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 $$(K_B : K_A)$$ is :

A stationary particle breaks into two parts. We need the ratio $$K_B : K_A$$.

By conservation of momentum, the initial momentum is zero since the particle is stationary, so $$m_A v_A = m_B v_B$$.

The kinetic energies of the parts are given by $$K_A = \frac{1}{2}m_A v_A^2, \quad K_B = \frac{1}{2}m_B v_B^2$$.

Thus, the ratio of the kinetic energies is $$\frac{K_B}{K_A} = \frac{m_B v_B^2}{m_A v_A^2} = \frac{m_B v_B}{m_A v_A} \times \frac{v_B}{v_A}$$. Since $$m_B v_B = m_A v_A$$ from conservation of momentum, the first factor equals 1, giving $$\frac{K_B}{K_A} = 1 \times \frac{v_B}{v_A} = \frac{v_B}{v_A}$$.

Therefore $$K_B : K_A = v_B : v_A$$.

The correct answer is Option (1): $$v_B : v_A$$.