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???
PLEASE HELP!!!
You are wrong in your math.Look at part second.
how am i wrong in my math? can you please show me which parts i did wrong?
I still can't get it. Oh well, it's due in an hour.
I apologize for the confusion. Let's go through the calculations step by step to make sure we get the correct answer.
First, let's calculate the change in momentum in the horizontal direction (Vx) and the vertical direction (Vy).
Given:
Mass of the softball (m) = 0.30 kg
Initial velocity in the horizontal direction (Vx1) = -10.39 m/s
Final velocity in the horizontal direction (Vx2) = 17 m/s
Change in momentum in the horizontal direction (Δp_x) = (Vx2 - Vx1) * m
Substituting the values, we get:
Δp_x = (17 - (-10.39)) * 0.30
Δp_x = 27.39 * 0.30
Δp_x = 8.217 kg·m/s
Next, let's calculate the change in momentum in the vertical direction (Vy).
Given:
Initial velocity in the vertical direction (Vy1) = 6.0 m/s
Final velocity in the vertical direction (Vy2) = -17 m/s
Change in momentum in the vertical direction (Δp_y) = (Vy2 - Vy1) * m
Substituting the values, we get:
Δp_y = (-17 - 6.0) * 0.30
Δp_y = -23 * 0.30
Δp_y = -6.9 kg·m/s
Now, let's calculate the magnitude of the change in momentum (Δp).
Magnitude of Δp = √ [ (Δp_x)^2 + (Δp_y)^2 ]
Substituting the values, we get:
Magnitude of Δp = √ [ (8.217)^2 + (-6.9)^2 ]
Magnitude of Δp = √ [ 67.542489 + 47.61 ]
Magnitude of Δp = √115.152489
Magnitude of Δp = 10.74 kg·m/s
So, the magnitude of the change in momentum of the ball while it is in contact with the bat is 10.74 kg·m/s.
I hope this helps! Let me know if you have any further questions.