JD wrote:
 . . .
 The statement of the problem is that the components collide and *stick*.
 Sticking, I assume, means that contribution (a) goes to zero. . . .
No. The objects can stick together and have the KE transferred to internal
degrees of freedom (contribution a), such as a spinning part. Your assumed
constraint is not necessarily so  neither is it required for the pursuit of
your conclusion, which is still valid.
Bob Sciamanda
Physics, Edinboro Univ of PA (Em) http://www.winbeam.com/~trebor/
trebor@winbeam.com
 Original Message 
From: "John Denker" <jsd@AV8N.COM>
To: <PHYSL@LISTS.NAU.EDU>
Sent: Thursday, December 08, 2005 10:36 AM
Subject: Re: Momentum Agina
 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 oftenuseful theorem that for a compound
 system, the KE in the lab frame is equal to
 a) the KE in the centerofmass frame, i.e. the KE of the components
 relative to the CM
 b) plus the blackbox KE in the lab frame, i.e. .5 M V^2 where M is
 the total mass of the system and V is the centerofmass 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 halfloss (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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