1- Force of a Baseball Swing. A baseball has mass 0.139 k g . A- If the velocity

Question

1-Force of a Baseball Swing. A baseball has mass 0.139

kg .

A-If the velocity of a pitched ball has a magnitude of 41.0

m/s and the batted ball's velocity is 53.0m/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.

B-If the ball remains in contact with the bat for 2.3

ms , find the magnitude of the average force applied by the bat.

2-On a frictionless, horizontal air table, puck A (with mass 0.250

kg ) is moving toward puck B (with mass 0.355kg ), that is initially at rest. After the collision, puck A has a velocity of 0.122m/s to the left, and puck B has velocity 0.655m/s to the right.

A-What was the speed of puck A before the collision?

B-Calculate the change in the total kinetic energy of the system that occurs during the collision

Explanation / Answer

1)

A)change in momentum = 0.139 X 41 - (-53)
= 13.06 kg m/s

impulse = change in momentum = 13.06 kg m/s

average force = change in momentum / time

= 13.06 / (2.3 X 10^-3) = 5913 N.

2)

Before the collision:
mv + MV = (.250kg)(v) + (.355kg)(0m/s) = v(.250kg)

KE1 = 1/2(.250kg)(v)^2 + 1/2(.355kg)(0m/s)^2 = 1/2(.250kg)(v)^2

After the collision:
mv + MV = (.250kg)(-0.122m/s) + (.355kg)(0.655m/s)

KE2 = 1/2(.250kg)(-0.122m/s)^2 + 1/2(.355kg)(0.655m/s)^2

But momentum is conserved, so:
(.250kg)(-0.122m/s) + (.355kg)(.655 m/s) = v(.255kg)

v = 0.7922 m/sec = speed of puck before collision

KE1 = 0.6275

KE2 = 0.0780

change in KE = -0.54948

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