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a 0.050 - kg lump of clay moving horizontally at 12 m/s strikes and sticks to a stationary 0.15 - kg cart that can move on a frictionless air track. determine the speed of the cart and clay after the collision.
zero m/s
12 m/s
9 m/s
6 m/s
3 m/s

Explanation:

Step1: Apply the law of conservation of momentum

The law of conservation of momentum states that \(m_1v_1 + m_2v_2=(m_1 + m_2)v\). Here, \(m_1 = 0.050\space kg\), \(v_1=12\space m/s\), \(m_2 = 0.15\space kg\), and \(v_2 = 0\space m/s\) (since the cart is stationary).
Substituting the values into the formula: \((0.050\times12)+(0.15\times0)=(0.050 + 0.15)v\)

Step2: Solve for \(v\)

First, calculate the left - hand side: \(0.050\times12=0.6\). The right - hand side is \((0.050 + 0.15)v=0.2v\).
So, \(0.6 = 0.2v\). Then, \(v=\frac{0.6}{0.2}=3\space m/s\)

Answer:

3 m/s