Re: [Phys-L] elastic collisions

[Chuck Britton <britton@ncssm.edu>]

Back in the ‘60s my freshman physics book was written by Hazen and Pidd (sp?) and contained one equation/comment that stuck with me for some reason.

They showed/claimed that for an elastic collision of two objects we find that the Velocity at which they approach each other will always equal (but opposite)  the velocity at which they separate from each other.

V(app) = -V(sep)

So we used this when we taught Momentum First and could do elastic as well as inelastic collisions without needing KE.

(Using the CM, which we had already covered in Static cases.)


> On Dec 4, 2016, at 8:48 PM, Jeffrey Schnick <JSchnick@Anselm.Edu> wrote:
>
> Energy remains energy.  Some of the kinetic energy of translation is transformed into kinetic energy of rotation.  The linear momentum of the system after the collision is still the same as the linear momentum of the system before the collision.  The angular momentum with respect to any fixed axis is the same after the collision as it is before the collision too.
>
> > -----Original Message-----
> > From: Phys-l [mailto:phys-l-bounces@mail.phys-l.org] On Behalf Of bernard
> > cleyet
> > Sent: Saturday, December 03, 2016 9:39 PM
> > To: Phys-L@Phys-L.org
> > Subject: Re: [Phys-L] elastic collisions
> >
> > My confusion is thinking some of the energy of the incident object is
> > transferred into angular momentum.   Linear momentum is still conserved?
> >
> >
> > The animation works well, but is of two point masses, so no spin.
> >
> >
> > bc
>
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