Question

A golf club exerts an average force of 1000 N on a 0.045-kg golf ball which is initially at rest. The club is in contact with the ball for 1.8 ms. What is the speed of the golf ball as it leaves the tee?

Answer 1

Given:

$F=1000\:\rm N$

$m=0.045\:\rm kg$

$\Delta t=0.0018\:\rm s$

$v_i=0\:\rm m/s$

The Newton's second law says

$$\frac{m\Delta v}{\Delta t}=F$$

Hence, the final speed of a ball

$$v_f=\frac{F\Delta t}{m}=\frac{1000*0.0018}{0.045}=40\:\rm m/s$$