Questions: Question Post-3: An automobile of mass 1500 kg moving at 25.0 m / s collides with a truck of mass 4500 kg at rest. The bumpers of the two vehicles lock together during the crash. a) Compare the force exerted by the car on the truck with the force exerted by the truck on the car during the collision. Is one force larger than the other, or are they equal in magnitude to each other? b) What is the final velocity of the car and truck just after the collision? Show your calculations.

Solution Steps

According to Newton's Third Law of Motion, for every action, there is an equal and opposite reaction. This means that the force exerted by the car on the truck is equal in magnitude and opposite in direction to the force exerted by the truck on the car during the collision.

The law of conservation of momentum states that the total momentum of a closed system remains constant if no external forces act on it. In this collision, the system consists of the car and the truck, and we assume no external forces are acting on them.

The initial momentum of the system is the sum of the momentum of the car and the truck. The truck is initially at rest, so its initial momentum is zero.

\[ p_{\text{initial}} = m_{\text{car}} \cdot v_{\text{car}} + m_{\text{truck}} \cdot v_{\text{truck}} = 1500 \, \text{kg} \cdot 25.0 \, \text{m/s} + 4500 \, \text{kg} \cdot 0 \, \text{m/s} \]

\[ p_{\text{initial}} = 37500 \, \text{kg} \cdot \text{m/s} \]

After the collision, the car and truck move together with a common velocity \(v_f\). The final momentum of the system is:

\[ p_{\text{final}} = (m_{\text{car}} + m_{\text{truck}}) \cdot v_f \]

By conservation of momentum:

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

\[ 37500 \, \text{kg} \cdot \text{m/s} = (1500 \, \text{kg} + 4500 \, \text{kg}) \cdot v_f \]

\[ 37500 \, \text{kg} \cdot \text{m/s} = 6000 \, \text{kg} \cdot v_f \]

Solving for \(v_f\):

\[ v_f = \frac{37500 \, \text{kg} \cdot \text{m/s}}{6000 \, \text{kg}} = 6.250 \, \text{m/s} \]

Final Answer