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

Answer:

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

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:  

$a = \frac{v_{f} - v_{i}}{t}$ (3)

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

$F = m(\frac{v_{f} - v_{i}}{t})$ (4)

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

$1.05ms . \frac{1s}{1000ms}$ ⇒ $0.00105s$

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

$148g . \frac{1kg}{1000gs}$ ⇒ $0.148Kgs$

Replacing all those values in equation 4 it is gotten:  

$F = (0.148Kg)(\frac{(45.0m/s - 40.7m/s)}{0.00105s})$

$F = (0.148Kg)(\frac{4.3m/s}{0.00105s})$

$F = (0.148Kg)(\frac{4.3m/s}{0.00105s})$

$F = 606.09Kg.m/s^{2}$

But $1N = 1Kg.m/s^{2}$, therefore:

$F = 606.09N$

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