My bolding
What about the energy required to decelerate the water in a hydroelectric plant.
There is a drop in velocity through the turbine, if deceleration requires energy how does the plant produce energy?
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Hardfire: Physics of 9/11
- Thread starter R.Mackey
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My bolding
What about the energy required to decelerate the water in a hydroelectric plant.
There is a drop in velocity through the turbine, if deceleration requires energy how does the plant produce energy?
Gravy
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Let's hope he's not the Killtown of physics.
C'mon, femr2, this concept is so simple that a science dummy like me can easily understand it. Something is holding you back. Don't feel bad...yet. Gordon Ross is a mechanical engineer and he apparently has the same mental block. He has to be right. Therefore the laws of physics as understood by everyone else have to be wrong.
You can do better than that. If for some reason you feel trickery is being played on you by people here, I encourage you to run this problem by a physics teacher or professor at a local school.
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Furcifer
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I think I know what the trickery is.
Rubber bands. Or elastics as we like to call them. Elastics have a fairly high coefficient or restitution. You can do pretty much what you like with an elastic without it breaking. This is carrying over into thinking about the collision. Imagine shooting a rubber band off your finger. As you finger collides with an elastic, the inelastic part is where the rubber band stretches, the momentum of your finger is now being stored in the deformation of the rubber band. If you continue to hold on to the elastic and pull back until it breaks that's essentially a perfectly inelastic collision. Both your hand and the elastic, after this perfectly inelastic collision are now moving with the same velocity, but slower than when you started to pull on the elastic. The loss in speed of your finger is due to the stretching of the elastic in the inelastic collision.
As you can see, thinking about elastics, when talking about elastic or inelastic collisions is very confusing. I had to read this post a couple of times to make sure I got it right.
When you conclude that there is no deformation of materials, isn't that a good time to start over?
No. The purpose of the discussion is to justify the separation of KE used during inelastic collision of two bodies of mass from subsequent KE usage to overcome structural resistance.
In the example presented, there is 13.04e8 J KE remaining following the collision available for further actions.
(Makes you wonder why these guys are trying so hard to fudge a simple conservation of momentum calculation really)
I see the same pattern here. If you could just admit to making an error, you wouldn't look nearly as ridiculous as you do when you hang on to your mistake come hell or high water.
Furcifer
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For the last time, where do you think the energy "used" goes? Do you know that energy cannot be created or destroyed? I'm sure you've heard this, and yet you're claiming it gets used? And accusing me of lying?
1 sentence. Where does this energy go when it is used up?
You messed up; big time. Confess you don't do physics. Where did you learn to do such a bad job at understanding physics?
It is perfectly clear and simple.
The kinetic energy is consumed by deceleration of the initial mass, and acceleration of the impacted mass.
The sum of these two KE sinks is exactly the amount of KE consumed during the inelastic collision, as is clearly shown in my calcs.
Your prior error, which I pointed out, was that you omitted the KE consumed decelerating the initial mass.
You then compounded that error by stating a KE equivalence equation (0.5mv^2), rather than a momentum equivalence equation (mv).
I have no need to discuss this point further.
iAi caramba!My response to you in post #888 is correct. So are the various objections raised by others in this thread. I'll show you, using your numbers.
So we've got two objects, call them M1 and M2, with masses 38.67 Gg and 2.47 Gg respectively. Before impact, M1 is moving at velocity V1 = 8.47 m/s, and M2 is stationary, V2 = 0.
We specify the collision is fully inelastic, so after collision, the masses have the same velocity, which we'll call V3. To find this velocity, we apply conservation of momentum -- momentum is conserved in any type of collision, whether elastic, partially inelastic, or fully inelastic.
Conservation of momentum simply states that the momentum before and after impact is the same. Momentum before is equal to M1 V1 + M2 V2. Since V2 = 0, the momentum before is just M1 V1 = 38.67 Gg x 8.47 m/s = 327.53 Gg m/s.
This is the same as the momentum afterward. Mass is also conserved, so the mass after collision is just M1 + M2, and we already said the new velocity was V3, so we have (M1 + M2) V3 = 327.53 Gg m/s. Since M1 + M2 = 41.14 Gg, this means V3 = 327.53 Gg m/s / (41.14 Gg) = 7.96 m/s.
So now let's talk about energy.
Before the collision, mass M1 had a kinetic energy equal to M1 V12 / 2 = 38.68 Gg x (8.47 m/s)2 / 2 = 1387 Gg m2/s2 = 1387 MJ.
After the collision, the combined mass (M1 + M2) has kinetic energy equal to (M1 + M2) V32 / 2, or we can rewrite this as M1 V32 / 2 + M2 V32 / 2. In other words, we can either look at the kinetic energy of the whole blob, or we can look at the energy of each now mashed-together piece. Either way, we get the same answer.
Let's do it the latter way. The new kinetic energy of mass from M1 is 38.68 Gg x (7.96 m/s)2 / 2 = 1225 MJ. Likewise, the new kinetic energy of mass M2 is 2.47 Gg x (7.96 m/s)2 / 2 = 78 MJ.
So let's now follow the path of the energy. Before collision, mass M1 had 1387 MJ, and mass M2 had none.
After collision, mass M1 has 1225 MJ left. It has lost a total of 162 MJ of energy.
Mass M2 has acquired 78 MJ. Some of that came from M1. The remainder that M1 lost, or 84 MJ, is unaccounted for.
What this means is we require an additional 84 MJ energy sink, and if we don't, then the two pieces have no reason to stay together. If M1 keeps that energy, it will move at a different velocity, and we won't have a fully inelastic collision anymore. If we had instead specified an elastic collision, then kinetic energy as well as momentum would be conserved, and it would mean M1 and M2 would exit the collision at different speeds, i.e. not stuck together at all. There is no other solution to the problem.
The most common energy sink is physical deformation. The reason for this is quite simple: In this problem, we've exchanged momentum between two objects. The flow of momentum between objects is a force, and the flow of momentum within objects is stress.
Now, whether we have an elastic or inelastic collision depends on how this stress affects the objects. If we have very strong objects, such that the strain is small, then the stress is elastic and very little energy can be dissipated -- the objects will bounce apart again after collision. This is because below the plastic limit, most solids behave much like springs.
Above the plastic limit, the object absorbs energy through deformation. That's the expected result. You can hit two objects harder and harder and harder, and they'll keep bouncing back, until you hit them so hard that the stress they feel at impact is no longer "small" and they start to deform as a result. At that point, you can start absorbing large amounts of energy. That is the transition from elastic collisions to inelastic collisions. It is no coincidence that the word "elastic" appears both in collisions and in material behavior.
If you smash them together harder still, not only will you deform the objects, but you will eventually cause them to yield, either causing rupture or fracture. At this point you are not only deforming the objects, but actually destroying them -- this is the mechanism responsible for most of the concrete crumbling during collapse, for instance. As you crumble or fracture the objects, they absorb more and more energy, as more and more of the initial mass is stressed to its yield point.
If you smash together even harder, you start to see new mechanisms arise. After things crumble, they start to fuse, friction between particles raising the temperature. You can also have fluid dissipation mechanisms come into play here. Keep going and eventually you can start causing static electrical effects or more radical molecular change. Keep cranking it up and you will start to see vaporization. Even more and you start making plasma. With still more you can start shedding UV or X-rays, then gamma rays, and finally you start in with relativistic effects like spontaneous pair creation. There really is no limit.
Fortunately for us, our problem is much slower and simpler. For us, mere deformation and fracture will dominate. There will be some attendant heating, but not much, mostly through compression of materials and hysteresis of solids. Friction is there too, but really only affects the contact surface, not the whole body until we've already caused widespread failure. But to say that there is no deformation in an inelastic collision is absolutely ludicrous. This is practically a contradiction in terms.
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Now, femr2, the next thing I want from you, here or anywhere, is an acknowledgment that you understand the above. I'm not interested in anything else. I won't sugar-coat this -- if you cannot understand the above, and you continue to fail at basic conservation of momentum and energy, then you have no hope whatsoever to correctly apply and interpret your own model. Your options at that point will be limited to two: Either acquire the expertise, which you should do on your own time; or stand aside and let those who understand the physics have the floor.
Sorry it turned out this way, but you really should have listened to me and kept this private.
ETA: Typo -- I had kJ instead of MJ throughout. The long-hand units were correct. Sorry for the confusion.
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Gravy
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No! It's time to redouble one's efforts to convince people who work in applied physics that they are numbskulls!
Lawdy I hope this ends with some learning.
This is a serious question. Do any of these people ever take a backward step? Do they ever admit that they might have gotten something wrong?
It's like they try to swim holding an anvil and they never think to let go of it.
A few have, and they have my respect. But not many. The omens are poor with this one.
Furcifer
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This sucks, I really wanted to play with his simulator. It has potential (no pun intended). At the very least it shows Heiwa and psi how wrong they are about collpase. femr2 has at least discovered that the collapse progresses. Even with his lack of understanding conservation of momentum, he is light years ahead of those two, and many others. He is possibly at a turning point, not so much in regards to the events of 9/11, but at least in regards to listening to authority. If he learns from this that people are not being deceitful maybe he will open up to listening. Listening to authority is the biggest obstacle many of them face. At least with science, there is no making things up, there is no bending of the rules, there really is no grey area. Both sides of this debate should be able to come to an agreement on the science. At least that's what I thought.
Let's stand back and look at the other side of the equation.
We begin with 13.87e8 J KE (m1 prior to impact)
We end with 13.04e8 J KE (m1+m2 post impact)
83.33e6 J KE difference.
You are suggesting that the 83.33e6 J is unaccounted for.
Yet we have accelerated a mass of 2.47e6 Kg from rest to 7.96m/s, and also decelerated a mass of 38.67e6 Kg from 8.47m/s to 7.96m/s.
You are suggesting that the changes in velocity are 'free' and consumed no KE in themselves. They are not free.
The deceleration of the initial mass (38.67e6 Kg) consumes 5.01e6 J
The acceleration of the impacted mass (2.47e6 Kg) consumes 78.32e6 J
5.01e6 + 78.32e6 = 83.33e6 J
If this is the end of conversation, so be it.
We begin with 13.87e8 J KE (m1 prior to impact)
We end with 13.04e8 J KE (m1+m2 post impact)
83.33e6 J KE difference.
You are suggesting that the 83.33e6 J is unaccounted for.
Yet we have accelerated a mass of 2.47e6 Kg from rest to 7.96m/s, and also decelerated a mass of 38.67e6 Kg from 8.47m/s to 7.96m/s.
You are suggesting that the changes in velocity are 'free' and consumed no KE in themselves. They are not free.
The deceleration of the initial mass (38.67e6 Kg) consumes 5.01e6 J
The acceleration of the impacted mass (2.47e6 Kg) consumes 78.32e6 J
5.01e6 + 78.32e6 = 83.33e6 J
If this is the end of conversation, so be it.