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???
To find the change in momentum of the ball, you need to find the differences in velocity in the horizontal and vertical directions before and after the collision.
Let's start by calculating the change in momentum in the horizontal direction (Δp_horizontal).
The initial x-component of the velocity (Vx1) is given as -10.39 m/s.
The final x-component of the velocity (Vx2) is given as +17 m/s.
To find the change in momentum in the horizontal direction, you subtract the initial velocity from the final velocity:
ΔV_horizontal = Vx2 - Vx1 = 17 m/s - (-10.39 m/s) = 27.39 m/s
Now, multiply the change in velocity in the horizontal direction by the mass of the softball (0.30 kg) to find the change in momentum in the horizontal direction (Δp_horizontal):
Δp_horizontal = ΔV_horizontal * mass
Δp_horizontal = 27.39 m/s * 0.30 kg = 8.217 kg·m/s
So, the magnitude of the change in momentum in the horizontal direction is 8.217 kg·m/s.
Now, let's move on to calculating the change in momentum in the vertical direction (Δp_vertical).
The initial y-component of the velocity (Vy1) is given as 6 m/s.
The final y-component of the velocity (Vy2) is given as -17 m/s.
To find the change in momentum in the vertical direction, you subtract the initial velocity from the final velocity:
ΔV_vertical = Vy2 - Vy1 = -17 m/s - 6 m/s = -23 m/s
Now, multiply the change in velocity in the vertical direction by the mass of the softball (0.30 kg) to find the change in momentum in the vertical direction (Δp_vertical):
Δp_vertical = ΔV_vertical * mass
Δp_vertical = -23 m/s * 0.30 kg = -6.9 kg·m/s
Note that the negative sign indicates that the change in momentum is in the opposite direction.
So, the magnitude of the change in momentum in the vertical direction is 6.9 kg·m/s.
Therefore, the magnitude of the total change in momentum while the ball is in contact with the bat is the sum of the magnitudes of the changes in momentum in the horizontal and vertical directions:
Δp_total = √(Δp_horizontal^2 + Δp_vertical^2)
Δp_total = √((8.217 kg·m/s)^2 + (6.9 kg·m/s)^2)
Δp_total ≈ 10.5 kg·m/s
So, the magnitude of the total change in momentum while the ball is in contact with the bat is approximately 10.5 kg·m/s.