A stationary body explodes into two fragments of masses $m_{1}$ and $m_{2}$ . If momentum of one fragment 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 A $\frac{p^{2}}{2 (m_{1} + m_{2})}$
Initially body is stationary (zero velocity)
using conservation of momentum
$(A) O = P + P_{2}, P_{2}$ is momentum of mass $m_{2}$ after explosion
$\begin{matrix} P_{2} = - P E n e r g y = E_{1} \Rightarrow \frac{P^{2}}{2 m_{1}} d E_{2} = \frac{{(- P)}^{2}}{2 m_{2}} n e t e n e r g y \Rightarrow E_{1} + E_{2} = \frac{P^{2}}{2 m_{2}} + \frac{P^{2}}{2 m_{2}} \Rightarrow \frac{P^{2}}{2 (m_{1} + m_{2})} \end{matrix}$

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

The correct option is A p22(m1+m2)Initially body is stationary (zero velocity)using conservation of momentum(A)O=P+P2,P2 is momentum of mass m2 after explosionP2=−PEnergy=E1⇒P22m1dE2=(−P)22m2netenergy⇒E1+E2=P22m2+P22m2⇒P22(m1+m2)

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