Answer:
The average force between ball and bat, F = 5236.11 N
Explanation:
Given that,
Mass of the baseball, m = 0.145 kg
Initial speed of the baseball, u = 29 m/s
Final speed of the ball, v = -36 m/s (negative as it rebounds)
Time, [tex]t=1.8\ ms=1.8\times 10^{-3}\ s[/tex]
Let J is the impulse imparted by the ball. It is equal to the change in momentum as :
[tex]J=m(v-u)=Ft[/tex]
[tex]F=\dfrac{m(v-u)}{t}[/tex]
[tex]F=\dfrac{0.145\times (-36-29)}{1.8\times 10^{-3}}[/tex]
F = -5236.11 N
|F| = 5236.11 N
So, the magnitude of the average force between ball and bat is 5236.11 N. Hence, this is the required solution.
