Question

Two masses $m$ and $2m$ are placed in a fixed horizontal circular smooth hollow tube of radius $r$ as shown. The mass $m$ is moving with speed $u$ and the mass $2m$ is stationary. If the collison between the masses is elastic collison, then choose the correct option(s).

• A. The velocity of the mass $2m$ after collision will be $\frac{2u}{3}$
• B. The velocity of the mass $m$ after collision will be $\frac{2u}{3}$
• C. Time elapsed for next collision after their first collision will be $\frac{2\pi r}{u}$
• D. All the above.

Answer 1

Answer: (A), (C)

Time elapsed for next collision after their first collision will be $\frac{2\pi r}{u}$

Let, after collision, the velocity of particle of mass $m$ be $u_1$ and for mass $2m$ be $u_2$.
There is no external force, hence momentum will be conserved.
$mu=m{u}_1+2m{u}_2$
$u=u_1+2u_2.......\left(i\right)$
Since the collision is elastic, the kinetic energy is also conserved in the collision.
Hence, $K{E}_i=K{E}_f$
$\frac{1}{2}m{u}^2=\frac{1}{2}m{u}_1^2+\frac{1}{2}\left(2m\right)u_2^2$
$\Rightarrow u^2=u_1^2+2u_2^2$
$\Rightarrow (u-u_1)(u+u_1)=2u_2^2$
Using $\left(i\right)$,
$\Rightarrow 2u_2(u+u_1)=2u_2^2$
$\Rightarrow u_2=u_1+u....\left(ii\right)$

From eq $\left(i\right)$ & $\left(ii\right)$,
$u_2=\frac{2u}{3}$ and $u_1=\frac{-u}{3}$
Now, time taken to collide again will be
$T=\frac{2\pi r}{u_2-u_1}$
$\Rightarrow T=\frac{2\pi r}{\frac{2u}{3}-\left(\frac{-u}{3}\right)}\Rightarrow T=\frac{2\pi r}{u}$
Hence, (A) & (C) are correct.