# The Three Stooges

```Physics Challenge for
Teachers and Students
Boris Korsunsky, Column Editor
Weston High School, Weston, MA 02493
[email protected]
Solution to March 2010 Challenge
w The Three Stooges
Two small spheres are attached to the ends of a long light nonconducting rod. A third, “middle” sphere can slide along the rod
without friction as shown. All three spheres are nonconducting,
have identical masses m, and a positive charge q is distributed evenly on the surface of each sphere. The whole system
is placed on a horizontal frictionless nonconducting surface.
Initially, all three spheres are at rest and the “middle” sphere
is located a distance 3d from one of the ends of the rod and a
distance d from the other. Find the maximum speed v of the
“middle” sphere after the system is released.
3d
d
Solution:
The solutions to the March Challenge, The Three
Stooges, came from all over the world. It was great to
see quite a few students among the solvers—although
not all of the contributors were successful. The most
common error was not recognizing the kinetic energy of
the “end” spheres. Here is one of the numerous correct
solutions:
“after” pictures and taking right as positive and left as
negative, conservation of momentum gives:
∑ pb = ∑ pa
0 = 2 mv2 − mv1 ,which gives
v2 =
v1
.
2
The maximum velocity of the middle charge will occur
when the middle charge is at…well, the middle. Beyond
the middle point there will be a net force to the right
that will tend to slow the charge down. The situation
will be the same for the middle charge even if the two
end charges are moving.
Applying the law of conservation of energy to the system and using the middle charge position of maximum
velocity as the “after” position, one gets:
Eb = Ea
PEb = KE + PEa
kq 2 kq 2 1 2 1
2 kq 2
+
= mv1 + (2 m )v22 +
.
3d
d
2
2
2d
Now using the conservation of momentum result above
for v2 and simplifying:
v2
kq 2 1 2
= mv1 + m 1 , upon further simplification,
3d 2
4
2
2
mv
3
kq
1
, finally solving for the maximum velocity:
=
3d
4
v1 =
2q k
.
3 md
(Contributed by Dave Bittel, Bristol Eastern High School, Bristol, CT)
Since the charge is evenly distributed, the charges can
be treated as point charges. Looking at the “before” and
The Physics Teacher ◆ Vol. 48, 2010
We are also pleased to recognize the following contributors:
Bruce Barnett (Johns Hopkins University, Baltimore, MD)
R. R. Bukrey (retired, Loyola University, Evanston, IL)
Phil Cahill (Lockheed Martin Corporation, North
Yorkshire, United Kingdom)
Dominic Calabrese (Sierra College, Rocklin, CA)
Daniel Cartin (Naval Academy Preparatory School,
Middetown, RI)
Craig Caylor (Westminster College, New Wilmington, PA)
R. C. Dhandhania (KalraShukla, Mumbai, India)
Hasan Fakhruddin (The Indiana Academy for Science,
Mathematics, and Humanities, Ball State University,
Muncie, IN)
Michael C. Faleski (Delta College, Midland, MI
Fernando Ferreira (Universidade da Beira Interior,
Covilh&atilde;, Portugal)
Art Hovey (retired, Milford, CT)
Mark Lenfestey (Homestead High School, Fort Wayne, IN)
John Mallinckrodt (Cal Poly Pomona, Pomona, CA)
Jeff Melmed (Eastern Maine Community College, Bangor,
ME)
Daniel Mixson (Naval Academy Preparatory School,
Newport, RI)
Eugene P. Mosca (U.S. Naval Academy, Annapolis, MD)
Carl E. Mungan (U. S. Naval Academy, Annapolis, MD)
Clark M. Neily, Jr. (Hermon’s of Alaska Christian School,
Allston, MA)
Frank Noschese (John Jay High School, Cross River, NY)
Pascal Renault (John Tyler Community College, Chester,
VA)
Jason L. Smith (Richland Community College, Decatur, IL)
Clint Sprott (University of Wisconsin – Madison, WI)
Ryan Yancey (Lancaster High School, Lancaster, CA)
Many thanks to all contributors and we hope to hear from
you in the future!