Ans.

A particle of mass $m_1$ moving with velocity $u_1$ collides head-on collision with the particle of mass $m_2$ moving with the velocity $u_2$.

Let us consider $v_1$ and $v_2$ be the velocities of mass $m_1$ and $m_2$ respectively.

According to the conservation of linear momentum,

$m_1u_1+m_2u_2=m_1v_1+m_2v_2 \tag{1}$

If $e$ be the coefficient of restitution, then

$\displaystyle{e=\frac{v_2-v_1}{u_1-u_2}}\tag{2}$

The loss of kinetic energy due to the impact is

=Kinetic energy before collision – Kinetic energy after collision

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

= $\frac{1}{2}\left[(m_1u_1^2+m_2u_2^2)-(m_1v_1^2+m_2v_2^2)\right]$

= $\frac{1}{2(m_1+m_2)}\left[(m_1+m_2)(m_1u_1^2+m_2u_2^2)-(m_1+m_2)(m_1v_1^2+m_2v_2^2)\right]$

= $\frac{1}{2(m_1+m_2)}\left[\left(m_1^2u_1^2+m_2^2u_2^2+m_1m_2(u_1^2+u_2^2)\right)\\-\left(m_1^2v_1^2+m_2^2v_2^2+m_1m_2(v_1^2+v_2^2)\right) \right]$

= $\frac{1}{2(m_1+m_2)}\left[\left({(m_1u_1+m_2u_2)}^2-2m_1m_2u_1u_2+m_1m_2(u_1^2+u_2^2)\right)\\-\left({(m_1v_1+m_2v_2)}^2-2m_1m_2v_1v_2+m_1m_2(v_1^2+v_2^2)\right) \right]$

= $\frac{1}{2(m_1+m_2)}\left[\left({(m_1u_1+m_2u_2)}^2+m_1m_2(u_1^2+u_2^2-2u_1u_2)\right)\\-\left({(m_1v_1+m_2v_2)}^2+m_1m_2(v_1^2+v_2^2-2v_1v_2)\right) \right]$

= $\frac{1}{2(m_1+m_2)}\left[{(m_1u_1+m_2u_2)}^2+m_1m_2{(u_1-u_2)}^2\\-{(m_1v_1+m_2v_2)}^2-m_1m_2{(v_1-v_2)}^2\right]$

= $\frac{1}{2(m_1+m_2)}\left[m_1m_2{(u_1-u_2)}^2-m_1m_2{(v_1-v_2)}^2\right]$
[ using equation (1)]

= $\frac{1}{2(m_1+m_2)}\left[m_1m_2{(u_1-u_2)}^2m_1m_2e^2{(u_1-u_2)}^2\right]$
[using equation (2)]

= $\displaystyle{\frac{m_1m_2}{2(m_1+m_2)}{(u_1-u_2)}^2(1-e^2)}$

CASE-I:
In case of a perfectly inelastic collision $e=0$,
So the loss of the kinetic energy will be
$\displaystyle{\frac{m_1m_2}{2(m_1+m_2)}{(u_1-u_2)}^2}$

CASE-II:
In case of perfectly elastic collision $e=1$,
So the loss of the kinetic energy will be
$0$
This means that after collision the total kinetic energy will remain unchanged.