A body of mass 2m is split into two equal parts by an internal explosion which generates a kinetic energy E. If after the explosion, the parts move in the same line as before, then what is their relative speed?

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$\sqrt{\frac{2 E}{m}}$

$\sqrt{\frac{E}{m}}$

$\sqrt{\frac{4 E}{m}}$

answer is C.

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Detailed Solution

Both parts will move with equal velocity in opposite direction to follow law of conservation of momentum.

So, $E = \frac{1}{2} {mv}^{2} + \frac{1}{2} {mv}^{2} = {mv}^{2} \Rightarrow v = \sqrt{\frac{E}{m}}$
Relative speed $= 2 v = 2 \sqrt{\frac{E}{m}} = \sqrt{\frac{4 E}{m}} .$

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