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Conservation of linear momentum and inelastic collision

  1. Apr 20, 2008 #1
    [SOLVED] Conservation of linear momentum and inelastic collision

    1. The problem statement, all variables and given/known data
    I have a homogeneous beam of length L and mass M, attached with a frictionless hinge at one of the endpoints O. The beam is affected by gravity, g, in the negative y-direction. Initially the beam hangs straight down in the negative y-direction. The moment of inertia of the beam for rotations around O in the direction given by the hinge is I.

    A small bullet of mass m is fired along the x-axis, hitting the beam at the bottom. The bullet has an initial velocity of v0 and remains lodged in the beam after the collision.

    I have to show that the linear momentum of the system is generally not conserved
    during the collision.

    2. Relevant equations

    Not quite sure.

    3. The attempt at a solution

    I just can't figure out where to start with this one. Only thing I got so far is that I can't shake the feeling that the hinge must exert a force (centripetal force?) on the beam to prevent it from flying off, so the sum of external forces on the beam is non-zero, ergo no conservation.

    Any help would be most appreciated.
     
  2. jcsd
  3. Apr 20, 2008 #2

    Doc Al

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    Staff: Mentor

    Yes, the horizontal force of the hinge prevents linear momentum of the system (beam + bullet) from being conserved.

    But what is conserved? Figure out the linear momentum of the system after the collision and show that it's not conserved.
     
  4. Apr 20, 2008 #3
    I guess that's my problem, I can't seem to figure out how to calculate the linear momentum after the collision... Should I use the formula for inelastic collision [tex]m_1v_1 + m_2v_2 = (m_1 + m_2)v_f[/tex]? If so, what would v_f be? The tangential velocity of the center of mass of the beam-with-bullet?

    Sorry for being dense here :) I just can't seem to find anything in my textbook describing a similar setup.
     
  5. Apr 21, 2008 #4

    Doc Al

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    Since that equation assumes conservation of linear momentum, we know it doesn't apply. And since we're trying to show that it doesn't, we certainly can't use it here.

    But another quantity is conserved. That's what you need to apply. Hint: The beam is free to rotate about the hinge.
     
  6. Apr 21, 2008 #5

    tiny-tim

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    Hail, My Lord Crc! :smile:

    Doc Al gave you two hints. You seem to have missed the first one …
    Do that first … then figure out the linear momentum. :smile:
     
  7. Apr 21, 2008 #6
    Ah yes, I know angular momentum is conserved, and I've already used it to compute the angular velocity of the beam-with-bullet after the collision.

    The problem is that I don't know how the linear momentum is defined in my post-collision setup. I guess it's still mv, but what's v? Is it the tangential velocity of the center of mass? Thats the only thing I can think of, but what do I know :) In that case I guess I could write v as a function of theta, and thus show that the components of the momentum aren't conserved?
     
  8. Apr 21, 2008 #7

    Doc Al

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    You're moving in the right direction.

    Use the angular speed to calculate the velocity of the center of mass of the beam-with-bullet. Then show that its linear momentum (right after the collision) does not equal the initial momentum of the bullet.
     
  9. Apr 21, 2008 #8
    Ok, lets see

    distance to center of mass is [tex]r = \frac{L(M + 2m)}{2(M + m)}[/tex]
    angular velocity is [tex]\omega = \frac{mv_0L}{I + mL^2}[/tex]

    So, putting it all together...
    [tex]p_0 = mv_0[/tex]
    [tex]
    p_1 = (m+M)v_1 = (M + m)r\omega = \frac{L(M+2m)}{2}\frac{mv_0L}{I + mL^2} = \frac{(M+2m)L^2}{2(I + mL^2)}p_0
    [/tex]

    So, unless [tex]\frac{(M+2m)L^2}{2(I + mL^2)} = 1[/tex], that is, [tex]\frac{ML^2}{2I} = 1[/tex], then linear momentum is not conserved. Or?

    At least this would neatly answer the second part of the question, in which special case is linear momentum conserved.
     
  10. Apr 21, 2008 #9

    Doc Al

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    Looks OK to me.
     
  11. Apr 21, 2008 #10
    Great! Thank you very much to both of you!
     
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