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.