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
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. .
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???
Based on the given information, you correctly calculated the initial and final velocities in the x-direction (Vx) and y-direction (Vy). However, before calculating the change in momentum, we need to convert the velocities to their respective components in the x and y directions.
For the initial velocity:
Vx1 = -12 m/s * cos(30°) = -10.392 m/s
Vy1 = -12 m/s * sin(30°) = -6.0 m/s
For the final velocity:
Vx2 = 17 m/s * cos(0°) = 17 m/s
Vy2 = -17 m/s * sin(90°) = -17 m/s
Now, let's calculate the change in momentum separately for the x and y directions:
Δp in the x-direction:
ΔVx = Vx2 - Vx1 = 17 m/s - (-10.392 m/s) = 27.392 m/s
Δp_x = ΔVx * mass = 27.392 m/s * 0.30 kg = 8.2176 kg·m/s
Δp in the y-direction:
ΔVy = Vy2 - Vy1 = (-17 m/s) - (-6.0 m/s) = -11 m/s
Δp_y = ΔVy * mass = (-11 m/s) * 0.30 kg = -3.3 kg·m/s
Note: Since the ball is moving vertically downward, the change in momentum in the y-direction is negative.
Therefore, the correct results are:
Δp_x = 8.2176 kg·m/s (change in momentum in the horizontal direction)
Δp_y = -3.3 kg·m/s (change in momentum in the vertical direction)