Question

A particle of mass $m$ moving with a speed $v$ hits elastically another stationary particle of mass $2m$ on a smooth horizontal circular tube of radius $r$. The time in which the next collision will take place is equal to

• A. $\frac{2\pi r}{v}$
• B. $\frac{4\pi r}{v}$
• C. $\frac{\pi r}{v}$
• D. $\frac{3\pi r}{v}$

Answer 1

The correct option is A $\frac{2\pi r}{v}$

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 $2m{v}_2$.

$mv=m{v}_1+2m{v}_2$

We can cancel out m and have:

$v=v_1+2v_2$

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

$\frac{1}{2}m{v}^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}2m{v}_2^2$

$v^2=v_1^2+2v_1^2$

The two equations can be rewritten in the form

$v-v_1=2v_2,v_2-v_1^2=2v_2^2$

Dividing both sides of the second equation by $v-v_1$

gives

$v+v_1=\frac{2v_2^2}{2v_2}=v_2$

(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 are

$v_2-v_1=v$

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

The time taken for this must be

$t=\frac{2\pi r}{v_2-v_1}=\frac{2\pi r}{v}$