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
