Discuss the completely inelastic collision in one dimension.

(N/A) Consider a particle of mass $m_{1}$ moving with velocity $v_{1i}$ colliding with a stationary particle of mass $m_{2}$. After the collision,both particles move together in the direction of $v_{1i}$ with a common final velocity $v_{f}$.
From the law of conservation of linear momentum,the total momentum before the collision equals the total momentum after the collision:
$m_{1} v_{1i} + m_{2} v_{2i} = (m_{1} + m_{2}) v_{f}$
Since the second particle is initially at rest,$v_{2i} = 0$:
$m_{1} v_{1i} = (m_{1} + m_{2}) v_{f}$
$v_{f} = \frac{m_{1}}{m_{1} + m_{2}} v_{1i}$
Kinetic energy is not conserved in an inelastic collision. The loss in kinetic energy $\Delta K$ is given by:
$\Delta K = K_{i} - K_{f}$
$\Delta K = \frac{1}{2} m_{1} v_{1i}^{2} - \frac{1}{2} (m_{1} + m_{2}) v_{f}^{2}$
Substituting $v_{f}$:
$\Delta K = \frac{1}{2} m_{1} v_{1i}^{2} - \frac{1}{2} (m_{1} + m_{2}) \left( \frac{m_{1}}{m_{1} + m_{2}} \right)^{2} v_{1i}^{2}$
$\Delta K = \frac{1}{2} m_{1} v_{1i}^{2} \left( 1 - \frac{m_{1}}{m_{1} + m_{2}} \right)$
$\Delta K = \frac{1}{2} \frac{m_{1} m_{2}}{m_{1} + m_{2}} v_{1i}^{2}$
Since $\Delta K > 0$,kinetic energy is dissipated during the collision.