Suppose two balls A and B of masses m₁ and m₂ are moving initially along the same straight line with velocities u₁ and u₂ respectively.

When u₁ > u₂ relative velocity of approach before collision,

= u₁ - u₂

Hence the two balls collide, let the collision be perfectly elastic. after collision, suppose v₁ is velocity of A and v₂ is velocity of B along the same straight lien,

(a) when v₂ > v₁, the bodies separate after collision.

Relative velocity of separation after collision

= v₂ - v₁

Linear momentum of the two balls before collision.

= m₁u₁ + m₂u₂

Linear momentum of the two balls after collision.

= m₁v₁ + m₂v₂

As linear momentum is conserved in an elastic collision, therefore

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ .............(i)

or m₂(v₂ − u₂) = m₁(u₁ − v₁) ….........(ii)

Total K.E. of the two balls before collision

= \(\frac{1}{2}m_1u^2_1+\frac{1}{2}m_1u^2_2..........(iii)\)

Total K.E. of the two balls after collision

= \(\frac{1}{2}m_1v^2_1+\frac{1}{2}m_1v^2_2..........(iv)\)

As K.E. is also conserved in an elastic collision, there from (iii) and (iv),

\(\frac{1}{2}m_1v^2_1+\frac{1}{2}m_1v^2_2=\frac{1}{2}m_1u^2_1+\frac{1}{2}m_1u^2_2\)

\(\frac{1}{2}m_2(v^2_2-u^2_2)=\frac{1}{2}m_1(u^2_1-v^2_1)\)

\(m_2(v^2_2-u^2_2)=m_1(u^2_1-v^2_1)......(v)\)

Dividing (v) by (ii),

\(\frac {m_2(v^2_2-u^2_2)}{m_2(v_2-u_2)}=\frac{m_1(u^2_1-v^2_1)}{m_1(u_1-v_1)}\)

\(\frac{(v_2+u_2)(v_2-u_2)}{(v_2-u_2)}=\frac{(u_1+v_1)(u_1-v_1)}{(u_1-v_1)}\)

or v₂ + u₂ = u₁ + v₁

or v₂ − v₁ = u₁ − u₂ …(iv)

Hence is one dimensional elastic collision relative velocity of separation after collision is equal to relative velocity of approach before collision.

From \(\frac{v_2-v_1}{u_2-u_1}=1\)

by definition, \(\frac{v_2-v_1}{u_2-u_1}=1\) = e =1