Question

A baseball with a mass of 148 g is thrown horizontally with a speed of 40.7 m/s (91 mi/h) at a bat. The ball is in contact with the bat for 1.05 ms and then travels straight back at a speed of 45.0 m/s (101 mi/h). Determine the average force exerted on the ball by the bat.

Answer 1

The average force exerted on the ball by the bat is 606.09N.

Step-by-step explanation:

The average force can be found using Newton's second law:

`F = ma` (1)

Where F is the force, m is the mass and a is the acceleration.

The acceleration can be determined using the equations for a Uniformly Accelerated Rectilinear Motion:

`v_(f) = v_(i) + at` (2)

Where
`v_(f)` is the final velocity,
`v_(i)` is the initial velocity, a is the acceleration and t is the time.

Then a will be isolated from equation 2:

`a = (v_(f) - v_(i))/(t)` (3)

Replacing equation (3) in equation (1) it is gotten:

`F = m((v_(f) - v_(i))/(t))` (4)

The time will be expressed from milliseconds to seconds before using it in equation (4):

`1.05ms . (1s)/(1000ms)` ⇒
`0.00105s`

The mass will be expressed from grams to kilograms before using it in equation (4):

`148g . (1kg)/(1000gs)` ⇒
`0.148Kgs`

Replacing all those values in equation 4 it is gotten:

`F = (0.148Kg)(((45.0m/s - 40.7m/s))/(0.00105s))`

`F = (0.148Kg)((4.3m/s)/(0.00105s))`

`F = (0.148Kg)((4.3m/s)/(0.00105s))`

`F = 606.09Kg.m/s^(2)`

But
`1N = 1Kg.m/s^(2)`, therefore:

`F = 606.09N`

So, the average force exerted on the ball by the bat is 606.09N.