You should hit a white ball by red one to sand the white ball directly (without reflection from the sidewalls) into a packet. How many choices you have to do that if: 1. The balls have the same weight. 2. The red ball is twice heavier than a white one.
So, my friends, nobody likes this problem, as I can see… But actually this simple kinematics problem lies in the foundation of the most principal branches of Physics. Let us return to the beginning… 1. As you know, if you have to send white ball in the direction BB (see the picture) due to hit of black ball A (the Initial disposition in the picture), you have the only one chance. And this chance is to aim ball A at the direction AA to hit white ball in point C (marked blue). 2. If the masses of both balls are the same, it is easy to prove that proposed solution is a unique: there is no other chance to send white ball along BB by hitting it with ball A. [Please, somebody should reveal (post) a proof of that!] 3. But what about case if mass of ball A is twice more than mass of ball B? When we will find out the answer, we will see amazing possibilities to reconsider a lot of our principal assumptions about many of the most fundamental postulates of Physics…
You aren't getting any responses because this doesn't really make a whole lot of sense! What do you mean by "how many choices"? Hitting at different angles? If you are thinking of the pocket as exactly fitting the ball, then there is only one (without rebounds). If the pocket is at all larger than the ball, then there are an infinite number of angles that will work. Different speeds? If there is no friction, any speed will work. If there is friction, then there will be a lower limit on how fast you can hit the cue ball to make the object ball go into the pocket but there will still an infinite number of possible speeds that will work. I must be misunderstanding. I don't see how any of this will cause us to "reconsider a lot of our principal assumptions".
Very, very good, guys! Both of you have noticed good points of the problem. But I have designed this problem exactly to show that Physics is a quantitative theory of Nature. That there are lots of very important problems that can not be solved ... by intuition, or any verbal logical analysis. That mathematics is the only way, which solves the problems and brings on light the possibilities, some of which can not be recognized ... without Math. So, let us go slowly and accurately. Can somebody post the prove that there indeed "if you are thinking of the pocket as exactly fitting the ball, then there is only one" solution? (Everything is considered non-relativistic, so there is no way to blame ...SRT!) When we will have this prove we will be able to go further...
Yuriy, are you serious? I do not have time for your trivial and petty problems. Do you actually believe that people will discuss insignificant "difficulties" with you that will result in Nothing. Start working on something that actually matter!
I am not sure what you mean by "how many choices?" Are you referring to how many pockets there are? For one set up assuming 6 pockets [the white ball spotted, the red ball on, or near, a miorror reflection of the white ball - the opposite "spot" would work. You may cut the white ball in one of the two nearest pockets, or you may bank the red ball on the back rail (nearest the white ball) back and then cut the white ball into one of two side pocket or one of two remaining corner pockets. Of course you may use variations on banking the red ball with as many sidewalls that do not diminish the momentunm of the collision process such that the white ball is not driven sufficiently far to enter a pocket. With sufficient velocity of the red ball struck with the cue stick the white each ball can be sent into the pocket on a staight line or with a "cut". The difference in weight of the ball can be compensated for with velocity variations of the moving red ball (AKA the cue ball). Now, what if the heavier red ball strikes the lighter white ball with the balls aligned perfectly? Will the heavier red ball follow the lighter one - angle of collision = 0 - into the pocket because of the excess momentum of the red ball? Not necessarily. By using low English generated backspin of the heavier red will prevent a scratch. Or this may be another choice, scratching that is. Or if you are playing "position" (still straight line redball-whiteball-hole) you might try using variations of English. The safest technique, depending on the relative distance balls and pocket, would be a variation on low English, either left or right. Geistkiesel
Try to give the mathematical proof to all your intuitive statements and you immediately will find out what I mean.... Remember? This problem is specially designed to show that there are the scientific problems that can not be apprehended ... due to intuition and "magic ... logical analysis" without application of Math. For MacM this problem is ... none resolvable, because it requires a math a little higher than a^2 +b^2=c^2, but you did not proclaim (yet) that you ...forgot even elementary Math! You can do it! So, DO IT!
What a petty pitty. I'm really proud that I have stuck in your craw. Please Register or Log in to view the hidden image!
Because our "experts" from crank-Museum are silent about solution of this classic problem, I will show you mine. See it here.
