1. Given data

• Mass of $A = m$
• Mass of $B = 2m$
• Mass of $C = m$
• Initially, $A$ moves with speed $9\,\text{m/s}$
• $B$ and $C$ are initially at rest
• First collision: elastic between $A$ and $B$
• Second collision: completely inelastic between $B$ and $C$

We need the final speed of object $C$.

2. First collision: elastic collision between $A$ and $B$

Let initial velocities be:

$u_A = 9, \qquad u_B = 0$

For a 1D elastic collision, final velocity of the second body is:

$v_B = \frac{2m_A}{m_A + m_B}u_A + \frac{m_B - m_A}{m_A + m_B}u_B$

Here,

$m_A = m, \qquad m_B = 2m, \qquad u_B = 0$

So,

$v_B = \frac{2m}{m+2m}\cdot 9 = \frac{2}{3}\cdot 9 = 6\,\text{m/s}$

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

3. Second collision: completely inelastic collision between $B$ and $C$

Now $B$ of mass $2m$ moves with speed $6\,\text{m/s}$ and collides with $C$ of mass $m$ at rest.

Since the collision is completely inelastic, they move together after collision with common velocity $v$.

Using conservation of momentum:

$2m \cdot 6 + m \cdot 0 = (2m + m)v$

$12m = 3mv$

$v = 4\,\text{m/s}$

Since $B$ and $C$ stick together, the final speed of object $C$ is also

$\boxed{4\,\text{m/s}}$

4. Comparison with stored answer

Stored correct answer = $4$

Our derived answer = $4$

So they agree.