Asked by Anonymous
A catcher “gives” with the ball when he catches .15 kg baseball moving at 25 m/s. (a) If he moves his glove a distance of 2.0 cm, what is the average force acting on his hand? (b) Repeat for the case in which his glove and hand more 10 cm.
Answers
Answered by
drwls
The average velocity of the ball between first contact with the glove and when it stops is 25/2 = 12.5 m/s. The time that the ball is decelerated is T = D/12.5 m/s, where d is the distance that the glove moves backwards.
The average force is given by the momentum law
(avg. Force) * T = initial momentum
avg. Force = M V*(V/2)/D = K.E./D
This is another way of saying that the ball does work on the backwards-moving glove that equals the initial kinetic energy of the ball.
The average force is given by the momentum law
(avg. Force) * T = initial momentum
avg. Force = M V*(V/2)/D = K.E./D
This is another way of saying that the ball does work on the backwards-moving glove that equals the initial kinetic energy of the ball.
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