Question

Prove that in an elastic collision in one dimension of two bodies, the relative velocity
of approach before collision is equal to the relative velocity of separation after the collision​

Answer 1

Explanation:

in elastic collision , linear momnetum and energy both are conserved .

let A body of mass m moving with u and B body of mass M moving with U velocity . after elastic collision their velocity v and V respectively .

use law of conservation of linear momentum ,

Fext = 0

so,

Pi = Pf

mu + MU = mv + MV

m( u -v) = -M(U - V) ------------(1)

now,

from conservation of energy ,

KEi = KEf

1/2mu² + 1/2MU² = 1/2mv² + 1/2MV²

1/2m(u² -v²) = 1/2M(V² -U²)

m(u² -v²) = - M(U²- V²)

m(u -v)(u + v) = - M( U - V)( U + V)

from equation (1)

(u + v) = ( U + V)

u - U = V - v

1 = ( V - V)/(u - U)

-1 = (V - v)/(U - u) =(v - V)/(u -U)

here ( v-V)/( u - U) is known as coefficient of restitution, where ( v-V) velocity of seperation and (u -U) is velocity of approach .

according to coefficient of restitution

- e = (v - V)/(u - U)

compare above to this ,

-e = -1

e = 1

hence , proved that velocity of separation equal to velocity of approach