A stationary body explodes into two fragments of masses $m_{1}$ and $m_{2}$ . If momentum of one fragments 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})}{4 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 D $\frac{p^{2} (m_{1} + m_{2})}{4 m_{1} m_{2}}$
applying the concept of momentum conservation
before collision
mass=M
velocity=V
after collision
mass of first piece will be $m_{1}$
mass of second piece will be $m_{2}$
velocity of first piece will be $v_{1}$
velocity of second piece will be $v_{2}$
now

$\frac{1}{2}$ $M V^{2}$ = $\frac{1}{2}$ $m_{1} v_{1}^{2}$ + $\frac{1}{2}$ $m_{2} v_{2}^{2}$

as we know that momentum (p)=mass $\times$ velocity
hence multiply by both numerator and denominator by m we get the final relation,

$\frac{1}{2}$ $\frac{p^{2}}{M}$ = $\frac{1}{2}$ ${\frac{p_{1}^{2}}{m}}_{1}$ + $\frac{1}{2}$ ${\frac{p_{2}^{2}}{m}}_{2}$

where,
$p_{1} = p_{2} = p$
finally we get

$E = \frac{P^{2} (m_{1} + m_{2})}{4 m_{1} m_{2}}$

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

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