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[Phys-L] Re: Aristotelian thinking among modern students



David T. Marx wrote:
....

The first question on the exam was the following:
A rock is thrown straight up from the Earth's surface. Which one of the following statements
concerning the net force acting on the rock at the top of its path is true?
(a) The net force is instantaneously equal to zero newtons.
(b) The net force is greater than the weight of the rock.
(c) The net force is less than the weight of the rock, but greater than zero newtons.
(d) The direction of the net force changes from up to down.
(e) The net force is equal to the weight of the rock.

Actually there is a reasonable non-Aristotelian interpretation that
favors answer (a).

It's a terminology issue. The word "acceleration" has at least two
meanings:
-- The vector acceleration is dv/dt. (This is favored in physics classes,
but there is no way a naive student could know that.)
-- The scalar acceleration is d(speed)/dt which is also equal to the
component of the vector acceleration in the direction of travel.
This meaning is very common in a host of nontechnical and semi-technical
settings. I use it myself:
http://www.av8n.com/how/htm/motion.html#sec-two-accel

Except on a set of measure zero, when a rock is thrown, the scalar acceleration
is zero at the top the parabolic arc (neglecting air friction). The exception
is for a rock thrown _straight_ up, as DTM specified, in which case the scalar
acceleration is undefined at the top of the trajectory, due to a discontinuity.
This is not one of the offered answers, but one could make a strong case that
zero is the 'principal value' and therefore answer (a) is better than the others.

You may choose to put a big poster on the classroom wall that says "in this
class acceleration always means vector acceleration" but unless/until that is
established, it seems unfair to penalize answer (a). I don't see it as a
dumb answer at all.


OTOH if the justification for answer (a) is

"The force pushing the rock up equals the force of gravity pulling it down so at the top its
acceleration is zero and the net force is zero."

then that's not good.

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

Pedagogical remarks:

-- Having two notions of acceleration is a medium-sized problem.
-- Having two notions _that go by the same name_ is a big problem. It is
a recipe for endless confusion.

One of the hardest things about learning and teaching is _unlearning_ something.
IMHO it is a pedagogical mistake to try to stamp out the notion of scalar
acceleration. Avoiding a fight is soooo much better than winning a fight.
My strategy is to say "scalar acceleration is not wrong, its just different
from vector acceleration, and here is a chart comparing and contrasting the
two notions. You will find that for physics, engineering, and other
quantitative work, the vector acceleration is more useful."

This also upholds the pedagogical principle that learning proceeds from the
known to the unknown (since scalar acceleration is familiar from driving et
cetera).