[Phys-L] Re: Momentum Agina

[John Denker <jsd@AV8N.COM>]

Bernard Cleyet wrote:

> > Very simply KE is dissipated until both are moving at the
> > same speed whether it's thru gross inelasticity (one ball is
> > sticky putty) or it's plastic (or not) deformation of a steel
> > railroad car coupling, or whether it's completed slowly or quickly.

Robert Cohen wrote:

> This is the key and I don't think it is obvious.  KE is dissipated until
> it is no longer necessary to dissipate it.

Perhaps the following will make this idea easier to understand ... and in
particular make it easier to accept that the result is independent of
mechanism.

We start by invoking the often-useful theorem that for a compound
system, the KE in the lab frame is equal to
  a) the KE in the center-of-mass frame, i.e. the KE of the components
   relative to the CM
  b) plus the black-box KE in the lab frame, i.e. .5 M V^2 where M is
   the total mass of the system and V is the center-of-mass velocity.
   (The name comes from the idea that you know the total mass of the
   black box and its overall velocity, but you can't look inside the
   black box to see whether there is any relative motion of the
   internal components.)

The statement of the problem is that the components collide and *stick*.
Sticking, I assume, means that contribution (a) goes to zero.  In the
CM frame, the KE does not decrease by half; it decreases by 100%.

This is independent of mechanism, because it involves nothing but the
definition of "sticking".  Perhaps it is easier to explain that total
loss (in the CM frame) is independent of mechanism, easier than
explaining half-loss (in the lab frame).

To finish the analysis, all you need to do is convince yourself that
the initial KE in the CM frame is half of the initial KE in the lab
frame.  That's true under the stated conditions (two objects of equal
mass, one initially at rest).

As I previously explained, this /half/ comes from the fact that the
energy/momentum relationship is a /second/-order equation.  In the
CM frame, each object has half the momentum.
   Lab frame:  KE =  1^2 +  0^2
   CM frame:   KE = .5^2 + .5^2 = half as much as in the lab frame.
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