Top Martin Gardner Physics Stumpers | Hacker News Reader

Top Martin Gardner Physics Stumpers

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59 pointsColinWright46 comments

One interesting common theme through many of these (helium balloon in a car; ice melting; bolts in a boat) is how counterintuitive buoyancy is. You can see why, when he had it figured out, Archimedes went all 'eureka' down the street naked [citation needed].

Similarly, the heated metal donut and gravity of a spherical shell questions challenge easy assumptions you may make. If your mental model of a torus is that it's a bent cylinder, you might conceive that when it expands the dominant behavior is that the cylinder gets fatter; if you mentally model gravitation of solid bodies as if it is equivalent to gravitation from point masses at their center of gravity, you'll assume a net attraction toward the center of a hollow shell.

The lesson to take from these puzzles is to modify your mental model so that the answers to all these questions become intuitive too. And that's why I love 'puzzles' like this - they help you deepen your understanding by focusing you on the places your assumptions might catch you out.

>> Archimedes went all 'eureka' down the street naked [citation needed]

Here you go:

http://www.thenakedscientists.com/HTML/articles/article/the-...

Two points here:

1 - "Whenever Archimedes is pictured in his watery moment of inspiration, he is typically alone in a tub in what appears to be his home. But the Greeks of antiquity, like the Romans after them, bathed in a public facility with attendants to receive and store patrons’ garments. If the legend is true, Archimedes would have been at the public baths when he had his Eureka! moment and might have rushed home to pursue the idea, perhaps neglecting to retrieve his clothes. Even then, history’s premier naked scientist might not have been completely bare. According to classicist Lydia Lake, “naked” in antiquity – Greek gumnos, Roman nudus – had a dual meaning: either stripped of all clothing or else lacking an outer garment, such as a toga or scarf-like wrap (chlamys). Archimedes might have sprinted through the streets of Syracuse in his underwear – what we would today recognize as a tunic"

2 - "The Eureka! story – fact or fable – is the touchstone of creative epiphany. It represents the culmination of Archimedean effort: that still mysterious cognitive path that scientists tread toward solutions, fundamental truths, and new ideas."

Let's go I'm doing each of these in 30 seconds or less:

1. You're rotating the image about a plane I think I remember

2. Larger

3. 600 miles

4. They're ferromagnetic so magnetize them both by keeping them near each other. Boom we're done here

5. Stay the same

6. Fall since it doesn't have to displace a mass of water (less dense) equivalent to the nuts and bolts (more dense)

7. Assume the pigeons begin at the top of the cab with zero initial momentum and we allow them to fall while the truck begins to accelerate. The truck would feel lighter and accelerate more easily up until the point where the free-falling pigeons contact the back of the truck and begin to be accelerated by the truck.

8. It moves in the direction of the car since the air in the vehicle is a fluid mass that piles up in the car opposite the direction of acceleration.

9. 30 seconds isn't enough to fully explain, but there's a separate solution to the field equation inside versus outside. Inside it should float where it is as attractive forces balance themselves out.

10. Mass of the liquid = mass of the can let's go with that

No guarantee I read the questions correctly in the 30 seconds I gave myself for each

I got 500 for 3. and for 4. I'd just touch the end of one rod to the middle of the other.

Woops, typo yeah it is 500 miles. The relative speed of one to the other is 30,000mph so it's 500 miles. For 4 I gave a cop-out answer since I didn't think of one within the time I gave myself

They travel 500 miles. They start 1317 miles apart. The distance afterward is 1317-500.

For 1) it's a flip along the axis of the mirror, not a rotation at all.

Pretty sure I remember it being possible to do a 4 dimensional rotation that's equivalent to a 3D mirror image. I'm at work now though and can't do the math for a while

Everything gets easier when you add dimensions - just ask the string theorists.

Sure. Add an extra dimension with coordinate w to 3d space.

Then just rotate around the (x,y)-plane until z -> -z and w -> -w. Everything in the (x,y)-plane stays the same and z has been inverted.

Thank you, that looks exactly right!

I think you are mostly right.

Some corrections and further explanations (no time limit):

1. A mirror flips the direction perpendicular to the mirror. Left is still left and right is still right, e. g., your left hand will still be on your left side in the mirror. But, your mirror image will be looking in the opposite direction compared to you.

7. The pigeons are flying around inside the truck. Doing this they exert a force on the air around them which in turn will exert a force on the truck containing the air. On average this force will balance the weight of the pigeons. Also, 200 pigeons at maybe a 200 grams each is just 40 kg to be compared to the weight of a truck of several tonnes.

9. Using spherical symmetry and the divergence theorem we can conclude that the gravitational field inside a spherical shell is zero.

I think number 10 is 1/4 of the can is full of soda, so 3 ozs of soda. There's a link here with the problem http://people.missouristate.edu/lesreid/Adv71.html.

1. Feynman explains it best https://www.youtube.com/watch?v=msN87y-iEx0

3. They approach one another at 30000 mph = 500 miles per minute, so two minutes before they collide they are 1000 miles apart, and who cares that they start out 1317 miles apart?

8. The balloon rushes forward, and gets pulled toward the inside of the curve. The best reasoning is via the equivalence principle: acceleration and gravity feel the same, and the fact that the balloon is buoyant in air means that it should rush away from whatever direction gravity is pointing.

9. It floats at the same location, by Gauss' theorem.

> 3. They approach one another at 30000 mph = 500 miles per minute, so two minutes before they collide they are 1000 miles apart, and who cares that they start out 1317 miles apart?

It was clearly just a distraction, and apparently an effective one, as the question was "calculate how far apart they are _one_ minute before they collide".

The version I heard that truly leads a calculus student down the wrong path is along the lines of:

Train a is going 40 mph, train b 60 mph. They are 200 miles apart on the same track, headed directly for each other. There is a bird flying between the two at 100 mph, starting from train a. Each time he reaches a train he turns around and heads back to the other train. How far does the bird travel before he is crushed between the trains?

I don't see what's so hard about that one, it's a simple infinite sum.

There's an easier way than an infinite sum.

40mph + 60mph = 100mph, so they will travel for 2 hours before they hit each other. The bird flying at 100 mph for 2 hours will travel 200 miles.

anon4 is referencing one of the tellings of this story where famous scientist [usually Von Neumann] is given this problem as a brainteaser and responds instantly having done it the `hard way'.

http://mathforum.org/dr.math/faq/faq.fly.trains.html

"...calculate how far apart they are one minute before they collide"

So, the answer is 500 miles (not 1000 miles).

Ah, on my screen the text wraps right where it says the number of minutes, and my brain parsed the Two from the beginning of the question.

A couple of these have been Mythbusters edutainment TV show topics. I am almost certain both the pigeons and the balloon have been on the show. I'm not sure how they'd use explosives to test the pigeon one, but I'm sure it would be interesting to watch. (edited to add, I just figured it out, they'll put frozen chicken parts from the supermarket in a really big box on top of some explosive, and put the works on a scale...)

The iron bars is more fun when analogized to the famous barometer building height challenge where the goal isn't someones abstract idea of simplest but to come up with the most unusual way to determine the answer (for the barometer I like the idea of tossing it off the roof and timing the impact, or bribing the building superintendent with a really nice barometer). For the iron bars I think a funny way to solve the problem is to heat one bar above the curie point and after it anneals back to room temp, if the bars don't stick to each other I guess the heat treated bar WAS a magnet. Another fun trick if you don't trust your string suspension to not bias the hanging position WRT north pole, simply spin each bar in each axis and one bar will have a funkier deceleration curves for various axes.

Depending on how solid the iron bars are, you could try breaking one in half. The two halves will behave like separate magnets if it was a magnet.

I think there is a better answer than yet given to the iron bar question. Lets assume the bar is magnetized so the ends are north and south (as opposed to the weird thing, where it is magnetized through its thickness). Position the bars near to each other and perpendicular, like the letter 'T'. Now switch the bars' positions, so the other bar is the cross and the other the stem. When the magnet is the stem of the 'T' there will be a strong attraction between the pieces, while when the magnet is the bar of the 'T' there will be almost no attraction of the pieces. It is just like sticking horseshoe magnet to a piece of metal with its tips is much stronger than sticking it with its back, just now the horseshoe is straightened. The attraction is to the poles, not the equator.

Number 10 is tricky. I solved it by using the fact that when you remove mass the center of mass moves away from the mass you removed, so the minimum will occur when the height of soda matches the height of the center of mass. This gives the equation:

    (1/2 c + (1/2 h) (h f)) / (c + h f) = h

Where c is the mass of the can, the can has height 1, h is the height of soda remaining in the can, and f is the mass of the fluid when the can is full. (Incidentally, I wrote this equation wrong like eight times because I kept missing a factor of h in the fluid's center of mass height.)

Working the equation:

    c + h^2 f = 2 h (c + h f)
    f h^2 + 2 c h - c = 0

Oh hey this will get simpler if we just look at the ratio:

    let r = c/f
    h^2 + 2 r h - r = 0
    h = -r +- sqrt(r^2 + r)
    h = sqrt(r^2 + r) - r

Well that's kind of gross. I'm actually surprised it's in the right range based on how it looks (but plotting it shows it is in fact in [0, 0.5]). Maybe the values from the problem are nice:

    r = 1.5/12 = 3/24
    h = sqrt((3/24)^2 + 3/24) - 3/24
    h = 1/4

I guess that's okay..

#10 seems to have confused a number of people, so I'll share my thoughts.

The reason that decreasing the amount of liquid lowers the center of gravity is that you're removing liquid from above the center of gravity. Likewise, the center of gravity rises again when you remove liquid from below the center of gravity. Thus, the center of gravity is lowest when the center of gravity is exactly at the level of the liquid.

Without the distribution of the weight of the can, I don't think we can get an actual number. The ratio of the weights of the top/bottom to the weight of the sides is important.

A final unrelated point: cans of soda are measured in fluid ounces, so the proposed numbers are somewhat less accurate than the questioner seems to think.

> Without the distribution of the weight of the can, I don't think we can get an actual number.

Because the distribution of weight in the can isn't changing, you can treat it as a point mass as far as the center of mass is concerned. The fluid's distribution is changing, but we know how.

Hrmmm. I did a full write-up assuming that worked, and then managed to convince myself it didn't. Looks like you're right, though. Here it is:

Let's call C the weight of the can, L the weight of the liquid, and x the proportion of the can that's full. Then the center of mass is (C/2 + xL/2)/(C+L). We want where this is equal to x. Solving for x gives us x=C/(2C+L).

This passes the sniff test: for C>>L, x->1/2, and for L>>C, x->0, as we'd expect. For the given weights, we get x=0.1, so when the can is 10% full.

The liquid's center of mass isn't xL/2, it's x L^2 /2. One factor of L for getting lighter, and one for getting lower.

I made the same mistake like five times.

In my formulation, x is the proportion and L the total weight, so I believe it would be x^2 L/2.

I also just realized that the denominator ought to contain xL, not L. My solution is a mess, clearly. The original idea is right, though.

The point of a fluid ounce is that it's the amount of water which weighs an ounce - roughly (allowing for various failures of standardization and definitions of water purity and measurement circumstances). And most beverages have a specific gravity close enough to 1.0 that you can consider them more or less water. So yes, the weight of the liquid in a 12oz beverage is, for puzzle purposes, pretty much 12oz.

That was the point hundreds of years ago, but its now fairly inaccurate. For the purposes of food labeling, a fluid ounce is defined as 30mL, which makes a fluid ounce of water weigh almost 5% more than an ounce. Sodas are typically more dense - Coke is 10% more dense than water, for example, for a total weight of over 13.5 ounces for a 12 fl oz can. Seems like a pretty big difference to me.

My favorite answer for #1:

There are two ways without a mirror that you could end up vertical and looking in the opposite direction: turn in place, or do a handstand. If you assume the former is what happened, then the mirror looks like if flipped left and right; if you assume the latter, then the mirror really flipped top-to-bottom. If you were to flip the image you get in the mirror upside down, you will see yourself upside-down with your hands not reversed.

10. when the soda's weight approaches the weight of the part of the can remaining above it (equality being the limit.) My initial thought was "...as half the can" but that's inaccurate.

... I think

I believe the answer is exactly half the can. Why do you think that is inaccurate?

That is inaccurate. Consider the case when the fluid is very, very much more dense than the can, and then consider the case when the fluid is very, very much less dense than the can.

You're right. Thanks.

An ice cube occupies a greater volume than the water that went into forming it, so I think the answer for question 5 would be the level of water in the beaker decreases.

You also need to remember that the ice cube is sticking out of the water. In fact the two effects cancel each other perfectly, so the water level stays the same.

To think about it in a more clear light, consider that the ice cube has the same weight frozen or melted. Thus by http://en.wikipedia.org/wiki/Archimedes%27_principle it will displace the same volume of water frozen or melted.

Ah touché, that's absolutely mind blowing. A run through of the maths here for anyone interested: http://physics.stackexchange.com/questions/110645/why-does-i...

This is almost the answer to the next question too.

Nope! :^) In the boat question, when the bolts are in the boat they are displacing water equal to their weight. After they drop to the bottom, they are displacing water less than their weight. So less water is displaced, i.e. total water level drops.

It's a buoyancy/displacement question. Similar to the one that follows. ;-)

Another one: what keeps the train on the tracks?

Feynman explains it well [1].

[1]: https://www.youtube.com/watch?v=y7h4OtFDnYE

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