A cart of mass $m$ moving at $12 \text{ m/s}$ to the right collides elastically with a cart of mass $4.0 \text{ kg}$ that is originally at rest. After the collision, the cart of mass $m$ moves to the left with a velocity of $6.0 \text{ m/s}$. Assuming an elastic collision in one dimension only, what is the velocity of the center of mass ($v_{\text{cm}}$) of the two carts before the collision?

$\textbf{(A) } v_{\text{cm}} = 2.0 \text{ m/s}\\
\textbf{(B) }  v_{\text{cm}}=3.0 \text{ m/s}\\
\textbf{(C) } v_{\text{cm}}=6.0 \text{ m/s}\\
\textbf{(D) } v_{\text{cm}}=9.0 \text{ m/s}\\
\textbf{(E) } v_{\text{cm}}=18.0 \text{ m/s}$
To find the velocity of the center of mass (\(v_{\text{cm}}\)) of the two carts before the collision, we can use the formula for the center of mass velocity in a system of particles:

\[
v_{\text{cm}} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}
\]

where:
- \(m_1\) and \(m_2\) are the masses of the two carts,
- \(v_1\) and \(v_2\) are their respective velocities.

Given:
- The mass of the first cart is \(m\),
- The velocity of the first cart is \(12 \text{ m/s}\) to the right,
- The mass of the second cart is \(4.0 \text{ kg}\),
- The velocity of the second cart is \(0 \text{ m/s}\) (since it is initially at rest).

1. **Calculate the total momentum of the system before the collision:**

\[
\text{Total momentum} = m \cdot 12 \text{ m/s} + 4.0 \text{ kg} \cdot 0 \text{ m/s} = 12m
\]

2. **Calculate the total mass of the system:**

\[
\text{Total mass} = m + 4.0 \text{ kg}
\]

3. **Calculate the velocity of the center of mass before the collision:**

\[
v_{\text{cm}} = \frac{12m}{m + 4.0 \text{ kg}}
\]

To find the value of \(m\), we use the information given about the collision. After the collision, the cart of mass \(m\) moves to the left with a velocity of \(6.0 \text{ m/s}\). Let \(v\) be the velocity of the \(4.0 \text{ kg}\) cart after the collision.

4. **Apply conservation of momentum:**

\[
12m = -6m + 4v
\]

Solving for \(v\):

\[
12m + 6m = 4v \implies 18m = 4v \implies v = \frac{9m}{2}
\]

5. **Apply conservation of kinetic energy (since the collision is elastic):**

\[
\frac{1}{2} m (12 \text{ m/s})^2 = \frac{1}{2} m (6 \text{ m/s})^2 + \frac{1}{2} \cdot 4 \text{ kg} \cdot v^2
\]

\[
72m = 18m + 2v^2
\]

Substitute \(v = \frac{9m}{2}\) into the kinetic energy equation:

\[
72m = 18m + 2 \left(\frac{9m}{2}\right)^2
\]

\[
72m = 18m + 2 \left(\frac{81m^2}{4}\right)
\]

\[
72m = 18m + \frac{162m^2}{4}
\]

\[
72m = 18m + 40.5m^2
\]

\[
54m = 40.5m^2
\]

\[
m = \frac{54}{40.5} = \frac{4}{3}
\]

6. **Substitute \(m = \frac{4}{3}\) back into the center of mass velocity equation:**

\[
v_{\text{cm}} = \frac{12 \cdot \frac{4}{3}}{\frac{4}{3} + 4} = \frac{16}{\frac{4}{3} + 4} = \frac{16}{\frac{4 + 12}{3}} = \frac{16}{\frac{16}{3}} = 3 \text{ m/s}
\]

The final answer is \( \boxed{3.0 \text{ m/s}} \)