Question

13. A bomb of mass M at rest explodes into two
fragments of masses m, and m. The total energy
released in the explosion is E. If E, and E2
represent the energies carried by masses m, and
m, respectively, then which of the following is
correct?
(1) E, = m2 E
(3) E,
(4) E, = "2E​

Answer 1

Explanation:

I suppose this is the answer... pls correct me if I have made mistakes

Answer 2

Question:-

A bomb of mass $\displaystyle\sf{M}$ at rest explodes into two  fragments of masses $\displaystyle\sf{m_1}$ and $\displaystyle\sf{m_2.}$ The total energy  released in the explosion is $\displaystyle\sf{E.}$ If $\displaystyle\sf{E_1}$ and $\displaystyle\sf{E_2}$ represent the energies carried by masses $\displaystyle\sf{m_1}$ and  $\displaystyle\sf{m_2}$ respectively, then find out the relationship between $\displaystyle\sf{E_1}$ and $\displaystyle\sf{E_2}$ connecting $\displaystyle\sf{m_1}$ and $\displaystyle\sf{m_2.}$

Solution:-

Let the velocities attained by $\displaystyle\sf{m_1}$ and $\displaystyle\sf{m_2}$ after collision be $\displaystyle\sf{v_1}$ and $\displaystyle\sf{v_2}$ respectively.

Since no external force is acting on the system, the linear momentum is conserved here. Hence,

$\displaystyle\longrightarrow\sf{m_1v_1+m_2v_2=0}$

$\displaystyle\longrightarrow\sf{v_2=-\dfrac{m_1v_1}{m_2}\quad\quad\dots(1)}$

Zero is because the system (the mass $\displaystyle\sf{M)}$ was initially at rest.

After collision the energies stored in each fragment provide kinetic energies for each. Therefore,

$\displaystyle\longrightarrow\sf{E_1=\dfrac{1}{2}m_1(v_1)^2\quad\quad\dots(2)}$

$\displaystyle\longrightarrow\sf{E_2=\dfrac{1}{2}m_2(v_2)^2\quad\quad\dots(3)}$

Dividing (2) by (3),

$\displaystyle\longrightarrow\sf{\dfrac{E_1}{E_2}=\dfrac{\frac{1}{2}m_1(v_1)^2}{\frac{1}{2}m_2(v_2)^2}}$

$\displaystyle\longrightarrow\sf{\dfrac{E_1}{E_2}=\dfrac{m_1(v_1)^2}{m_2(v_2)^2}}$

From (1),

$\displaystyle\longrightarrow\sf{\dfrac{E_1}{E_2}=\dfrac{m_1(v_1)^2}{m_2\left(-\dfrac{m_1v_1}{m_2}\right)^2}}$

$\displaystyle\longrightarrow\sf{\dfrac{E_1}{E_2}=\dfrac{m_1(m_2)^2(v_1)^2}{(m_1)^2m_2(v_1)^2}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{\dfrac{E_1}{E_2}=\dfrac{m_2}{m_1}}}}$

Or,

$\displaystyle\longrightarrow\sf{\underline{\underline{E_1m_1=E_2m_2}}}$