Question

1. two cars are traveling in opposite directions on a highway when they collide head on. the chevy has a mass of 1200 kg and is traveling north with a speed of 35 m/s. the ford has a mass of 1400 kg and is traveling south with a speed of 28 m/s. the two cars stick together after the collision.

a) draw a picture showing the situation described before and after the collision. draw arrows showing the velocities of each car.
b) calculate the total momentum of the system before the collision.
c) calculate the total momentum of the system after the collision.
d) calculate the velocities of the cars after the collision.

Explanation:

Step1: Define the positive direction

Let's take the north - direction as positive.

Step2: Calculate the momentum of each car before collision

The momentum of an object is given by \(p = mv\).
For the Chevy: \(m_1=1200\space kg\), \(v_1 = 35\space m/s\), so \(p_1=m_1v_1=1200\times35 = 42000\space kg\cdot m/s\)
For the Ford: \(m_2 = 1400\space kg\), \(v_2=- 28\space m/s\) (since it is moving south), so \(p_2=m_2v_2=1400\times(-28)=-39200\space kg\cdot m/s\)

Step3: Calculate the total momentum before collision

The total momentum before collision \(p_{total - before}=p_1 + p_2\)
\(p_{total - before}=42000-39200=2800\space kg\cdot m/s\)

Step4: Calculate the total momentum after collision

According to the law of conservation of momentum \(p_{total - after}=p_{total - before}\)
So \(p_{total - after}=2800\space kg\cdot m/s\)

Step5: Calculate the velocity after collision

Since the two cars stick together after collision, \(m = m_1 + m_2=1200 + 1400=2600\space kg\)
Using \(p = mv\), we can find \(v=\frac{p}{m}\)
\(v=\frac{2800}{2600}=\frac{14}{13}\approx1.08\space m/s\)

Answer:

b) The total momentum of the system before the collision is \(2800\space kg\cdot m/s\)
c) The total momentum of the system after the collision is \(2800\space kg\cdot m/s\)
d) The velocity of the cars after the collision is approximately \(1.08\space m/s\) (in the north - direction)