1) Uppercase alphabet letters split into four following groups. Explain the principle of separation.

?  1. A,H,I,M,Y,T,U,V,W
?  2. B,C,D,E,K
?  3. S,X,Z,N,O
?  4. F,G,J,L,P,Q,R


2) There is two balls: one made of gold and hollow, other is not hollow and made of titanium. Both balls are identical in size and weight, and covered with a shell made of non-transparent, electricity/heat/cold non conducting material. Find a way to determine which ball is made of what, without damaging the shell (you are not allowed to burn it, place in liquid, etc.) and without using any external measurement equipment.


3) A stick with length a suspended by the edges horizontally on two parallel wires with the same length b. If the stick is rotated up to phi angle around the Y axis, (the Y axis is in center of the stick). Up to how much the stick will lift?


4) Lets suppose a man flying with a speed of light holding a flashlight emitting light in the direction of movement. What will happen with the light emitted from the flashlight?
Posted on 2006-03-10 12:19:09 by arafel

2) There is two balls: one made of gold and hollow, other is not hollow and made of titanium. Both balls are identical in size and weight, and covered with a shell made of non-transparent, electricity/heat/cold non conducting material. Find a way to determine which ball is made of what, without damaging the shell (you are not allowed to burn it, place in liquid, etc.)
Measure the Sound Velocity through each.?  Gold is dense and slower.?  Titanium would be quicker every time.?  The shell being constant between the test objects.

Regards,?  P1?  8)
Posted on 2006-03-10 13:15:47 by P1
Using any measurement equipment is not allowed. There is more elegant way, which doesn't require any external instruments.
Posted on 2006-03-10 13:29:21 by arafel
Using any measurement equipment is not allowed. There is more elegant way, which doesn't require any external instruments.
Where was that in the question above?  ;)

Seeing we are having some fun at our expense, and I will wait for the answers.  :)

Regards,  P1  8)
Posted on 2006-03-10 13:38:42 by P1

Using any measurement equipment is not allowed. There is more elegant way, which doesn't require any external instruments.
Where was that in the question above??  ? ;)

Sorry, forgot to include that condition in the question.

Fixed.


Seeing we are having some fun at our expense, and I will wait for the answers.  :)

I'll post the answers bit later..
Posted on 2006-03-10 14:44:29 by arafel
2) There is two balls: one made of gold and hollow, other is not hollow and made of titanium. Both balls are identical in size and weight, and covered with a shell made of non-transparent, electricity/heat/cold non conducting material. Find a way to determine which ball is made of what, without damaging the shell (you are not allowed to burn it, place in liquid, etc.) and without using any external measurement equipment.
Thump it with your finger, a hollow ball with have an echo.  The solid ball would dissipate the sound quickly.

Well, I figured, I had nothing to lose for a second try.

Regards,  P1  8)
Posted on 2006-03-10 17:02:19 by P1

4) Lets suppose a man flying with a speed of light holding a flashlight emitting light in the direction of movement. What will happen with the light emitted from the flashlight?

According to the Theory of Relativity, the man holding the flashlight won't notice anything particular (he will see everything 'normally'). A 'static', 'external' observer (watching that man with flashlight) will 'see' that the flashlight doesn't emit any light, or something like that (depending on their relative difference in speed vectors). Of course this is very theoretical and requires abstract thinking.
Posted on 2006-03-10 17:17:08 by ti_mo_n
arafel,
好bsp; 好bsp; 好bsp;I've been busy working on my taxes, but I immediately knew the answer for the second problem, so I will quickly dash out an answer.好bsp; That problem brought back memories of my college physics classes.好bsp; You can't really tell what the material is directly, but you can determine the mass distribution.好bsp; How?好bsp; By rolling both balls down a inclined plane and observing which is fastest/slowest.好bsp; At the bottom of the incline, the balls are going to have rotational energy and translational energy.好bsp; The sum of these two kinds of energy for each ball are determined by the mass of the ball, the gravity field, and the height of the incline, which is constant for both balls.好bsp; The ball with more of the mass distributed on the outside will have a higher rotational inertia, also called moment of inertia.好bsp; It will hog more of the available energy leaving less energy for translational motion.好bsp; Any good physics book will explain this.好bsp; Anyway, the bottom line is that the hollow ball of the same mass and rolled from the same height within the same gravity field will take longer to get to the bottom of the incline.好bsp; Ratch

P.S.

    The hollow ball can only be gold because less dense hollow titanium can never be the same size as solid gold if both are equal mass.  Ratch
Posted on 2006-03-10 17:36:25 by Ratch
Ratch,
That is correct.

P1, your alternative answer has been accepted as well  ;), although I don't think it will work for relatively small balls (also don't forget the shell which may compensate the sound). It will be impossible to hear the difference.
Posted on 2006-03-10 18:47:28 by arafel
regarding the second problem :

The gold (hollow) ball will float in water, regardless that it is made of gold - in order for it to weigh the same as the titanium ball, it must have a very thin wall section, and therefore contains a large volume of air relative to its volume of metal.

This is the same principle which allows steel ships to float.
Posted on 2006-03-10 23:08:29 by Homer
EvilHomer2k,

The gold (hollow) ball will float in water, regardless that it is made of gold - in order for it to weigh the same as the titanium ball, it must have a very thin wall section, and therefore contains a large volume of air relative to its volume of metal.


    I am afraid not.  Here's why.  Both balls are the same size, shape, and weigh the same.  Therefore they both displace the same amount of liquid.  The amount of liquid displaced is equal to the amount of lifting (floating) force.  So if one ball floats or sinks, so will the other.  Since the solid titanium ball will surely sink, so will the identically shaped and equally heavy gold ball.  Ratch
Posted on 2006-03-11 00:03:16 by Ratch
1)
group 1 are all symmetrical about the vertical axis
group 2 are all symetrical about the horizontal axis
group 3 have rotational symmetry about the axis out of the plane of the letter, rotate by 180deg and you get the same back again
group 4 don't fit the above groups



2) the gold ball has a greater moment of inertia and will pick up spin less easily


3) b-sqrt(b^2 - (a*sin(phi/2))^2) for -180deg < phi < 180deg. Insufficient data for other angles.

4) Depends on who is observing.
The man holding the light will see the light emerge as normal and as if he were standing still
An external, stationary observer will see no light emitted.
The difference is explained by time dilation, to the external observer the travelling man is frozen in time.

edit: 3) above was not right... corrected now.
Posted on 2006-03-13 16:03:28 by pdixon
pdixio,
All correct!

By the way, full solution for the third problem will be:

x^2+y^2 = (a/2)^2

x = (a/2)*cos(phi)  ==>

y^2 = (a/2)^2-((a/2)*cos(phi))^2
y^2 = (a/2)^2(1-cos(phi)^2)
y^2 = (a/2)*sin(phi)^2

((a/2)-x)^2 + y^2 + (b-h)^2 = b^2
((a/2)-((a/2)*cos(phi)))^2 + ((a/2)*sin(phi))^2 + (b-h)^2 = b^2
(a/2)^2 - 2(a/2)(a/2)*cos(phi) + (a/2)^2*cos(phi)^2 + (a/2)^2*sin(phi)^2 + (b-h)^2 = b^2
(a/2)^2(1-2cos(phi)+cos(phi)^2+sin(phi)^2) + (b-h)^2 = b^2
(a/2)^2(2-2cos(phi)) + (b-h)^2 = b^2
((a^2)/2)*(1-cos(phi)) + (b-h)^2 = b^2

(b-h) = sqrt(b^2-(a^2)*(1-cos(phi))/2)

h = b-sqrt(b^2-(a^2)*(1-cos(phi))/2)



P.S. Guys if you know any cool puzzles/problems - start posting them. Don't be shy :)
Posted on 2006-03-13 20:57:02 by arafel
Don't be shy

OK, try this one.
You have 12 identical balls but one has a fault which makes it very slightly different in weight to the others. You don't know if it's too heavy or too light.

The only tool you have to check is an accurate balance with a pan on the left, a pan on the right and it tilts in the direction of the heavier pan, like this: http://www.thebest3d.com/smart3d/Libra.jpg

What is the minimum number of weighings you need to make to guarantee to identify the faulty ball and whether it is heavy or light?
.. of course, you have to explain how your method works.

You could compare every single ball with every other, that would take 66 attempts, but there are a lot of redundant weighings in there.
You could choose 1 ball and compare it to all others one at a time, this take from 2 (if your lucky) to 11 (if unlucky) weighings, so you would need 11 weighings to guarantee to find the answer with that method.

I'll give you a clue, the answer is below 11.

Paul.
Posted on 2006-03-14 07:34:27 by pdixon
There is a similar problem. (but much simpler)

9 balls. And you know that one of them is lighter than the others.

(3 or less steps)

Posted on 2006-03-14 09:44:08 by Azrim
Split the balls into 4 equal group. Lets call it A,B,C,D

weight A against B
weight A against C

We have seven possible outcomes now:

I) All three groups are equal. Means the faulty ball is in fourth group. Take two balls out of fourth group and weight them:

  a) If no balance, replace the lighter ball with the third ball. Weight again. If libra balances, then the ball we removed is lighter than other, if stays as before then other ball is heavier than others. (total 4 weightings)

  b) If balance. Replace one of the balls with third ball which we have left outside. Weight again to check if the ball is heavier or lighter (total 4 weightings)

II) group A heavier/lighter than B and C. Or only B heavier/lighter than A. Or only C heavier/lighter than A. For example B was lighter (the following method is correct for all six outcomes so I'll refer only to B group case) so we know that there is a lighter ball in group B and all other groups are of correct weight.
Split B group (that is two balls and one). Lets calls group with one ball H and other J. Add to H ball from A,C or D. And weight H against J.

  a) If H lighter than J. Then original ball from H is what we are looking for (total 3 weightings)

  b) If J is lighter. Weight balls from J against each other to determinate the faulty one (total 4 weightings)


So the minimum required number of weightings is either four or three. Depends on the case.
Posted on 2006-03-14 10:27:54 by arafel
Arafel,
  so your answer is 4.

.. but there is a better method which gives a lower answer.

Paul.
Posted on 2006-03-14 11:19:06 by pdixon
Ok, I give up.

12 balls and 3 weightings means checking 27 different outcomes (3*3*3 possible scenarios, with 3 initial groups with four balls in each ). Anyhow that's just too much  :)
Posted on 2006-03-14 13:43:14 by arafel
For the first,

- Consideration 1: There are 2 posible results when weighting the same number of balls in the balance at each side equal or not equal E, NE.
- consideration 2: if a == b, c != d, then c or d equals one of a or b, that is b == c XOR b == d is true. In other words a == b and b == c, or the other words a == b and b == d but not both of them.


Take groups of 1-1 and you will have 6 groups thus 6 weights, 5 of them are E and the other is NE.
You already know wich one is more weight than the other in the case of NE.
Take a ball of a group that result in E and compare to one ball of the result of NE, watching if they are balanced or unbalanced will say wich one of them is equal and because you already know the comparation, also say wich one is weight.


The other?
Take groups of 3, you will have 3 groups.
First compare two of them, if they are equal, then the other group of 3 contain a ball that is less weight than the others, if not, then you already have the group in the side of the balance that is more high.

Take the group of 3 that you already know that contain the less weight ball.

do a secont comparation and you will see if they are equal, then the other (the one that havent been balanced) is the ball with less weight, if not, you already have the one in the side that is more high of the balance.
Posted on 2006-03-14 13:44:14 by rea
Also for the first answer, you can reduce the number if you take first groups of 4 balls, compare first 2 of them, and you already have a group of 4 that contain the defective ball.

Then you will have 2 groups of two one of them is E and the other NE inside the group of 4 that you already know that contain a defective ball, then weight one ball of E with the other of NE, and you will have reduced the initial 6 weights + 1 to...

1 weight + 2 (for get wich one of the two groups of 2 balls is) + 1 weight for the final weight.


----------------
Correction, take 3 groups of 4 A,B,C, compare AB and BC, you already know if the ball is light or weight, then you need 2 extra balance for get the defective ball.

I dont see how you can doit in 3 movements, because you need at less 1 for know if the ball is weight or lighter, the only way to do it in 3 movements (at less like I see) is dont say if the ball is lighter or weight, and you delete 1 movement.
Posted on 2006-03-14 14:03:49 by rea