Explanation:
Given that,
Average force exerting on the golf club, F = 1000 N
Mass of the ball, m = 0.045 kg
Initial speed of the ball, u = 0
The time of contact between the ball and the club, [tex]t=1.8\ ms=1.8\times 10^{-3}\ s[/tex]
1. Let v is the speed of the golf ball as it leaves the tee. The product of force and time is equal to the change in its momentum as :
[tex]Ft=m(v-u)[/tex]
[tex]Ft=mv[/tex]
[tex]v=\dfrac{Ft}{m}[/tex]
[tex]v=\dfrac{1000\times 1.8\times 10^{-3}}{0.045}[/tex]
v = 40 m/s
2. Force applied by the batter, F = 8000 N
Time, [tex]t=1.1\ ms=1.1\times 10^{-3}\ s[/tex]
Let J is the magnitude of the impulse delivered to the baseball. It is equal to the product of force and time as :
[tex]J=F\times t[/tex]
[tex]J=8000\ N\times 1.1\times 10^{-3}\ s[/tex]
J = 8.8 kg-m/s