While being thrown, a net force of 132 N acts on a baseball (mass = 140 g) for a period of 4.5 X 10^-2 seconds.
The balls initial velocity is zero before leaving the pitchers hand.
When the batter hits the ball, a net force of 1320 N, opposite the direciton of the balls initial motion, acts on the ball for 9 X 10^-3 seconds during hte hit.

What is the magnitude of the change in momentum of the as it is a)thrown b)hit back.

What is the velocity of the ball after it is a)thrown b) hit back.

Solution Preview

We know that Impulse = change in momentum = mass * dV
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<br>Also impulse = force * time
<br>(Let the force act in the +ve x direction)
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<br>From data, Impulse = 132N * 4.5* 10^-2 second = 5.94 N Sec
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<br>This impulse would cause a momentum change of 5.94 kg*m/s
<br>
<br>Thus a) change in momentum = +5.94 Kg m/s ...

Solution Summary

Solution is given with all required mathematical steps.

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