received by the particle during the time from zero to 8 s is

(a) 24 Ns (b) 20Ns (c) 12Ns (d) 6Ns (e) zero

The area under the force-time graph gives the impulse imparted to the particle. Impulse is a vector quantity and so you must consider its sign while adding the areas. The impulse received from zero to 2 seconds is positive and is equal to the area of the triangle, which is 6 Ns. The impulse received during the time from 2 s to 4 s is the area of the rectangle, which is – 6 Ns. The impulse received during the time from 4 s to 8 s is the area of the larger rectangle, which is 12 Ns.

Hense the net impulse received during the time from zero to 8 seconds is 6 –6 +12 = 12 Ns.

Since the impulse is equal to the change of momentum, the correct option is (c).

Now consider the following question:

A boy caught a ball of mass 200g moving with a speed of 30 ms^–1. If the catching process be completed in 0.1 s, the force of impact exerted by the ball on the hands of the boy is

(a) 60 N (b) 40 N (c) 30 N (d) 20 N (e) 10 N

We have force F = dp/dt where ‘dp’ is the change in linear momentum during the time dt.

The initial momentum of the ball is 0.2×30 = 6 kgms^–1 and the final momentum is zero so that dp = 6.

Therefore, F = 6/0.1 = 60 N.

The following MCQ which appeared in Karnataka CET 2002 question paper is meant for checking your grasp of the law of conservation of linear momentum:

A projectile is moving at 20 ms^–1at its highest point, where it breaks into two equal parts due to an internal explosion. One part moves vertically up at 30 ms^–1 with respect to the ground. Then the other part will move at

(a) 30 ms^–1 (b) 50 ms^–1 (c) 10√3 ms^–1 (d) 20 ms^–1

If the mass of the projectile is ‘m’, the mass of each fragment after the explosion is m/2. The momentum of the part which moved upwards is (m/2)×30 =15m. This is shown as vector OA in the figure.

The momentum of the projectile at the highest point of its path is 20m and its direction is horizontal. This is shown as vector OC.

The momentum of the other part (after the explosion) is (m/2)×v where ‘v’ is its velocity. This is shown as vector OB in the figure.

Since the momentum is conserved, the total final momentum, which is the vector sum of the momentum vectors OA and OB must be equal to the initial momentum, represented by the vector OC.

Evidently, OB = √(OA² + OC²).

Or, mv/2 = √[(15m)² + (20m)²] = 25m.

This gives v = 50 ms^–1.