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
​

Answer 1

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 2mv

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

2

1

mv

2

=

2

1

mv

1

2

+

2

1

2mv

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

=

2v

2

2v

2

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 2πr.

The time taken for this must be

t=

v

2

−v

1

2πr

=

v

2πr

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