3.7 Collisions

A collision is an interaction between two objects over a short time interval. To solve these problems, we can use the concept of momentum conservation and energy conservation that we have been studying in the last two chapters. Consider two blocks sliding along a frictionles surface towards each other (1-dimensional problem). The blocks have masses $m_{1}$ , $m_{2}$ and velocities $v_{1}$ , $v_{2}$ respectively. What we want to find is the velocities of the blocks after the collision. To do this, we write the total momentum and kinetic energy before and after as

Before:	$\displaystyle P=m_{1}v_{1}+m_{2}v_{2}$	(3.101)
	$\displaystyle K=\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}$	(3.102)
After:	$\displaystyle P^{\prime}=m_{1}v_{1}^{\prime}+m_{2}v_{2}^{\prime}$	(3.103)
	$\displaystyle K^{\prime}=\frac{1}{2}m_{1}v_{1}^{\prime 2}+\frac{1}{2}m_{2}v_{2% }^{\prime 2}.$	(3.104)

Total momentum is always conserved in collisions. On the other hand, depending on the forces involved during the collision, total kinetic energy may or may not be conserved. We call the case where it is conserved “elastic” and the case where it is not “inelastic”.

3.7.1 Elastic Collisions

In the case of elastic collisions, where total kinetic energy is conserved, we can write

	$\displaystyle m_{1}v_{1}+m_{2}v_{2}$	$\displaystyle=m_{1}v_{1}^{\prime}+m_{2}v_{2}^{\prime}$		(3.105)
	$\displaystyle\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}$	$\displaystyle=\frac{1}{2}m_{1}v_{1}^{\prime 2}+\frac{1}{2}m_{2}v_{2}^{\prime 2}$		(3.106)

This is a system of two equations for two unknowns, $v_{1}^{\prime}$ and $v_{2}^{\prime}$ . We can solving this system using algebra, and the solution is

	$\displaystyle v_{1}^{\prime}$	$\displaystyle=\frac{m_{1}-m_{2}}{m_{1}+m_{2}}v_{1}+\frac{2m_{2}}{m_{1}+m_{2}}v% _{2}$		(3.107)
	$\displaystyle v_{2}^{\prime}$	$\displaystyle=\frac{2m_{1}}{m_{1}+m_{2}}v_{1}+\frac{m_{2}-m_{1}}{m_{1}+m_{2}}v% _{2}.$		(3.108)

Does this make sense? To examine whether this answer makes physical sense we can take some limits and see what happens to the solution. Set $v_{2}=0$ and then consider the limit where $m_{1}\gg m_{2}$ . In this case, $v_{1}^{\prime}\to v_{1}$ and $v_{2}^{\prime}\to 2v_{1}$ . This is like a bowling ball colliding with a ping-pong ball, the bowling ball keeps on going and the ping ball gets deflected in the same direction with twice the speed. If the two masses are equal, $v_{1}^{\prime}=0$ and $v_{2}^{\prime}=v_{1}$ , which is like a perfect billiard ball collision. On the other hand, if $m_{1}\ll m_{2}$ , $v_{1}^{\prime}\to-v_{1}$ and $v_{2}^{\prime}\to 0$ . This corresponds to a ping-pong ball hitting a bowling ball at rest. It bounces off with the same speed in the opposite direction while the bowling ball stays still.

If we transform the velocities into the centre of mass frame we get

$\displaystyle v_{1,\text{COM}}$	$\displaystyle=\frac{m_{2}(v_{1}-v_{2})}{m_{1}+m_{2}}$	(3.109)
$\displaystyle v_{2,\text{COM}}$	$\displaystyle=\frac{m_{1}(v_{2}-v_{1})}{m_{1}+m_{2}}=-\frac{m_{1}}{m_{2}}v_{1,% \text{COM}}$	(3.110)
$\displaystyle v_{1,\text{COM}}^{\prime}$	$\displaystyle=-v_{1,\text{COM}}$	(3.111)
$\displaystyle v_{2,\text{COM}}^{\prime}$	$\displaystyle=-v_{2,\text{COM}}.$	(3.112)

So in the centre of mass frame, the two objects approach each other from opposite directions with velocities antiproportional to thei masses. After the collision, the magnitude of the velocities remains the same but they switch sign.

3.7.2 Inelastic Collisions

In an inelastic collision, we only have conservation of momentum since some energy is lost to non-conservative forces in the collision. To solve the system, we need another constraint on the velocities after the collision. In the case where the maximum kinetic energy is lost, which is when the objects stick together and move as a single body with velocity $v^{\prime}=v_{1}^{\prime}=v_{2}^{\prime}$ . This reduces the two equations for two unknowns that we had to solve before to one equation for one unknown.

	$\displaystyle m_{1}v_{1}+m_{2}v_{2}$	$\displaystyle=m_{1}v^{\prime}+m_{2}v^{\prime}$		(3.113)
	$\displaystyle\implies v^{\prime}$	$\displaystyle=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{1}+m_{2}}.$		(3.114)

Notice that $v^{\prime}$ is simply the centre of mass velocity. So, if we transform into the centre of mass frame the final velocity is 0

v_{\text{COM}}^{\prime}=v^{\prime}-v_{\text{COM}}=0.

(3.115)

This means that in a perfectly inelastic collision seen from the centre of mass frame, the objects approach each other with the same velocities as in the elastic case, but then come together at rest at the origin.

Example 3.9.

Golf ball on a basketball.