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Change in momentum

  1. Nov 9, 2008 #1
    1. The problem statement, all variables and given/known data
    There are two 'gliders' X and Y on an air track. The mass of X is 0.2kg and it's velocity is 1.5m/s to the right. The mass of Y is 0.3kg and it is stationary. When 'glider' X collides with trolley Y they move off together.

    Calculate the velocity of the 'gliders' after the collision and give their direction.


    2. Relevant equations
    mass x velocity = momentum
    mass Y x velocity Y = -(mass X x velocity X)

    3. The attempt at a solution
    mass Y x velocity Y = -(mass X x velocity X)
    0.3 x velocity Y = -(0.2 x 1.5)
    [tex]Velocity Y = \frac{-0.3kgm/s}{0.3kg} = -1[/tex]

    So it's moving 1m/s to the left? I'm not sure whether that's the velocity of both of the gliders, or just velocity of Y because that's what my equation shows.
     
  2. jcsd
  3. Nov 9, 2008 #2

    Doc Al

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    Staff: Mentor

    This is incorrect. What would be true is that the change in Y's momentum will be equal and opposite to the change in X's momentum.

    Treat this as a completely inelastic collision (the carts stick together and thus end up with the same final velocity) and write an equation for momentum conservation.
     
  4. Nov 10, 2008 #3
    My book says that's the equation for conservation of momentum.
     
  5. Nov 10, 2008 #4

    Doc Al

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    Staff: Mentor

    What book are you using?

    I suspect that the equation in your book is more like:
    [tex]M_y \Delta V_y = - M_x \Delta V_x[/tex]

    For inelastic collisions, I recommend this version:
    [tex]m_1v_1 + m_2v_2 = (m_1 + m_2)v_f[/tex]
     
  6. Nov 10, 2008 #5
    My book is AQA Science GCSE Physics

     
  7. Nov 10, 2008 #6
    You have to learn to distinguish an equation that is generally valid (such as conservation of momentum when there is no external force), and equations that apply to specific situations (such as the one you are using, even though that equation follows from the more general law).
     
  8. Nov 10, 2008 #7
    There is no other equations to do with conservation of energy in my book though :(

    I hate my stupid Physics/Chemistry teachers, he teaches us jack all. He told us out loud he's not going to teach us the theory. WTF is he going to teach us then.
     
  9. Nov 10, 2008 #8
    Wow, that's sad....I suppose he is "teaching" you how to mindlessly put numbers into equations.

    In any case, the general equation for conservation of momentum (assuming the masses stay constant!) is that:
    [tex]
    \sum_i m_i \vec{v}_i^{{\rm b}} = \sum_i m_i \vec{v}_i^{{\rm a}}
    [/itex]
    thus relating velocities before (b) and after (a) a collision between particles with masses m_i.
     
  10. Nov 10, 2008 #9
    I thought mass was constant O.O

    What's that big E thing mean?
     
  11. Nov 10, 2008 #10
    I was just having troubles with LaTeX: it's fine now. There's no big E anymore.
     
  12. Nov 10, 2008 #11
    It's still there, this thing: [tex]\sum[/tex]
     
  13. Nov 10, 2008 #12
    Oh, that's a summation sign: you sum over all particles in your system. In your case you have two gliders, X and Y, so you'd sum over i taking the values X and Y.
     
  14. Nov 10, 2008 #13
    Written in a different way, you have:

    [tex]

    m_X v_X^b+m_Yv_Y^b=m_X v_X^a+m_Yv_Y^a

    [/tex]
     
  15. Nov 10, 2008 #14
    Now that, I understand apart from the ^a and ^b, where has a and b come from or is that what I need to work out?
     
  16. Nov 10, 2008 #15
    a=after
    b=before
     
  17. Nov 10, 2008 #16
    So:
    [tex]0.2\times1.5 + 0.3\times0 = (m_Xv_X + m_Yv_Y)^a[/tex]
    [tex]\frac{0.3}{1.5 + 0.3} = (v_X + v_Y)^a[/tex]

    and that gets me the final velocity?
     
  18. Nov 10, 2008 #17
    Not quite, but you're on the right track:

    first you should note that after the collision X and Y end up traveling with the same velocity, so v_X^a=v_Y^a.

    Second, I don't know why you divided by 1.5+0.3: the masses are 0.2 and 0.3, not 1.5 and 0.3!
     
  19. Nov 10, 2008 #18

    Doc Al

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    Staff: Mentor

    Don't forget that after the collision the trolleys "move off together"--that means they have a single velocity after the collision.

    It might be simpler for you to use the equation I gave in post #4. (The 2nd equation. The left side represents the momentum before the collision; the right side, after the collsion.)
     
  20. Nov 10, 2008 #19
    I think I've got it now:
    [tex]Velocity = \frac{0.3}{0.5}[/tex]

    So the velocity is 0.15m/s to the left?
     
  21. Nov 10, 2008 #20

    Doc Al

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    Staff: Mentor

    Good.
    No. (That fraction doesn't equal this.)
    Why to the left?
     
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