Force of a Baseball Swing. A baseball has mass 0.150 kg . Part A If the velocity

Question

Force of a Baseball Swing. A baseball has mass 0.150 kg .
Part A
If the velocity of a pitched ball has a magnitude of 44.0 m/s and the batted ball's velocity is 56.5m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.
Express your answer to three significant figures and include the appropriate units.

If the ball remains in contact with the bat for 1.5 ms, find the magnitude of the average force applied by the bat.
Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

change in momentum is the subtraction from "after momentum" of the "before momentum" as momentum is a vector quantity this is a vector subtraction. m1v1 = initial momentum of ball = (0.145)(44.0) = 6.38 kg-m/s m1v2 = after momentum = (0.145)(64.0) = 9.28 kg-m/s since these momentums are 180° opposite one must be called negative so their difference = 15.67 kg-m/s change in momentum = 15.67 kg-m/s ANS a1 Impulse = change in momenutm = 15.67 ANS a2

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