# A particle of mass m moving with speed v hits elastically another stationary particle of mass 2m inside a smooth horizontal circular tube of radius r. The time after which the second collision will happen?

Mar 5, 2018

The circumference of the circle is $2 \pi r$ units, so that is the net distance the particles will travel after the collision. The time taken will be given by $t = \frac{2 \pi r}{v}$

#### Explanation:

The momentum before the collision will be $m v$. Momentum is conserved, so the momentum after the collision will be the same. It will be distributed between the two objects.

Call the velocity of the particle of mass $m$ after the collision ${v}_{1}$ and the velocity of the particle of mass $2 m$ ${v}_{2}$.

$m v = m {v}_{1} + 2 m {v}_{2}$

We can cancel out $m$ and have:

$v = {v}_{1} + 2 {v}_{2}$

Since the collision is elastic, kinetic energy is conserved in the collision

$\frac{1}{2} m {v}^{2} = \frac{1}{2} m {v}_{1}^{2} + \frac{1}{2} 2 m {v}_{2}^{2}$

Canceling out $\frac{1}{2} m$ gives us

${v}^{2} = {v}_{1}^{2} + 2 {v}_{1}^{2}$

The two equations can be rewritten in the form

$\textcolor{red}{v - {v}_{1} = 2 {v}_{2}} , q \quad {v}^{2} - {v}_{1}^{2} = 2 {v}_{2}^{2}$

Dividing both sides of the second equation by $v - {v}_{1}$
gives

$\textcolor{red}{v + {v}_{1} =} \frac{2 {v}_{2}^{2}}{2 {v}_{2}} = \textcolor{red}{{v}_{2}}$

(Note that the last equation could have been directly obtained by using the alternative definition of elastic collisions - relative speed of approach equals that of separation - that would considerably shorten the answer!)

This implies that the relative speed of separation of the two objects is

${v}_{2} - {v}_{1} = v$

When the two bodies collide again, the distances traveled by them must differ by $2 \pi r$.

The time taken for this must be

$t = \frac{2 \pi r}{{v}_{2} - {v}_{1}} = \frac{2 \pi r}{v}$