## Saturday, May 05, 2007

### Two Multiple Choice Questions on Elastic Collision

You should remember that momentum and kinetic energy are conserved in elastic collisions where as momentum alone is conserved in inelastic collisions. Let us consider two questions (MCQ) involving elastic collision in one dimension:

(1) A system consisting of two identical blocks A and B, each of mass ‘m’, connected by a light spring of force constant ‘k’ is resting on a smooth horizontal surface. A third identical block C of mass ‘m’ moving with a velocity ‘v0’along the direction of the line joining A and B collides elastically with A and compresses the spring. The maximum compression of the spring is

(a) √(mv0/k) (b) √(mv0/2k) (c) √(m/v0k) (d) v0√(m/k) (e) v0√(m/2k)

At the instant of collision, you need consider the two identical colliding masses C and A only. As the collision is elastic, the entire momentum and kinetic energy of C are transferred to A and C comes to rest. The block A then moves towards B, compressing the spring. When the compression is maximum, both A and B move with the same velocity.

Since the momentum is always conserved,

mv0 = (m + m)v where ‘v’ is the common velocity with which the masses A and B move. Therefore, v = v0/2.

The entire kinetic energy of C is transferred to A, which then compresses the spring and pushes B so that we have

½ mv02 = ½ (m + m)(v0/2)2 + ½ kx2, where ‘x’ is the maximum compression.

This gives x = v√(m/2k).

(2) An α-particle of mass ‘m’ suffers a one-dimensional elastic collision with a nucleus of unknown mass at rest. After the collision the α-particle is scattered directly backwards, losing 75% of its kinetic energy. Then, the mass of the nucleus is

(a) m (b) 2m (c) 3m (d) (3/2)m (e) 5m

This MCQ appeared in the Kerala Engineering Entrance 2003 question paper.

The total momentum of the system is equal to the initial momentum p1 of the α-particle. Equating the total initial and final momenta,

p1 = P – p2 where P and p2 are respectively the final momenta of the nucleus and the α-particle. (The negative sign for p2 is because of the backward motion of the α-particle after the collision).

But the magnitude of p2 is p1/2 since the final kinetic energy of the α-particle is ¼ of its initial kinetic energy. [Remember that K.E. = p2/2m and hence the momentum p is directly proportional to the square root of the kinetic energy].

Therefore, P = p1 + (p1/2) = (3/2) p1.

Since the collision is elastic, kinetic energy also is conserved.

Therefore, K.E. lost by the α-particle = K.E gained by the nucleus so that

P2/2M = (3/4) p12 /2m where ‘M’ is the mass of the nucleus.

Substituting for P, [(3/2)p1]2/ 2M = [(3/4)p12]/2m

This gives M = 3m.

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