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question 9
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 change in momentum

The formula for momentum is \(p = mv\). The initial velocity \(v_i=15\ m/s\) (right, so positive), final velocity \(v_f = - 15\ m/s\) (left, so negative). Mass \(m = 0.065\ kg\).
Change in momentum \(\Delta p=m(v_f - v_i)\)
\(\Delta p=0.065\times(-15 - 15)\)
\(\Delta p=0.065\times(-30)=- 1.95\ kg\cdot m/s\)

Step2: 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 for \(F_{avg}\) gives \(F_{avg}=\frac{\Delta p}{\Delta t}\)
\(F_{avg}=\frac{-1.95}{0.020}=-97.5\ N\)

The magnitude of the force is \(|F_{avg}| = 98\ N\) (rounded to two significant figures)

Answer:

98 N