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Two balls A and B of masses m and 2m respectively are lying on a smooth surface as shown in the figure. Ball A hits the ball B (which is at rest) with a velocity v = 16 m/s. Such that B just reaches the highest point of inclined plane. $(g = 10 m / s^{2})$
where e is the coefficient of restitution and $v_{A}$ is velocity of the ball A after collision.

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e = 3/4

e = 7/8

$v_{A} = - 4 m / s$

$v_{A} = 8 m / s$

answer is C, B.

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Detailed Solution

$P_{1} = m \times 16 + 2 m \times 0 = 16 m$

$P_{1} = m V_{A} + 2 m V_{B}$

$∴ \frac{1}{2} \times 2 m V_{B}^{2} = 2 m g h \Rightarrow V_{B} = 10 m / s 16 m = m V_{A} + 2 m \times 10 \Rightarrow V_{A}^{'} = - 4 m / s e = \frac{V_{B}^{'} - V_{A}^{'}}{V_{A} - V_{B}} = \frac{10 - (- 4)}{16 - 0} = \frac{14}{16} = \frac{7}{8}$

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