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Homework Help: Linear Momentum of Particle and Rod

  1. Dec 7, 2011 #1
    I don't need a solution to this problem, I just need some help understanding a part of it.

    1. The problem statement, all variables and given/known data
    The figure below shows a small mass, m, moving at an initial speed, v0 , colliding with a stick with
    length, L, and mass, M. Both the mass and the stick lie on top of a table. The collision happens at the
    tip of the stick. After the collision the mass continues in the same direction, but now with the speed, v.
    For a specific value of the ratio m/M, the stick will collide with the small mass a second time. What is
    this value, and how far will the small mass have traveled between collisions? Assume that the
    collision is elastic.

    2. Relevant equations
    Initial Angular Momentum: (L/2)mv0
    Angular Momentum After Collision: (1/12)mL2ω + (L/2)mv

    3. The attempt at a solution
    I know that the rod and the particle must be moving at same linear velocities for the second collision to happen. What I don't 100% understand is why the angular momentum for the particle after the collision is still (L/2)mv. Initially, the origin is at the center of the rod, but after the collision both the particle and the rod move to the right. I don't understand why, for the particle, the distance for the angular momentum is just (L/2); doesn't this place the origin at the center of mass of the rod, which is now moving and thus a noninertial frame?

    Thanks for the help!
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Dec 8, 2011 #2


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    Homework Helper

    The reference system has the origin at the centre of the rod in rest and stays the same after impact. With respect to that point, the angular momentum of the particle is mv0L/2 before the impact and mvL/2 after the impact (supposing that the velocity of the particle is perpendicular to the rod). Do not forget that the angular momentum is the vector product of the position vector with the linear momentum, so it is
    mv*r*sin(angle between r and v)=mvL/2.

    It is more convenient to use a frame of reference in rest then one connected to the centre of mass.

    This is a very nice problem, really. It took me some time to imagine how the rod can hit the particle again. :smile:

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