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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: <puzzles/archive/physics_745653851@questrel.com> 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: <puzzles/archive/Instructions_745653851@questrel.com> 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 r<l: The milk pot is larger than your coffee cup. (E.g, it really is a pot.) r>l: 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). 
