March: I`ve got two worlds on a string

Physics Challenge for
Teachers and Students
Boris Korsunsky, Column Editor
Weston High School, Weston, MA 02493
[email protected]
Solution to March 2015 Challenge
w I’ve got two worlds on a string
Two small metal spheres are connected by a light flexible
metal wire of length l. The spheres have equal radii r (r<<l)
but different masses: m and 2m. Initially, the spheres are held
at rest on a large insulating frictionless table at a distance l/2
from each other in a uniform electrostatic field of magnitude
E, directed along the straight line connecting the spheres.
The spheres are released simultaneously. Find the maximum
speed v attained by the sphere of mass m.
Answer:
Since the electric field is uniform, the direction it points
along the line joining the spheres is irrelevant. I will arbitrarily draw the field pointing from the smaller mass
toward the larger mass. The electric field will cause a
charge separation such that the larger mass has charge
+q and the smaller mass has charge –q. Making the assumption that the masses of the spheres are orders of
magnitude greater than the mobile electrons, we can say
that the charges are at all times in equilibrium. When
the spheres are a distance s apart, the potentials due to
the separation of the charges in the field and the potentials due to the charge concentration on the spheres
must balance.
2s/3
E
m
-q
x
s/3
the magnitude
(2)
The force on the charged sphere is
(3)
The force that the separated charges exert on each other
is ignored. It is small compared to the force caused by the
external field (this is ensured by the geometry, r << l. Yes, I
did check).
Relating the kinetic energy of the sphere to the work done
on the charge results in
(4)
Since x = 2s/3, Eq. (4) can be rewritten as
(5)
Evaluation of the integral gives
2m
+q
(6)
After substitution of the limits and rearranging terms, Eq.
(6) gives the value of v as
(7)
(Submitted by Don Easton, Lacombe, Alberta, Canada)
In the diagram, the position of the smaller mass is
marked as x, the distance from the center of mass. This
is to remind us that the distance the smaller mass moves
must always be twice what the larger mass moves. The
balancing of electric potentials requires that
(1)
where k is the Coulomb constant.
From Eq. (1), we see that the charge on each sphere has
The Physics Teacher ◆ Vol. 53, 2015
We also recognize the following successful contributors:
Gerald E. Hite (TAMUG, Galveston, TX)
Carl E. Mungan (U. S. Naval Academy, Annapolis,
MD)
Pascal Renault (John Tyler Community College,
Midlothian, VA)
Joseph Rizcallah (School of Education, Lebanese
University, Beirut, Lebanon)
Many thanks to all contributors and we hope to hear
from many more of you in the future!
Guidelines for contributors:
– We ask that all solutions, preferably in Word format,
be submitted to the dedicated email address
[email protected] Each message will receive an
automatic acknowledgment.
– The subject line of each message should be the same
as the name of the solution file.
– The deadline for submitting the solutions is the last
day of the corresponding month.
– Each month, a representative selection of the successful solvers’ names will be published in print and
on the web. – If your name is—for instance—Brian May, please
name the file “May15May” (do not include your
first initial) when submitting the solution.
– If you have a message for the Column Editor, you
may contact him at [email protected];
however, please do not send your solutions to this
address.
As always, we look forward to your contributions
and hope that they will include not only solutions
but also your own Challenges that you wish to submit for the column. Boris Korsunsky, Column Editor
The Physics Teacher
◆ Vol. 53, 2015
`