Re: [Phys-l] momentum and energy

["John Ertel" <jpe@usna.edu>]

I would posit that while I was never taught the "Conservation of Kinetic
Energy" principle and indeed some books go out of their way to deride the
use of the expression "Conservation of Kinetic Energy" relating to any of
the conservation principles, when there is no change in potential energy and
no losses in/by the system, Kinetic Energy must stay constant.
 
Additionally, one illuminating way of looking at the losses in a two-body
collision is to monitor the Kinetic Energy as seen within the CoM reference
frame (as compared to the Lab Reference Frame.
 
As an example, for those with access to Interactive PhysicsC (in any of its
versions), just try a collinear two-body collision between equal masses.
Run the simulation monitoring KE (both System and CoM parts) making the
elasticities exactly ONE(1).
 
Now, re-run the simulation adjusting the elasticity and monitor KE's.
 
Finally, rerun the simulation making the elasticities exactly ZERO(0) and
monitor KE's.
 
My students claim this really gives them a clearer view of what the words
"Totally Elastic" and "Totally Inelastic" mean.
 
Of course, for your Physics Majors, this works even better when you show
them what happens in the CoM Reference Frame when the masses are not equal
and possibly very different.
 
Try it, you'll like it!
 
Keep Fizzing,
Just John.
 
+=================================+
 
-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu
[mailto:phys-l-bounces@carnot.physics.buffalo.edu] On Behalf Of Bernard
Cleyet
Sent: Monday, November 13, 2006 11:19 AM
To: Forum for Physics Educators
Subject: Re: [Phys-l] momentum and energy
 
"... note that you do not need the elasticity condition if you conserve 
kinetic energy
                   regards,
                        Jack"
 
 
Are they not analytically the same?
 
 
bc
 
 
 
Jack Uretsky wrote:
 
> Hi-
>    consider a 1-dimensional elastic collision between a mass m and a 
> mass M, M>m, M at rest before the collision, so that m reverses direction 
> as a result of the collision.  Conserve energy and momentum.  How much of 
> the energy does M acquire as M become very, very, very large?
>        note that you do not need the elasticity condition if you conserve 
> kinetic energy
>                  regards,
>                       Jack
>
>
>
>
>
> On Sun, 12 Nov 2006, Spencer, Rob wrote:
>
>  
>
> > Greetings,
> > I am a high school teacher and I have a question regarding an assumed
elastic collision.  Suppose a golf ball strikes a massive wall and recoils
with a velocity equal in magnitude to its initial velocity.  If one were to
evaluate the situation with conservation of energy, then energy is conserved
(defined to be perfectly elastic).  All of the (kinetic) energy remains in
the golf ball.  None is transferred to the wall.  However, the wall must
acquire momentum due to the fact the system momentum must be conserved.  So
the wall has final momentum when it initially had none but has no kinetic
energy.  I am having a hard time with an object having momentum but no
energy conceptually...any guidance would be appreciated.
> > Regards,
> > Rob Spencer
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