Received: from MIT.EDU (PACIFIC-CARRIER-ANNEX.MIT.EDU [18.69.0.28]) by bloom-picayune.MIT.EDU (8.6.13/2.3JIK) with SMTP id OAA03924; Sat, 20 Apr 1996 14:57:27 -0400
Received: from [199.164.164.1] by MIT.EDU with SMTP
id AA15891; Sat, 20 Apr 96 14:13:25 EDT
Received: by questrel.questrel.com (940816.SGI.8.6.9/940406.SGI)
for news-answers-request@mit.edu id LAA25282; Sat, 20 Apr 1996 11:14:20 -0700
Newsgroups: rec.puzzles,news.answers,rec.answers
Path: senator-bedfellow.mit.edu!bloom-beacon.mit.edu!spool.mu.edu!howland.reston.ans.net!europa.eng.gtefsd.com!uunet!questrel!chris
From: chris@questrel.questrel.com (Chris Cole)
Subject: rec.puzzles Archive (physics), part 27 of 35
Message-Id:
Followup-To: rec.puzzles
Summary: This is part of an archive of questions
and answers that may be of interest to
puzzle enthusiasts.
Part 1 contains the index to the archive.
Read the rec.puzzles FAQ for more information.
Sender: chris@questrel.questrel.com (Chris Cole)
Reply-To: archive-comment@questrel.questrel.com
Organization: Questrel, Inc.
References:
Date: Wed, 18 Aug 1993 06:06:29 GMT
Approved: news-answers-request@MIT.Edu
Expires: Thu, 1 Sep 1994 06:04:11 GMT
Lines: 470
Xref: senator-bedfellow.mit.edu rec.puzzles:25014 news.answers:11534 rec.answers:1934
Apparently-To: news-answers-request@mit.edu
Archive-name: puzzles/archive/physics
Last-modified: 17 Aug 1993
Version: 4
==> physics/balloon.p <==
A helium-filled balloon is tied to the floor of a car that makes a
sharp right turn. Does the balloon tilt while the turn is made?
If so, which way? The windows are closed so there is no connection
with the outside air.
==> physics/balloon.s <==
Because of buoyancy, the helium balloon on the string will want to move
in the direction opposite the effective gravitational field existing
in the car. Thus, when the car turns the corner, the balloon will
deflect towards the inside of the turn.
==> physics/brick.p <==
What is the maximum overhang you can create with an infinite supply of bricks?
==> physics/brick.s <==
You can create an infinite overhang.
Let us reverse the problem: how far can brick 1 be from brick 0?
Let us assume that the brick is of length 1.
To determine the place of the center of mass a(n):
a(1)=1/2
a(n)=1/n[(n-1)*a(n-1)+[a(n-1)+1/2]]=a(n-1)+1/(2n)
Thus
n 1 n 1
a(n)=Sum -- = 1/2 Sum - = 1/2 H(n)
m=1 2m m=1 m
Needless to say the limit for n->oo of half the Harmonic series is oo.
==> physics/bubbles.p <==
In a universe with the same physical laws, but which is mostly water
with little bubbles in it, do the bubbles attract, repel, or what?
==> physics/bubbles.s <==
A bubble should produce a gravitational field that is the negative of
that produced by an equal volume of water in an empty universe. This is
because a point in space would be affected only by the water in the
symmetric image of the bubble with respect to that point. The effect
on another bubble in that field would be to attract that bubble since
it would be pushing the water around it away. Therefore, the bubbles
should attract.
==> physics/cannonball.p <==
A person in a boat drops a cannonball overboard; does the water level change?
==> physics/cannonball.s <==
The cannonball in the boat displaces an amount of water equal to the MASS
of the cannonball. The cannonball in the water displaces an amount of water
equal to the VOLUME of the cannonball. Water is unable to support the
level of salinity it would take to make it as dense as a cannonball, so the
first amount is definitely more than the second amount, and the water level
drops.
==> physics/magnets.p <==
You have two bars of iron. One is magnetized along its length, the
other is not. Without using any other instrument (thread, filings,
other magnets, etc.), find out which is which.
==> physics/magnets.s <==
Take the two bars, and put them together like a T, so that one bisects the
other.
___________________
bar A ---> |___________________|
| |
| |
| |
| |
bar B ------------> | |
| |
| |
|_|
If they stick together, then bar B is the magnet. If they don't, bar A is
the magnet. (reasoning follows)
Bar magnets are "dead" in their centers (i.e., there is no magnetic force,
since the two poles cancel out). So, if bar A is the magnet, then bar B
won't stick to its center.
However, bar magnets are quite "alive" at their edges (i.e., the magnetic
force is concentrated). So, if bar B is the magnet, then bar A will stick
nicely to its end.
==> physics/milk.and.coffee.p <==
You are just served a hot cup of coffee and want it to be as hot as
possible later. If you like milk in your coffee, should you add it
when you get the cup or just before you drink it?
==> physics/milk.and.coffee.s <==
Normalize your temperature scale so that 0 degrees = room temperature.
Assume that the coffee cools at a rate proportional to the difference
in temperature, and that the amount of milk is sufficiently small that
the constant of proportinality is not changed when you add the milk.
An early calculus homework problem is to compute that the temperature
of the coffee decays exponentially with time,
T(t) = exp(-ct) T0, where T0 = temperature at t=0.
Let l = exp(-ct), where t is the duration of the experiment.
Assume that the difference in specific heats of coffee and milk are
negligible, so that if you add milk at temperature M to coffee at
temperature C, you get a mix of temperature aM+bC, where a and b
are constants between 0 and 1, with a+b=1. (Namely, a = the fraction
of final volume that is milk, and b = fraction that is coffee.)
If we let C denote the original coffee temperature and M the milk
temperature, we see that
Add milk later: aM + blC
Add milk now: l(aM+bC) = laM+blC
The difference is d=(1-l)aM. Since l<1 and a>0, we need to worry about
whether M is positive or not.
M>0: Warm milk. So d>0, and adding milk later is better.
M=0: Room temp. So d=0, and it doesn't matter.
M<0: Cold milk. So d<0, and adding milk now is better.
Of course, if you wanted to be intuitive, the answer is obvious if you
assume the coffee is already at room temperature and the milk is
either scalding hot or subfreezing cold.
Moral of the story: Always think of extreme cases when doing these puzzles.
They are usually the key.
Oh, by the way, if we are allowed to let the milk stand at room
temperature, then let r = the corresponding exponential decay constant
for your milk container.
Add acclimated milk later: arM + blC
We now have lots of cases, depending on whether
rl: The milk pot is smaller than your coffee cup.
(E.g., it's one of those tiny single-serving things.)
M>0: The milk is warm.
M<0: The milk is cold.
Leaving out the analysis, I compute that you should...
Add warm milk in large pots LATER.
Add warm milk in small pots NOW.
Add cold milk in large pots NOW.
Add cold milk in small pots LATER.
Of course, observe that the above summary holds for the case where the
milk pot is allowed to acclimate; just treat the pot as of infinite
size.
==> physics/mirror.p <==
Why does a mirror appear to invert the left-right directions, but not up-down?
==> physics/mirror.s <==
Mirrors invert front to back, not left to right.
The popular misconception of the inversion is caused by the fact that
a person when looking at another person expects him/her to face her/him,
so with the left-hand side to the right. When facing oneself (in the
mirror) one sees an 'uninverted' person.
See Martin Gardner, ``Hexaflexagons and other mathematical
diversions,'' University of Chicago Press 1988, Chapter 16. A letter
by R.D. Tschigi and J.L. Taylor published in this book states that the
fundamental reason is: ``Human beings are superficially and grossly
bilaterally symmetrical, but subjectively and behaviorally they are
relatively asymmetrical. The very fact that we can distinguish our
right from our left side implies an asymettry of the perceiving
system, as noted by Ernst Mach in 1900. We are thus, to a certain
extent, an asymmetrical mind dwelling in a bilaterally symmetrical
body, at least with respect to a casual visual inspection of our
external form.''
Martin Gardner has also written the book ``The Ambidextrous Universe.''
==> physics/monkey.p <==
Hanging over a pulley there is a rope, with a weight at one end.
At the other end hangs a monkey of equal weight. What happens if
the monkey starts to ascend the rope? Assume that the mass of the
rope and pulley are negligible, and the pulley is frictionless.
==> physics/monkey.s <==
The monkey is pulling down on the rope hard enough to pull itself up. This
increases the tension in the rope just enough to cause the weight to rise at
the same rate as the monkey, since they are of equal mass.
==> physics/pole.in.barn.p <==
Accelerate a pole of length l to a constant speed of 90% of the speed of
light (.9c). Move this pole towards an open barn of length .9l (90%
the length of the pole). Then, as soon as the pole is fully inside the
barn, close the door. What do you see and what actually happens?
==> physics/pole.in.barn.s <==
What the observer sees depends upon where the observer is, due to
the finite speed of light.
For definiteness, assume the forward end of the pole is marked "A" and
the after end is marked "B". Let's also assume there is a light source
inside the barn, and that the pole stops moving as soon as end "B" is
inside the barn.
An observer inside the barn next to the door will see the following
sequence of events:
1. End "A" enters the barn and continues toward the back.
2. End "B" enters the barn and stops in front of the observer.
3. The door closes.
4. End "A" continues moving and penetrates the barn at the far end.
5. End "A" stops outside the barn.
An observer at the other end of the barn will see:
1. End "A" enters the barn.
2. End "A" passes the observer and penetrates the back of the barn.
3. If the pole has markings on it, the observer will notice the part
nearest him has stopped moving. However, both ends are still
moving.
4. End "A" stops moving outside the barn.
5. End "B" continues moving until it enters the barn and then stops.
6. The door closes.
After the observers have subtracted out the effects of the finite speed
of light on what they see, both observers will agree on what happened:
The pole entered the barn; the door closed so that the pole was
completely contained within the barn; as the pole was being stopped it
elongated and penetrated the back wall of the barn.
Things are different if you are riding along with the pole. The pole
is never inside the barn since it won't fit. End A of the pole penetrates
the rear wall of the barn before the door is closed.
If the wall of the barn is impenetrable, in all the above scenarios insert
the wording "End A of the pole explodes" for "End A penetrates the barn."
==> physics/resistors.p <==
What is the resistance between various pairs of vertices on a lattice
of unit resistors in the shape of a
1. Cube,
2. Platonic solid,
3. N dimensional Hypercube,
4. Infinite square lattice,
and
5. between two small terminals on a continuous sheet?
==> physics/resistors.s <==
1. Cube
The key idea is to observe that if you can show that two
points in a circuit must be at the same potential, then you can
connect them, and no current will flow through the connection and the
overall properties of the circuit remain unchanged. In particular, for
the cube, there are three resistors leaving the two "connection
corners". Since the cube is completely symmetrical with respect to the
three resistors, the far sides of the resistors may be connected
together. And so we end up with:
|---WWWWWW---| |---WWWWWW---| |---WWWWWW---|
| | |---WWWWWW---| | |
*--+---WWWWWW---+-+---WWWWWW---+-+---WWWWWW---+---*
| | |---WWWWWW---| | |
|---WWWWWW---| |---WWWWWW---| |---WWWWWW---|
|---WWWWWW---|
This circuit has resistance 5/6 times the resistance of one resistor.
2. Platonic Solids
Same idea for 8, 12 and 20, since you use the symmetry to identify
equi-potential points. The tetrahedron is a hair more subtle:
*---|---WWWWWW---|---*
|\ /|
W W W W
W W W W
W W W W
| \ / |
\ || |
\ | /
\ W /
\ W / <-------
\ W /
\|/
+
By symmetry, the endpoints of the marked resistor are equi-potential. Hence
they can be connected together, and so it becomes a simple:
*---+---WWWWW---+----*
| |
+-WWW WWW-+
| |-| |
|-WWW WWW-|
3. Hypercube
Think of injecting a constant current I into the start vertex.
It splits (by symmetry) into n equal currents in the n arms; the current of
I/n then splits into I/n(n-1), which then splits into I/[n(n-1)(n-1)] and so
on till the halfway point, when these currents start adding up. What is the
voltage difference between the antipodal points? V = I x R; add up the voltages
along any of the paths:
n even: (n-2)/2
V = 2{I/n + I/(n(n-1)) + I/(n(n-1)(n-1)) + ... + I/(n(n-1) )}
n odd: (n-3)/2
V = 2{I/n + I/(n(n-1)) + I/(n(n-1)(n-1)) + ... + I/(n(n-1) )} (n-1)/2
+ I/(n(n-1) )
And R = V/I i.e. replace the Is in the above expression by 1s.
For the 3-cube: R = 2{1/3} + 1/(3x2) = 5/6 ohm
For the 4-cube: R = 2{1/4 + 1/(4x3)} = 2/3 ohm
This formula yields the resistance from root to root of
two (n-1)-ary trees of height n/2 with their end nodes identified
(-when n is even; something similar when n is odd).
Coincidentally, the 4-cube is such an animal and thus the answer
2/3 ohms is correct in that case.
However, it does not provide the solution for n >= 5, as the hypercube
does not have quite as many edges as were counted in the formula above.
4. The Infinite Plane
For an infinite lattice: First inject a constant current I at a point; figure
out the current flows (with heavy use of symmetry). Remove that current. Draw
out a current I from the other point of interest (or inject a negative current)
and figure out the flows (identical to earlier case, but displaced and in the
other direction). By the principle of superposition, if you inject a current I
into point a and take out a current I at point b at the same time, the currents
in the paths are simply the sum of the currents obtained in the earlier two
simpler cases. As in the n-cube, find the voltage between the points of
interest, divide by I and voila`!
As an illustration, in the adjacent points case: we have a current of I/4 in
each of the four resistors:
^ |
| v
<--o--> -->o<--
| ^
v |
(inject) (take out)
And adding the currents, we have I/2 in the resistor connecting the two points.
Therefore V=(1 ohm) x I/2 and effective resistance between the points = 1/2 ohm.
We do not derive it, but the equivalent resistance between two nodes k
diagonal units apart is (2/pi)(1+1/3+1/5+...+1/(2k-1)); that, plus
symmetry and the known equivalent resistance between two adjacent
nodes, is sufficient to derive all equivalent resistances in the
lattice.
5. Continuous sheet
I think the answer is (rho/dz)log(L/r)/pi where rho is the resistivity,
dz is the sheet thickness, L is the separation, r is the terminal radius.
cf. "Random Walks and Electric Networks", by Doyle and Snell, published by the
Mathematical Association of America.
==> physics/sail.p <==
A sailor is in a sailboat on a river. The current is 3 knots with respect
to the land. The wind (air velocity) is zero, with respect to the
land. The sailor wants to proceed downriver as quickly as possible,
maximizing his downstream speed with respect to the land.
Should he raise the sail, or not?
==> physics/sail.s <==
Depends on the sail. If the boat is square-rigged, then not, since
raising the sail will simply increase the air resistance.
If the sailor has a fore-and-aft rig, then he should, since he can then
tack into the wind. (Imagine the boat in still water with a 3-knot head
wind).
==> physics/shoot.sun.p <==
If you are standing at the equator at sunrise, where must you point a laser
cannon to hit the Sun dead center? Assume that the Sun is stationary and
that the Earth's orbit around it is circular.
==> physics/shoot.sun.s <==
You aim it at the horizon. The sun is exactly in the place where it
appears to be. It is true that the sun wasn't on the horizon 8 minutes
ago (the specific number is 2 degrees), when it emitted the light you
are now seeing. However, "the sun wasn't on the horizon" doesn't mean
the sun moved; it means the horizon moved.
==> physics/skid.p <==
What is the fastest way to make a 90 degree turn on a slippery road?
==> physics/skid.s <==
For higher speeds (measured at a small distance from the point of initiation
of a sharp turn) the fastest way round is to "outside loop" - that is, steer
away from the curve, and do a skidding 270.
This technique is taught in advanced driving schools.
References:
M. Freeman and P. Palffy, American Journal of Physics, vol 50, p. 1098, 1982.
P. Palffy and Unruh, American Journal of Physics, vol 49, p. 685, 1981.
==> physics/spheres.p <==
Two spheres are the same size and weight, but one is hollow. They are
each made of uniform material, though of course not the same material.
With a minimum of apparatus, how can I tell which is hollow?
==> physics/spheres.s <==
Since the balls have equal diameter and equal mass, their volume and
density are also equal. However, the mass distribution is not equal,
so they will have different moments of inertia - the hollow sphere has
its mass concentrated at the outer edge, so its moment of inertia will
be greater than the solid sphere. Applying a known torque and observing
which sphere has the largest angular acceleration will determine which
is which. An easy way to do this is to "race" the spheres down an
inclined plane with enough friction to prevent the spheres from sliding.
Then, by conservation of energy:
mgh = 1/2 mv^2 + 1/2 Iw^2
Since the spheres are rolling without sliding, there is a relationship
between velocity and angular velocity:
w = v / r
so
mgh = 1/2 mv^2 + 1/2 I (v^2 / r^2) = 1/2 (m + I/r^2) v^2
and
v^2 = 2mgh / (m + I / r^2)
From this we can see that the sphere with larger moment of inertia (I) will
have a smaller velocity when rolled from the same height, if mass and radius
are equal with the other sphere. Thus the solid sphere will roll faster.
==> physics/wind.p <==
Is a round-trip by airplane longer or shorter if there is wind blowing?
==> physics/wind.s <==
It will take longer, by the ratio (s^2)/(s^2 - w^2) where s is the
plane's speed, and w is the wind speed. The stronger the wind the
longer it will take, up until the wind speed equals the plane's speed,
at which point the plane will run out of fuel before too long.
Math:
s = plane's speed
w = wind speed
d = distance in one direction
d / (s + w) = time to complete leg flying with the wind
d / (s - w) = time to complete leg flying against the wind
d / (s + w) + d / (s - w) = round trip time
d / (s + w) + d / (s - w) = ratio of flying with wind to
------------------------- flying with no wind (bottom of
d / s + d / s equation is top with w = 0)
this simplifies to s^2 / (s^2 - w^2).