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Resited motion question (1 Viewer)

shindig

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Hi there
I have been going through the examples in different books and I found same question being repeated in few books but they had different values for the acceleration of the particle and I am not sure which one I should follow. I will type up the question from every book

Cambridge
A particle is projected vertically upwards in a medium where the resistance is proportional to the square of the velocity. The velocity of projection is V, mass is m and initial resistance is mkV2(why is m present in the resistance??
F = ma = -mg -mkv2 and so a = -g -kv2

Terry Lee
A particle is projected vertically upwards from the ground with the initial velocity U. It experiences an air resistance which is proportional to the square of its speed
ma = -mg – kv2 (no m present with the kv2.. why?) and so a = -g -kv2/m

Fitzpatrick
A particle is projected vertically upwards with a velocity of u m/s in a revisiting medium. Assuming that the retardation sue to this resistance is equal to kv2 (why is this different form the Cambridge assumption???), find greatest height and time taken to reach it.
ma = -mg –mkv2 and so a = -g –kv2

Excel book has ma= -mg – R (where R is the resistance)

So which one should I follow ?? could it be that Terry lee’s one is just a typo.
Anyway are you supposed to work the acceleration first and then think about ma? Because I always draw a diagram and from that what I would have got what terry lee has as I do not understand why the extra m is present with the kv2

I would be glad for any help if you could make sense of all that
 

vafa

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the question always mentions what format of r you need to use.
 

Wackedupwacko

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no they are all correct HOWEVER for convinience's sake when u come to algebra use the mkv² form since the m disappears when u convert it to a=-g-kv². and no you are not suppose to work out acceleration first then deal with it. what happens is you are suppose to draw a force diagram. force by definition is ma (newtons 2nd law) then by dividing through by "m" to get "a". Normally when the question states : x is proportional to y you would write it as x = ky. so terry lees book is right in writing it as R as kv² hence:
F= ma = -mg-kv²
a =-g-kv²/m
of course it is more convinient if u used mkv² because it becomes:
a=-g-kv²

if u still dont get it... basically k is a constant. m is a constant. so mk is a constant that can also be called c or in terry lees case k therefore all those forms are correct.
 
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Riviet

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I remember being confused with this earlier in the year, just remember ultimately that m and the constant of proportionality are constants, hence mk is a constant, and in your answer, you just have to specify what your k value represents.
 

shindig

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yeah i get it now. m and k are both constants. Thankyou for clearing that up
 

Naylyn

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resistance is a force, while retardation is an acceleration so of the retardation is kv^2 the resistance will become mkv^2. Which is what is done in the fitzpatrick text book.
 

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