Question

question 12 of 28
a 1450 kg car drives toward a 60 kg shopping cart that has a velocity of
-1.2 m/s toward the car. the two objects collide, giving the car a final velocity
of 5.13 m/s, and the shopping cart a velocity of 11.75 m/s. what was the
initial velocity of the car?
a. 5.36 m/s
b. 5.67 m/s
c. -5.67 m/s
d. -5.36 m/s

Explanation:

Step1: Apply the law of conservation of momentum

The law of conservation of momentum states that \(m_1u_1 + m_2u_2=m_1v_1 + m_2v_2\), where \(m_1 = 1450\space kg\) (mass of the car), \(m_2=60\space kg\) (mass of the shopping - cart), \(u_1\) is the initial velocity of the car (to be found), \(u_2=- 1.2\space m/s\), \(v_1 = 5.13\space m/s\), and \(v_2 = 11.75\space m/s\).
Substitute the values into the formula: \(1450u_1+60\times(-1.2)=1450\times5.13 + 60\times11.75\)

Step2: Simplify the equation

First, calculate the right - hand side: \(1450\times5.13+60\times11.75=1450\times5.13 + 705\)
\(1450\times5.13=(1400 + 50)\times5.13=1400\times5.13+50\times5.13=7182+256.5 = 7438.5\)
\(1450\times5.13 + 705=7438.5+705 = 8143.5\)
The left - hand side is \(1450u_1-72\)
So the equation becomes \(1450u_1-72 = 8143.5\)

Step3: Solve for \(u_1\)

Add \(72\) to both sides of the equation: \(1450u_1=8143.5 + 72=8215.5\)
Then \(u_1=\frac{8215.5}{1450}\)
\(u_1 = 5.67\space m/s\)

Answer:

B. \(5.67\space m/s\)