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question 16
question 16 of 22
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a 3.0 - kg cart moving to the right with a speed of 1.0 m/s has a head - on collision with a 5.0 - kg cart that is initially moving to the left with a speed of 2.0 m/s. after the collision, the 3.0 - kg cart is moving to the left with a speed of 1.0 m/s. what is the final velocity of the 5.0 - kg cart?
0.80 m/s to the left
zero m/s
2.0 m/s to the right
2.0 m/s to the left
0.80 m/s to the right

Explanation:

Step1: Apply the law of conservation of momentum

The law of conservation of momentum states that \(m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}\). Let the right - hand direction be positive. So, \(m_1 = 3.0\space kg\), \(v_{1i}=1.0\space m/s\), \(m_2 = 5.0\space kg\), \(v_{2i}=- 2.0\space m/s\), \(v_{1f}=-1.0\space m/s\).
Substitute the values into the formula: \((3.0\times1.0)+(5.0\times(- 2.0))=(3.0\times(-1.0))+(5.0\times v_{2f})\)

Step2: Simplify the equation

First, calculate the left - hand side: \(3.0-10.0=-7.0\). Then, calculate the right - hand side: \(-3.0 + 5v_{2f}\).
So, the equation becomes \(-7.0=-3.0 + 5v_{2f}\).

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

Add \(3.0\) to both sides of the equation: \(-7.0 + 3.0=5v_{2f}\), i.e., \(-4.0 = 5v_{2f}\). Then \(v_{2f}=\frac{-4.0}{5}=-0.8\space m/s\)

Answer:

A. \(0.80\space m/s\) to the left