Question

Two masses m and 2m are placed in 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. After their first collision, the time elapsed for next collision. (coefficient of restitution e = 1/2)

• A. $\frac{2\pi r}{u}$
• B. $\frac{4\pi r}{u}$
• C. $\frac{3\pi r}{u}$
• D. $\frac{12\pi r}{u}$

Answer 1

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

Let the velocity of the particle of mass $m$, after the collision is $u_1$ and the velocity of the particle of mass $2m$ is $u_2$.

From conservation of momentum, it can be written as,

$mu=m{u}_1+2m{u}_2$

$u=u_1+2u_2$

Since, the collision is elastic, the kinetic energy is conserved in the collision and it can be written as,

$\frac{1}{2}m{u}^2=\frac{1}{2}m{u_1}^2+\frac{1}{2}2m{u_2}^2$

$u^2={u_1}^2+2{u_2}^2$

$u^2-{u_1}^2=2{u_2}^2$

$\left(u-u_1\right)\left(u+u_1\right)=2{u_2}^2$

$\left(u+u_1\right)=u_2$

$u=u_2-u_1$

The time taken is given as,

$t=\frac{2\pi r}{u_2-u_1}$

$t=\frac{2\pi r}{u}$

Thus, the time elapsed for the next collision is $\frac{2\pi r}{u}$.