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Validating the kinetic energy for the two particle collision problem

  1. Dec 29, 2012 #1
    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): [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.

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


    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:

    [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 dont see will cancel out to give eq. 3
     
  2. jcsd
  3. Dec 29, 2012 #2
    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.
     
  4. Dec 29, 2012 #3

    TSny

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    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##.
     
  5. Dec 29, 2012 #4
    I guess it can be done that way, althought I had something else in mind:

    [tex]\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 [/tex]

    [tex]\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) [/tex]
     
  6. Dec 29, 2012 #5

    TSny

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    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.:redface:)
     
  7. Dec 29, 2012 #6
    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.
     
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