Two bodies of same mass are connected with an inelastic string of length $l$ are placed together on a horizontal surface. One of them is projected vertically upwards from the surface with a velocity $u = \sqrt{6 g l} .$ The maximum height up to which the centre of mass of the system of the two masses rises is

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$l$

$\frac{3 l}{4}$

$\frac{3 l}{2}$

$\frac{l}{2}$

answer is D.

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Detailed Solution

The velocity of B just before the string is taut $v_{1} = \sqrt{u^{2} - 2 g l} = \sqrt{6 g l - 2 g l} = \sqrt{4 g l}$

After string is taut both move with same speed $v_{2} = \frac{m_{1} v_{1}}{m_{1} + m_{2}} = \frac{v_{1}}{2} = \sqrt{g l}$

By the time centre of mass is at $h = \frac{l}{2}$

Further rise in centre of mass $= \frac{v_{2}^{2}}{2 g} = \frac{l}{2}$

Total maximum $H = \frac{l}{2} + \frac{l}{2} = l$

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