Question

A bomb of mass $M$ at rest explodes into two fragments of masses $m_1$ and $m_2$. The total energy released in the explosion is $E$. If $E_1$ and $E_2$ represent the energies carried by masses $m_1$ and $m_2$ respectively, then which of the following is correct?

• A. $E_2=\frac{m_1}{M}E$
• B. $E_1=\frac{m_1}{m_2}E$
• C. $E_1=\frac{m_1}{M}E$
• D. $E_1=\frac{m_2}{m_1}E$

Answer 1

The correct option is D $E_1=\frac{m_2}{m_1}E$

As no external force act on the system during explosion, lineare momentum is conserve.

$\Rightarrow P_{initial}=P_{final}$

Now,

$P_{initial}=0$ (Man M was at rust )....(2)

Let, after explosion velocity of $M_1$ and $M_2$ be $V_1$ and $V_2$ respectively such that

$E_1=\frac{1}{2}{m}_1{v}_1^2$ and $E_2=\frac{1}{2}{m}_2{v}_2^2$,

$\Rightarrow P_{final}=m_1{v}_1+m_2{v}_2$

from (1), using (2) and (3)

$m_1{v}_1+m_2{v}_2=0\Rightarrow m_1{v}_2=-m_2{v}_2$

$\Rightarrow (m_1{v}_1{)}^2=(m_2{v}_2{)}^2$ (squaring both sides)

$\Rightarrow \left(\frac{1}{2}{m}_1{v}_1\right)m_1=\left(\frac{1}{2}{m}_2{v}_2^2\right)m_2$ (multiplying wiht $\frac{1}{2}$ on both sides)

$\Rightarrow E_1{m}_1=E_2{m}_2$

$\Rightarrow E_1=\frac{m_2}{m_1}{E}_2$