Question

A big league hitter attacks a fastball! The ball has a mass of 0.16 kg. It is pitched at 38 m/s. After the player hits the ball, it is now traveling 44 m/s in the opposite direction. The impact lasted 0.002 seconds. How big of a force did the ballplayer put on that ball?

Answer 1

The ballplayer put 6560 newtons of force onto the ball.

Step-by-step explanation:

List out all the information the question provides:

Ball mass: 0.16

Initial Velocity: 38 m/s

Final velocity: -44 m’s (Remember the negative sign!! The ball is going the OPPOSITE direction.)

Time: 0.002

Change in velocity: 82 (difference of 38 and -44)

This is the formula:

F∆t=m∆v

(∆t= time, m is mass, ∆v is the change in velocity.)

then solve.

F•(0.002)=(0.16•82)

F•(0.002)=(13.12)

divide 13.12 by ∆t

F=6560

The ballplayer put 6560 newtons of force onto the ball.

Hope this helps! It's pretty straightforward once you get the hang of it.

Answer 2

Momentum = (mass) x (velocity)
Original momentum before the hit =
(0.16 kg) x (38 m/s) this way <==
= 6.08 kg-m/s this way <==
Momentum after the hit =
(0.16) x (44 m/s) that way ==>
= 7.04 kg-m/s that way ==>
Change in momentum = (6.08 + 7.04) = 13.12 kg-m/s that way ==> .-----------------------------------------------
Change in momentum = impulse.
Impulse = (force) x (time the force lasted)
13.12 kg-m/s = (force) x (0.002 sec)
(13.12 kg-m/s) / (0.002 sec) = Force
6,560 kg-m/s² = 6,560 Newtons = Force
( about 1,475 pounds ! ! ! )
Hoped this helped!! ☺