Mechanics question (1 Viewer)

[vds700]

froim Kings 2006 Trial

A particle is released so that it falls vertically , under the influence of gravity, in a medium whose retardation varies as the square of the velocity. Show that the terminal velocity of the particle is given by:

V = sqrt(g/k)

i am not sure about setting up the force equation. Is it:
1. R = ma = mg - mkv^2 or
2. R = ma = mg - kv^2?

Once i know this im fine with the rest of the question, just to find v as a function of t and the terminal velocity occurs when t->infinity etc.

Thanks for any help.

 

[vds700]

ok thanks...but how can u be surre in an exam, aren't they supposed to tell u the equation so theres no confusion???


Or do u just assume the constant is always mk unless they tell u otherwise??? Im confused

 

[vds700]

they'll probably tell you in an exam. here its obvious though as they give you the terminal velocity which is independent of m. if it was ma = mg - kv^2 the terminal velocity would be sqrt(mg/k)

ah right that makes sense. Thanks mate!

 

[vds700]

ask your teacher if you can just do:

the velocity will increase until the magnitude of the resistive force approaches that of the gravitational force, hence
mg - mkV^2 = 0
V = sqrt (g/k)

wow thats a much better way of doing it, yeah i will find out if u can do it this way....much quicker than the way I did it.


EDIT: The question is only worth 2 marks so I think thats the way its meant to be done.

 

[vds700]

ask your teacher if you can just do:

the velocity will increase until the magnitude of the resistive force approaches that of the gravitational force, hence
mg - mkV^2 = 0
V = sqrt (g/k)

i asked my teacher and he said that way is fine