Solving Catching a Fastball Problem | Force, Mass & Time

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SUMMARY

The discussion focuses on calculating the average force acting on a baseball catcher's mitt when catching a fastball thrown at 43 m/s with a mass of 0.15 kg. The initial calculations by the user, Mastac741, incorrectly assumed no acceleration acting on the ball, leading to erroneous results for both the time interval and the average force. By recognizing the need to account for the deceleration of the ball as it comes to rest, Mastac741 corrected the approach, ultimately leading to the correct calculation of the average force.

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mastac741
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Here's the problem I'm having trouble with:

A baseball catcher is catching a fastball that is thrown at 43 m/s by the pitcher. If the mass of the ball is .15kg and if the catcher moves his mitt backward toward his body by 8.0 cm as the ball lands in the glove, what is the magnitude of the average force acting on the catcher's mitt? Estimate the time interval required for the catcher to move his hands.

OK, so this is how I solved it(tried to solve it). I got it wrong, so something's wrong with my logic, so please try to point it out.

Vo = 43 m/s
m = .15 kg

p = mv = (.15) (43) = 6.45 N/s = Fav (Delta) t

I need to find t, so I use x-x0 = Voxt (no acceleration on the ball, and no acceleration acting in the x direction). I find that t = .0186 sec (which is wrong). I divide 6.45 by .0186 sec to get Fav to = 346.8 N (which also is wrong).

Some help would be appreciated.

Mastac741
 
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mastac741 said:
I need to find t, so I use x-x0 = Voxt (no acceleration on the ball, and no acceleration acting in the x direction).

But there obviously is an acceleration on the ball in the x-direction. If there weren't, the ball would never come to rest in the glove.
 
gotcha

Right you are. If I account for the acceleration in the -x direction I get the right answer. Thanks for filling in my logic gaps.

Mastac741
 

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