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A 2kg ball moving at $6 m s^{- 1}$ collides head on with a $3 k g$ ball moving in the opposite direction at $12 m / s$ . The velocity of each ball after the impact, if the coefficient of restitution is 1/2, is

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$- 1 0. 2m / s; - 1 . 2 m / s$

$- 10 m / s; - 2 m / s$

$10 m / s; 14 m / s$

$1 0. 2m / s, - 2 m / s$

answer is A.

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Detailed Solution

$v_{1} = \frac{m_{1} - e m_{2}}{m_{1} + m_{2}} u_{1} - \frac{(1 + e) m_{2}}{m_{1} + m_{2}} u_{2}$ $= \frac{2 - \frac{1}{2} \times 3}{2 + 3} \times 6 - \frac{(1 + \frac{1}{2}) 3}{2 + 3} \times 12$

= $\frac{3}{5} - \frac{9}{10} \times 12$ = $0.6 - 10.8 = - 10.2 m s^{- 1}$

$v_{2} = \frac{(1 + e) m_{1}}{m_{1} + m_{2}} u_{1} - (\frac{m_{2} - e m_{1}}{m_{1} + m_{2}}) u_{2}$ = $- 1.2 m s^{- 1}$

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