Posted by winterWx on Saturday, February 24, 2007 at 12:28am.
A 0.30-kg softball has a velocity of 12 m/s at an angle of 30° below the horizontal just before making contact with the bat. What is the magnitude of the change in momentum of the ball while it is in contact with the bat if the ball leaves the bat with a velocity of
17 m/s, vertically downward, Δp = __ kg·m/s, and 17 m/s, horizontally back toward the pitcher? Δp = __ kg·m/s
drwls, Friday, February 23, 2007 at 5:40am
All you have to do is calculate the change in momentum in the horizontal and vertical directions, separately. Initially, you have
Vx = -12 cos 30 = -10.39 m/s
Vy = -12 sin 30 = 6.0 m/s
if Vx is defined as positive in the direction of the pitcher and Vy is postive upward.
After the hit, Vx = +17 m/s and Vy = -17 m/s. Just do the subtractions to get delta V, and multiply by the mass of the softball to get delta p.
winterWx, Saturday, February 24, 2007 at 12:27am
Hmmm...This is what I did:
Vx1-Vx2=-10.39-17=-27.39 m/s
(-27.39)(.3 kg)=-8.217 kg*m/s
Vy1-Vy2=-6-(-17)=11 m/s
(11)(.3)=3.3 kg*m/s
But both of those are incorrect. What am I doing wrong???
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