Question

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.)

Answer 1

(4273.7
`\hat{j}` - 3281.6
`\hat{i}`)

Step-by-step explanation:

`\underset{v_(i)}{\rightarrow}` = initial velocity of the baseball before collision = 43
`\hat{i}` m/s

`\underset{v_(f)}{\rightarrow}` = final velocity of the baseball after collision = 56
`\hat{j}` 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

`\underset{F_(avg)}{\rightarrow}` = Average force vector

Using Impulse-change in momentum equation

`\underset{F_(avg)}{\rightarrow}` t = m (
`\underset{v_(f)}{\rightarrow}` -
`\underset{v_(i)}{\rightarrow}` )

`\underset{F_(avg)}{\rightarrow}` (0.0019) = (0.145) (56
`\hat{j}` - 43
`\hat{i}`)

`\underset{F_(avg)}{\rightarrow}` = (4273.7
`\hat{j}` - 3281.6
`\hat{i}`)