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Michael Watson
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Message 1452362 - Posted: 10 Dec 2013, 15:32:10 UTC

We're all familiar with the formula for the area of a circle: A=Pi . r2.
Who can give a formula for the area of a circle that does not mention it's radius?

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Message 1452365 - Posted: 10 Dec 2013, 15:44:54 UTC - in response to Message 1452362.
Last modified: 10 Dec 2013, 15:45:45 UTC

Circumference squared divided by 4pi

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Message 1452367 - Posted: 10 Dec 2013, 15:49:12 UTC - in response to Message 1452362.
Last modified: 10 Dec 2013, 15:51:10 UTC

Can you prove :in a purely elastic collision between two bowling balls of the same mass that are approaching head on at two different speeds: That they simply rebound and when they do their speeds are exchanged??

Hint: Both momentum and kinetic energy are preserved in an elastic collision.

It might help to pick two specific speeds since you will get two answers, one of which you must show is to be discarded.

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Message 1452404 - Posted: 10 Dec 2013, 20:43:05 UTC
Last modified: 10 Dec 2013, 20:44:37 UTC

To make the problem a little more challenging, how about an area-of-a-circle formula that makes no reference to either the radius, or the circumference?

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Message 1452517 - Posted: 11 Dec 2013, 0:19:08 UTC

a = pi * (diameter/2)^2
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Message 1452739 - Posted: 11 Dec 2013, 12:39:23 UTC - in response to Message 1452367.

Can you prove :in a purely elastic collision between two bowling balls of the same mass that are approaching head on at two different speeds: That they simply rebound and when they do their speeds are exchanged??

Hint: Both momentum and kinetic energy are preserved in an elastic collision.

It might help to pick two specific speeds since you will get two answers, one of which you must show is to be discarded.


The bottom expression is for v2', it is either v1 or v2. v2'=v2 is the trivial solution (no impact happened). Back subbing will show v1'=v2.

(ignore the detour I took to remember how the quadratic equation goes)

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Message 1453064 - Posted: 12 Dec 2013, 1:34:49 UTC

I don't think that you quite have it. Try it with V1 =5 and V2=2 say. I think it will be much easier. You need to state what the new V1 and V2 are. You will get another incorrect solution which you must show is extraneous.

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Message 1453091 - Posted: 12 Dec 2013, 3:32:58 UTC - in response to Message 1453064.

I don't think that you quite have it. Try it with V1 =5 and V2=2 say. I think it will be much easier. You need to state what the new V1 and V2 are. You will get another incorrect solution which you must show is extraneous.

Admittedly my chicken scratches don't exactly flow, but the result is:

Pre-impact velocity of balls: v1, v2
Post-impact velocity of balls: v1', v2'

There are two solutions

Solution 1: v1' = v1, v2' = v2
Solution 2: v1' = v2, v2' = v1

Solution 1 says that neither ball changes velocity. That conserves kinetic energy and momentum but is not the solution we are interested in by inspection because it does not describe an impact event.

Solution 2 is therefore what we want. It shows the swapping of velocities between the balls.

I leave it as an exercise for the reader to plug in specific numbers.

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Message 1453556 - Posted: 13 Dec 2013, 15:38:11 UTC - in response to Message 1453091.
Last modified: 13 Dec 2013, 15:43:17 UTC

The math must show that the exchange solution is correct. The math must show what happens to the velocities and therefore that the other solution does not preserve momentum and by induction kinetic energy and is consequently invalid. You cannot use the apriori idea that the collision must exchange velocities since that is what you are trying to prove.

I am always amazed that Math accurately describes what physically happens in nature--it's the ultimate truth.

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Message 1453557 - Posted: 13 Dec 2013, 15:42:53 UTC

Can you prove that there are two solutions for firing angles to achieve the same range in ballistics. Assume the same muzzle velocity; Ignore air friction and the Coreolis effect.

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Message 1453563 - Posted: 13 Dec 2013, 15:52:03 UTC

Thank you to everyone who responded to the topic of an alternate formula for the area of a circle. Dividing the diameter in half in a formula for the area of a circle seems to refer to the radius. It is possible to get the area without invoking either the radius, diameter, or circumference of a circle. clue: It should take either ten minutes or two hours, as measured by the clock.

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Message 1453620 - Posted: 13 Dec 2013, 21:00:49 UTC - in response to Message 1453556.
Last modified: 13 Dec 2013, 21:01:13 UTC

... I am always amazed that Math accurately describes what physically happens in nature--it's the ultimate truth.

Yes and no...

Yes in that Maths is designed to be self-consistent and so it is it's own "Truth", by design.

With regards to nature, it just so happens that we can accurately match some of the mathematical truths to what we observe in Nature.

Meanwhile, Nature will do it's own thing regardless of what our Maths might describe...


Having said/typed all that, so far we have found some of our Maths to describe and predict what Nature does, right down to some fantastic levels of precision and accuracy...

Keep searchin',
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Message 1453875 - Posted: 14 Dec 2013, 12:00:44 UTC

Having said/typed all that, so far we have found some of our Maths to describe and predict what Nature does, right down to some fantastic levels of precision and accuracy...

Sshhhh silly! If ID hears you he'll start banging on about intelligent design again .......

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Message 1454191 - Posted: 15 Dec 2013, 14:33:07 UTC

OK, so not exactly a mathematical approach...
Take a cylinder of known height, measure its volume using a displacement technique, and so since
volume = area x height
and we now know both height and area we can very simply calculate the area as being
area = volume / height



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Message 1454493 - Posted: 16 Dec 2013, 15:12:50 UTC - in response to Message 1454465.

You could probably use Archimedes "Method of Exhaustion" as well

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