1. Use conservation of momentum

Since the particle was initially stationary, its initial momentum was zero. After breaking into two parts,

$m_A v_A = m_B v_B$

(in opposite directions; for magnitudes we use the above relation).

2. Write kinetic energies of the two parts

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

So,

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

3. Use the momentum relation

From

$m_A v_A = m_B v_B$

we get

$m_B = \frac{m_A v_A}{v_B}$

Substitute into the ratio:

$\frac{K_B}{K_A} = \frac{\left(\frac{m_A v_A}{v_B}\right) v_B^2}{m_A v_A^2} = \frac{v_B}{v_A}$

Hence,

$K_B : K_A = v_B : v_A$

4. Check options

• A: $v_B : v_A$ ✅
• B: $1:1$ ❌
• C: $m_B v_B : m_A v_A$ = $1:1$ from momentum conservation, so ❌
• D: $m_B : m_A$ ❌

Therefore, the correct option is A.