March 01, 2004

• You are given 2 identical looking spheres. They have the same mass and have the same diameter. Physically, they look the same, and have the same surface texture. (ie you can't visually pick them apart) They are both hard, thus they won't bounce and they won't have any 'give'. They both have perfectly smooth surface. One is made of less dense material and is soild and uniform through out. The other is made of higher density material, but since having the same mass and volume as the other, it is hollow at its centre (assume a spherical cavity with the centre of cavity and centre of the whole sphere at the same point). With a minimum of instruments, how can you determine which one is hollow and which one is solid? My second thought: Break it open. Also: very cool site. Thanks, Gyan.
• Cna we attempt to aswers some of them here? or that would be unfair. I would like to discuss some solutions.
• I'm for it (well, obviously, since I wrote in my answer to one). The easy ones are, well, pretty easy, but I hit the medium ones and apparently I'm going to have to do some work to answer those. Except the manhole one, which I already knew the answer to.
• Go right ahead. Note that particular tricky ones as well.
• Some of the easy ones are not so easy. Lateral thinking required.
• If you weight the spheres on shallow water (not covering the spheres entirely) the one with the cavity should weight less because of the low density of the air in the cavity compared with the sorrounding water. The difference could be also tested by how they produce sound when hit by a little hammer. The one with a cavity should sound hollow I though. :) So much time lapsed since my last physics course... Memory hurts...
• ... I though. I though, therefore I are...
• Oh! by the way... [banana] O love this kind of puzzles. Although, I can rarely solve them.
• O love ... I'm felling sharp today.
• The mother is 21 years older than the child. In 6 years from now, the mother will be 5 times as old as the child. Question: Where's the father? Any thoughts? Zemat, I'm not seeing how that works. From the liquid's perspective, the ball is still a closed system, with a consistent weight and density, so the properties shouldn't change based on the depth of the water. If it's not buoyant enough to float, then it'll displace water based on the volume of the sphere and not the density of the submerged part; if it's too buoyant to sink, then you won't be able to weigh it as an independent item. But I admit I may be missing something. The tapping method should work, if the material is acoustically alive enough.
• I thought a lot about it too, Sandspider. I'm not totally convinced myself. But my reasoning is that if you weight the to spheres in the air they show the same weight regardless that the sphere with the cavity has more mass because of the air enclosed in it. That's because there's no difference between the air pressure inside the sphere and outside. So the pressures negate each other. But if you put the spheres half sunken in water then you create a difference between the pressures inside and outside the hollow sphere. since the pressure of the water push the sphere uppward. If the sphere is completely sunken it doesn't work out because the pressure from the water acts in every direction toward the center of the sphere. I hope to reach my hose to try to reach a prof on paper.
• Obviously all that I said doesn't work if the spheres are in outher space, and the hollowed sphere is also devoid from air inside.
• by hose I mean house. chezz...
• Question: Where's the father? Obviously, in Michigan. Q.E.D.
• On second thought, I'm mistaken about having no distinction in weight inside of water. The air inside the hollow sphere would act as added mass. Since the material of both spheres occupy the same volume and have the same overall density (as sphere-mass/sphere-volume), the bubble of air would increase the density of the hollow sphere slightly making it heavier than the filled sphere.
• But if the surface is solid, smooth, and hard, would the pressure on the outside be enough to affect the pressure on the inside? It's been forever since I've dealt with pressure, so I don't really know the answer to this. But what I do know the answer to is: "Proof" that 2 = 1: a = b a2 = ab a2 - b2 = ab-b2 (a-b)(a+b) = b(a-b) a+b = b b+b = b 2b = b 2 = 1 Does this argument make sense? And we're the only ones on this thread, aren't we?
• Floating the spheres in water would make no difference, assuming that they have the same mass. (I'm pretty sure they meant to say that the space in the hollow sphere was filled with a vacuum, rather than air; otherwise, you could just weigh the two spheres with a normal balance.) The law of buoyancy states that the water will push upward with a force equal to the weight of the displaced water. The distribution of mass in the sphere is irrelevant in this case, as long as the mass, volume, and shape are the same. What you can do, though, is spin them. The solid sphere has a lesser moment of inertia than the hollow sphere, so you could probably tell which one is which by how much it resists being turned by hand; the hollow sphere should resist more. I'd like to figure out a more quantifiable test, though; you might be able to attach a motor to the ball and measure how much energy it takes to spin it up to some angular speed, and use that to calculate its moment of inertia. And Sandpiper: you can't divide by zero, silly!
• Wow, I got that one wrong. Sorry, Sandspider. *checks glasses*
• This reminds me of the
• And, of course, I meant to finish that with rec.puzzles archive Go HTML skills!
• Exactly, pmdboi. That one shoulda been in the easy category. Also not hard is: Sheila and He-Man are twins; Sheila is the OLDER twin. Assume they were born immediately after each other, an infinitesimally small - but nonzero - amount of time apart. During one year in the course of their lives, Sheila celebrates her birthday two days AFTER He-Man does. How is this possible? Note: For both Sheila and He-Man, these birthday celebrations happen on the actual birthday date -- it cannot be a celebration that occurs at a date earlier or later than the actual birthday date for whatever reasons of convenience. Also, the solution has nothing to do with the theory of relativity or any other over complicated nonsense like that. Though there may be cognitive reasons why it's a pretty easy one.
• This is quite a lot of fun (and an excellent diversion from my homework)... The mother is 21 years older than the child. In 6 years from now, the mother will be 5 times as old as the child. Question: Where's the father? I believe that he's currently, *ahem*, getting it on with the mother.
• Sandspider: being a leap year baby myself, I hold a special place in my heart for that one. Unfortunately, my parents have never been outside of North America and I'm not a twin.
• pmdboi: floating the spheres in water would measure volume. Here's a semi-helpful thingie on volume, mass, density and weight.
• pmbdoi - but they weren't leap year babies?
• Thanks for the correction and the correct solution pmdboi. I'm gonna stick to the very easy ones (those that desn't involve physics, mathematics or logic) from now on. So, what... is the air-speed velocity of an unladen swallow?
• Great post! Will likely spend alot of time at this site. Thanks!
• I had to quit before I cried in frustration. Did anyone work out why Jeremiah didn't die?
• I figured that the poison needed to be a certain temperature before it would work. So when Jeremiah kept dunking those drinks, it was too cold. Whereas after the few minutes of talking, the ice melted, the scotch got warm, and Wallace drank it when the poison was active. But the most popular solution was that the poison was in the ice. Which doesn't really make sense, cos Wallace would try to refill the same glass (full of ice) with scotch hoping that the ice would melt right? And after three glasses, wouldn't it?
• path: The solution I've heard is the following: The mother flies to some Pacific island on the east side of the International Date Line on some non-leap year, then gets in flight and gives birth to the children such that He-Man is born before midnight on February 28 to the east of the line, and Sheila is born after midnight on March 2 to the west of the line (crossing the line from east to west adds a day, give or take an hour). Then on any leap year, their birthdays are separated by two days: February 29 and March 1. I don't know; I guess I feel a kind of kinship to anything that exploits our calendrical quirks, being a victim of them myself. And yeah, you can measure the volume of the spheres by immersing them (Archimedes and "Eureka!" and all that), but what good will it do you? No water will get into the inner cavity, so the apparent volume of the two spheres will be the same as long as their outer forms are the same. I personally think the problem could have been more elegantly stated by filling the center of the hollow sphere with cork or styrofoam or some other substance less dense than that used in the solid sphere, and making it so that the masses of the two spheres are still equal. That way the volume of the spheres, in terms of the part of them that isn't air, is the same, and the only thing that changes between the two is the distribution of mass. Then again, one might consider the air in the cavity in the present problem to be that less dense substance. But, meh. I should be coding right now. Stupid garbage collectiion.
• pmdboi, regarding your solution about spinning the spheres. It wouldn't be easier just to let them roll down an slope and see what sphere reaches the ground first? The one with the cavity should take longer since it has a greater rotational inertia. Am I right?
• Zemat: Yeah, that would probably be a lot easier, certainly easier than attaching a motor to the ball. So, if we assume bananas are spherical, and we have a fresh one and one that has been eaten but has a really heavy peel... *pines for spherical bananas*
• Zemat, so you want to know about swallows, do ya?
• Re the twins, if they were born during a non-leap year, one before midnight Feb. 28 and the other after, their birthdays would be 2 days apart in a leap year. (Or, did everyone else get it already and I'm just late for the party?)