Questions: A 1500 kg car traveling at 15.0 m / s to the south collides with a 4500 kg truck that is initially at rest at a stoplight. The car and truck stick together and move together after the collision. What is the final velocity of the two-vehicle mass?

Solution Steps

• Mass of the car, \( m_1 = 1500 \, \text{kg} \)
• Initial velocity of the car, \( v_1 = 15.0 \, \text{m/s} \) (south)
• Mass of the truck, \( m_2 = 4500 \, \text{kg} \)
• Initial velocity of the truck, \( v_2 = 0 \, \text{m/s} \) (at rest)

In a perfectly inelastic collision, the total momentum before the collision is equal to the total momentum after the collision.

The total initial momentum \( p_{\text{initial}} \) is: \[ p_{\text{initial}} = m_1 v_1 + m_2 v_2 \]

Substitute the given values: \[ p_{\text{initial}} = (1500 \, \text{kg} \times 15.0 \, \text{m/s}) + (4500 \, \text{kg} \times 0 \, \text{m/s}) \] \[ p_{\text{initial}} = 22500 \, \text{kg} \cdot \text{m/s} \]

Let \( v_f \) be the final velocity of the combined mass after the collision. The total mass after the collision is \( m_1 + m_2 \).

Using the conservation of momentum: \[ p_{\text{initial}} = (m_1 + m_2) v_f \]

\[ 22500 \, \text{kg} \cdot \text{m/s} = (1500 \, \text{kg} + 4500 \, \text{kg}) v_f \] \[ 22500 \, \text{kg} \cdot \text{m/s} = 6000 \, \text{kg} \cdot v_f \] \[ v_f = \frac{22500 \, \text{kg} \cdot \text{m/s}}{6000 \, \text{kg}} \] \[ v_f = 3.75 \, \text{m/s} \]

The final velocity of the two-vehicle mass is \( 3.75 \, \text{m/s} \) to the south.

Final Answer