Multiple Choice Questions on Elastic and Inelastic Collisions

Today we will discuss two questions on collisions. The first question pertains to elastic collision in one dimension and the second question pertains to inelastic collision in two dimensions:

(1) A particle A of mass m moving along the positive x-direction with kinetic energy K suffers an elastic head-on collision with a stationary particle B of mass M. After the collision the particle A moves along the negative x-direction with kinetic energy K/9. What is the mass of particle B?

(a) 9 m

(b) 6 m

(c) 3 m

(d) 2 m

(e) m/3

Since the kinetic energy of A after collision is one-nineth of its initial kinetic energy, the momentum of A after collision is one-third of its initial momentum.

Since the momentum is to be conserved, we have

p = p’ – p/3 where p is initial momentum of A and p’ is the momentum of B after the collision.

[The final momentum of particle A is negative since its direction is reversed].

Therefore, p’ = 4p/3

The kinetic energy gained by particle B due to the collision is p’²/2M where M is the mass of particle B.

The kinetic energy lost by particle A due to the collision is (8/9)×p²/2m.

[Note that the initial kinetic energy of particle A is p²/2m and its final kinetic energy is (1/9) p²/2m].

Since the kinetic energy too is conserved in elastic collisions, the kinetic energy gained by particle B is equal to the kinetic energy lost by particle A. Therefore, we have

p’²/2M = (8/9) p²/2m

Substituting for p’ = 4p/3, we have

(16/9) (p²/2M) = (8/9) p²/2m from which M = 2 m.

(2) A particle of mass m moving westwards with speed v collides with an identical particle moving northwards with speed v. If they stick together after the collision, what is their common speed after the collision?

(a) 4v

(b) 2√2 v

(c) 2v

(d) √2 v

(e) v/√2

The total momentum before collision is the vector sum of the initial momenta mv and mv at right angles (since one is westwards and the other is northwards) and is equal to √2 mv acting along the north-west direction (Fig.).

The combined mass is 2m. If V is the common speed, the total momentum after collision is 2m V, acting along the north-west direction.

Since the momentum is conserved, we have

2m V = √2 mv so that V = v/√2

IIT-JEE 2007 Assertion-Reason Type MCQ on Elastic Collisions

The following simple Assertion-Reason Type multiple choice question which appeared in IIT-JEE 2007 question paper is quite simple and is meant for checking whether you have mastered the fundamentals of elastic and inelastic collisions:

STATEMENT-1

In an elastic collision between two bodies, the relative speed of the bodies after collision is equal to the relative speed before the collision

because

Statement-2

In an elastic collision, the linear momentum of the system is conserved.

(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct

explanation for Statement-1

(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct

explanation for Statement-1

(c) Statement-1 is True, Statement-2 is False

(d) Statement-1 is False, Statement-2 is True

The important points you have to remember here are:

(i) momentum is conserved in elastic as well as inelastic collisions

(i) in elastic collisions, kinetic energy also is conserved.

[Remember, we are talking of kinetic energy. Note that the total energy is conserved in all collisions in accordance with the law of conservation of energy].

In an elastic collision, the relative speed of the bodies after collision is equal to the relative speed before the collision and this is a consequence of the conservation of kinetic energy.

Therefore, statement-1 is true. Statement-2 also is true, but it is not the reason for statement-1. So, the correct option is (b).

[Note that the ratio of the relative speed of the bodies after collision to the relative speed before collision is called the coefficient of restitution, e. The value of ‘e’ lies between 0 and 1. For perfectly elastic collisions, e =1 and for perfectly inelastic collisions, e = 0]