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Re: momentum



At 21:28 7/1/99 -0700, you wrote:
The nearest equivalence I can think of is a head on collision
between two cars moving at 35 and a head on collision between
two cars, one of which is moving at 70 and the other at rest
*with brakes off*. You may supply whatever velocity units you
like above.

Herb's comment that energy is conserved is either trivial and
potentially confusing, or else it is wrong. Such statements
can only distract students from a correct line of reasoning,
and they already have enough distractions from their raging
hormones so that we need not provide gratuitous distractions.

Leigh


Bearing in mind Leigh's dictum that the truth that confuses must
be concealed, I proceed like this:

The car at 35 mph of mass x kg has speed of 1 in my speed units
and mass one in my mass units.
The car at 70 mpg has speed 2 and mass one, the stationary car
has speed 0, mass 1

Momentum before:
car1 1 car2 -1
fast car1 2 slow car2 0

KE before:
car1 0.5 car2 0.5
fast car1 2 slow car2 0


Momentum after:
car1 0 car2 0
fast car1 1 slow car2 1

KE After:
fast car1 0 slow car2 0

Loss of momentum:
car1 + car2 (before) - (car1 + car2 after) = 0
fast car1 + slow car2 - (fastcar1 + slow car2) = 0

Loss of Energy:
car1 + car2(before) - (car1 + car2 after) = 1
fast car1 + slow car2 - (fast car1 + slow car2) = 2

I conclude that here, momentum is conserved, and energy is not
conserved. Energy is expended in cold forming the vehicle
structures to novel shapes with some structural heating.
This is the general outcome of an inelastic collision.

I see that the delta KE is greater for the fast vehicle.

(Desperately trying to regain my self respect after wrestling with an
inelastic collision question in Science100 a while back)

brian whatcott <inet@intellisys.net>
Altus OK