Re: Newton's second law

[Justin Parke <FIZIX29@AOL.COM>]

Conceptually I understand quite clearly the difference between sand with no initial horizontal component of velocity falling *into* a rolling train car and sand leaking *out* of a train car.  My question has to do with my apparent misunderstanding of the statement of NII and what "vdm/dt" means.

John Mallinckrodt has provided me with the most satisfying answer so far, but I welcome further input.

Basically, I do not want to tell my students something like, "this term is equal to zero just because you know it must be."  This perpetuates the idea of physics as a mysterious topic rather than a straightforward one.

In a message dated 11/6/2003 4:12:12 PM Eastern Standard Time, rlamont@POSTOFFICE.PROVIDENCE.EDU writes:

> Maddening, isn't it? :-)
>
> Think of the difference between dropping stationary sand onto the moving
> flatcar (basically the long discussion just concluded on this list about
> rain falling into a flatcar) and having that same sand slowly leaking
> out a hole at the bottom of the flatcar.
>
> In one case you are accelerating the falling sand from a horizontal
> speed of zero to a horizontal speed that is the same as the flatcar. You
> are in essence creating a momentum, v*dm, for the falling sand each time
> interval dt. If the flatcar is rolling freely and is not part of a
> train, there are no external forces. The force creating the momentum of
> the the fallen sand is counterbalanced by a force of the fallen sand
> back on the flatcar. The term m*dv/dt represents the resulting
> deceleration of the flatcar which has an instantaneous mass m (as
> someone has already pointed out.) The sum of the two terms is zero.
>
> In the second case, the sand leaks out, but keeps its horizontal speed
> v. There is no horizontal creation or destruction of momentum for the dm
> lost in each time interval, so the v*dm/dt term is not included when
> evaluating dmv/dt.
>
> The key here is that you are looking at the rate of change of momentum.
> If dm has an associated momentum change then use the v*dm/dt term,
> otherwise not.
>
> Bob at PC
>
> Justin Parke wrote:
> > In a message dated 11/6/2003 3:08:09 PM Eastern Standard Time, rlamont@POSTOFFICE.PROVIDENCE.EDU writes:
> >
> > > The v in
> > > this case is the sideways speed of the sand, not the speed of the
> > > flatcar down the track - which remains constant because
> > > there is no
> > > force in this direction.
> > >
> > > Bob at PC
> >
> > How do we know that "in this case" v is the sideways speed of the sand?  Sorry to sound so stupid...just trying to
> completely straighten out my thinking on this.
> > Justin Parke
> > Oakland Mills High School
> > Columbia, MD

Justin Parke
Oakland Mills High School
Columbia, MD