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Finding the final velocity using momentum and KE

  1. Oct 28, 2013 #1
    Before the collision, there are two masses. Mass 1 is at rest and mass 2 is moving. No friction. It is an elastic collision and mass 2 is moving in positive x direction.

    After the collision, is it correct that mass 1 moves in positive x direction but mass 2 rebounds and moves in negative x direction? If so, is this the equation for momentum:

    m[itex]_{2}[/itex]v[itex]_{2i}[/itex]=m[itex]_{1}[/itex]v[itex]_{1f}[/itex]-m[itex]_{2}[/itex]v[itex]_{2f}[/itex]

    And the equation for KE:

    [itex]_{2}[/itex](v[itex]_{2i}[/itex])^2=m[itex]_{1}[/itex](v[itex]_{1f}[/itex])^2+m[itex]_{2}[/itex](v[itex]_{2f}[/itex])^2

    And if solving for v[itex]_{1f}[/itex] :

    v[itex]_{1f}[/itex]=[itex]\sqrt{\frac{m_{2}(v_{2i})^2}{m_{1}}}[/itex]

    And the velocity of this mass should be in the positive x direction.

    Does everything look correct?
     
  2. jcsd
  3. Oct 28, 2013 #2

    haruspex

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    Does the RHS represent total momentum after collision? Doesn't look like it.
     
  4. Oct 28, 2013 #3
    Thanks Haruspex... Does mass 2 then not change direction and not travel along negative x? Mass 2 is twice the mass as mass 1.
     
    Last edited: Oct 28, 2013
  5. Oct 28, 2013 #4

    haruspex

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    What do the equations tell you? You shouldn't need to be taking any square roots, thereby introducing ambiguity in sign. If you handle the conservation equations correctly you can obtain a simple ratio between relative velocities. Combined with the mass ratio you can then determine all velocities in terms of the initial velocity of mass 2.
     
  6. Oct 29, 2013 #5
    Thanks for pointing out the proper method. So since mass 1 has half the mass of mass 2, mass 2 does not rebound and change directions? What if the masses were switched? If mass 1 were moving and collided with the more massive mass 2 at rest, would mass 1 change direction in this case?
     
  7. Oct 29, 2013 #6

    haruspex

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    Is that what the equations tell you? Please post your working.
     
  8. Oct 29, 2013 #7

    HallsofIvy

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    It is not a matter of "changing direction" or "not changing direction". Total momentum is always [itex]m_1v_1+ m_2v_2[/itex]. Whether or not a mass changes direction is signaled by the sign of its v. If mass 2 is twice mass 1 and has initial velocity [itex]v_i[/itex] then "conservation of momentum" is [itex]2mv_i= mv_1+ 2mv_2[/itex] where [itex]v_1[/itex] is the velocity after the collision of mass 1 and [itex]v_2[/itex] is the velocity after the collision of mass 2.
    "Conservation of kinetic energy" is [itex](1/2)2mv_i^2= (1/2)mv_1^2+ (1/2)(2m)v_2^2[/itex]. Of course,, you can divide both equations by "m" eliminating that. You will have [itex]v_1+ 2v_2= 2v_i[/itex] and [itex]v_1^2+ 2v_2^2= 2v_i^2[/itex]. You could then, for example, solve the first equation for [itex]v_1= 2v_i- 2v_2[/itex] and replace [itex]v_1[/itex] in the second equation by that to get a quadratic equation for [itex]v_2[/itex].
     
  9. Oct 29, 2013 #8
    Thanks Haruspex and HallsofIvy, I appreciate your help!

    Haruspex has asked me more than once "Is that what the equation tells you?" Now I realize what you are saying (thanks HallsofIvy for your explanation).... you are saying that we do not know if the mass will rebound or not until we solve the equation. So, we can't choose a direction for the final velocity before we have solved for it. But if the final velocities are in opposite directions, then we would conclude that the moving mass rebounded. Is this right, Haruspex?
     
  10. Oct 29, 2013 #9

    haruspex

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    Yes. It often happens when trying to find a vector (force, velocity, whatever) that you don't know in advance which direction it will be in. So you just choose a direction as positive, and if the equations give you a negative answer then you know it was the other way.
     
  11. Oct 30, 2013 #10
    Awesome!!! Thanks for your help, Haruspex... This concept is very important!
     
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