Question

a professional golfer is examining a video of a practice swing. the high - speed footage shows that his club is in contact with the ball (which was initially at rest on the tee) for only $delta t = 0.871$ ms, and the radar gun clocks the speed of the ball as $v_{b}=146$ mph after it comes off the club. the golf ball has a mass of $m_{b}=45.9$ g.
what is the magnitude $j$ of the impulse imparted to the ball by the club?
$j=\frac{kgcdot m}{s}$
what is the magnitude $f_{avg}$ of the average force of contact between the club and the ball?
$f_{avg}=n$

Explanation:

Step1: Convert the speed unit

We know that \(1\ mph=\frac{1609.34}{3600}\ m/s\). So \(v_{b}=146\ mph = 146\times\frac{1609.34}{3600}\ m/s\approx65.27\ m/s\)
The mass of the ball \(m_{b}=45.9\ g = 45.9\times10^{- 3}\ kg\)
The initial velocity of the ball \(u = 0\ m/s\)

Step2: Calculate the impulse

According to the impulse - momentum theorem \(J=\Delta p=m_{b}(v - u)\)
Substitute \(m_{b}=45.9\times10^{-3}\ kg\), \(v = 65.27\ m/s\) and \(u = 0\ m/s\) into the formula
\(J=m_{b}v=(45.9\times 10^{-3})\times65.27\ kg\cdot m/s\approx2.99\ kg\cdot m/s\)

Step3: Calculate the average force

We know that \(J = F_{avg}\Delta t\), where \(\Delta t=0.871\ ms=0.871\times10^{-3}\ s\)
From \(F_{avg}=\frac{J}{\Delta t}\), substitute \(J = 2.99\ kg\cdot m/s\) and \(\Delta t=0.871\times 10^{-3}\ s\)
\(F_{avg}=\frac{2.99}{0.871\times10^{-3}}\ N\approx3433\ N\)

Answer:

\(J\approx2.99\ \frac{kg\cdot m}{s}\)
\(F_{avg}\approx3433\ N\)