Question

Prove that relative velocity of approach before collision is equal to relative velocity of seperation after elastic collision.

Answer 1

u1-u2 => relative velocity of separation
v1-v2=> relative velocity of approach
u1-u2=v1-v2
relative velocity of separation=relative velocity of approach

Answer 2

Answer:

To prove:

The realtive velocity of approach is equals to the realative velocity of sapration.

To derive :

Pf = Pi

Derivation:

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