A stationary body explodes into two fragments of masses $m_{1}$ and $m_{2}$ . If momentum of one fragment is p, the energy of explosion is :

\frac{p^{2}}{2 (m_{1} + m_{2})}

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\frac{p^{2}}{2 \sqrt{m_{1} m_{2}}}

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\frac{p^{2} (m_{1} + m_{2})}{2 m_{1} m_{2}}

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\frac{p^{2}}{2 (m_{1} - m_{2})}

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Solution

The correct option is B $\frac{p^{2} (m_{1} + m_{2})}{2 m_{1} m_{2}}$
Mass of the fragments be $m_{1}$ and $m_{2}$
Momentum of one particle is $p$ .
As initial momentum of the body is zero. Thus final momentum of the system must also be zero.
$⟹$ Magnitude of momentum of second fragment $| P_{2} | = p$
Energy of the explosion must be equal to sum of kinetic energy of the two fragments.
Thus $E = K . E_{1} + K . E_{2}$

$∴$ $E = \frac{p^{2}}{2 m_{1}} + \frac{p^{2}}{2 m_{2}} = \frac{p^{2} (m_{1} + m_{2})}{2 m_{1} m_{2}}$

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A stationary body explodes into two fragments of masses m1 and m2. If momentum of one fragment is p, the energy of explosion is :

A stationary body explodes into two fragments of masses $m_{1}$ and $m_{2}$ . If momentum of one fragment is p, the energy of explosion is :