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Orlov [11]
3 years ago
9

A pitcher throws a 150g baseball so that it crosses home plate with a speed of 40 m/s. The ball hit straight back towards the pi

tcher at 45 m/s. If the contact time between the ball and the bat is 2x10^-3 seconds, what is the magnitude of the force on the ball?
Physics
1 answer:
Fed [463]3 years ago
4 0

Momentum = (mass) x (velocity)

On the way to the plate, the momentum of the ball was

(0.15 kg) x (40 m/s this way <==) = 6 kg-m/s this way <==

After the hit, the momentum of the ball was

(0.15 kg) x (45 m/s that way ==>) = 6.75 kg-m/s that way ==>

The CHANGE in momentum was 12.75 kg-m/s that way ==>

That's the momentum that the bat gave it.

Change in momentum = Impulse

Impulse = (force) x (length of time the force lasts)

12.75 kg-m/s = (force) x (0.002 second)

Force = (12.75 kg-m/s) / (0.002 second)

<em>Force = 6,375 Newtons, that way ==></em>

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