Question

A 0.150-kg baseball traveling in a horizontal direction with a speed of 21.0 m/s hits a bat and is popped straight up with a speed of 15.0 m/s. If the bat was in contact with the ball for 51.0 ms, what was the average force of the bat on the ball

Answer 1

Answer:

17.65N

Explanation:

We know that the relationship for the impulse and momentum is given as

Ft=mv

given data

mass m=0.15kg

t= 51ms= 26.5/1000= 0.051seconds

velocity v1= 21m/s

Velocity v2=15m/s

substituting in the expression we have

Ft=m(v1-v2)

make F subject of formula we have

F=m(v1-v2) /t

F=0.15(21-15)/0.051

F=0.15*6/0.051

F=0.9/0.051

F=17.65N

Answer 2

This question involves the concepts of momentum and impulse.

The average force of the bat on ball is "17.65 N".

What is the average force?

According to Impulse Equation the change in momentum is equal to the product of applied force and time of contact. Mathematically:

$\Delta P=F\Delta t\\\\F=\frac{m\Delta v}{\Delta t}$

where,

• F = Average force = ?
• m = mass of baseball = 0.15 kg
• Δv = change in speed = 21 m/s - 15 m/s = 6 m/s
• Δt = time of contact = 51 ms = 0.051 s

Therefore,

$F=\frac{(0.15\ kg)(6\ m/s)}{0.051\ s}$

F = 17.65 N

Learn more about momentum here:

https://brainly.com/question/24030570

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