How does the floor contribute to the ball's energy during a bounce?

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Homework Help Overview

The discussion revolves around the energy transfer during the bouncing of a ball dropped from a height. Participants examine the work done by the floor on the ball and the implications of energy conservation in the context of elastic and inelastic collisions.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the calculation of work done by the floor, questioning whether the floor does any work if it does not move. There are discussions about the nature of forces during the collision and the energy transformations involved.

Discussion Status

The conversation is ongoing, with various interpretations being explored regarding the work done by the floor and the energy lost by the ball. Some participants suggest that the problem is underspecified, leading to uncertainty about the energy dissipation mechanisms involved.

Contextual Notes

There are assumptions about the rigidity of the floor and the ball, as well as the nature of the collision (elastic vs. inelastic) that are being debated. Participants note that real-world factors, such as deformation and energy loss, complicate the analysis.

  • #31
verty said:
If the wall does deflect but returns to its starting position, is there work done? Can we say that the ball does work deflecting the wall and the wall does work reversing that deflection?
To be thorough, it may be that the floor deforms elastically while the ball deforms inelastically (or better, imperfectly elastically). In that case the floor does no net work. Or it could be the other way around, or anything in between.
 
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  • #32
darkxponent said:
Never heard of this "the movement of point of application of Force". What i knew was "Work is done when Force is applied to an abject and it moves some distance under the application of Force.". Do not understand by what you mean by "point of application of Force" or maybe i might not be familier to it.
Instead of a uniform ball, consider a point mass attached atop a massless spring landing on the floor. In the frame of reference of an undeflected floor, the point mass does work on the spring, then, during rebound, the spring does work on the point mass. The spring is not moved by the floor, so the floor does no work.
You can extend this to a vertical chain of point masses joined by springs, thereby approximating a ball.
 
  • #33
Tanya Sharma said:
But that does give ∫F.dx = -25 J .Can we say that the center of mass work(not the "real" work done by that force) is equal to -25J ?
Absolutely.

You invariably make complete sense.Thanks for the nice explanation.:smile:
You are most welcome. :wink:
 
  • #34
darkxponent said:
Interna thermal energy you said Doc Al!.

What i am interested in is know what happens after the ball comes at rest(see my attachment).

During the collision, there is a time when ball is at rest. Now, what happens is that the ball starts moving in the opposite direction and finally comes to velocity v/2. That is momentum = mv/2 and K.E = mvv/8
I have to questions in mind.

1) What increases the momentum of the ball?- my answer would be the normal force by the ground. I came to this conclusion from Newtons second law.

2) What increases K.E of the ball?- My answer would be the Normal force, as it is responsible for increase in momentum so it is responsible for increase in KE.

Where am i wrong here? Is this not a mechanics question, rather a thermodynamics question. Do tell me
There's nothing wrong with applying dynamics to this problem. That's exactly what I did with the equation:
∫F.dx = ΔKE + ΔGPE

But you'll be missing something essential if you don't also consider energy. Since the floor does no real work on the ball, as I explained above, for the ball at rest to rise up from the floor there must be some stored elastic energy within the ball.

(I am assuming the floor doesn't move for this exercise. Imagine a rubber ball bouncing on a concrete floor.)
 

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