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Collisions in Two Dimensions

  1. Mar 21, 2006 #1
    After a completely inelastic collision, two objects of the same mass and same initial speed are found to move away together at 1/4 their initial speed. Find the angle between the initial velocities of the objects.

    I figured.. 2m(v/4)
    but then i just get lost... can someone help! thanks!
     
  2. jcsd
  3. Mar 21, 2006 #2

    nrqed

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    Pick a direction for the first object, let's say East. Let's say the second object is coming from an angle [itex] \theta [/itex] North of East. Then Decompose their momenta into x and y components, apply conservation of momentum along x and y directions. You will find the final x and y velocities along x and y in terms of the initial speed v and the angle [itex] \theta [/itex]. Impose that [itex] {\sqrt { v_{x final}^2 + v_{y final}^2}} = v/4 [/itex] and that will give you a single equation for [itex] \theta [/itex].


    Patrick
     
  4. Mar 22, 2006 #3
    OK, so the collision is in 2 dimensions, so you know you're going to be using vectors. So let me start you off with what you know.

    [tex]m(\underline{v}_1 + \underline{v}_2 ) = 2m \underline{v}_{final}[/tex]

    Do you know how to make vectors loose their directional components?

    (Hint: Multiply both sides by a vector you know)

    Hope this Helps, Sam
     
  5. Mar 22, 2006 #4
    OK, pick the vector carefully.

    I have just done this question in 5 lines (instead of 10) by picking another vector. If you don't choose your vector well, you'll have to use some trig. identities (which I don't like if I can avoid it).

    Draw a diagram and note what the angles have in relation to each other due to symmetry. If you want, I'll put up a diagram of what I'm trying to say... just ask if you want it.

    Regards,
    Sam
     
  6. Mar 22, 2006 #5
    hey thanks. this is what i did so far. but im still have trouble.

    M1V1i + M2V2i = M1V1f + M2V2f

    initial for 1
    X = mv cos Theta
    y = mv sin theta

    final for 1
    X = mv
    y=0

    Initial for 2
    X = mv cos theta
    Y = mv sin theta

    final for 2
    X =mv
    y = 0


    mv cos theta + mv sin theta = MVi
    Mv (cos theta + sin theta) = MVi

    ???? then i get lost.
     
  7. Mar 22, 2006 #6
    The components of their initial velocities perpendicular to the direction of their final velocity adds to zero. And the components along this direction must thus be same and add upto v/4.
     
  8. Mar 22, 2006 #7
    Sorry. The second sentence is - the component of their initial velocities along the direction of final velocity must thus be same and add upto v/4
     
  9. Mar 22, 2006 #8
    im still confused.

    can someone show me the steps?
     
  10. Mar 22, 2006 #9
    Let a be the angle made with their initial velocities along the direction of final veloctitis.
    Now,
    v*cosa + v*cosa = v/4.
    Thus cosa = 1/8
     
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