N identical balls are placed on a smooth horizontal surface. An another ball of same mass collides elastically with velocity u with first ball of N balls. A process of collision is thus started in which first ball collides with second ball and the second ball with the third ball and so on. The coefficient of restitution for each collision is e. Find speed of Nth ball.

After collision, first ball starts moving with velocity u

Now, $\mathrm{e}=\frac{{\mathrm{V}}_2-{\mathrm{V}}_1}{\mathrm{u}}$

and $\mathrm{mu}={\mathrm{mV}}_1+{\mathrm{mV}}_2$

or ${\mathrm{V}}_1 = \mathrm{u}-{\mathrm{V}}_2$

so, $\mathrm{e}=\frac{{\mathrm{V}}_2-\mathrm{u}+{\mathrm{V}}_2}{\mathrm{u}}$

or ${\mathrm{V}}_2 = \frac{\mathrm{u}(1+\mathrm{e})}{2}$

Incoming velocity of second ball is ${\mathrm{V}}_2 \mathrm{and} \mathrm{after} \mathrm{collision}, \mathrm{velocity} \mathrm{of} \mathrm{third} \mathrm{ball} \mathrm{is} {\mathrm{V}}_3 \mathrm{and} \mathrm{velocity} \mathrm{of} \mathrm{second} \mathrm{ball} \mathrm{is} {{\mathrm{V}}_2}^{\text{'}}$

 $\mathrm{e} = \frac{{\mathrm{V}}_3-{\mathrm{V}}_2\text{'}}{{\mathrm{V}}_2}$

and ${\mathrm{mV}}_2 = {\mathrm{mV}}_3+{\mathrm{mV}}_2\text{'}$

or ${\mathrm{V}}_2\text{'} = {\mathrm{V}}_2-{\mathrm{V}}_3=\frac{\mathrm{u}(1+\mathrm{e})}{2}-{\mathrm{V}}_3$

So, $\mathrm{e} = \frac{{\mathrm{V}}_3-{\displaystyle \left[\frac{(\mathrm{u}+\mathrm{ue}-2{\mathrm{V}}_3)}{2}\right]}}{{\displaystyle \left[\frac{\mathrm{u}(1+\mathrm{e})}{2}\right]}}$

On solving ${\mathrm{V}}_3 = \frac{\mathrm{u}{(1+\mathrm{e})}^2}{2^2}$ 

So, velocity of N^th ball = $\frac{\mathrm{u}{(1+\mathrm{e})}^{\mathrm{N}-1}}{2^{\mathrm{N}-1}}$