mechanics q (1 Viewer)

[D94]

Wouldn't the initial equation where the particle is vertically projected be ma = -mg-kv^2 ?

Well, I thought we have the particle moving up in the positive direction, it has a weight force in the downwards direction, and air resistance acts against the moving particle, so in the downwards direction. Sum of the forces in the y direction yields the result: ma-mg-kv²=0, then rearrange that in terms of 'a'.

However, I think it depends on how you define g, and which direction you take g as... or, I could just be completely wrong.

 

[Carrotsticks]

I don't think of it as 'net forces' or 'forces add to zero' or anything like that. I think of it as:

"Everything versus MG"

So much more simple.

 

[Carrotsticks]

 

[john-doe]

View attachment 26187

legend

i still dont understand when resistance is force or acceleration...some questions require it as force while others dont! and it can alter the final solution!

 

[Carrotsticks]

legend

i still dont understand when resistance is force or acceleration...some questions require it as force while others dont! and it can alter the final solution!

I don't quite get what you mean here about how it can alter the final solution...

 

[Sindivyn]

I think he means how resistance is sometimes mkv^2 or mkv versus kv^2 or kv.

 

[Carrotsticks]

I think he means how resistance is sometimes mkv^2 or mkv versus kv^2 or kv.

Oh yes I understand now. Usually, the question in the HSC tells you what expression to use by having the first part "Prove that acceleration is XXXX".

In actual fact, mkv^2 or kv^2 should NOT make a difference in the final result numerically, because m is a constant and k is a constant too, so think about it as m being 'a part of k' when just written as kv^2.

As for whether it's kv^2 or k/v or k/v^2 etc etc, the question will say something like "Resistance is proportional to the reciprocal of the velocity, squared" or 'Resistance is proportional to the velocity squared" etc.

 

[Sindivyn]

Ah, yeah, would you have to specify that k incorporates both a constant as well as the mass though?

 

[Carrotsticks]

Ah, yeah, would you have to specify that k incorporates both a constant as well as the mass though?

If I were doing a relatively ambiguous question, then yes.

But otherwise I would look at the final result. If it has M in it, it means your resistance shouldn't have M in it, to avoid them all cancelling out.

If it doesn't have M in it, that implies you have ma = mg - mkv^2 or something like that, such that all the M's cancel out.

 

[Sindivyn]

If I were doing a relatively ambiguous question, then yes.

But otherwise I would look at the final result. If it has M in it, it means your resistance shouldn't have M in it, to avoid them all cancelling out.

If it doesn't have M in it, that implies you have ma = mg - mkv^2 or something like that, such that all the M's cancel out.

That makes sense, made the mistake of not cancelling m out in my trial when the final answer had no m - didn't even think to go back to change it. Wont make that mistake again , thanks.

 

[D94]

View attachment 26187

the answer for the velocity at the point of projection is

u^2/ {sqrt [1+ (ku^2/mg)]} note the k in this answer doesnt include mass

?

 

[Carrotsticks]

I didn't read any prior posts, I just answered the question straight out without knowing where to aim.

So if that's the case, then I guess the original resistance was intended to be kv^2 as opposed to what I did mkv^2 to simplify things.

 

[jyu]

http://www.itute.com/wp-content/uploads/rn_kinematics_mechanics.pdf
p8, 9

 

[Amundies]

http://www.itute.com/wp-content/uploads/rn_kinematics_mechanics.pdf
p8, 9

How long does it usually take for a class to cover all of that?

 

[john-doe]

If I were doing a relatively ambiguous question, then yes.

But otherwise I would look at the final result. If it has M in it, it means your resistance shouldn't have M in it, to avoid them all cancelling out.

If it doesn't have M in it, that implies you have ma = mg - mkv^2 or something like that, such that all the M's cancel out.

what if they dont give u the final result..eg finding the answer rather than showing it?? u wouldnt know then i guess

 

[D94]

Resistance is a force, you can't really dispute that in these types of basic questions. F = ma, therefore, even if you only have kv², that equals 'ma', not 'a'. There is nothing ambiguous about it, no assumptions are needed - just think resistance is a force, then equate whatever they've given you with the other terms.

 

[Fus Ro Dah]

Resistance is a force, you can't really dispute that in these types of basic questions. F = ma, therefore, even if you only have kv², that equals 'ma', not 'a'. There is nothing ambiguous about it, no assumptions are needed - just think resistance is a force, then equate whatever they've given you with the other terms.

I think the confusion came from the fact that suppose we have ma=kv^2, then k can be considered to be mpv^2, where p is some constant, in which case we have a=pv^2. It's just notation, which should be specified in the question.