Question

A stick of length $L$ and mass $M$ lies on a frictionless horizontal surface on which it is free to move in any way, A ball of mass is $m$ and speed $v$ as shown in fig. What must be the mass of the ball so that it remains at rest immediately after the collision?

Answer 1

Given: a stick of length L and mass M lies on a frictionless horizontal surface on which it is free to move in any way, A ball of mass is m and speed v as shown in fig.

To find the mass of the ball so that it remains at rest immediately after the collision

Solution:

Applying the conservation of momentum

$mv=MV.......\left(i\right)$

By conservation of angular momentum,

$mv\times \frac{L}{2}=\frac{M{L}^2}{12}\times \omega ............\left(ii\right)$

eqn(i) in eqn(ii), we get

$MV\times \frac{L}{2}=\frac{M{L}^2}{12}\times \omega ⟹\omega =\frac{6V}{L}...\left(iii\right)$

As the collision is elastic, we have

$\frac{1}{2}m{v}^2=\frac{1}{2}M{V}^2+\frac{1}{2}I{\omega }^2$

Substituting the value of I and $\omega$ we get

$\frac{1}{2}m{v}^2=\frac{1}{2}M{V}^2+\frac{1}{2}\times \frac{M{L}^2}{12}\times \frac{(6V{)}^2}{L^2}⟹\frac{1}{2}m{v}^2=\frac{M{V}^2}{2}+\frac{3M{V}^2}{2}⟹\frac{1}{2}m{v}^2=\frac{4M{V}^2}{2}$

From eqn(i), $V=\frac{mv}{M}$

So the above equation becomes,

$m{v}^2=4M\times \frac{(mv{)}^2}{M^2}⟹M=4m⟹m=\frac{M}{4}$

is the mass of the ball so that it remains at rest immediately after the collision