Need help with momentum related questions

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In summary, a bat applies an average force of 0.25 kg to a ball that is moving at 45 m/s after being struck by the bat.
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kenny243
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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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  • #2
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...
 
  • #3
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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  • #4
I think I see why now. They give the time that the force was applied, not the distance. Thanks haruspex! :-)
 
  • #5
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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Related to Need help with momentum related questions

1. What is momentum?

Momentum is a measurement of an object's motion, and it is calculated by multiplying an object's mass by its velocity.

2. How is momentum related to force?

According to Newton's second law of motion, force is equal to the rate of change of an object's momentum. This means that an object with a larger momentum will require a larger force to change its motion.

3. Can momentum be conserved?

Yes, momentum is a conserved quantity, meaning that it remains constant in a closed system where there is no external force acting on the system.

4. How does momentum relate to collisions?

In collisions, the total momentum of the objects involved before and after the collision remains the same. This is known as the law of conservation of momentum.

5. How does momentum relate to impulse?

Impulse is the change in an object's momentum over time. It is equal to the force applied to an object multiplied by the time it is applied. This relationship is described by the impulse-momentum theorem.

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