Question

a 57 g tennis ball is traveling at 45 m/s to the right, when it hits a racket. the ball reverses direction and travels at 33 m/s. the ball is in contact with the racket for 0.0085 s. what is the magnitude of the force that was exerted on the ball? 1.8 n 12 n 81 n 520 n

Explanation:

Step1: Convert mass to SI - unit

First, convert the mass of the tennis - ball from grams to kilograms. Given $m = 57\ g=0.057\ kg$.

Step2: Define initial and final velocities

Let the initial velocity $v_0 = 45\ m/s$ (to the right) and the final velocity $v = - 33\ m/s$ (to the left).

Step3: Calculate the change in momentum

The change in momentum $\Delta p=m(v - v_0)$. Substitute the values: $\Delta p=0.057\times(-33 - 45)=0.057\times(-78)=-4.446\ kg\cdot m/s$. The magnitude of the change in momentum is $|\Delta p| = 4.446\ kg\cdot m/s$.

Step4: Use the impulse - momentum theorem

The impulse - momentum theorem states that $J=\Delta p$ and $J = F_{avg}\Delta t$, where $F_{avg}$ is the average force and $\Delta t$ is the time of contact. We know $\Delta t = 0.0085\ s$. Rearranging for $F_{avg}$, we get $F_{avg}=\frac{\Delta p}{\Delta t}$. Substitute $|\Delta p| = 4.446\ kg\cdot m/s$ and $\Delta t = 0.0085\ s$ into the formula: $F_{avg}=\frac{4.446}{0.0085}\approx523\ N\approx520\ N$.

Answer:

$520\ N$