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Re: car acceleration



John Barrer wrote:

I'm uncomfortable with the "momentum transfer but no
energy transfer" model described in earlier posts.

I still think it's a good description of what's going
on. Indeed I think that the car-acceleration problem
is confusing precisely because students don't clearly
appreciate the distinction between energy and momentum,
and can't figure out which one the problem is asking
about.

Consider a similar situation, the hypothetical
perfectly elastic collision between a fall ing ball
and the floor. Model the floor as an ideal spring.
During the first half of the interaction, as the
"floor spring" is being compressed, the floor does
"negative work" on the ball, thereby reducing both its
KE and momentum to zero (in the floor ref frame).
During the second half of the interaction, as the
"floor spring" is recovering it is doing "positive
work" on the ball, increasing both its KE and momentum
to their original absolute values.

Even though this is not a particularly close
analogy to the car-acceleration problem, it
actually does serve as a good illustration of
the point I was making about energy versus
momentum. When you consider the two halves
of the interaction together, the net effect on
the energy is very small. It's certainly not
positive; it's slightly less than zero if we
account for dissipation. Meanwhile ..... The
contribution to the momentum is large and positive
during both halves of the interaction.

To repeat: the trampoline contributes nothing
(or slightly less than nothing) to the net energy,
but contributes hugely to the net momentum.

So, suppose you want to include road-deformation
in the analysis of the car-acceleration problem
(which I do _not_ recommend). The question still
must be broken down into two pieces:
-- what does road-deformation contribute to
the momentum budget?
-- what does road-deformation contribute to
the energy budget?

These are two different questions! They have
different answers (unless you decide that they're
both negligible).

======================

Are people shocked by the notion that there
can be a large momentum transfer without a
large energy transfer? They'd better get used
to it!

In general, the energy E associated with momentum
p is:
E = p^2 / (2m)
and in the case where m is the mass of the earth,
E is very very small. (If m is the mass of Scott's
cardboard on rollers, the story changes a bit!)

================

Carpenters deal with wood.
Dairy farmers deal with milk.
Physicists deal with energy and momentum.