1. First collision: A with B (elastic)

Masses:

• $m_A = m$
• $m_B = 2m$
• $m_C = 2m$

Initial velocities:

• $u_A = 9\,\text{m/s}$
• $u_B = 0$
• $u_C = 0$

For a 1D elastic collision, the final velocity of $B$ is $v_B = \frac{2m_A}{m_A+m_B}u_A + \frac{m_B-m_A}{m_A+m_B}u_B$ Since $u_B=0$, $v_B = \frac{2m}{m+2m}\cdot 9 = \frac{2}{3}\cdot 9 = 6\,\text{m/s}$

(Also, $A$ slows down to $v_A = \frac{m_A-m_B}{m_A+m_B}u_A = \frac{m-2m}{3m}\cdot 9 = -3\,\text{m/s}$ though this is not needed further.)

So after the first collision, $B$ moves toward $C$ with speed $6\,\text{m/s}$.

2. Second collision: B with C (completely inelastic)

Now $B$ of mass $2m$ moving at $6\,\text{m/s}$ collides with $C$ of mass $2m$ at rest, and they stick together.

Use conservation of momentum: $2m\cdot 6 + 2m\cdot 0 = (2m+2m)V$ $12m = 4mV$ $V = 3\,\text{m/s}$

Since $B$ and $C$ stick together, the final speed of $C$ is also $3\,\text{m/s}$

3. Option check

• A: $6\,\text{m/s}$ ❌
• B: $9\,\text{m/s}$ ❌
• C: $4\,\text{m/s}$ ❌
• D: $3\,\text{m/s}$ ✅

Therefore, the correct answer is D.