Suppose that a narrow and straight tunnel is made passing through the centre of the earth. Two small balls $A$ and $B$ are released simultaneously at the two ends of tunnel. After collision, $A$ comes to rest and $B$ escapes earth’s gravitational field. If mass of $B$ is $(\sqrt{3} - 1) k g .$

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$e = \frac{1}{\sqrt{2}}$

$m_{A} = 2 k g$

$m_{A} = (\sqrt{3} + 1) k g$

$e = \frac{\sqrt{3}}{2}$

answer is A, D.

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Detailed Solution

They collide at centre of earth let $V_{0}$ be speed just before collision

$\frac{- G M m}{R} = \frac{1}{2} m v_{0}^{2} - \frac{3 G M m}{2 R}$
$V_{0} = \sqrt{\frac{G M m}{2 R}}$
Escape velocity $\frac{- 3 G M m}{2 R} + \frac{1}{2} m v^{2} = 0$
$v = \sqrt{\frac{3 G M}{R}}$
$e = \frac{v_{r e l}}{u_{r e l}} = \frac{\sqrt{\frac{3 G M}{R}}}{2 \sqrt{\frac{G M}{R}}} = \frac{\sqrt{3}}{2}$
$(m_{A} - m_{B}) V_{0} = m_{B} . v$
$(m_{A} - \sqrt{3} + 1) \sqrt{\frac{G M}{R}} = (\sqrt{3} - 1) \sqrt{\frac{3 G M}{R}}$
$(m_{A} - \sqrt{3} + 1) = (3 - \sqrt{3})$
$m_{A} = 2 k g$