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 - units

The mass of the ball $m = 57\ g=0.057\ kg$. Initial velocity $v_1 = 45\ m/s$ (right - direction positive), final velocity $v_2=- 33\ m/s$ (reversed direction), and time of contact $\Delta t = 0.0085\ s$.

Step2: Calculate the change in momentum

The change in momentum $\Delta p=m(v_2 - v_1)$. 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 $|\Delta p| = 4.446\ kg\cdot m/s$.

Step3: Use the impulse - momentum theorem

According to the impulse - momentum theorem $F_{avg}\Delta t=\Delta p$. So, $F_{avg}=\frac{\Delta p}{\Delta t}$. 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