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Energy, momentum and elastic collisions

  1. May 23, 2015 #1
    1. The problem statement, all variables and given/known data
    A particle (of mass m velocity v) makes a perfect elastic collision with a stationary particle. After the collision both particles travel 30 degrees from original path. Use conservation of momentum/energy to obtain 3 equations relating the masses/velocities.

    2. Relevant equations


    3. The attempt at a solution
    I can get two equations, one from using initial momentum = final momentum, and the other using initial kinetic energy = final kinetic energy, but I am unsure of where the third equation comes from. Any help would be appreciated.
     
  2. jcsd
  3. May 23, 2015 #2

    Orodruin

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    Initial momentum = final momentum is a vector equation and should give you two independent relations.
     
  4. May 23, 2015 #3
    Thanks for your response, so would it be,

    Initial momentum = final momentum
    m1v = (m1 + m2)(v(y component)/sin(30))
    and
    m1v = (m1 + m2)(v(x component)/cos(30)) ?
     
  5. May 23, 2015 #4

    Orodruin

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    No. First of all the x and y components of the original velocity are not equal. Second, the particles are not moving with the same velocity after the collision.

    Please expand on your argumentation why you think something. It will help us understand how you think and where you go wrong - and it is usually enlightening to listen to your own arguments as well.
     
  6. May 23, 2015 #5
    Since its an elastic collision and the masses are constant then the initial velocity v = final velocity of particle 1(call it u) + velocity of particle 2(call it s). Resolving those into the x and y components would therefore give for particle 1 u(x component)cos(30) and u(y component)sin(30) and for particle 2 s(x component)cos(30) and s(y component)sin(30).
    Would the answer then be for the momentums
    m1v = m1u(x component)cos(30) + m2s(x component)cos(30)
    m1v = m1u(y component)sin(30) + m2s(y component)sin(30)
    Or am I completely thinking about it wrong?
     
  7. May 23, 2015 #6

    Orodruin

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    You need to split the initial velocity into components too.

    The particles are also not going in the same direction after impact.

    Edit: In addition, you are both taking the components of the velocities and multiplying with a trigonometric function. This is doing the same thing twice, which is one too many.
     
  8. May 23, 2015 #7
    Okay would it then be

    m1(√ [(vcos30)^2+(vsin30)^2] ) = m1ucos30 + m2scos30
    m1(√ [(vcos30)^2+(vsin30)^2] ) = m1usin30 + m2ssin30
     
  9. May 24, 2015 #8

    Orodruin

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    No. Again, you have the same initial velocity component in both directions. This is not the case. In which direction is the particle originally moving? On top of that, you are still working with the assumption that the particles move in the same direction after impact, they are not - one is going at a 30 degree angle to one side and the other in a 30 degree angle to the other side.
     
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