Two masses $m_{1}$ and $m_{2}$ are connected by a string of length l. They are held in a horizontal plane at a height H above two heavy plates A and B made of different materials placed on the floor. Initially, the distance between the two masses is $a < l$ . When the masses are released they fall freely and collide with the plates simultaneously. The coefficient of restitution for the collision of the first ball-plate pair is 0.8. and for the second ball-plate pair is 0.4. The time after the collision when the string becomes taut is

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$t = \frac{5}{2} \sqrt{\frac{l^{2} - a^{2}}{2 gH}}$

$t = \frac{5}{4} \sqrt{\frac{l^{2} - a^{3}}{2 gH}}$

$t = \frac{5}{2} \sqrt{\frac{l_{a}}{2 gH}}$

$t = \frac{5}{4} \sqrt{\frac{l_{a}}{2 gH}}$

answer is A.

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Detailed Solution

The string will become taut when
$\begin{array}{l} h_{1} - h_{2} = \sqrt{l^{2} - a^{2}} \\ h_{1} = 0 .8 vt - \frac{1}{2} {gt}^{2} \\ h_{2} = 0 .4 vt - \frac{1}{2} {gt}^{2} \\ h_{1} - h_{2} = 0 .4 vt \\ 0 .4 vt = \sqrt{l^{2} - a^{2}}, where v = \sqrt{2 gh} \end{array}$
$t = \frac{5}{2} \sqrt{\frac{l^{2} - a^{2}}{2 gH}}$