Question

A golf club exerts an average horizontal force of 1000 n on a 0.045 -kg golf ball that is initially at rest on the tee. the club is in contact with the ball for 1.8 ms. what is the speed of the golf ball just as it leaves the tee?

Answer 1

The impulse (the variation of momentum of the ball) is related to the force applied by
$\Delta p = F \Delta t$
where $\Delta p$ is the variation of momentum, F is the intensity of the force and $\Delta t$ is the time of application of the force. 
Using F=1000 N and $\Delta t = 1.8 ms=1.8 \cdot 10^{-3}s$, we can find the variation of momentum:
$\Delta p = (1000 N)(1.8 \cdot 10^{-3} s)=1.8 kg m/s$

This $\Delta p$ can be rewritten as
$\Delta p = p_f - p_i = mv_f - mv_i$
where $p_f$ and $p_i$ are the final and initial momentum. But the ball is initially at rest, so the initial momentum is zero, and
$\Delta p = mv_f$
from which we find the final velocity of the ball:
$v_f = \frac{\Delta p }{m}= \frac{1.8 kg m/s}{0.045 kg}= 40 m/s$