T.R | Title | User | Personal Name | Date | Lines |
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902.1 | Binding Energy | HACKIN::YOUNG | Really DWOVAX::YOUNG | Sat Jul 16 1988 14:57 | 22 |
| The indestructable, white body particles/objects you describe would
not 'stick' together after the collision.
The 'lost' energy of a 'perfectly inelastic' collision goes into
holding the 2 objects together. You see, their individual energy
and momentum exchanges result in different velocity vectors for
the two objects. Which is to say that they want to go to different
places. If you want them to go to the same place, you have to expend
energy to keep them together.
In the microscopic world, what happens to this energy is tied up
in quantum mechanical explanantions, wave equations, "binding energy"
etc. In the macroscopic world this energy is tied up in deforming
the two objects into one mass and intermixinbg their component
particles so that they will 'stick' together (see micrscopic world
above).
If you have objects made up of some theoretical particles that cannot
somehow absorb the energy, then they will not stick together, or
become one object.
-- Barry
|
902.2 | Yes, but..... | GIDDAY::ALLEN | | Sun Jul 17 1988 01:25 | 35 |
|
Thanks for the reply, Barry.
Yes, I can appreciate that things are different in the microscopic world,
quantum physics then takes over.
My experiment takes place in the macroscopic world. I used the word 'particle'
for want of a better word, but really they could be of masses measured in
tonnes. To get the 'objects' to bind together I envisaged use of a frictionless
docking system- with no means of vibration/oscilation of the combined objects
through the mechanism after impact.
I understand what you say about energy required to prevent the objects from
going in the directions they would go in if the collision were elastic. But
surely this is a fixed amount of energy required to do this work? If this is
so and it accounts for all the 'lost' energy in the collision (because
there is no means for energy absorbtion and dissipation in my setup), then
what if we change the experiment so that we CAN absorp and dissipate
energy - more like a real experiment. Now I have the particles surrounded by
a sticky substance that allow them to stick together on impact, the particles
have an average heat capacity so they can dissipate heat sound etc.
The law of conservation of momentum states that the final velocity of the
combined mass will be the same as that in my first experiment. However now
I will definitely be losing energy to the system (and the universe in the form
of heat). But if I still need the same energy to prevent the objects from
travelling in the directions they would otherwise go, where does this
heat energy, etc, come from? It seems to me that depending on the physical
setup different amounts of energy become involved.
I must be missing something here?
Any further explanations would be welcome. Thanks.
Mike
|
902.3 | Its still in there... | HACKIN::YOUNG | Really DWOVAX::YOUNG | Sun Jul 17 1988 14:14 | 51 |
| What I was trying to say was:
1) Some macrosopic effects that seem 'invisible' are so because
they have been translated into microscopic effects that are
governed by quantum mechanics. Such things are heat, sound,
strength of material bonding, etc.
2) If you postulate 2 objects that have no way to absorb and
/or dissipate the energy to be lost in the exchange then
they CANNOT stick together.
Lets take your example of a 'linking' or 'docking' mechanism between
2 large objects that cannot dissipate the lost energy of the exchange.
Lets say that this docking mechanism is just 2 large hooks made
of some super-strong material that will link together to hold the
2 objects together. Now when the collision (and link-up) occur
there will be a huge amount of momentum and energy exchange. But
because the two objects are now hooked into one, they will move
as one object. However, the excess energy will try to pull the
two objects apart, so that they will be straining against the hooks
holding them together.
Now, by definition, the hooks are strong enough to resist, so they
still stay together. Now the two objects will rebound against their
constraints and collide (or just push if there is no 'play' in the
hooks) into each other. They will exchange energy and momenta
again and pull away against the hooks again. This cycle will repeat
endlessly. All the while the 2 objects are rebounding against each
other like this, the system as a whole will be moving along in space
with the combined mass and momentum of the 2 objects originaly,
and with an 'apparrent' loss of energy.
The loss of energy is on apparent however, because the energy is
contained in the constant oscillation or rebounding of the two original
objects. Now you might think that this oscillation should dampen
and eventually just dissapear. This normally happens because the
energy is being dissapted by other activities (sound, friction,
heat, etc.) that either radiates the energy off into other objects
or else internalizes it at the microscopic level.
Instead what we have is a system where the excess energy has been
internalized at the macroscopic level. Because all of that oscillation
is just a macroscopic form of heat. But the energy IS STILL THERE.
If you where to go in at some later point and release the hooks,
the two objects would suddenly fly apart with the velocityies that
would have been predicted for the original collision if it where
perfectly elastic. IE. the proper momentum AND energy would be
imparted to both as they flew apart.
-- Barry
|
902.4 | a question | ZFC::DERAMO | Supersedes all previous personal names. | Sun Jul 17 1988 17:19 | 13 |
| If you redo your calculations in a different frame of
reference, moving with constant velocity with respect
to your "stationary" mass B, you get different values
for the kinetic energies of the two. You find the "after"
velocities by applying conservation of momentum in any
such frame of reference. Then you calculate the "after"
kinetic energies from those velocities, and in general
get different answers in different frames.
Question: Is the ratio of kinetic energy after to kinetic
energy before unchanged between reference frames?
Dan
|
902.5 | This ain't hard or do I miss something? | HPSTEK::XIA | | Sun Jul 17 1988 18:48 | 15 |
| re .4
In general, no. Counterexample:
Suppose we have two balls of mass m, and made a plastic collision.
Suppose we are doing it in R, and suppose that the initial velocities
are v1 and v2, and the final velocity is v3 in an enertia frame
F1. Then before the collision the kinetic energy is
1/2*m*(v1^2 + v2^2). After the collision, it is m*v3^2. Hence,
the ratio is 2*v3^2/(v1^2+v2^2). Now suppose
an arbitrary enertia frame F is moving with respect to F1 with
velociy v0. Then the new ratio is 2*(v3-v0)^2/((v1-v0)^2+(v2-v0)^2).
There is no way this mess gonna be a constant when you change v0.
As you can see, you do not even need conservation of momentum and
energy to do that.
Eugene
|
902.6 | you may stick a star on your forehead | ZFC::DERAMO | Supersedes all previous personal names. | Mon Jul 18 1988 11:51 | 6 |
| Re: .4, .5:
No, you didn't miss anything, it just wasn't difficult, that's
all. :-)
Dan
|
902.7 | heat, I think | EAGLE1::BEST | R D Best, sys arch, I/O | Mon Jul 18 1988 12:08 | 43 |
| >
>This loss of energy, according to physics text books, is given up as internal
>energy to the system - heat, sound etc. This is understandable in the
>situation described in one book - where particle A is a bullet, and B is
>a block of wood hanging on the end of a piece of string. Depending on the
>height achieved by the block of wood after the bullet has impacted with it, the
>velocity of the bullet can be calculated. The internal energy aquired by the
>system is the damage to the block of wood - heat, sound, energy required to
>tear wood fibres etc. etc.
In this case, the energy is probably converted to heat; the temperature of
the mass rises initially. As time passes, the temperature will drop until
the rate at which heat is radiated is equal to the rate at which it is
absorbed from the environment.
>
>In my experimental setup, however, I have decided to use indestructible
>particles. They are perfect white bodies (can't radiate heat) they have zero
>heat capacity, it occurs in outer-space (no sound generated). I.E I have
^
You may have decided this, but I suspect nature won't allow it. It's always
possible to over-constrain a problem by positing too many requirements to
get a solution. But nature isn't fooled. It KNOWs there's an answer
(it's the behavior you observe).
If both initial objects are white bodies, then when they merge, they can't
reach thermal equilibrium. I think this violates one of the laws of
thermodynamics.
>decided that they will have no medium, through which to absorb the internal
>energy or in which to dissipate it. Possibly a physical impossibity, but
>THEORETICALLY possible, surely?
You can postulate any set of ideal conditions you want about some physical
situation, but you will only get results that correlate to real world
observations if the things you postulate are realistic approximations to the
physical situation, and the set of conditions you postulate are
self-consistent.
>So, in this case, where does the lost energy go?
I would say heat (but you have to throw out the notion of perfect white
bodies).
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