Chronology Current Month Current Thread Current Date [Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

# [Phys-L] Re: Momentum Agina

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: <PHYS-L@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 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.
|
_______________________________________________
Phys-L mailing list
Phys-L@electron.physics.buffalo.edu
https://www.physics.buffalo.edu/mailman/listinfo/phys-l