Question

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

Moderate

Solution

From conservation of linear momentum.

$\large 0 = mv\hat j + mv\hat i + 3m{\vec v_1}$
$\large {\vec v_1} = - \frac{v}{3}\left( {\hat i + \hat j} \right);{v_1} = \frac{{\sqrt 2 }}{3}v$
$\large K{E_i} = 0$
$\large K{E_f} = \frac{1}{2}m{v^2} + \frac{1}{2}m{v^2} + \frac{1}{2}\left( {3m} \right){\left( {\frac{{\sqrt 2 }}{3}} \right)^2}{v^2}$
$\large = m{v^2} + \frac{{m{v^2}}}{3} = \frac{4}{3}m{v^2}$
$\large \Delta KE = K{E_f} - K{E_i} = \frac{4}{3}m{v^2}$

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