Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper

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

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

  4. Apr 6, 2005 #3

    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


    User Avatar
    Science Advisor
    Homework Helper

    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.

  6. Apr 6, 2005 #5
    *****http://img.photobucket.com/albums/v225/Fanman22/c7f601bb.jpg [Broken]*****

    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 by a moderator: May 2, 2017
  7. Apr 6, 2005 #6


    User Avatar
    Science Advisor
    Homework Helper

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

  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

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook