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A particle of mass 5m which is at rest explodes into 3 fragments. Two equal fragments each of mass 2m are found to move with a speed v each in opposite directions. The energy released in the process of explosion is

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$3 m v^{2}$

$2 m v^{2}$

$\frac{3}{2} m v^{2}$

$\frac{5}{2} m v^{2}$

answer is A.

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Detailed Solution

From conservation of momentum,

2mv-2mv +mv'= 0;

hence v'=0,

total kinetic energy = $2 (\frac{1}{2} 2 {mv}^{2}) = 2 {mv}^{2}$

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