Question

A body initially at rest explodes into $2$ pieces of mass $2M$ and $3M$ respectively having a total kinetic energy of a piece of mass $2M$ after the explosion is?

Answer 1

In order to solve this question, we will first formulate the equation for velocities of two bodies using conservation of linear momentum and then use the sum of kinetic energies of both body equal to given total energy, hence will solve for energy of body of mass $2M.$

Complete step by step answer:
According to the question, we have given that
${m_1} = 2M$ mass of one body and let its velocity be ${v_1}$
${m_2} = 3M$ mass of another body and let its velocity be ${v_2}$
Since, the body which exploded was initially at rest so its momentum will be zero.
Using, law of conservation of momentum we have,
Initial momentum $=$ Final momentum.
As, initial momentum of the system is zero.
Final momentum of system is ${m_1}{v_1} + {m_2}{v_2}$ so we have,
${m_1}{v_1} + {m_2}{v_2} = 0$ on putting the value of parameters we get,
$2M{v_1} + 3M{v_2} = 0$
Now, given that total energy of these two bodies is E. $2{v_1} + 3{v_2} = 0 \to (i)$
Let ${E_1},{E_2}$ be the Kinetic energy of these two bodies after collision then we have,
$E = {E_1} + {E_2}$
Now, ${E_1} = \dfrac{1}{2}2M{v_1}^2$
and ${E_2} = \dfrac{1}{2}3M{v_2}^2$
on putting the values in equation $E = {E_1} + {E_2}$ we get,
$E = \dfrac{1}{2}2M{v_1}^2 + \dfrac{1}{2}3M{v_2}^2$
$\dfrac{E}{M} = {v_1}^2 + \dfrac{3}{2}{v_2}^2 \to (ii)$
On solving for ${v_1},{v_2}$ from equations (i) and (ii) we get,
$\dfrac{E}{M} = {v_1}^2 + \dfrac{3}{2}{( - \dfrac{2}{3}{v_1})^2}$
$\dfrac{E}{M} = {v_1}^2(1 + \dfrac{2}{3})$
$\dfrac{E}{M} = {v_1}^2(\dfrac{5}{3})$
on rearranging above equation we can write it as,
$\dfrac{{3E}}{5} = \dfrac{1}{2}2M{v_1}^2$
and since we have, ${E_1} = \dfrac{1}{2}2M{v_1}^2$ kinetic energy of first body, so
${E_1} = \dfrac{3}{5}E$
Hence, the kinetic energy of a body having mass of $2M$ is $\dfrac{3}{5}$ times of total energy E.

Note: It should be remembered that, negative sign of velocity indicates that both bodies move in opposite direction after explosion and its obviously in order to balance the law of conservation of momentum, a body which is at rest will always have zero momentum because its velocity is zero.