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Inelastic collision of equal masses and velocities

  1. Apr 6, 2005 #1
    After a completely inelastic collision between two objects of equal mass, each having initial speed, v, the two move off with speed v/5.5. What was the angle between their initial directions?

    Well, inelastic collision so it looks like I'll be using the momentum equation in here. My professor loves these "ratio-type" problems and I believe that I'll have to use something of the sort on this.

    Momentum in = Momentum out
    mv + mv = m(v/5.5) + m(v/5.5)

    Not sure where to go from there, and I don't see how I will find an angle out of all of this. Anyone have any sugestions as to where I should start?
     
  2. jcsd
  3. Apr 6, 2005 #2

    dextercioby

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    No,no,conservation of momentum is a typical example of vector relation/equation...

    Sides,in the final state,there's only one particle...:wink:

    Daniel.
     
  4. Apr 6, 2005 #3
    Fanman22,

    Have you tried drawing a picture showing the particles and their directions of travel before and after the collision? If not, it might help.
     
  5. Apr 6, 2005 #4

    dextercioby

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    Yeah,you'd see an isosceles triangle there.It would help you with the projection of the vector equation on some nicely chosen axis of coordinates.

    Daniel.
     
  6. Apr 6, 2005 #5
    *****Drawing*****

    Is that even remotely correct? I'm not sure the isosceles triangle idea makes sense to me. I don't understand how it represents the motion of the 2 particles before the collision and how it represents the velocity of the total mass afterwards.
     
    Last edited: Apr 6, 2005
  7. Apr 6, 2005 #6

    dextercioby

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    Nope.It should have been more like an Y.Actually exactly like an Y...

    Daniel.
     
  8. Apr 6, 2005 #7
    I can see the "Y-shape" now.....What I did was take the components (in the direction of the final velocity) of each V. So...

    Vsin(theta) + Vsin(theta) = (v/5.5) = 2Vsin(theta)
    Theta = 5.216

    To find the angle between the particles...
    180 - 5.216 - 5.216 = the middle angle = 169.6

    But of course, I got it wrong again :grumpy:

    Where did I go wrong?
     
  9. Apr 7, 2005 #8
    Fanman,

    You almost have it. But what does your equation:

    Vsin(theta) + Vsin(theta) = (v/5.5)

    say is conserved? What's really conserved?
     
  10. Apr 7, 2005 #9
    yes what jdavel said
    Edit* didn't look too closely myself
     
    Last edited: Apr 7, 2005
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