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 :

<p>A stationary particle breaks into two parts with masses $m_A$ and $m_B$, which then move with velocities $v_A$ and $v_B$, respectively. We need to determine the ratio of their kinetic energies $K_A$ and $K_B$.</p> <p>Since the initial momentum of the particle is zero, the momentum of the two parts must be equal and opposite to conserve momentum:</p> <p>$m_A v_A = m_B v_B$</p> <p>Here, the ratio of kinetic energies is given by:</p> <p>$$ \frac{K_A}{K_B} = \frac{\frac{1}{2} m_A v_A^2}{\frac{1}{2} m_B v_B^2} = \frac{m_A v_A^2}{m_B v_B^2} $$</p> <p>However, using the momentum relationship, we can substitute $m_A v_A = m_B v_B$ into the kinetic energy ratio, leading us to simplify:</p> <p>$\frac{K_A}{K_B} = \frac{v_A}{v_B}$</p> <p>Therefore, the ratio of their kinetic energies $\left(K_B: K_A\right)$ is:</p> <p>$\frac{K_B}{K_A} = \frac{v_B}{v_A}$</p>