[geocentrism] Re: Angular momentum
- From: "Dr. Neville Jones" <ntj005@xxxxxxxxxxx>
- To: geocentrism@xxxxxxxxxxxxx
- Date: Wed, 27 Oct 2004 01:17:13 +0100 (BST)
Dear Mike,
I have tried to demonstrate that the inequality you yourself came up with
amounts to the same thing, but for the "negative" cases of particles moving
right to left. This is nothing more than a convention (in the same way that one
has to be careful to keep to convention when considering ray optics).
The labelling of the particles is totally arbitrary and either way round you
care to look at it, you are left with an inequality which cannot be supported.
Furthermore, the physics argument is perfectly straightforward. If some kinetic
energy is lost in an inelastic collision, then either some mass or some
velocity, or both, has been lost. But if it's been lost, then some linear
momentum (which is itself a product of nothing more than mass and velocity) has
also been lost. It doesn't just pop back in again, so as to save a so-called
"law."
Neville.
Mike <mboyd@xxxxxxxxxxxx> wrote:
Dear Neville,
The fact is that if you divide an inequality equation by -1 you change
the sense of the inequality.
10 + -5 > 0
-10 + 5 < 0
When you divided your inequality you *assumed* that what you were
dividing by was positive because you retained the '>'.
For any given set of values for which the rest of you analysis is valid,
whether this *assumption* is true or not depends entirely on which way
round you label the particles. So whether (u1+v1) > (u2+v2) or (u1 +
v1) < (u2 + v2) is also entirely *dependant* on your choice of labelling
contrary to your conclusion that:
"This [ (u1+v1) > (u2+v2) ] is a preposterous result, since even the
decision to label them this way round in the first instance was totally
arbitrary".
That sentence was the entirity of your demonstration that momentum
cannot be conserved in an inelastic collision.
Regards,
Mike.
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