Question 7.SP.37: Rework Problem 7.36 for the impact of purely elastic bodies ......

For elastic impact, substitute $e=1$ in the equations for $v_{2}$ and $v_{1}$ derived in Problem 7.36. There results

$\begin{array}{l}v_{1}=\frac{2 m_{2} u_{2}  +  u_{1}\left(m_{1}-m_{2}\right)}{m_{2}  +  m_{1}} \qquad \qquad &(1) \\  \\v_{2}=\frac{2 m_{1} u_{1}  +  u_{2}\left(m_{2}-m_{1}\right)}{m_{2}  +  m_{1}}\qquad \qquad &(2)\end{array}$

Of special interest is the case when $m_{1}=m_{2}=m$. Then the equations become

$\begin{array}{l}v_{1}=\frac{2 m u_{2}  +  0}{m+m}=u_{2}\qquad \qquad &(3) \\  \\v_{2}=\frac{2 m u_{1}  +  0}{m  +  m}=u_{1}\qquad \qquad &(4)\end{array}$

The above equations explain what takes place when a moving coin strikes an identical stationary coin on a frictionless surface. The final speed of the moving coin will be the initial speed of the other coin (in this case zero), while the final speed of the formerly stationary coin will be the initial speed of the moving coin. In other words, the moving coin stops and the stationary one assumes its speed. The same thing is true of a straight row of coins. The first one (farthest away from the moving coin) moves away while the others remain stationary.