Question

A stationary body explodes into three fragments of masses $m_1,m_2$ and $m_3$. If momentum of one fragment is $p$ and one fragment of mass $m_3$ remains at rest, the energy of explosion is:

• A. $\frac{p^2}{2(m_1+m_2)}$
• B. $\frac{p^2}{2\sqrt{m_1{m}_2}}$
• C. $\frac{p^2(m_1+m_2)}{2m_1{m}_2}$
• D. $\frac{p^2}{2(m_1-m_2)}$

Answer 1

Answer: (C)

$\frac{p^2(m_1+m_2)}{2m_1{m}_2}$

Let, $p$ be the momentum of one fragment with mass $m_1$ and $p^{\prime }$ be the momentum of other fragment with mass $m_2$,
The energy of explosion is

$KE=\frac{1}{2}{m}_1(v_1{)}^2+\frac{1}{2}{m}_2(v_2{)}^2$

$KE=\frac{1}{2m_1}\left(m_1{)}^2\right(v_1{)}^2+\frac{1}{2m_2}\left(m_2{)}^2\right(v_2{)}^2$

$KE=\frac{1}{2m_1}{p}_1^2+\frac{1}{2m_2}{p}_2^2$

$KE=\frac{2(m_2{p}_1^2+m_1{p}_2^2)}{4m_1{m}_2}$

Since, after explosion momentum will be conserved hence, $p=p_1=p_2$

$KE=\frac{m_2{p}^2+m_1{p}^2}{2m_1{m}_2}$

$KE=\frac{p^2(m_1+m_2)}{2m_1{m}_2}$