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

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Aicharya · Accepted Answer

To solve this problem, a diagram of the problem will be helpful. Along with that, the conservation of linear mornentum, which is giver by. $P_{i}=p_{f}. As per Newton's second law their montentum is conserved. The explosion occurs purely due to internal forces. Step by step salution: Let's start by making a diagram of the problem. As per the problem, a stationary ball of mass (m) and initial velocity $v_0$ after an explosion breaks into two masses $m_{1}$ and $m_{2}$ - These two 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, $\frac{E_{1}}{E_{2}}=\frac{\frac{1}{2} m_{1}\left(v_{1}\right)^{2}}{\frac{1}{2} m_{2}\left(v_{2}\right)^{2}}=\frac{m_{1}\left(v_{1}\right)^{2}}{m_{2}\left(v_{2}\right)^{2}}$. Now, 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 imerse of the masses. So, the correct answer is option A. Note: Another way of solving this problem is by only using the moenemtums of the masses. From the conservation of momentum before 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 momentum of the masses $m_{1}$ and $m_{2}$ respectively. We also know that kinetic energy is given try $E=\frac{p^{2}}{2 m}$ - Again, we will remove the negative sign as it only states the direction, therefore $\frac{E_{1}}{B_{2}}=\frac{\frac{p_{1}^{2}}{2 m_{1}}}{\frac{p_{2}^{2}}{2 m_{2}}}=\frac{\frac{\left(-p_{2}\right)^{2}}{m_{1}}}{\frac{p_{2}^{2}}{m_{2}}}=\frac{\frac{p_{2}^{2}}{m_{1}}}{\frac{p_{2}^{2}}{m_{2}}}=\frac{m_{2}}{m_{1}}$ .

A stationary body explodes into two parts of masses ' $\mathrm{M}_{1}$ ' and ${ }^{\prime} \mathrm{M}_{2}{…

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