Ans.

Let us consider two particles of masses \( m_1 \) and \( m_2 \) travelling along the same straight line with velocities \( u_1 \) and \( u_2 \) respectively impinge on each other. Let \( v_1 \) and \( v_2 \) be their respective velocities after perfectly elastic collision in the direction of \( u_1 \) and \( u_2 \).

According to the principle of conservation of linear momentum we can write,

\( m_1u_1+m_2u_2=m_1v_1+m_2v_2\tag{1} \)

and according to the conservation of kinetic energy we can write,

\( \frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2 \)

or, \( m_1(u_1^2-v_1^2)=m_2(v_2^2-u_2^2)\tag{2} \)

Deviding equation (2) by equation (1) we get

\( u_1+v_1=u_2+v_2 \)

or, \( v_2=u_1+v_1-u_2\tag{3} \)

From equations (1) and (3) we get,

\( m_1(u_1-v_1)=m_2(u_1+v_1-2u_2) \)

or, \( (m_1+m_2)v_1=(m_1-m_2)u_1+2m_2u_2 \)

or, \( \displaystyle{v_1=\frac{(m_1-m_2)u_1+2m_2u_2}{m_1+m_2}}\tag{4} \)

From equations (2) and (4) we get,

\( \displaystyle{v_2=u_1-u_2+\frac{(m_1-m_2)u_1+2m_2u_2}{m_1+m_2}} \)

= \( \displaystyle{\left(1+\frac{m_1-m_2}{m_1+m_2}\right)u_1+\left(\frac{2m_2}{m_1+m_2}\right)u_2} \)

= \( \displaystyle{\frac{2m_1u_1}{m_1+m_2}+\frac{-m_1+m_2}{m_1+m_2}u_2} \)

or, \( \displaystyle{v_2=\frac{2m_1u_1+(m_2-m_1)u_2}{m_1+m_2}}\tag{5} \)

Special cases for perfectly elastic collision:

Case I :
When \( m_1=m_2 \), from equations (4) & (5) we get,
\( v_1=u_2 \) & \( v_2=u_1 \),
For perfectly elastic collision of two bodies of equal masses, they exchange their velocities after collision.

Case II :
When \( u_2=0 \), then from equations (4) & (5) we get
\( \displaystyle{v_1=\frac{m_1-m_2}{m_1+m_2}u_1} \) & \( \displaystyle{v_2=\frac{2m_1}{m_1+m_2}u_1} \)
In this condition, there are three cases \( \cdots \)

1. If \( m_1=m_2 \) then
   \( v_1=0 \) & \( v_2=u_1 \)
   So after collision the first body stops and the second body moves with the velocity which the first one originally had before collision.
2. If \( m_1>>m_2 \) then
   \( v_1\simeq{u_1} \) & \( v_2\simeq{2u_1} \)
   So when a heavy body collide with the much lighter body which is at rest, then velocity of the heavy body remains unchanged after collision, but after collision the lighter body moves with the velocity which is twice the velocity of the heavy body.
3. If \( m_1<<m_2 \) then
   \( v_1\simeq{-u_1} \) & \( v_2\simeq{0} \)
   So when a lighter body collides with the much heavier body which is at rest, then after collision the heavier body remains approximately at rest and the velocity of lighter body approximately reversed.