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Conservation of Momentum (Elastic Collision)

  1. Sep 4, 2008 #1
    1. The problem statement, all variables and given/known data
    A particle with mass m and speed v collides with another particle with
    mass m, which is initially stationary. The collision is elastic. Show that
    the particles travel in perpendicular directions after the collision.


    2. Relevant equations
    1)initial momentum = final momentum
    2)initial kinetic energy = final kinetic energy

    3. The attempt at a solution
    Assume that B is the moving particle and A is the stationary particle.

    By conservation of momentum,
    For the horizontal component of the motion
    mvi=mvfcos[tex]\theta[/tex]+mufcos[tex]\phi[/tex]

    For the vertical component of the motion
    0=mvfsin[tex]\theta[/tex]+mufsin[tex]\phi[/tex]

    where
    vi = initial velocity of B
    vf = final velocity of B
    uf = final velocity of B
    [tex]\theta[/tex] = angle between vector A and horizontal axis
    [tex]\phi[/tex] = angle between vector B and horizontal axis

    By conservation of energy,
    [tex]\frac{1}{2}[/tex]mvi2=[tex]\frac{1}{2}[/tex]m(vf2+uf2)
    So, vi2=vf2+uf2

    These are all the equations (3 equations 5 variables) I can get,
    But it seems that I have too many variable so I cant solve for [tex]\theta[/tex] and [tex]\phi[/tex]. Is there any more assumption I need to make in order to solve the question?
     
  2. jcsd
  3. Sep 4, 2008 #2

    LowlyPion

    User Avatar
    Homework Helper

    Looks like a good start. You look armed and dangerous.
     
  4. Sep 5, 2008 #3
    In these sorts of problems I find it useful to draw of diagram of the vectors. You know that the sum of all the x- and y-velocity vectors before and after the collison are equal which gives you the following options:

    y-velocity:

    uf*sin(phi) = - vf*sin(theta) Equation (1)
    [equal and opposite final velocities of the two particles in the y-direction]

    a) phi = 0, theta = 180 (in degrees)
    b) phi = 180, theta = 0

    c) 0 <= phi <= 90, -90 <= theta <= 0
    d) 0 <= theta <= 90, -90 <= phi <= 0

    From conservation of kinetic energy, you can eliminate options a) and b) for these particles of mass 'm'. You can then choose either c) or d) without loss of generality.

    You then need to sub in Equation (1) to the equations you have for conservation of x-momentum and conservation of kinetic energy. It will require some funky algebra to solve the equation, but the trick is to place certain boundaries on what the final angles can be.

    Hope this helps
     
  5. Sep 6, 2008 #4
    Thank you for your help.
     
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