Question

1. a tennis ball with a mass of 0.025 kg is moving from left to right and 2.5 m/s. it is hit by a squash racket, which applies a force for 0.005 s so that the ball leaves the racket at 7.5 m/s moving from right to left. use the impulse - momentum theorem to calculate the average force on the ball. - 75 n 25 n - 5 n - 50 n

Explanation:

Step1: Calculate the initial and final momentum

The initial velocity \(v_{i}=2.5\ m/s\) (right - direction, assume positive \(x\) - axis). The final velocity \(v_{f}=- 7.5\ m/s\) (left - direction). The mass \(m = 0.025\ kg\).
The initial momentum \(p_{i}=mv_{i}\), so \(p_{i}=0.025\times2.5=\frac{0.025\times2.5}{1}=0.0625\ kg\cdot m/s\)
The final momentum \(p_{f}=mv_{f}\), so \(p_{f}=0.025\times(-7.5)=- 0.1875\ kg\cdot m/s\)

Step2: Calculate the change in momentum

The change in momentum \(\Delta p=p_{f}-p_{i}\)
\(\Delta p=-0.1875 - 0.0625=-0.25\ kg\cdot m/s\)

Step3: Use the impulse - momentum theorem \(J = F\Delta t=\Delta p\)

We know \(\Delta t = 0.005\ s\) and \(\Delta p=-0.25\ kg\cdot m/s\)
From \(F=\frac{\Delta p}{\Delta t}\), substituting the values: \(F=\frac{-0.25}{0.005}=- 50\ N\)

Answer:

\(-50\ N\)