Question

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?

Answer 1

The circumference of the circle is `2pir` units, so that is the net distance the particles will travel after the collision. The time taken will be given by `t=(2pir)/v`

Explanation:

The momentum before the collision will be `mv`. 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 `2m` `v_2`.

`mv=mv_1+2mv_2`

We can cancel out `m` and have:

`v=v_1+2v_2`

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

`1/2mv^2 = 1/2mv_1^2+1/2 2mv_2^2`

Canceling out `1/2m` gives us

`v^2 = v_1^2+2v_1^2`

The two equations can be rewritten in the form

`color(red)(v-v_1 = 2v_2),qquad v^2-v_1^2 = 2v_2^2`

Dividing both sides of the second equation by `v-v_1`
gives

`color(red)(v+v_1 =) {2v_2^2}/(2v_2)=color(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 `2pi r`.

The time taken for this must be

`t = {2pi r}/{v_2-v_1} = {2pi r}/v`