Question

a 2,800 kg truck traveling forward at 12 m/s collides with a 1,200 kg parked vehicle. the park vehicle moves forward at 17 m/s. what is the final velocity of the truck? (1 point)
11.9 m/s
-11.7 m/s
4.7 m/s
8.4 m/s

Explanation:

Step1: Apply the law of conservation of momentum

The law of conservation of momentum states that \(m_1u_1 + m_2u_2=m_1v_1 + m_2v_2\). Here, \(m_1 = 2800\space kg\), \(u_1=12\space m/s\), \(m_2 = 1200\space kg\), \(u_2 = 0\space m/s\) (since the second vehicle is parked), and \(v_2=17\space m/s\). Substitute these values into the formula:
\(2800\times12+1200\times0 = 2800v_1+1200\times17\)

Step2: Simplify the equation

First, calculate \(2800\times12 = 33600\) and \(1200\times17=20400\). The equation becomes \(33600=2800v_1 + 20400\)

Step3: Solve for \(v_1\)

Subtract \(20400\) from both sides: \(33600 - 20400=2800v_1\). So, \(13200 = 2800v_1\). Then, \(v_1=\frac{13200}{2800}\approx4.7\space m/s\)

Answer:

\(4.7\space m/s\)