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The Proof is Trivial!

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Original post by bananarama2
F=12mVCdv2A F= \frac{1}{2}\frac{m}{V}C_d v^2 A


Yes I second this.
Original post by und
The force will be the same for any given velocity, but the acceleration will be different since F=ma.


I promise you mass doesn't come in.

mg−12mVCdv2A=ma mg- \frac{1}{2}\frac{m}{V}C_d v^2 A =ma

An I missing something?
(edited 10 years ago)
Original post by Jkn
Hahahaha :lol:
Quite the contrary! The nature of assumption is, in this case, such that if the problem does not suggest air resistance is being ruled out, we must assume that it is not! Lack of assumption in Physics forces you towards more sophisticated theories whilst lack of assumption is mathematics often renders a problem un-doable! :smile:


But the lack of assumptions does not mean you can assume whatever you wish to make the problem solvable. I could equally validly assume things that makes the question still unsolvable, such that bananarama always stack the coins up before putting them in the bag such that the surface which air resistance acts on is the same. Unless you name the particular assumptions you want to hear about, what you are saying only justifies you receiving an 100-page document exploring all the different possibilities under each possible assumption. :wink:

Also I think you've been caught in a corner with your question and are just flailing. :tongue:
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und
OP
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Original post by bananarama2
Drag F=12mVCdv2A F= \frac{1}{2}\frac{m}{V}C_d v^2 A

Weight mg mg

Ratio is independent of mass...

I think you're confusing the density of the air with the density of the object.
Reply 1544
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Jkn
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Original post by bananarama2
Drag F=12mVCdv2A F= \frac{1}{2}\frac{m}{V}C_d v^2 A

Weight mg mg

Ratio is independent of mass...

Are you sure that is right? You are asserting that, with all things constant, the mass of an object is directly proportional to it's drag?
Original post by ukdragon37
But the lack of assumptions does not mean you can assume whatever you wish to make the problem solvable. I could equally validly assume things that makes the question still unsolvable, such that bananarama always stack the coins up before putting them in the bag such that the surface which air resistance acts on is the same. Unless you name the particular assumptions you want to hear about, what you are saying only justifies you receiving an 100-page document exploring all the different possibilities under each possible assumption. :wink:

Also I think you've been caught in a corner with your question and are just flailing. :tongue:

Hahaha, perhaps that's what I'm expecting :colone:

Nope! :tongue:

Edit: With lack of assumption, we fall to the most valid theories. From these valid theories (special relatively, etc...) we then use simplified models based on the requirements of the question though we must be sure of the validity of the things we assume are negligible. In this case, whilst it is safe to assume 'Bananarama 2' isn't capable of throwing coins at speeds close to the speed of light, etc... it is not safe to assume there is no air resistance as, having a different dependency on mass than weight does, it is crucial in measuring a difference between the timings of the drops. :tongue:
(edited 10 years ago)
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Jkn
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Original post by und
I think you're confusing the density of the air with the density of the object.

Precisely!

The correct equation: FD=12ρv2CDAF_D=\frac{1}{2} \rho v^2 C_D A with ρ\rho being the density of the fluid :smile:
Problem 227 **/***

Solve y′′+αy=u(x) y'' + \alpha y = u(x) , for any function u(x).
Reply 1547
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und
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Of course, since this is "The Proof if Trivial!", I will accept no solution that doesn't model for the changing density of the air.
Original post by Jkn
Precisely!

The correct equation: FD=12ρv2CDAF_D=\frac{1}{2} \rho v^2 C_D A with ρ\rho being the density of the fluid :smile:


It's lateish :tongue:
Original post by und
Of course, since this is "The Proof if Trivial!", I will accept no solution that doesn't model for the changing density of the air.


I think we should go for a general solution to Navier Stokes equation :tongue:
Original post by Lord of the Flies
Yes I second this.


:facepalm: Backing the wrong person there:tongue:
Original post by Jkn
Edit: With lack of assumption, we fall to the most valid theories. From these valid theories (special relatively, etc...) we then use simplified models based on the requirements of the question though we must be sure of the validity of the things we assume are negligible. In this case, whilst it is safe to assume 'Bananarama 2' isn't capable of throwing coins at speeds close to the speed of light, etc... it is not safe to assume there is no air resistance as, having a different dependency on mass than weight does, it is crucial in measuring a difference between the timings of the drops. :tongue:


You are utilising a circular argument here. You say "based on the requirements of the question" as if one of the requirements is that the question is solvable, when this is not stated. But clearly an assumption of "the question is solvable" is not negligible. :tongue: I.e. what you said would be valid had you also stated in the question "assume there is a solution".

EDIT: In fact, not only are you predisposing towards the assumption that the question is solvable, the question itself asks whether the task could be done. Sounds like a large circle to me if to solve it you have to assume the positive answer first and make assumptions such as air resistance not negligible.
(edited 10 years ago)
Original post by bananarama2
:facepalm: Backing the wrong person there:tongue:


LOL I don't even know what you guys are talking about.

Original post by james22
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Original post by Mladenov
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Original post by Jkn
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Original post by henpen
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The numbers are mixed up, there appears to be two problems 214. Starting from the second 214 (your series Jkn), could you +1 to each question/solution number? I have amended mine.
Original post by Lord of the Flies
LOL I don't even know what you guys are talking about.


I gathered :colone:



I'll argue with you some more after M3 tomorrow guys. I'm off to read some of Catch 22.
I haven't posted on here in ages, but my first offering will be the following, which is a slight adaptation of a question from the cambridge entrance exam in 1981 (so roughly STEP level)...

How many sixth powers are there of form 2013n+9,n∈N2013n+9, n \in N?

It requires AS knowledge at the most, and I know of no solution that uses more complex methods.

EDIT: the original problem was flawed. It works now.

Hint:

Spoiler

(edited 10 years ago)
Original post by TheMagicMan
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How was first year? :colone:
Original post by ukdragon37
How was first year? :colone:


Difficult. Didn't do as much work as I probably should have in the first two terms but I somewhat pulled it together for the exams.
Reply 1557
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Jkn
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Original post by ukdragon37
You are utilising a circular argument here. You say "based on the requirements of the question" as if one of the requirements is that the question is solvable, when this is not stated. But clearly an assumption of "the question is solvable" is not negligible. :tongue: I.e. what you said would be valid had you also stated in the question "assume there is a solution".

Well the question must have a solution by nature because either you can consistently predict the coin is counterfeit or you cannot :tongue: Though, as I write this, it occurs to me that you are referring to the question within the question:

In Physics you are permitted to take information from the natural world in order to form theories and so, if I asked a question like "where will a coin land if I drop it off the Eiffel Tower?", even without telling you what to assume etc.. a method to find the solution would exist by simply dropping coins off the Eiffel Tower and, if the effect is large enough, you will be able to do a probabilistic analysis as to where it lands. In this case you may deduce that the information will not be retained. If I attached to this the assumption that air resistance is negligible, we could build a tube beside the Eiffel Tower and drop it in the tube. Obviously you do not necessarily need experiments to deduce answer to questions (as we may deduce our answers from what have been observed in previous experiments) and so as answer must always exist.

Note that, on a philosophical level (detached from day-to-day Physics), this would most-certainly be up for debate :tongue:

EDIT: In fact, not only are you predisposing towards the assumption that the question is solvable, the question itself asks whether the task could be done. Sounds like a large circle to me if to solve it you have to assume the positive answer first and make assumptions such as air resistance negligible.

You do not assume it is negligible for that very reason :tongue:
Original post by bananarama2
x

You have to solve it now.. :colone:
Reply 1558
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Jkn
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Original post by Lord of the Flies
LOL I don't even know what you guys are talking about.

****s sake :lol: Hahahahahahah!
The numbers are mixed up, there appears to be two problems 214. Starting from the second 214 (your series Jkn), could you +1 to each question/solution number? I have amended mine.

Actually I believe ukdragon withdrew his problem :tongue: Hence why is numbered mine 214 :smile:
Original post by FireGarden
Problem 227 **/***

Solve y′′+αy=u(x) y'' + \alpha y = u(x) , for any function u(x).


This technically isn't solvable for any function u(x). You need certain conditions on the integrability of u to hold (something like u continuous is sufficient but not necessary).

I believe a necessary and sufficient condition would be something along the lines of having a measure 0 set of discontinuities on the domain of y in the solution.

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