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Re: [Phys-L] Physics, Errors and Differenet Teaching Styles



-----Original Message-----
From: phys-l-bounces@mail.phys-l.org
[mailto:phys-l-bounces@mail.phys-l.org] On Behalf Of William Maddox
Sent: Friday, June 22, 2012 12:59 PM
To: phys-l@phys-l.org
Subject: [Phys-L] Physics, Errors and Differenet Teaching Styles

From: W.C. Maddox

Topic: Physics, Mistakes, Different Teaching Styles

Warning: This message may pose a quandary for fans of Mazur
in particular and PER in general but who do not like multiple
choice questions. You may wish to stop at this point.

These are being considered for use here as a "clicker
question"/demo combination (IDL). The questions below came
from a book by Eric Mazur called Peer Instruction. They came
from a collection of "concept test" questions for use in
introductory physics but are variations of common textbook
questions. The three questions are variations of the same
idea. Before giving the book's answer and some experimental
(demonstration) results, I would like to get your answers.

#1 " A lead weight is fastened on top of a large solid piece
of Styrofoam that floats in a container of water. Because of
the weight of the lead , the water level is now flush with
the top of the surface of the Styrofoam. If the piece of
Styrofoam is turned upside down so that the weight is now
suspended underneath it,

1. The arrangement sinks.

2. The water level is now below the top surface of the
Styrofoam.

3. The water level is still flush with the top surface of the
Styrofoam."


2. The water level is now below the top surface of the styrofoam. The
volume of that amount of water that would fit in that region below the
surface of the water that is occupied by stuff other than water hasn't
changed, it is that volume of water on which the gravitational force
would be equal to the gravitational force acting on the lead+Styrofoam
object (because the buoyant force has to be equal in magnitude to the
gravitational force acting on the lead+Styrofoam object in order for it
to be floating at rest). But now, the lead is occupying some space
below the surface of the water so for the total volume of water that
would fit in the space below the surface occupied by other stuff to be
the same the Styrofoam has to be occupying less space below the surface
of the water so the water level relative to the Styrofom is lower than
it was before.

#2 " A lead weight is fastened to a large solid piece of
Styrofoam that
floats in a container of water. Because of the weight of the
lead , the water level is now flush with the top of the
surface of the Styrofoam. If the piece of Styrofoam is turned
upside down so that the weight is now suspended underneath
it, the water level in the container

1) rises

2) drops

3) remains the same "

3) It remains the same in that the volume of water displaced is the
same, namely that volume that would experience a gravitational force
equal to the gravitational force acting on the lead+Styrofoam object.


#3. "A boat carrying a large rock Is floating on a lake. The
boulder is thrown overboard and sinks. The water level in the
lake (with respect to the
shore)

1) rises

2) drops

3) remains same"
2) The water level in the lake, relative to its banks, drops. (I am
assuming that the density of the rock is greater than that of the water
so the rock sinks to and comes to rest on the bottom where, in
equilibrium the bottom is exerting a non-zero normal force upward on the
rock--e.g. we are not talking about a pummice rock that might sink down
below the surface just after being dropped but would then bob back up to
the surface.) The gravitational force that would act on that amount of
water that would fit in the space occupied by the boat+rock below the
surface of the water (let's call this the displaced water) is now less
than the gravitational force on the boat+rock because the buoyant force
is no longer equal in magnitude to the gravitational force on the
boat+rock. Rather, the magnitude of the net upward force which is equal
to the magnitude of the total buoyant force on the boat+rock plus the
magnitude of the normal force exerted on the rock by the bottom of the
lake minus the magnitude of the gravitational force on the boat+rock is
zero, so the magnitude of the buoyant force on the boat+rock is equal to
the magnitude of the gravitational force on the boat+rock minus the
magnitude of the normal force exerted on the rock by the bottom of the
lake. Now if the gravitational force that would act on the displaced
water is less, that means the volume of the displaced water is less.
The net change is equivalent to taking a thin layer of water from the
top of the water in the lake and using it to fill in some of the
original displacement volume, just enough to reduce the displacement
volume to the new value. The new water level in the lake is reduced by
the thickness of that layer.

End Message

PS Remember some snarks may be boojums. If you are having a
really slow day this summer you may want to try answering
this non-physics question. One of these does not fit with the
others: Batt, Connolly, Daltrey, Howard, Hurt, Lennon,
Sensible, Ure, Williams. What could he have been called so
that he would fit in?





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