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Homework Help: Can this equation be solved?

  1. Apr 17, 2008 #1
    1. The problem statement, all variables and given/known data

    This problem is related to momentum and energy. I have set the mass to unity to simplfy things and reduce the variable count. u,v,w and x are velocities. The first equation is obtained from energy considerations and the second and third equations from conservation of momentum. I am trying to find velocity u in terms of w (or vice versa) and eliminate v and x or alternatively find v in terms of x and eliminate u and w.

    2. Relevant equations

    [tex]{1 \over \sqrt{1-u^2}}+{3 \over \sqrt{1-v^2}} = {1 \over \sqrt{1-w^2}} + {2 \over \sqrt{1-x^2}}[/tex]

    [tex]v = {u \over \sqrt{9-8u^2}[/tex]

    [tex]x = {w \over \sqrt{4-3 w^2}[/tex]

    3. The attempt at a solution

    I substituted the given values for v and x in the first equation and simplified to get an expression in terms of u and w only:

    [tex]{1 \over \sqrt{1-u^2}}+3{\sqrt{9-8u^2} \over \sqrt{9-9u^2}} = {1 \over \sqrt{1-w^2}} + 2 { \sqrt{4-3w^2}\over \sqrt{4-4w^2}}[/tex]

    By substituting y for [itex]9/u^2[/itex] and z for [itex]4/w^2[/itex] (hoping to extract u or w later) and further simplifying I get:

    [tex]{1 \over \sqrt{1-9/y}}+3{ \sqrt{y-{8}} \over \sqrt{y-9}} = {1 \over \sqrt{1-4/z}} + 2{ \sqrt{z-3}\over \sqrt{z-4}}[/tex]

    Despite the relatively simple appearance of the simplified equation the online solver I use still can not solve for z in terms of y (or vice versa). Any ideas?
    Last edited: Apr 17, 2008
  2. jcsd
  3. Apr 17, 2008 #2


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    It's my experience that in this type of problem, working with four-momenta instead of energies and three-momenta separately makes life much easier. Can you state the problem in terms of a scattering problem? It's two body to tow body? And you are trying to find the speed of which particle in terms of which other particle? They are all of equal mass?
  4. Apr 17, 2008 #3
    A bit more detail


    I will add a bit more detail about the thought experiment. Initially I thought it just a maths problem so I did not add to much of the details of the physics involved.

    Initially there is a cannon of mass 3M and a cannon ball of mass M. It can be considered an electrial rail gun so we do not have to consider gunpowder particles flying all over the place. The cannon is fired with a "charge" of stored energy E. A velocity for the cannon ball is assumed and the recoil velocity of the cannon is obtained from conservation of momentum.

    In the second part of the experiment the cannon is brought to rest. The mass of 3M for the cannon given in the first part included a spare cannon ball of mass M. The cannon is fired again with an identical "charge" of stored energy E. The velocity of the cannon ball (or the recoil velocity of the cannon that now has mass 2M) on this occasion is what I am trying to calculate. I have tried to do this by assuming the kinetic energy on each occasion is the same.

    Thinking about it, I have not allowed for the fact that kinetic energy is total energy minus the rest mass in relativity, so I think the equation I am trying to solve is now:

    [tex]{1 \over \sqrt{1-9/y}}+3{ \sqrt{y-{8}} \over \sqrt{y-9}} -4 = {1 \over \sqrt{1-4/z}} + 2{ \sqrt{z-3}\over \sqrt{z-4}} -3 [/tex]

    If you can solve the problem using 4 momenta then that would be welcome too ;)
  5. Apr 17, 2008 #4


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    Ah, sorry, I thought it was a relativistic problem because of all the square roots. This is purely classical so no need to use four-momenta. It would help a lot to see your steps in getting to this equation. I don't think it should be that complicated. Unfortunately, I have to go teach in a few minutes.
  6. Apr 17, 2008 #5
    It is a relativistic problem ;) If I can solve this stage of the problem I intend to transform it to a reference frame with relative motion and try and figure out what happens to the internal energy of the "charge" under Lorentz transformation.

    Under further consideration, momentum is not conserved independently in relativity so I guess I will have to rethink the thought eperiment more appropriately. I am still puzzled as to why the equation given in the first post is so difficult to solve from a purely mathematical point of view considering its simple appearance.
    Last edited: Apr 17, 2008
  7. Apr 17, 2008 #6
    if you factor out 9 and 4 from the square roots from left and right side respectively it will look much simplier. Now make the substitution a=1-u^2, b=1-w^2 and you end up with a much more comfortable equation:
    [tex] \frac{1+\sqrt{8a+1}}{\sqrt{a}}= \frac{1+\sqrt{3b+1}}{\sqrt{b}}[/tex]
    not sure it will help though
    edit: i think i got it. set the let side of this equation to c, then you can move 1/sqrt(a) to the right side, square, move around little more, and you can solve for a, then for u.
    Last edited: Apr 17, 2008
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