Question

A golf ball with a mass of 0.08 kg falls to the ground with a velocity of 8.2 m/s. It bounces upwards with a velocity of 5.9 m/s. If the total time of impact was 0.015 s, what was the average force the golfball was under during the impact?

Answer 1

Given data:

* The mass of the golf ball is m = 0.08 kg.

* The initial velocity of the golf ball is u = 8.2 m/s.

* The final velocity of the golf ball is v = 5.9 m/s.

* The time of impact is t = 0.015 s.

Solution:

According to Newton's second law, the average force on the golf ball in terms of the change in momentum with time is,

`F=(m(v-u))/(t)`

Substituting the known values,

`\begin{gathered} F=(0.08(5.9-8.2))/(0.015) \\ F=(-0.08*2.3)/(0.015) \\ F=12.27\text{ N} \end{gathered}`

Hencem