Validating the kinetic energy for the two particle collision problem

1. Dec 29, 2012

Andreasdreas

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
The relation between V and F: $$F(x)=-\frac{dV}{dx}$$

3. 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 dont see will cancel out to give eq. 3

2. Dec 29, 2012

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. xi.

3. Dec 29, 2012

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 $x_2$.

4. Dec 29, 2012

cosmic dust

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

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

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

5. Dec 29, 2012

TSny

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.)

6. Dec 29, 2012

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.