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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?

Respuesta :

memo213le1
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!! ☺

Answer:

The ballplayer put 6560 newtons of force onto the ball.

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.

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