
Top Martin Gardner Physics Stumpers - ColinWright
http://martin-gardner.org/Top10MGPhysics.html
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jameshart
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.

~~~
at-fates-hands
>> Archimedes went all 'eureka' down the street naked [citation needed]

Here you go:

[http://www.thenakedscientists.com/HTML/articles/article/the-...](http://www.thenakedscientists.com/HTML/articles/article/the-
original-naked-scientist/)

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."_

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_sword
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

~~~
anon4
I got 500 for 3. and for 4. I'd just touch the end of one rod to the middle of
the other.

~~~
_sword
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

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

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evanb
1\. Feynman explains it best
[https://www.youtube.com/watch?v=msN87y-iEx0](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.

~~~
JshWright
> 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".

~~~
LanceH
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?

~~~
anon4
I don't see what's so hard about that one, it's a simple infinite sum.

~~~
jameshart
There's an easier way than an infinite sum.

~~~
Khao
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.

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VLM
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.

~~~
FreeFull
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.

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anon1110
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.

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Strilanc
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..

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thedufer
#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.

~~~
jameshart
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.

~~~
thedufer
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.

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aidenn0
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.

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delinka
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

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

~~~
ColinWright
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.

~~~
avian
You're right. Thanks.

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Lorem678
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.

~~~
jeremysalwen
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](http://en.wikipedia.org/wiki/Archimedes%27_principle)
it will displace the same volume of water frozen or melted.

~~~
Lorem678
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...](http://physics.stackexchange.com/questions/110645/why-does-ice-
melting-not-change-the-water-level-in-a-container)

~~~
DougBTX
This is almost the answer to the next question too.

~~~
anon1110
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.

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imranq
Another one: what keeps the train on the tracks?

~~~
symmetricsaurus
Feynman explains it well [1].

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

