Answer :
(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:
[tex]I=\Delta p = m(v-u)[/tex]
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
[tex]I=(0.144 kg)(-47 m/s-(42 m/s))=-12.8 kg m/s[/tex]
And since we are interested in the magnitude only,
[tex]I=12.8 kg m/s[/tex]
(b) 2.78 kN
The impulse exerted on the ball is also equal to the product between the average force and the contact time:
[tex]I=F\Delta t[/tex]
where
F is the average force exerted on the ball
[tex]\Delta t=0.0046 s[/tex] is the contact time
Solving the formula for F, we find
[tex]F=\frac{I}{\Delta t}=\frac{12.8 kg m/s}{0.0046 s}=2783 N = 2.78 kN[/tex]
(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
[tex]W=mg[/tex]
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
[tex]W=(0.144 kg)(9.8 m/s^2)=1.4 N[/tex]
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):
[tex]\frac{F}{W}=\frac{2783 N}{1.4 N}=1988[/tex]