Need help with momentum related questions

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A user seeks help calculating the average force exerted by a bat on a baseball after it reverses direction and increases speed. They initially calculate the impulse based on the change in momentum and provide a result of 3542.67 J. Another participant clarifies that average force should be calculated using the formula for change in momentum over time, rather than energy. The discussion emphasizes the importance of using correct formulas and understanding the relationship between force, momentum, and energy in physics problems. The thread highlights common misconceptions in calculating average force in similar scenarios.
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kenny243 said:
A bat strikes a 0.25 kg baseball pitched at 40 m/s. The bat reverses the ball’s direction and gives it a speed of 45 m/s. What average force does the bat apply to the ball if they are in contact for 0.006 s?
I got 3542.67J.

I used Impulse=change in momentum.
F(0.006s)=0.25kg(45m/s-(-40ms))

Did I do it right?
 
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kenny243 said:
A bat strikes a 0.25 kg baseball pitched at 40 m/s. The bat reverses the ball’s direction and gives it a speed of 45 m/s. What average force does the bat apply to the ball if they are in contact for 0.006 s?
I got 3542.67J.

I used Impulse=change in momentum.
F(0.006s)=0.25kg(45m/s-(-40ms))

Did I do it right?

Welcome to the PF.

In the future, please use the Homework Help Template that you are provided when you start a new thread in the HH forums. It helps to organize the post into sections to make it easier for us to help you. :-)

To check your answer, you could calculate the average force a second way. Can you do it using the change in the kinetic energy and the time? Show us your work on that, and see if you get the same answer...
 
berkeman said:
Can you do it using the change in the kinetic energy and the time?
Not that I can see. Bear in mind that average force is ##\vec {\Delta p}/\Delta t##. Energy doesn't really enter into it.
kenny243 said:
Did I do it right?
Yes.
 
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I think I see why now. They give the time that the force was applied, not the distance. Thanks haruspex! :-)
 
berkeman said:
I think I see why now. They give the time that the force was applied, not the distance. Thanks haruspex! :)
There's a bit more to it than that. As I posted, average force is ##\vec{\Delta p}/\Delta t##, not ##\Delta E/\Delta s##. Using the latter formulation would lead to the bizarre result that the average force might be different from mass times average acceleration. They do give the same result when the force is constant, but rarely otherwise. Besides, taking the vector equation ##\vec F.\vec{ds} = dE## and dividing through by ds is not really on.
Sadly, questions from textbooks and teachers asking for average force given energy and distance are a regular occurrence on this forum. I do my best to enlighten the students that this is not actually valid, hoping that some will able to pass this on to their elders.
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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