A ball of mass m moving with a constant velocity u strikes against a ball of same mass at rest. If e is the coefficient of restitution, then what will be the ratio of velocity of two balls after collision?

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

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

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

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

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Solution

The correct option is C $\frac{1 - e}{1 + e}$
A. $\frac{1 - e}{1 + e}$

Given,

$m_{1} = m_{2} = m$ (say)

$u_{1} = u$ , $u_{2} = 0$

let, $v_{1} =$ velocity of ball 1 after collision

$v_{2} =$ velocity of ball 2 after collision

The coefficient of restitution,

$e = \frac{v_{2} - v_{1}}{u_{1} - u_{2}}$

$e u = v_{2} - v_{1}$ . . . . . . .(1)

By the conservation of Linear momentum,

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$

$u = v_{1} + v_{2}$ . . . . . . . . .(1)

By solving equation (1) and (2), we get

$v_{1} = \frac{(1 - e) u}{2}$

$v_{2} = \frac{(1 + e) u}{2}$

The ratio of velocity of two ball,

$\frac{v_{1}}{v_{2}} = \frac{1 - e}{1 + e}$

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