A golf ball of mass 0.050 kg is at rest on the tee. Just after being struck, it has a velocity of 102 m/s. If the club and ball were in contact for $0.81 \mathrm {~ms} ,$ what is the average force exerted on the ball by the club?

Question

Quizplus · Accepted Answer

The average force can be calculated using the impulse-momentum theorem: $ F \Delta t = m \Delta v $. Here, $ m = 0.050 \, \text{kg} $, $ \Delta v = 102 \, \text{m/s} $, and $ \Delta t = 0.81 \, \text{ms} = 0.00081 \, \text{s} $. Solving for $ F $, we get $ F = \frac{m \Delta v}{\Delta t} = \frac{0.050 \times 102}{0.00081} = 6,296.3 \, \text{N} \approx 6.3 \, \text{kN} $.

A Golf Ball of Mass 0 0.81 ms,0.81 \mathrm {~ms} ,0.81 ms, What Is the Average Force Exerted on the Ball by on the Tee

A Golf Ball of Mass 0 $0.81 \mathrm {~ms} ,$ What Is the Average Force Exerted on the Ball by on the Tee