A stationary body explodes into two parts of masses ' \(\mathrm{M}_{1}\) ' and \({ }^{\prime} \mathrm{M}_{2}{ }^{\prime}\). They move in opposite directions with velocities ' \(v_{1}\) ' and ' \(v_{2}\) '. The ratio of their kinetic energies is

To solve this problem, a diagram of the problem will be helpiul. Alorig with that, the corservation of linear mornentum, which is giver by. \(P_{i}=p_{f} \mathrm{~ . ~ A s ~ p e r ~ N e w t a n ' s ~ s e c a n d ~ l a}\) ther montentum is conserved. The explosion occurs purely due to internal forces.
Step ty step salution:
Let's sart by making a diagram of the problem.

As per the problem, a stationary ball of mass (m) arnd initial velocity cercolu-0) after an explosion breaks into two masses \(m_{1}\) and \(m_{2}\) - These twò masses move in apposite directions with velocities \(v_{1}\) and \(v_{2}\) respectively.
We will use the conservation of linear momentum now, given by, \(p_{i}=p_{f}\). The initial and linal mormenta will be equal. This means that, \(p_{i}=p_{f} \Rightarrow m_{i}(0)=m_{1}\left(-v_{1}\right)+m_{2}\left(v_{2}\right) \Rightarrow m_{1}\left(v_{1}\right)=\) \(m_{2}\left(v_{2}\right)\). Hence, the ratio of the velocities will be equal to, \(\frac{v_{1}}{v_{2}}=\frac{m_{2}}{m_{1}}\).
The Final kinetic energy of the system after the explosion is the sum of the kinetic energies of the two bodies \(E_{1}\) arnd \(B_{2}\).
Finding the ratio of these two kinetic entergies becomes, \(\frac{E_{1}}{E_{2}}=\frac{\frac{1}{2} m_{1}\left(v_{1}\right)^{2}}{\frac{1}{2} m_{2}\left(v_{2}\right)^{2}}=\frac{m_{1}\left(v_{1}\right)^{2}}{m_{2}\left(v_{2}\right)^{2}}\). Naw, substituting in the ratio of velocites that we found out earlier, the ratio of kinetic energies becomes, \(\frac{E_{1}}{E_{2}}=\left(\frac{m_{1}}{m_{2}}\right)\left(\frac{v_{1}}{v_{2}}\right)^{2}=\left(\frac{m_{1}}{m_{2}}\right)\left(\frac{m_{2}}{m_{1}}\right)^{2}=\frac{m_{2}}{m_{1}} .\) Hence it is equal to the ratio of the irnerse of the masses. Sa, the correct answer is option A.
Note:
Another way of solving this problem is by only using the moenemtums of the masses. From the corservation of momentum belore and after the explosion become,
\(p_{i}=p_{f} \Rightarrow m(0)=p_{1}+p_{2} \Rightarrow p_{1}=-p_{2}\), where \(p_{1}\) and \(p_{2}\) are the momencum of the masses m \(_{1}\) and TR2respectively.
We also know that kinetic energy is given try \(E=\frac{p^{2}}{2 m}\) - Again, we will remowe the negative sign as it anly states the direclion, therefore \(\frac{E_{1}}{B_{2}}=\frac{\frac{p_{1}^{2}}{2 m_{1}}}{\frac{p_{2}^{2}}{2 m_{2}}}=\frac{\frac{\left(-p_{2}\right)^{2}}{m_{1}}}{\frac{p_{2}^{2}}{m_{2}}}=\frac{\frac{p_{2}^{2}}{m_{1}}}{\frac{p_{2}^{2}}{m_{2}}}=\frac{m_{2}}{m_{1}}\)