- #1

Andreasdreas

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## Homework Statement

The following is a summaration of a piece of text in my textbook:

The setting is two particles moving in one dimension, under mutual forces, so that the equations of motion are

(1): [itex]m_1 a_1=F[/itex] and [itex] m_2 a_2=-F [/itex] (

*a*stands for acceleration)

[itex]F[/itex] is only a function of the relative distance between the particles [itex]x=x_1-x_2[/itex] : [itex]F=F(x)[/itex] so [itex]F[/itex] is conservative and a potential energy function [itex]V(x)[/itex] can be introduced.

The law of conservation of energy then takes the form

(2): [itex] T + V=E=constant [/itex]

with

(3):[itex] T=\frac{1}{2}m_1(v_1)^2+\frac{1}{2}m_2(v_2)^2 [/itex] (

*T*is the kinetic energy and

*v*stands for velocity, this is also the relative velocity of the two particles: [itex]v=v_1-v_2[/itex]).

The books then says that:

eq. 3 can easily be verified by differentiating eq. 2 and using eq. 1 and the realtion between

*V*and

*F*.

## Homework Equations

The relation between

*V*and

*F*: [tex] F(x)=-\frac{dV}{dx} [/tex]

## The Attempt at a Solution

I am quit unsure with what i should differentiate with respect to when differentiating eq. (2)

*V*is a function of

*x*and T is a function of

*v*. What i have tried is this:

[tex]T+V=constant \implies \frac{dT}{dv} \frac{dv}{dt}=-\frac{dV}{dx}=F(x)[/tex]

Then i am unsure what term i should use for

*F(x)*.

What i have done is to put it equal to [itex]ma [/itex] where

*a*is the acceleration of the relative position [itex]x=x_1-x_2[/itex] and i guess

*m*should be [itex]m_1+m_2[/itex] .

But integrating [itex]ma [/itex] dosent seem to yield eq. 3, I have integatred like this:

[tex]\int \int (ma) dt dv = \int (mv) dv =\frac{1}{2}mv^2=\frac{1}{2}(m_1+m_2)(v_1-v_2)^2[/tex]

[itex]\frac{1}{2}(m_1+m_2)(v_1-v_2)^2[/itex] must give a lot of terms which i don't see will cancel out to give eq. 3