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BabaBlast [244]
3 years ago
11

1. 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 ba

ll, 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?
Physics
1 answer:
MariettaO [177]3 years ago
8 0
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  ! ! ! )
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Explanation:

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To solve this problem, start with setting up the net force equations for both block A and B:

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a_B = a_A-0.5 \frac{m}{s^2}

Now we can solve the system of 3 equations for a_A, a_B and finally for F_fr:

30kg\cdot a_A = 75N - F_{fr}\\24kg\cdot a_B = F_{fr}\\a_B= a_A-0.5 \frac{m}{s^2}\\\implies \\a_A=\frac{87}{54}\frac{m}{s^2},\,\,\,a_B=\frac{10}{9}\frac{m}{s^2}\\F_{fr} = 24kg \cdot \frac{10}{9}\frac{m}{s^2}=\frac{80}{3}kg\frac{m}{s^2}\approx 26.7N

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This answer has been verified by multiple people and is correct for the provided values in your question. I recommend double-checking the text of your question for any typos and letting us know in the comments section.

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