Two identical balls, of equal masses A and B, are lying on a smooth surface as shown in Fig. Ball A hits ball B (which is at rest) with a velocity v = 16m/s. What should be the minimum value of coefficient of restitution between A and B so that B just reaches the highest point of inclined plane. $(g = 10 m s^{- 2})$

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$\frac{1}{3}$

$\frac{2}{3}$

$\frac{1}{4}$

$\frac{1}{2}$

answer is B.

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Detailed Solution

v = 16 m/s

Let velocity of B after collision be $v_{2}$ and that of (A) be $v_{1}$ then

$v_{2}$ = $\sqrt{2 gs} = \sqrt{2 x 10 x 5}$ = 10m/s

$m v_{1} + m v_{2} = m v \Rightarrow v_{1} + 10 = 16 \Rightarrow v_{1}$ = 6m/s

e = $v_{2} - v_{1} / v = \frac{10 - 6}{16} = \frac{1}{4}$

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