A shell of mass $(m_{1} + m_{2})$ is fined with a given velocity in a given direction. At the highest point of its path, the shell explodes into two fragments of mass $m_{1}, m_{2}$ . The explosion produces an additional kinetic energy $E$ and the fragments separate in a horizontal direction. Find the horizontal distance on the ground at which they hit the ground, if vertical component of velocity is $v_{0}$

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$x^{'} = \frac{v_{0}}{g} {\sqrt{E (\frac{1}{m_{1}} + \frac{1}{m_{2}})}}^{3}$

$x^{'} = \frac{v_{0}}{g} \sqrt{E (\frac{1}{m_{1}} + \frac{1}{m_{2}})}$

$x^{'} = \frac{v_{0}}{g} \sqrt{2 E (\frac{1}{m_{1}} + \frac{1}{m_{2}})}$

$x^{'} = \frac{2 v_{0}}{g} \sqrt{2 E (\frac{1}{m_{1}} + \frac{1}{m_{2}})}$

answer is D.

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Detailed Solution

Time taken by the shell to reach the highest point $T = \frac{v_{0}}{g}$

The fragments $m_{1}$ and $m_{2}$ take the same time $T$ to reach the ground. In this durations the horizontal component of relative speed of the fragments is $(v_{1} + v_{2})$ so

$x^{'} = (v_{1} + v_{2}) T = (u_{1} + u_{2}) \frac{v_{0}}{g} (i)$

if $u$ is the speed of shell before explosion, then

$(m_{1} + m_{2}) u = m_{1} u_{1} - m_{2} u_{2} (i i) \frac{1}{2} (m_{1} + m_{2}) u^{2} + E = \frac{1}{2} m_{1} {u^{2}}_{1} + \frac{1}{2} m_{2} {u^{2}}_{2} (i i i)$