Validating the kinetic energy for the two particle collision problem

[Andreasdreas]

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): $m_1 a_1=F$ and $m_2 a_2=-F$ (a stands for acceleration)

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

The law of conservation of energy then takes the form

(2): $T + V=E=constant$

with

(3):$T=\frac{1}{2}m_1(v_1)^2+\frac{1}{2}m_2(v_2)^2$ (T is the kinetic energy and v stands for velocity, this is also the relative velocity of the two particles: $v=v_1-v_2$).

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: $$F(x)=-\frac{dV}{dx}$$

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:

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

Then i am unsure what term i should use for F(x).
What i have done is to put it equal to $ma$ where a is the acceleration of the relative position $x=x_1-x_2$ and i guess m should be $m_1+m_2$ .
But integrating $ma$ dosent seem to yield eq. 3, I have integatred like this:
$$\int \int (ma) dt dv = \int (mv) dv =\frac{1}{2}mv^2=\frac{1}{2}(m_1+m_2)(v_1-v_2)^2$$
$\frac{1}{2}(m_1+m_2)(v_1-v_2)^2$ must give a lot of terms which i don't see will cancel out to give eq. 3

 

[cosmic dust]

When we say tha energy is conserved, we mean that is constant as time passes. So the derivative of energy w.r.t. time will be zero. Althought V depends explicitly on x = x₁-x₂, since x₁ and x₂ are functions of time, then V is actually a function of time. So try to differntiate T+V=E w.r.t. time and then use the equations of motions to get rid of the forces. Also, the force applied to "i" particle will be equall to minus the partial derivative of V w.r.t. x_i.

 

[TSny]

To elaborate a little on cosmic dust's good suggestion:

Note that $T$ is a function of $v_1$ and $v_2$: $T(v_1, v_2)$ and $V$ is a function of $x_1$ and $x_2$: $V(x_1, x_2)$. So, overall, the total energy is a function of four variables: $E(v_1, v_2, x_1, x_2)$.

The time derivative is $\dot E = \frac{\partial E}{\partial v_1} \dot v_1 + \frac{\partial E}{\partial v_2} \dot v_2 + \frac{\partial E}{\partial x_1} \dot x_1 + \frac{\partial E}{\partial x_2} \dot x_2$

In carrying this out, you’ll need to use $\frac{\partial V}{\partial x_1} = \frac{\partial V}{\partial x} \frac{\partial x}{\partial x_1}$ and similarly for $x_2$.

 

[cosmic dust]

I guess it can be done that way, althought I had something else in mind:

$$\frac{dE}{dt}=0\Rightarrow \frac{dT}{dt}+\frac{dV}{dt}=0\Rightarrow \frac{dT}{dt}+\frac{\partial V}{\partial {{x}_{1}}}{{v}_{1}}+\frac{\partial V}{\partial {{x}_{2}}}{{v}_{2}}=0\Rightarrow$$

$$\Rightarrow \frac{dT}{dt}-{{m}_{1}}{{\dot{v}}_{1}}{{v}_{1}}-{{m}_{2}}{{\dot{v}}_{2}}{{v}_{2}}=0\Rightarrow \frac{dT}{dt}=\frac{d}{dt}\left( \frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2} \right)$$

 

[TSny]

I guess it can be done that way, althought I had something else in mind:

$$\frac{dE}{dt}=0\Rightarrow \frac{dT}{dt}+\frac{dV}{dt}=0\Rightarrow \frac{dT}{dt}+\frac{\partial V}{\partial {{x}_{1}}}{{v}_{1}}+\frac{\partial V}{\partial {{x}_{2}}}{{v}_{2}}=0\Rightarrow$$

$$\Rightarrow \frac{dT}{dt}-{{m}_{1}}{{\dot{v}}_{1}}{{v}_{1}}-{{m}_{2}}{{\dot{v}}_{2}}{{v}_{2}}=0\Rightarrow \frac{dT}{dt}=\frac{d}{dt}\left( \frac{1}{2}{{m}_{1}}v_{1}^{2}+\frac{1}{2}{{m}_{2}}v_{2}^{2} \right)$$

Yes, that's much better. I lost sight of the question! You want to verify that T has the form given in OP's equation (3). Very good.
(I was assuming (3) and verifying E is constant. Not what was asked.)

 

[Andreasdreas]

Thanks for the responds, i see it clearly now.
I got confused in the different variables and also i was not proberly aware that a kinetic and potential energy is attached to each of the particles.