Friction/heat kinetic energy - reactive force

[TonyCross]

Yes. For that one question the answer is yes.

Just be aware that this is not a general principle and it cannot be applied in other contexts.

Thanks very much, I really must phrase my questions better in future...(:

 

[haruspex]

if the ball did not heat up would the ball hit the wall harder?

Again, you need to define hitting harder.
The peak force could be exactly the same, indeed, the whole force profile during compression could be the same, but the change in momentum will be greater since the forces during expansion will be greater.
Consider instead a spring mechanism. In one scenario the springs rebounds perfectly elastically; in another, a latch operates at maximum compression to inhibit any rebound.

 

[jbriggs444]

initial compression phase of the ball takes 8 milliseconds and the rebound takes 2 milliseconds

Those numbers suggest a rebound that is four times as fast as the impact. Is the ball coated with explosives?

 

[A.T.]

..., surely more velocity more force, less velocity less force.

No. You can have a very soft but elastic ball, that will have a lot of rebound but low peak force, because the impulse is spread out over a long time.

 

[haruspex]

I modeled it as a spring with two different constants, a higher one during compression and a lower during relaxation. As you can see, the rebound takes longer.

Read the next chart as clockwise around the loop. The enclosed area represents the net KE lost.

And for completeness:

 

[Physics guy]

Take a hammer and nail, I move the hammer at a constant speed with a certain mass this is kinetic energy or potential energy, it hits the nail with a Force (Newtons) the action of the nail has to be the reactive force causing
the nail to move into the wood with the transferred kinetic energy, or am I wrong.

First of all before giving my opinion you seem to be a heck lot confused with the concept of force and energy. Maybe because you think a little too much about it. Force and energy are concepts which help us understand and study the motion around us. They are compatible with each other but each helps us understand motion in a different way.

Energy can be of any type but one thing that we know for sure is that the total energy is conserved for an isolated system no matter in what situation. This has been strongly established. Kinetic energy is the energy due to the motion of the body and potential energy is the energy due to the position of a body with respect to another body. In ideal situations when there are no dissipative forces we can conclude that the total mechanical energy (sum of kinetic and potential energy) is conserved. The question of hammering a nail if you want to study it by forces then we apply an external force on the nail which causes the nail to move and it hits the wall bit stops because the wall exerts force causing the nail to stop (note that this has nothing to do with the reaction force due to the force of the nail on the wall). If we were to study it in terms of energy then we use the energy stored in the body to be converted to kinetic energy of the hammer which inturn transfers the energy to the nail causing it to enter the wall. But the dissipative forces exerted by the wall cause the kinetic energy to be converted to any other form of energy about which we are not concerned. I don't understand what potential energy has to do in this situation.

 

[TonyCross]

I modeled it as a spring with two different constants, a higher one during compression and a lower during relaxation. As you can see, the rebound takes longer.
View attachment 280488
Read the next chart as clockwise around the loop. The enclosed area represents the net KE lost.

View attachment 280489

And for completeness:
View attachment 280491

I have attempted to use your graph with some imaginary velocities, does this seem correct?

 

[A.T.]

I have attempted to use your graph with some imaginary velocities, does this seem correct?

To much imaginary stuff there, especially that "make it a triangle for the sake of ease" part.

Why would the rebound take longer, if the rebound force is the same as the compression force? You have now more impulse transmitted on the rebound, than on the compression. Also, if rebound force is the same as the compression force, then you have no energy dissipation, so why is the final speed smaller than the initial speed.

 

[TonyCross]

To much imaginary stuff there, especially that "make it a triangle for the sake of ease" part.

Why would the rebound take longer, if the rebound force is the same as the compression force? You have now more impulse transmitted on the rebound, than on the compression. Also, if rebound force is the same as the compression force, then you have no energy dissipation, so why is the final speed smaller than the initial speed.

All good points how would you approach the problem?

 

[A.T.]

All good points how would you approach the problem?

What is the question?

 

[Dale]

All good points how would you approach the problem?

I like the approach by @haruspex

 

[TonyCross]

I like the approach by @haruspex

So do I...

 

[Dale]

So do I...

OK, so just use that and don’t change it.

 

[TonyCross]

OK, so just use that and don’t change it.

Thanks Dale, you actually answered my initial question earlier in the thread, haruspex made a really good examination of the physics. I think we can now call this question well and truly answered. Thanks to everyone who gave their input to the discussion.