1. Set up the situation

Let mass $m_1$ move initially with speed $u$, and mass $m_2$ be at rest.

• Initial velocity of $m_1 = u$
• Initial velocity of $m_2 = 0$

After collision, they move with equal speeds in opposite directions. Let that common speed be $v$.

So we can take:

• Final velocity of $m_1 = -v$
• Final velocity of $m_2 = +v$

2. Apply conservation of linear momentum

Initial momentum: $p_i = m_1 u$

Final momentum: $p_f = m_2(v) + m_1(-v) = v(m_2 - m_1)$

Equating initial and final momentum: $m_1 u = v(m_2 - m_1) ag{1}$

3. Use coefficient of restitution for collision

Since the problem says they collide and then move apart, we take it as an elastic collision type standard JEE interpretation unless otherwise stated.

For a head-on elastic collision, $\text{relative speed of separation} = \text{relative speed of approach}$

Before collision: $u_1 - u_2 = u - 0 = u$

After collision: $v_2 - v_1 = v - (-v) = 2v$

Thus, $2v = u \quad \Rightarrow \quad v = \frac{u}{2}$

4. Substitute into momentum equation

From (1): $m_1 u = \frac{u}{2}(m_2 - m_1)$

Cancel $u$: $m_1 = \frac{m_2 - m_1}{2}$

So, $2m_1 = m_2 - m_1$ $m_2 = 3m_1$

Hence, $m_2 : m_1 = 3 : 1$

5. Check options

• A: $1:1$ ❌
• B: $3:1$ ✅
• C: $2:1$ ❌
• D: $1:2$ ❌

Therefore, the correct answer is B.