# Homework Help: Newtonian Mechanics - the movement of a particles

1. Mar 4, 2010

### bobey

(Moderator's note: thread moved from "Differential Equations")

A particle of mass m which moves along a horizontal straight line with a velocity of v encounters a resistance of av + b(v^3), where a and b are constants. If there is no other force beside the resistance acting on the particle and its initial velocity is U , show that the particle will stop after it has moved a distance of m(ab)^(-1/2)tan^(-1)(U(b/a)^1/2). Also show that the velocity is 1/2U after time (m/2a)ln(4a+b(U^2))/(a+b(U^2)))

to show the distance when the particle stops :

i use :

F = ma

=> m (dv/dt) = - (av+b(v^3))

i let v = y and t = x to make it less confusion ....(1)

thus m(dy/dx) = -(ay+b(y^3))

dy/dx = -(ay/m)-(b(y^3)/m)

dy/dx + (a/m)y = -(b/m)(y^2) ... (2)

(y^-3)(dy/dx) + (a/m)(y^-2) = (-b/m) ===> this is Bernoulli equation

let v = (y^-2)

thus, dv/dx = -2(y^-3) (dy/dx)

=> -1/2(dv/dx) + (a/m)v= -b/m

searching for integrating factor, miu(x) = exp(integrating(a/m)dx)

thus, miu (x) = exp(ax/m)

miu(x) x (2) : d/dx (exp(ax/m).v) = exp(ax/m).(-b/m)

exp(ax/m).v = (-b/m) integrating (exp(ax/m) dx)

v = (-b/a) + c(exp-(ax/m))

but v = (y^-2) = (v^-2)

thus,

v^-2 = (-b/a) + c(exp-(ax/m))

v^2 = (1/c(exp-(ax/m))) - (a/b)

v = sqrt ((1/c(exp-(ax/m))) - (a/b))

applying initial condition where v_0 = u, x=0, t=0

c = 1/u^2 + b/a

thus v = dx/dt = sqrt ((1/(1/u^2 + b/a)) - a/b)

when particle stop dx/dt = 0

then

0 = sqrt ((1/(1/u^2 + b/a)) - a/b)
.
.
.
x = m/a ln |(1/ab)+1|

why i can't get the answer given which is m(ab)^(-1/2)tan^(-1)(U(b/a)^1/2)???????

where i'm going wrong??????

p/s : i don't know to use SYNTAX. sorry =p

Last edited by a moderator: Mar 4, 2010
2. Mar 4, 2010

### gabbagabbahey

But this makes things more confusing, as does your unnecessary use of integrating factors and Bernoulli's equation. In fact, it's such confusing notation, that I can't be bothered to try and go through it and see where you are going wrong.

For example, here:

You are essentially defining $v=\frac{1}{v^2}$...recycling variables like this (using $v$ to represent more than one quantity) leads to confusion, and may have lead to your error(s).

$$m\frac{dv}{dt}=-(av+bv^3)\implies -m \int_{v_0}^{v} \frac{dv'}{av'+b(v')^3}=\int_{t_0}^{t} dt=t-t_0$$