# Mechanics: Chapter 16 – Collisions

If there is no external force acting on a system, the total momentum of a system is always conserved. In addition, if a collision is elastic, the total kinetic energy before a collision is equal to the total kinetic energy after the collision i.e kinetic energy is also conserved. In other words, in an elastic collision between two particles of mass $m_{1}$ and $m_{2}$,

• momentum is conserved: $m_{1}v_{1} + m_{2}v_{2} = m_{1}v_{1}^{'} + m_{2}v_{2}^{'}$
• kinetic energy is conserved: $\frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^{2} = \frac{1}{2}m_{1}v_{1}^{'2} + \frac{1}{2}m_{2}v_{2}^{'2}$

If the particle with mass $m_{2}$ is stationary, i.e $v_{2} = 0$, we can derive the following two expressions,

• $v_{1}^{'} = \frac{m_{1} - m_{2}}{m_{1} + m_{2}}v_{1}$
• $v_{2}^{'} = \frac{2m_{1}}{m_{1} + m_{2}}v_{1}$

If $m_{2} >> m_{1}$, we can neglect $m_{1}$ in our expressions and we get

• $v_{1}^{'} = - v_{1}$
• $v_{2}^{'} = 0$

Which makes sense. For example, imagine a tennis ball colliding with a wall.

If $m_{1} = m_{2}$, we get

• $v_{1}^{'} = 0$
• $v_{2}^{'} = v_{1}$

which also makes sense. Imagine two snooker balls colliding.

If a collision is inelastic its kinetic energy is not conserved (but the momentum is still conserved). However, if a collision is completely inelastic i.e the colliding particles get stuck together after the collision, we can still derive an easy expression to figure out their combined velocity simply from conservation of momentum:

• $v_{1}^{'} = v_{2}^{'} = v^{'} = \frac{m_{1}v_{1} + m_{2}v_{2}}{m_{1} + m_{2}}$