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
2πru
B
4πru
C
3πru
D
12πru

Solution

## The correct option is A $$\dfrac{2\pi r}{u}$$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, $$mu = m{u_1} + 2m{u_2}$$ $$u = {u_1} + 2{u_2}$$ Since, the collision is elastic, the kinetic energy is conserved in the collision and it can be written as, $$\dfrac{1}{2}m{u^2} = \dfrac{1}{2}m{u_1}^2 + \dfrac{1}{2}2m{u_2}^2$$ $${u^2} = {u_1}^2 + 2{u_2}^2$$ $${u^2} - {u_1}^2 = 2{u_2}^2$$ $$\left( {u - {u_1}} \right)\left( {u + {u_1}} \right) = 2{u_2}^2$$ $$\left( {u + {u_1}} \right) = {u_2}$$ $$u = {u_2} - {u_1}$$ The time taken is given as, $$t = \dfrac{{2\pi r}}{{{u_2} - {u_1}}}$$ $$t = \dfrac{{2\pi r}}{u}$$ Thus, the time elapsed for the next collision is $$\dfrac{{2\pi r}}{u}$$.Physics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More