Question

a 71.80 kg person holding a steel ball stands motionless on a frozen lake. the person then throws the ball, which propels the person at 0.15 m/s to the right and the ball 3.50 m/s to the left. if the initial momentum of the system is zero, what is the mass of the steel ball? a. 3.50 kg b. 23.33 kg c. 20.51 kg d. 3.08 kg

Explanation:

Step1: Apply the law of conservation of momentum

Since the initial momentum of the system \(p_{i}=0\), according to the law of conservation of momentum \(p_{i} = p_{f}\), where \(p_{f}=m_{person}v_{person}+m_{ball}v_{ball}\). Let the right - hand direction be positive. So \(m_{person} = 71.80\space kg\), \(v_{person}=0.15\space m/s\), \(v_{ball}=- 3.50\space m/s\).

Step2: Solve for the mass of the ball

We have \(0=m_{person}v_{person}+m_{ball}v_{ball}\). Then \(m_{ball}=-\frac{m_{person}v_{person}}{v_{ball}}\). Substitute the values: \(m_{ball}=-\frac{71.80\times0.15}{- 3.50}\).
Calculate \(\frac{71.80\times0.15}{3.50}=\frac{10.77}{3.50}\approx3.08\space kg\)

Answer:

D. \(3.08\space kg\)