Question

1. a 2 kg pumpkin flies through the air at 42 m/s (93 mph)! it strikes a 500kg volkswagen bug that has been abandoned in a corn field for quite some time. the final velocity of the pumpkin is -2.7 m/s (-6 mph). what is the final velocity of the car?

Explanation:

Step1: Apply the law of conservation of momentum

The law of conservation of momentum states that \(m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}\). Here, the initial velocity of the car \(v_{2i} = 0\) (since it is abandoned and initially at rest), \(m_1 = 2\space kg\), \(v_{1i}=42\space m/s\), \(v_{1f}=- 2.7\space m/s\), and \(m_2 = 500\space kg\).
Substituting the values into the formula: \(2\times42+500\times0=2\times(-2.7)+500\times v_{2f}\)

Step2: Simplify the equation

First, calculate the left - hand side: \(2\times42=84\)
The right - hand side: \(2\times(-2.7)=-5.4\), so the equation becomes \(84=-5.4 + 500v_{2f}\)

Step3: Solve for \(v_{2f}\)

Add \(5.4\) to both sides of the equation: \(84 + 5.4=500v_{2f}\)
\(89.4 = 500v_{2f}\)
Then \(v_{2f}=\frac{89.4}{500}\)

Step4: Calculate the value of \(v_{2f}\)

\(v_{2f}=0.1788\space m/s\)

Answer:

\(0.1788\space m/s\)