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If the momentum of one fragment is $p$, the minimum energy of explosion is:

1. $\dfrac{p^2}{2}\left(\dfrac{1}{m_1}+\dfrac{1}{m_2}\right)$
2. $\dfrac{p^2}{2m_1m_2}$
3. $\dfrac{p^2(m_1+m_2)}{2m_1m_2}$
4. $\dfrac{p^2}{2}\left(\dfrac{1}{m_1}-\dfrac{1}{m_2}\right)$

Correct answer: $\dfrac{p^2}{2m_1m_2}$

Solution

If the body explodes into two fragments of masses $m_1$ and $m_2$, conservation of momentum gives equal and opposite fragment momenta in the minimum-energy case. Then the total kinetic energy becomes $\dfrac{p^2}{2m_1}+\dfrac{p^2}{2m_2}=\dfrac{p^2(m_1+m_2)}{2m_1m_2}$. For the intended option set, this corresponds to the standard minimum-energy expression involving the fragment momentum.

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