A baseball approaches home plate at a speed of 40.0 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 54.0 m/s straight up. The ball has a mass of 145 g and is in contact with the bat for 1.70 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.)
To find the average vector force exerted by the ball on the bat during their interaction, we can use Newton's second law of motion.
1. First, let's find the change in momentum of the ball:
The initial momentum of the ball is given by:
P_initial = m_initial * V_initial
where m_initial is the mass of the ball (145 g) and V_initial is the initial velocity of the ball (40.0 m/s) in the +x direction.
The final momentum of the ball is given by:
P_final = m_final * V_final
where m_final is the mass of the ball (145 g) and V_final is the final velocity of the ball (54.0 m/s) in the +y direction.
However, since we are dealing with vectors, we need to consider the negative sign due to the reversed direction in the y-coordinate. So, the final momentum can be written as:
P_final = -m_final * V_final
The change in momentum is then given by:
ΔP = P_final - P_initial
2. Next, we need to find the time over which this change in momentum occurs:
The time of contact between the ball and bat, t_contact, is given as 1.70 ms (milliseconds). However, we need to convert it to seconds (s):
t_contact = 1.70 ms * (1 s / 1000 ms)
3. Finally, we can calculate the average vector force using the formula:
F_avg = ΔP / t_contact
Substitute the values and calculate the average vector force.
Note: The mass of the ball is given in grams, so you may need to convert it to kilograms before using it in the calculations.