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A bomb of mass 3m is initially at rest. Suddenly it explodes into two fragments, ratio of whose masses is 1 : 2. Energy released in the process is E and 75% of energy released appears as kinetic energy of the fragments, then kinetic energy of the smaller fragment is

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$\frac{3 E}{4}$

$\frac{E}{2}$

$\frac{E}{4}$

$\frac{2 E}{3}$

answer is A.

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Detailed Solution

If p be the magnitude of momentum of each fragment just after the explosion, Then

$\frac{p^{2}}{2 m} + \frac{p^{2}}{2 (2 m)} = \frac{75}{100} x E \Rightarrow \frac{3 p^{2}}{4 m} = \frac{3 E}{4} \Rightarrow \frac{p^{2}}{2 m} = \frac{E}{2}$

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