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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)
1079930_c644429635c0402296e7e762cc941579.png


A
2πru
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B
4πru
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C
3πru
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D
12πru
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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

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