What Is the Average Force Exerted on a Rebounding Steel Ball?

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SUMMARY

The average force exerted on a 56g steel ball during its elastic rebound from a steel plate is calculated using the principles of impulse and momentum. The ball is in contact with the plate for 0.5 ms, and the average force can be determined by calculating the change in momentum over the contact time. The discussion emphasizes the importance of understanding the momentum before and after the collision to accurately compute the average force, with potential answers ranging from 4390 N to 10,260 N.

PREREQUISITES
  • Understanding of impulse and momentum concepts
  • Knowledge of elastic collisions
  • Basic physics of free fall and gravitational acceleration
  • Ability to perform vector calculations
NEXT STEPS
  • Learn how to calculate momentum before and after a collision
  • Study the principles of elastic and inelastic collisions
  • Explore the relationship between force, time, and change in momentum
  • Practice solving impulse problems using different mass and height scenarios
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Physics students, educators, and anyone interested in understanding the dynamics of collisions and impulse in mechanics.

williams31
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A 56g steel ball is released from rest and falls vertically onto a steel plate. The ball strikes the plate and is in contact with it for .5ms. The ball rebounds elastically, and returns to its original height. The time interval for a round trip is 8.00s. In this situation, the average force exerted on the ball during contact with the plate is closest to:
A) 4390 N
B) 10,260 N
C) 5870 N
D) 7300 N
E) 8780 N

I was wondering what formula I should use to solve the problem.
 
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This is an impulse question right here. The formulas for impulse are your best friend in this case. Remember that it's not exactly straitfoward, but requires a little though outside of the definition of Impulse. Hope that gets you started.
 
Im not sure if I am following you.
 
williams31 said:
A 56g steel ball is released from rest and falls vertically onto a steel plate. The ball strikes the plate and is in contact with it for .5ms. The ball rebounds elastically, and returns to its original height. The time interval for a round trip is 8.00s. In this situation, the average force exerted on the ball during contact with the plate is closest to:
A) 4390 N
B) 10,260 N
C) 5870 N
D) 7300 N
E) 8780 N

I was wondering what formula I should use to solve the problem.
the average force during a collision is the change of momentum divided by the time interval during which the force acts. Here, by change of momentum, I mean the final momentum (after the collision) minu sthe initial momentum (before the collision). You have to be careful, this is a vector difference!
 
Im lost...but thanks for your guys help anyway.
 
williams31 said:
Im lost...but thanks for your guys help anyway.
First step: find the momentum of the ball *just* before it hits the surface (you may essentially treat its height as being zero). You know it takes 4 seconds to fall and it started from rest so you may find the velocity it has once it reaches the surface. The magnitude of the momentum will be the mass times the speed. (here, I am treating the time that the collision last negligible compared to the time of fall)

Second step: If the collision is elastic, what is the momentum just after the collision? If you take the momentum vector just after minus the momentum vector just before the collision, you get a difference which is not zero. What is its magnitude?

Third step: take this last result and divide by the time interval of the collision.
 
williams31 said:
Im lost...but thanks for your guys help anyway.
What is the height that the ball is dropped from? (You can work this out from the time of the round trip).
What is its velocity just before it strikes the plate?
What is it after?
What is the change in velocity?
What is the change in momentum?
What is the relationship between (average) Force, time, and change in momentum?

Answer those questions and you will have your answer.

AM
 

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