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?

1 Answer

Answer:

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#