A baseball approaches home plate at a speed of 43.0 m/s, moving horizontally just before being hit by a bat. The batter hits a pop-up such that after hitting the bat, the baseball is moving at 56.0 m/s straight up. The ball has a mass of 145 g and is in contact with the bat for 1.90 ms. What is the average vector force the ball exerts on the bat during their interaction? (Let the +x-direction be in the initial direction of motion, and the +y-direction be up.)

Respuesta :

Answer:

(4273.7 [tex]\hat{j}[/tex] - 3281.6 [tex]\hat{i}[/tex])

Explanation:

[tex]\underset{v_{i}}{\rightarrow}[/tex] = initial velocity of the baseball before collision = 43 [tex]\hat{i}[/tex] m/s

[tex]\underset{v_{f}}{\rightarrow}[/tex] = final velocity of the baseball after collision = 56 [tex]\hat{j}[/tex] m/s

m = mass of the ball = 145 g = 0.145 kg

t = time of contact of the ball with the bat = 1.90 ms = 0.0019 s

[tex]\underset{F_{avg}}{\rightarrow}[/tex] = Average force vector

Using Impulse-change in momentum equation

[tex]\underset{F_{avg}}{\rightarrow}[/tex] t = m ([tex]\underset{v_{f}}{\rightarrow}[/tex] - [tex]\underset{v_{i}}{\rightarrow}[/tex] )

[tex]\underset{F_{avg}}{\rightarrow}[/tex] (0.0019) = (0.145) (56 [tex]\hat{j}[/tex] - 43 [tex]\hat{i}[/tex])

[tex]\underset{F_{avg}}{\rightarrow}[/tex] = (4273.7 [tex]\hat{j}[/tex] - 3281.6 [tex]\hat{i}[/tex])