Question

question 9
5 points save answer
a 0.065 - kg tennis ball moving to the right with a speed of 15 m/s is struck by a tennis racket, causing it to move to the left with a speed of 15 m/s. if the ball remains in contact with the racquet for 0.020 s, what is the magnitude of the average force exerted on the ball?
hint: assign vector directions
right is positive +
left is negative -
98 n
320 n
160 n
240 n
zero newtons

Explanation:

Step1: Calculate the initial and final momentum

The formula for momentum is \(p = mv\).
Initial velocity \(v_{i}=15\ m/s\), mass \(m = 0.065\ kg\), so initial momentum \(p_{i}=mv_{i}=0.065\times15\ kg\cdot m/s = 0.975\ kg\cdot m/s\)
Final velocity \(v_{f}=- 15\ m/s\), so final momentum \(p_{f}=mv_{f}=0.065\times(-15)\ kg\cdot m/s=- 0.975\ kg\cdot m/s\)

Step2: Calculate the change in momentum

The change in momentum \(\Delta p=p_{f}-p_{i}\)
\(\Delta p=-0.975 - 0.975=-1.95\ kg\cdot m/s\)

Step3: Use the impulse - momentum theorem \(F_{avg}\Delta t=\Delta p\)

We know \(\Delta t = 0.020\ s\), and we want to find \(F_{avg}\). Rearranging the formula \(F_{avg}=\frac{\Delta p}{\Delta t}\)
\(F_{avg}=\frac{- 1.95}{0.020}=-97.5\ N\)

The magnitude of the average force \(|F_{avg}| = 98\ N\)

Answer:

A. 98 N