A stationary body explodes into two fragments of masses $m_{1}$ and $m_{2}$ . If momentum of one tangent is $p$ , the minimum 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 C $\frac{p^{2} (m_{1} + m_{2})}{2 m_{1} m_{2}}$
Since the initial momentum of the system is zero. So, according to the conservation of linear momentum, final momentum of the system is zero. Thus momentum of the second fragment is also $p$ .
Kinetic energy of first fragment $E_{1} = \frac{p^{2}}{2 m_{1}}$
Kinetic energy of second fragment $E_{2} = \frac{p^{2}}{2 m_{2}}$
Minimum energy of explosion $E = E_{1} + E_{2} = \frac{p^{2} (m_{1} + m_{2})}{2 m_{1} m_{2}}$

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

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