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A particle of mass 4 m which is at rest explodes into three fragments. Two of the fragments each of mass m are found to move with speed V each in mutually perpendicular directions. The minimum energy released in the process of explosion is

\[\begin {array} {1 1} (a)\;\frac{2}{3} mv^2 & \quad (b)\;\bigg(\frac{3}{2}\bigg)mv^2 \\ (c)\;\bigg(\frac{4}{3}\bigg)mv^2 & \quad  (d)\;\bigg(\frac{3}{4}\bigg)mv^2 \end {array}\]

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By momentum conservation
$2mv'=\sqrt 2 mV$
$v'=\large\frac{v}{\sqrt 2}$
$KE= 2\{\large\frac{1}{2}$$mv^2\}+\large\frac{1}{2}$$ (2m)\large\frac{v^2}{2}$
$\qquad= mv^2+\large\frac{mv^2}{2}$
$\qquad= \large\frac{3}{2}$$mv^2$
answered Nov 22, 2013 by meena.p
edited Jun 17, 2014 by lmohan717

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