# M4 - Motion of Particles in One DimensionWatch

#1
M4 - Ex3A Q.8

A small stone of mass m is projected vertically upwards with speed Vtan(a) where a is a constant. Air resistance varies as the square of the speed. Terminal speed is V.
Show that the total energy, kinetic plus potential, lost in its ascent is:

1/2 mV2[tan2(a) - 2 ln sec(a)]

This is what I thought:

At the top of its ascent, it will have zero kinetic energy, so the kinetic energy lost = 1/2 mV2tan2(a).

The potential energy it gains will be equal to mgh where h is the height it reaches, ie. the distance, x, it has travelled when v=0.

So, taking the upwards direction as positive:

ma = -mg - v2
=> a = - (g + v2/m)

v dv/dx = - (g + v2/m)
=> INT v/(g + v2/m) dv = INT -dx
=> m INT v/(mg + v2) dv = A - x
=> m/2 ln (mg + v2) = A - x

Using x=0, v = Vtan(a):

A = m/2 ln (mg + V2tan2(a))

So, x = m/2 ln [(mg + V2tan2(a))/(mg + v2)]

Subbing in v=0 for the height it reaches clearly isn't going to give the required value for x.

Can anyone see where I've gone wrong?
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12 years ago
#2
Mistake's in the first line: ma = -mg - v2

It should be: ma = -mg - kv2

There has to be a constant of proportionality
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12 years ago
#3
Lol - took me a while, but I've done it...

As you rightly say, the kinetic energy is the easy bit. So we have to work out mgh, and thus must find the height at which it comes to rest instantaneously.

As a preliminary thing, we're told terminal velocity is V. This means weight = air resistance at this speed i.e. mg = kV2, so V2 = mg/k

So...

v*dv/dx = -{g + (k/m)v2}

-∫0Vtan(a) v/{g + (k/m)v2} = ∫0h dx

[(m/2k) * ln{g + (k/m)v2]0Vtan(a) = h

ln{(g + (k/m)V2tan2(a))/g} = 2kh/m

ln{1 + (k/mg)*V2tan2(a)} = 2kh/m

Now using, V2 = mg/k on both the LHS and the RHS....

ln{1 + tan2(a)} = 2gh/V2

h = (1/2)*(V2/g)*ln{sec2(a)}

so PE gain = mgh = (1/2)*mV2*ln{sec2(a)}

which nearly fits your given answer, but I still think it's sec2(a)!

EDIT:(1/2)mV2*ln{sec2(a)}

can be simplified to

mV2*ln{sec(a)}

by just taking the power of 2 out of the log....
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12 years ago
#4
Arghh, I remember this question.
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#5
(Original post by Worzo)
Mistake's in the first line: ma = -mg - v2

It should be: ma = -mg - kv2

There has to be a constant of proportionality
I was confused by the question, it says "resistance varies as the square of the speed." I thought that if a constant of proportionality was necessary, it would say something like "resistance varies proportionally with the square of the speed". Oh, well.

(Original post by Worzo)
As a preliminary thing, we're told terminal velocity is V. This means weight = air resistance at this speed i.e. mg = kV2, so V2 = mg/k
Ahhh, I wasn't sure what to do with the terminal speed. Would never have got that. Thanks
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12 years ago
#6
When something varies "as" or "with" or "proportionally with" something else, you always need a constant of proportionality in order to keep the equation dimensionally consistent (even if that constant has no units, or is 1).

If a force, air resistance, varies with the square of the speed, you'd have:
F = k*v2

So the units are:
N = [k] * m2s-2

If k wasn't there, you'd have newtons = (metres per second)2, which doesn't make sense.

Looking at both sides of this equation, we call tell that the units of k, [k], have to be Nm-2s2 in order to make the RHS newtons as well.
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12 years ago
#7
(Original post by e-unit)
Ahhh, I wasn't sure what to do with the terminal speed. Would never have got that. Thanks
You could think of it as if it means acceleration = 0 (which will give what Worzo got). If we had things in terms of t, then usually taking t->inf will do the trick.
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#8
(Original post by dvs)
You could think of it as if it means acceleration = 0 (which will give what Worzo got). If we had things in terms of t, then usually taking t->inf will do the trick.
Yeah, it was the only question in the exercise where you had to think of it that way. All the rest would end up with something like v = k(1-e-t) and, as you say, you'd just let t-->inf.
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