[Phys-L] Re: collision question (revisited) (again)

[Hugh Haskell <hhaskell@MINDSPRING.COM>]

At 8:45 -0800 2/27/05, John Barrer wrote:
> Perhaps if I throw in some numbers my question will be
> better stated. All vector quantities are expressed in
> the lab frame. Let's assume the inelastic collision
> occurs in a vacuum chamber so there are no air
> molecules present to carry away some of the system KE.
> The system consists of two carts (a=.25kg, b=1.0kg) on
> a "frictionless" surface. At time
> 0-delta t, Va=0.5m/s, Vb=0.0m/s so that before the
> collision, system KE=.03125J and system p=.125kgm/s
> with 100% of both "resident" in the CoM motion of cart
> a. At time 0+delta t, Va=Vb=0.1m/s which gives the
> following:
> Cart a: p=.025kgm/s, KE=.00125J
> Cart b: p=.1kgm/s, KE=.005J
> Conservation of p, 80% dissipation of CoM KE.
>
> My question boils down to this: In the time interval
> from 0-delta t to 0+delta t, how were the original
> system momentum and system KE distributed?
Someone may have mentioned this before, but I'm going to mention it
again, because I think it is the best way to address the question you
have raised.

To start with, put yourself in the CofM reference frame, where, by
definition the momentum is always zero, but the kinetic energy can be
most anything (this because KE is always positive, and since both
objects can be moving, will not add to zero, unless both are at
rest). No matter what happens in the collision, the total momentum
will remain zero, even though the momentum of any of the parts
involved in the collision will not necessarily be zero. This is
required by the definition of the center of mass, and will be true
irrespective of the type of collision. If one object in the collision
suffers a momentum change of delta-p, then the other object must see
a change in its momentum of -delta-p. If one object has its outgoing
speed changed by a factor epsilon, then the other object has its
outgoing speed changed by the same factor epsilon. The process is a
bit more complicated if there is mass transfer between the two
objects during their collision, but the net momentum change during
the collision must be zero.

As JD has pointed out, what happens between the various 10^23 or so
atoms during the collision is much more complex, and difficult to
figure out. But until the time intervals you want to look at become
comparable to the time it takes to transmit the interaction signals
between the various individual particles, it is probably safe to
treat those interactions as the same as those between the two objects
as a whole, and not worry about the short time scale violations of
newton's third law. Those time scales will be typically much shorter
than the time scale for the interaction as a whole.

Now that what is going on in the CofM frame is settled, we can move
to an inertial frame moving WRT the CofM frame. The observed momentum
will be different, of course. Now you will see the momentum
associated with that of the motion of the CofM relative to your new
point of view. The energy will now also be different, but in a more
complex way, since we can now discuss the kinetic energy *of* the
CofM and the KE *relative to* the CofM (which is the same as the
energy measured while *in* the CofM.

Now when the collision occurs the total momentum cannot change, since
it is just the momentum of the CofM, which we saw is due entirely to
the motion of the objects from the POV of the observers frame. Since
we have tacitly assumed that the collision involves no interaction
with the surroundings, it's CofM motion cannot change. Since the
motion of the CofM doesn't change the total momentum, as seen by the
observer cannot change.

Similarly, the KE associated with the CofM motion also will not
change, but what can change is the KE relative to the CofM. Some of
it can be converted into thermal motion of the atoms and molecules,
some may be absorbed by the atoms and converted to internal energy of
the atoms and molecules, and some may be radiated away in various
forms of electromagnetic or acoustic radiation. But any such actions
will have to conserve the initial momentum in the objects involved in
these microscopic interactions, because those atoms can be considered
as collisions similar to the overall collision(however, with the
added complication that the atoms and molecules are interacting with
an external force as well as each other--an external force which is
internal to the interacting system as a whole, however, and thus one
whose effect must be such that when summed over the entire system,
its effects add to zero).

In short, if we can find a frame of reference such that the total
momentum must remain unchanged irrespective of the complexity of the
collision (even though the kinetic energy *can* change), then we can
say that (at least to the extent that the Galilean transformation is
valid) no momentum is lost to internal processes, and overall
momentum is conserved as seen from any inertial reference frame.
although momentum may be redistributed within an object, that
redistribution remains visible as external momentum of the center of
mass. Since the center of mass reference frame is one such frame, I
think the proposition is established.

I'm sure this is also true under Lorentz transformations, but I have
never worked that one out.

Hugh
--

Hugh Haskell
<mailto:haskell@ncssm.edu>
<mailto:hhaskell@mindspring.com>

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