Physics Challenges for
Teachers and Students
Solutions to January Challenges
◗ Out in the Field
Challenge: Two charged particles (M, +Q) and (m,
–q) are placed in a uniform electric field E. After the
particles are released, they stay at a constant
distance from each other. What is this distance (L)?
Substituting Eq. (2) into Eq. (1) yields
kQ |q|
(Q –q)E
ᎏ
ᎏ .
QE – ᎏᎏ
=
M
L2
(M + m)
冦
冧
(3)
Solving Eq. (3) for L gives the desired result:
Solution:
+Q, M
•
–q, m
L
L=
•
→ → → →E → → → →
→→→→→→→→
In order to maintain a constant separation, the
particles must have the same acceleration. To
find that acceleration, let us analyze the forces
acting on the particles. Consider the forces
acting on particle Q: the leftward force due to
E-field acting on Q equals QE; the rightward
force due to particle q acting on Q equals
kQ |q| L2.
Assume that Q > |q|; then the system of both
charges will accelerate to the left. Newton’s
second-law equation of motion for particle Q is
then
QE – kQ |q|/L2 = Ma.
(1)
Considering the system of both charges
immersed in an E-field and accelerating
together once they are the equilibrium distance
L apart, we find their common acceleration to
be
F et
(Q – q)E
= ᎏᎏ.
a = ᎏnᎏ
total mass (M + m)
THE PHYSICS TEACHER ◆ Vol. 41, April 2003
(2)
(M + m)kQq
冪ᎏ
莦E莦(q莦M莦+ᎏ
莦Q莦m莦) .
The same result can be obtained when charge
(–q, m) is analyzed and, of course, it makes no
difference which charge is assumed to have the
greater absolute value.
(Contributed by Kurt Fisher, Dowling College,
Oakdale, NY)
◗ In Hot Water
Challenge: Two identical light metal containers are
filled with equal amounts of water and placed in a
room with constant air temperature. A heavy ball is
submerged into the center of one of the containers
on a thin nonconducting string. The mass of the ball
equals to the mass of the water, and the density of
the ball is much greater than that of water.
Both containers are heated to the boiling point of
water and are then allowed to cool. The container
with the ball in it takes k times longer to cool down
to the room temperature than the container without
the ball. Find the specific heat of the material of the
ball c b in terms of k and the specific heat of water
cw.
Solution: The equation that relates the rate of
heat flow to the temperature difference is
dQ/dt = ␣(T – TR),
where T is the current temperature of the
A-1
container’s surface, TR is the room temperature, and the constant ␣ depends on the
conductivity of the materials and the geometry
of the container. The containers are identical so
they should have the same constant ␣. After all,
each container is made of metal and conducts
heat well, minimizing the discrepancy that
results from the greater surface area of the
slightly higher water level in the container with
the submerged ball.
tension force is approximately constant:
T = mg. The elongation of the spring is then
L = mg /k and its total length is 2L = 2 mg/k.
The linear mass density for the stretched spring
is then ␮ = m/(2L) = k/(2g). The speed of
苶苶
/␮ = g兹2苶m
苶/k
苶
waves on the spring is v = 兹T
and the time for the pulse to travel the length
苶/k
苶.
of the spring is t = 2L/v = 兹2苶m
(Contributed by Inge H. A. Pettersen, Arendal,
Norway)
dQ/dt = ␣(T – TR)
dQ = ␣(T – TR) dt
mc dT = ␣(T – TR) dt
mcdT /(T – TR) = ␣dt
First to Submit Correct Solutions
The above equation shows that the time to
cool to a temperature T is proportional to the
amount of heat released by the container
(Q = mc⌬T). Therefore,
Dylan Consla, student (Maine School of Science
and Mathematics, Limestone, ME)
Art Hovey (Milford, CT)
Carl E. Mungan (U.S. Naval Academy,
Annapolis, MD)
k = (mcw⌬T + mcb ⌬T)/mcw⌬T
Several other readers also sent us correct
solutions to the January Challenges. We would
like to recognize the following contributors:
k = (cw + cb)/cw
We appreciate you submissions and hope to
receive more solutions in the future.
and, finally,
Note to contributors:
cb = (k–1)cw.
(Contributed by Ken Wright, Ithaca High School,
Ithaca, NY)
◗ The World on a Spring
Challenge: A spring has a force constant k and
mass m. The spring hangs vertically, and a block of
an unknown mass is attached to its bottom end. It is
known that the mass of the block is much greater
than that of the spring. The hanging block stretches
the spring to twice its relaxed length. How long (t)
would it take for a low-amplitude transverse pulse to
travel the length of the spring stretched by the
hanging block?
Solution: Since the mass of the spring is small
compared to the mass m of the block, the
A-2
As the number of submissions grows, we
request that certain guidelines be observed, in
order to facilitate the process more efficiently:
• Please email the solutions as Word files.
• Please name the file as : “Apr03HSimpson” if
— for instance — your name is Homer
Simpson, and you are sending the solutions
to April 2003 Challenges.
• State your name, hometown, and professional affiliation in the file, not only in the email
message.
Many thanks!
Please send correspondence to:
Boris Korsunsky
444 Wellesley St.
Weston, MA 02493-2631;
korsunbo@gse.harvard.edu
THE PHYSICS TEACHER ◆ Vol. 41, April 2003