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A 4kg ball moving at $48 m s^{- 1}$ undergoes inelastic head-on collision with an $8 k g$ ball moving in the opposite direction at $96 m s^{- 1} .$ If the coefficient of restitution is $\frac{2}{3},$ their velocities in $m s^{- 1}$ after impact are

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$- 16, - 112$

$- 56, - 8$

$- 7, - 1$

$- 14, - 2$

answer is B.

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Detailed Solution

$4 \times 48 - 8 \times 96 = 4 \times v_{1} + 8 v_{2}$

$192 - 768 = 4 v_{1} + 8 v_{2}$

$4 v_{1} + 8 v_{2} = - 576$ -----(1)

$\frac{2}{3} = \frac{v_{2} - v_{1}}{u_{1} + u_{2}} \Rightarrow \frac{2}{3} = \frac{v_{2} - v_{1}}{48 + 96}$

$v_{2} - v_{1} = \frac{2}{3} \times 144 = 96$ -----(2)

Solving (1) and (2)

$v_{1} + 2 v_{2} = - 144$

$- v_{1} + v_{2} = 96$

$3 v_{2} = - 48$

$v_{2} = - 16 m / s$

$v_{1} = - 112 m / s$

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