A stationary particle breaks into two parts of masses $m_A$ and $m_B$ which move with velocities $v_A$ and $v_B$ respectively. The ratio of their kinetic energies $\left(K_B: K_A\right)$ is :

Question

Aicharya · Accepted Answer

Initial momentum is zero. $\begin{aligned} & 	ext { Hence }\left|\mathrm{P}_{\mathrm{A}}ight|=\left|\mathrm{P}_{\mathrm{B}}ight| \ & \Rightarrow \mathrm{m}_{\mathrm{A}} \mathrm{v}_{\mathrm{B}}=\mathrm{m}_{\mathrm{B}} \mathrm{v}_{\mathrm{B}} \ & \frac{(\mathrm{KE})_{\mathrm{A}}}{(\mathrm{KE})_{\mathrm{B}}}=\frac{\frac{1}{2} \mathrm{~m}_{\mathrm{A}} \mathrm{v}_{\mathrm{A}}^2}{\frac{1}{2} \mathrm{~m}_{\mathrm{B}} \mathrm{v}_{\mathrm{B}}^2}=\frac{\mathrm{v}_{\mathrm{A}}}{\mathrm{v}_{\mathrm{B}}} \ & \frac{(\mathrm{KE})_{\mathrm{B}}}{(\mathrm{KE})_{\mathrm{A}}}=\frac{\mathrm{v}_{\mathrm{B}}}{\mathrm{v}_{\mathrm{A}}}\end{aligned}$

A stationary particle breaks into two parts of masses $m_A$ and $m_B$ which move with velocities $v_A$ and…

Solution