(a) 12.8 kg m/s
The impulse delivered by the bat on the baseball is equal to the change in momentum of the baseball:
![I=\Delta p = m(v-u)](https://tex.z-dn.net/?f=I%3D%5CDelta%20p%20%3D%20m%28v-u%29)
where we have
m = 0.144 kg is the mass of the ball
v = -47 m/s is the final velocity of the ball
u = 42 m/s is the initial velocity of the ball
Substituting into the equation, we find
![I=(0.144 kg)(-47 m/s-(42 m/s))=-12.8 kg m/s](https://tex.z-dn.net/?f=I%3D%280.144%20kg%29%28-47%20m%2Fs-%2842%20m%2Fs%29%29%3D-12.8%20kg%20m%2Fs)
And since we are interested in the magnitude only,
![I=12.8 kg m/s](https://tex.z-dn.net/?f=I%3D12.8%20kg%20m%2Fs)
(b) 2.78 kN
The impulse exerted on the ball is also equal to the product between the average force and the contact time:
![I=F\Delta t](https://tex.z-dn.net/?f=I%3DF%5CDelta%20t)
where
F is the average force exerted on the ball
is the contact time
Solving the formula for F, we find
![F=\frac{I}{\Delta t}=\frac{12.8 kg m/s}{0.0046 s}=2783 N = 2.78 kN](https://tex.z-dn.net/?f=F%3D%5Cfrac%7BI%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B12.8%20kg%20m%2Fs%7D%7B0.0046%20s%7D%3D2783%20N%20%3D%202.78%20kN)
(c) The force exerted on the ball is much larger (1988 times more) than the weigth of the ball
The weight of the ball is given by
![W=mg](https://tex.z-dn.net/?f=W%3Dmg)
where
m = 0.144 kg is the mass of the ball
g = 9.8 m/s^2 is the acceleration due to gravity
Solving the equation for W, we find
![W=(0.144 kg)(9.8 m/s^2)=1.4 N](https://tex.z-dn.net/?f=W%3D%280.144%20kg%29%289.8%20m%2Fs%5E2%29%3D1.4%20N)
So as we see, the force exerted on the ball (2783 N) is almost 2000 times larger than the weight of the ball (1.4 N):
![\frac{F}{W}=\frac{2783 N}{1.4 N}=1988](https://tex.z-dn.net/?f=%5Cfrac%7BF%7D%7BW%7D%3D%5Cfrac%7B2783%20N%7D%7B1.4%20N%7D%3D1988)