A particle of mass 5m at rest suddenly breaks on its own into three fragments. Two fragments of mass each move along mutually perpendicular directions with speed each. The energy released during the process is: [2019]

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Solution

Q.
A particle of mass 5m at rest suddenly breaks on its own into three fragments. Two fragments of mass $m$ each move along mutually perpendicular directions with speed $v$ each. The energy released during the process is: [2019]

1 $\frac{3}{5} m v^{2}$

2 $\frac{5}{3} m v^{2}$

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

4 $\frac{4}{3} m v^{2}$

Ans.
(4)

Let the speed of the third fragment of mass $3 m$ be $v' .$

From the law of conservation of linear momentum,

$3 m v' = \sqrt{2} m v \Rightarrow v' = \frac{\sqrt{2} v}{3}$ ...(i)

$∴$ Energy released during the process is,

$K.E. = 2 (\frac{1}{2} m v^{2}) + \frac{1}{2} (3 m) v'^{2} = m v^{2} + \frac{1}{2} (3 m) \frac{2 v^{2}}{9} (Using eqn. (i))$

$= m v^{2} + \frac{m v^{2}}{3} = \frac{4}{3} m v^{2}$

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