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This problem involves a two-dimensional inelastic collision, where both objects stick together after the collision. To solve it, we can use the conservation of momentum in both the \(x\)-direction (east-west) and the \(y\)-direction (north-south).

### Step 1: Conservation of momentum in the \(x\)-direction (east-west)

- The truck (A) is moving east at 6.00 m/s with mass 1200 kg.
- The car (B) is moving north at 9.00 m/s with mass 600 kg.

We apply conservation of momentum to each direction separately.

The initial momentum in the \(x\)-direction is only due to the truck because the car has no velocity in that direction:

\[
p_{x_{\text{initial}}} = m_A \cdot v_{A_x} = 1200 \, \text{kg} \cdot 6.00 \, \text{m/s} = 7200 \, \text{kg} \cdot \text{m/s}
\]

After the collision, the combined object (A + B) moves with a final velocity \(v_f\) in the \(x\)-direction:

\[
p_{x_{\text{final}}} = (m_A + m_B) \cdot v_{f_x} = 1800 \, \text{kg} \cdot v_{f_x}
\]

By conservation of momentum in the \(x\)-direction:

\[
p_{x_{\text{initial}}} = p_{x_{\text{final}}}
\]

\[
7200 = 1800 \cdot v_{f_x}
\]

Solving for \(v_{f_x}\):

\[
v_{f_x} = \frac{7200}{1800} = 4.00 \, \text{m/s}
\]

### Step 2: Conservation of momentum in the \(y\)-direction (north-south)

The car is moving north, so it has initial momentum in the \(y\)-direction:

\[
p_{y_{\text{initial}}} = m_B \cdot v_{B_y} = 600 \, \text{kg} \cdot 9.00 \, \text{m/s} = 5400 \, \text{kg} \cdot \text{m/s}
\]

After the collision, the combined object moves with velocity \(v_f\) in the \(y\)-direction:

\[
p_{y_{\text{final}}} = (m_A + m_B) \cdot v_{f_y} = 1800 \, \text{kg} \cdot v_{f_y}
\]

By conservation of momentum in the \(y\)-direction:

\[
p_{y_{\text{initial}}} = p_{y_{\text{final}}}
\]

\[
5400 = 1800 \cdot v_{f_y}
\]

Solving for \(v_{f_y}\):

\[
v_{f_y} = \frac{5400}{1800} = 3.00 \, \text{m/s}
\]

### Step 3: Finding the final speed

The final velocity is the vector sum of the \(x\)- and \(y\)-components:

\[
v_f = \sqrt{v_{f_x}^2 + v_{f_y}^2}
\]

Substituting the values for \(v_{f_x}\) and \(v_{f_y}\):

\[
v_f = \sqrt{(4.00 \, \text{m/s})^2 + (3.00 \, \text{m/s})^2} = \sqrt{16 + 9} = \sqrt{25} = 5.00 \, \text{m/s}
\]

Thus, the final speed of the combined object is **5.00 m/s**.

The correct answer is option **(c)**.

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Would you like more details or have any questions? Here are 5 related questions:

1. What would happen if the collision were elastic instead of inelastic?
2. How would the result change if the truck were heavier than 1200 kg?
3. Why do we apply conservation of momentum in both directions separately?
4. How does the concept of center of mass relate to this problem?
5. What role does friction (or lack thereof) play in this scenario?

**Tip:** In inelastic collisions, kinetic energy is not conserved, but momentum always is!

A truck (mass 1200 kg) is traveling due east at 6.00 m/s, and a car (mass 600 kg) is traveling due north at 9.00 m/s. The horizontal roads are icy and frictionless. The two vehicles collide and stick together. What is the speed of the combined 1800 kg object immediately after the collision?

Learn how to solve a 2D inelastic collision problem where a truck and car collide and stick together. The final speed is calculated using momentum conservation principles and vector addition. This problem involves understanding momentum, two-dimensional motion, and applying the Pythagorean theorem. Suitable for high school students.

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