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 $$K_B$$.

Since the initial momentum of the particle is zero, the momentum of the two parts must be equal and opposite to conserve momentum:

$$ m_A v_A = m_B v_B $$

Here, the ratio of kinetic energies is given by:

$$ \frac{K_A}{K_B} = \frac{\frac{1}{2} m_A v_A^2}{\frac{1}{2} m_B v_B^2} = \frac{m_A v_A^2}{m_B v_B^2} $$

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:

$$ \frac{K_A}{K_B} = \frac{v_A}{v_B} $$

Therefore, the ratio of their kinetic energies $$\left(K_B: K_A\right)$$ is:

$$ \frac{K_B}{K_A} = \frac{v_B}{v_A} $$