A ball of mass $m$ moving with a speed $2 v_{0}$ collides head-on with an identical ball at rest. If $e$ is the coefficient of restitution, then what will be the ratio of velocity of two balls after collision?

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$\frac{1 - e}{1 + e}$

$\frac{1 + e}{1 - e}$

$\frac{e - 1}{e + 1}$

$\frac{e + 1}{e - 1}$

answer is A.

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Detailed Solution

According to the law of conservation of linear momentum, we get

$m (2 v_{0}) + m \times 0 = m v_{1} + m v_{2}$

where $v_{1}$ and $v_{2}$ are the velocities of the balls after collision.

$v_{1} + v_{2} = 2 v_{0} . . . (i)$

By definition of coefficient of restitution

$e = \frac{v_{2} - v_{1}}{u_{1} - u_{2}} = \frac{v_{2} - v_{1}}{2 v_{0}} (∵ u_{1} = 2 v_{0}, u_{2} = 0) v_{2} - v_{1} = 2 e v_{0} . . . (i i)$

Adding $(i)$ and $(i i)$ ,we get $2 v_{2} = 2 v_{0} + 2 e v_{0}; v_{2} = (1 + e) v_{0}$

Subtract $(i i)$ from $(i)$ , we get

$2 v_{1} = 2 v_{0} - 2 e v_{0}; v_{1} = v_{0} (1 - e)$

Their corresponding ratio is $\frac{v_{1}}{v_{2}} = \frac{1 - e}{1 + e}$

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