Head-On Collision

The initial momentum is $\vec{p} = m_{1} \vec{v}_{1} + m_{2} \vec{v}_{2} = m_{1} (\vec{v}_{1} - \vec{v}_{1}) = 0$ . The final momentum is 0 by conservation of momentum, so the final energy is also 0.

Doing it the hard way, we can plug into the general equation for the energy of an inelastic collision

$E_f = \frac{1}{2} (\frac{m_1^2}{(m_1 + m_2)} v_1^2 + \frac{m_2^2}{(m_1 + m_2)} v_2^2 + \frac{m_1 m_2}{(m_1 + m_2)} v_1 v_2 \text{cos} \theta)$

Setting $m_1 = m_2 = m$ , $v_1 = v_2= v$ and $\theta = \pi$ gives

$E_f = \frac{1}{2} (\frac{m^2}{2m} v^2 + \frac{m^2}{2m} v^2 + 2 \frac{m^2}{2m} v^{2} (-1))$ $E_f = 0$

For inelastic collision V= m2u2/m1+ m2 +m2u2/m1+m2 After putting the value V=0 So KE (f) = 0