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*’ = 4*p*/3

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

The kinetic energy lost by particle A due to the collision is (8/9)×*p*^{2}/2*m*.

[Note that the initial kinetic energy of particle A is* p*^{2}/2*m* and its final kinetic energy is (1/9)* p*^{2}/2*m*].

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}/2*M = *(8/9)* p*^{2}/2*m *

Substituting for *p*’ = 4*p*/3, we have

(16/9)* *(*p*^{2}/2*M*) = (8/9)* p*^{2}/2*m* 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) 4*v*

(b) 2√2 *v*

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

Since the momentum is conserved, we have

2*m* *V = *√2 *mv* so that *V = v*/√2

beautiful job sir ji..........

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