A baseball (m = 149 g) approaches a bat horizontally at a speed of 39.8 m/s (89.0 mph) and is hit straight back at a speed of 45.8 m/s (102 mph). If the ball is in contact with the bat for a time of 1.27 ms, what is the average force exerted on the ball by the bat? Neglect the weight of the ball, since it is so much less than the force of the bat. Choose the direction of the incoming ball as the positive direction.
To find the average force exerted on the ball by the bat, you can use Newton's second law of motion:
Force = mass × acceleration
In this case, the acceleration can be calculated using the equation:
Acceleration = (Change in velocity) / (Time)
First, we need to find the change in velocity of the ball. Since the initial velocity (before hitting the bat) is 39.8 m/s and the final velocity (after hitting the bat) is 45.8 m/s, the change in velocity is:
Change in velocity = Final velocity - Initial velocity
= 45.8 m/s - 39.8 m/s
= 6 m/s
Now we can calculate the acceleration:
Acceleration = (Change in velocity) / (Time)
Given that the time of contact is 1.27 ms, we need to convert it to seconds:
Time = 1.27 ms / 1000
= 0.00127 s
Using the formula:
Acceleration = (6 m/s) / (0.00127 s)
= 4716.54 m/s^2
Now, we can calculate the force using Newton's second law:
Force = mass × acceleration
Given that the mass of the ball is 149 g, we need to convert it to kilograms:
mass = 149 g / 1000
= 0.149 kg
Using the formula:
Force = (0.149 kg) × (4716.54 m/s^2)
= 703.27 N
Therefore, the average force exerted on the ball by the bat is 703.27 Newtons.