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Momentum and Energy

  1. Mar 30, 2009 #1
    1. The problem statement, all variables and given/known data
    Two air track gliders of mass 0.30kg and 0.20 kg are moving towards each other with speeds of 0.50 m/s and 1.0m/s respectively. Take the direction of the more massive glider as positive.

    A) if the collision is elastic, find the velocity of each glider after the collision.

    Answer is: [ -0.70m/s , 0.80m/s ]


    2. Relevant equations
    m1v1 + m2v2 = (m1 + m2 ) v'

    Ek = 1/2 mv2

    for.. elastic equations i think..
    Ek = Ek'

    3. The attempt at a solution

    I've tried using the v1' and v2' equations and subbing in the values, but the answer was not correct O_O. Uhm.. I just need a little hand on where to go next! (Note: I've also tried drawing the diagrams and I think they are correct xP) okay I've got one of the answers.. -0.7m/s but with that 1 equation.. how do I get the other answer lol...
     
    Last edited: Mar 30, 2009
  2. jcsd
  3. Mar 30, 2009 #2
    Your first two equations are not correct. How did you get those equations? Remember that in an eleastic collision, both momentum and energy are conserved.
     
  4. Mar 30, 2009 #3
    1 quick question, how do I know if the system is a conserved momentum or... not? if it is conserved momentum.. I believe I would use 1/2 Mv^2 + 1/2 mv^2 = 1/2 mv'^2 + 1/2 mv'^2

    (didn't put the little signs =P)
     
  5. Mar 30, 2009 #4
    As I understand it, in nearly all collisions, momentum is conserved. In inelastic collisions, energy is not conserved.
     
  6. Mar 30, 2009 #5
    mmm k that helped ^^

    ok just want to double check here...

    so basically.. for all or most questions that involve a collision other then inelastic, it's associated with the formula...

    m1v1 + m2v2 = (m1 + m2 ) v'

    EDIT: okay I just used that, and only got the final velocity of 1 of the objects lol...
    errr.. still trying x.x
    think i just got it ..
     
    Last edited: Mar 30, 2009
  7. Mar 30, 2009 #6
    blaghh nvm I didn't get the second final velocity..
     
  8. Mar 30, 2009 #7
    No. That equation would be used for what are called perfectly inelastic collisions. In this situation, energy is NOT conserved, and the objects stick together. That's why the right hand side of the equation has the (m1+m2) term.

    However, in this problem we are dealing with elastic collisions and so we would use this equation to represent the conservation of momentum.

    m1v1 + m2v2 = m1V1f + m2V2f where the f subscript represents the velocity of the mass after the collision. Hopefully that helps!
     
  9. Mar 30, 2009 #8
    hm.. I think that cleans up some bit of stuff rofl thanks.. if I run into any trouble I'll post back in about 3 - 5 minutes
     
    Last edited: Mar 30, 2009
  10. Mar 30, 2009 #9
    If I use that equation.. wouldn't I have 2 unknown variables? V1f and V2f?

    ok well, I used that equation and I ended up with

    0.35 = 0.3v1' + 0.2v2'
     
  11. Mar 30, 2009 #10
    You would. However, you can also use the fact that kinetic energy is conserved to formulate a second equation. With two equations and two unknown variables, you should be able to solve.

    Remember, K.E. = 1/2mv^2
     
  12. Mar 30, 2009 #11
    uhmm I used the 1/2 mv12 + 1/2mv22 = 1/2 mv12 + 1/2 mv22

    ... Ek1 + Ek2 = Ek1' + Ek2'....?

    and I ended up with a weird number for V12'
     
  13. Mar 30, 2009 #12
    What number did you end up with?

    You need to use a combination of the conservation of energy equation and conservation of momentum equation. You have two unknown variables. You can solve for one of these unknown's in terms of the other unknown variable, and then plug that into the second equation
     
  14. Mar 30, 2009 #13
    is the formula I typed right above your post correct?
     
  15. Mar 30, 2009 #14
    ok well I got Square Root of ( ( 0.1375 - 0.1v22')/ 0.15) = V1'
     
  16. Mar 30, 2009 #15


    The first equation is wrong ( or you forgot your subscripts ). You forgot to label the masses m1 and m2, and in your equation the velocities are the same on both sides. A more precise equation would be

    1/2m1v1^2 + 1/2m2v2^2 = 1/2m1v1f^2 + 1/2m2v2f^2.

    I know it's kind of confusing, especially typed out. Hopefully it makes sense to you!
     
  17. Mar 30, 2009 #16
    BUUMP still need help =/ I gots a test tomorrow
     
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