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3 years ago

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A pitcher throws a 0.144-kg baseball toward the batter so that it crosses home plate horizontally and has a speed of 42 m/s just

before it makes contact with the bat. The batter then hits the ball straight back at the pitcher with a speed of 47 m/s. Assume the ball travels along the same line leaving the bat as it followed before contacting the bat. (a) What is the magnitude of the impulse delivered by the bat to the baseball? 12.816 N · s (b) If the ball is in contact with the bat for 0.0046 s, what is the magnitude of the average force exerted by the bat on the ball? kN (c) How does your answer to part (b) compare to the weight of the ball? Fav mg =

1 answer:

Solnce55 [7]3 years ago

8 0

(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:

$I=\Delta p = m(v-u)$

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

$I=(0.144 kg)(-47 m/s-(42 m/s))=-12.8 kg m/s$

And since we are interested in the magnitude only,

$I=12.8 kg m/s$

(b) 2.78 kN

The impulse exerted on the ball is also equal to the product between the average force and the contact time:

$I=F\Delta t$

where

F is the average force exerted on the ball

$\Delta t=0.0046 s$ is the contact time

Solving the formula for F, we find

$F=\frac{I}{\Delta t}=\frac{12.8 kg m/s}{0.0046 s}=2783 N = 2.78 kN$

(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

$W=mg$

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

$W=(0.144 kg)(9.8 m/s^2)=1.4 N$

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):

$\frac{F}{W}=\frac{2783 N}{1.4 N}=1988$

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