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Inelastic Collision and Angular Momentum.

  1. Oct 20, 2012 #1
    1. The problem statement, all variables and given/known data

    The professor very generally created a very simple conceptual problem as a basis for harder ones, but I don't understand how to answer it.

    A piece of clay, with mass (m) and speed (v) collides with a motionless stick of length L (with uniform mass density and total mass M). The stick is aligned along the Z-axis with its center at z=0. The clay is initially moving in the x-direction. The clay collides at the stick's end at z = L/2. Assume there is no gravity or external forces.

    Questions:
    What is the momentum in the y direction after the collision? (the clay completely sticks to the stick upon collision).

    a. 1/2(m+M)v
    b. 1/2(m/M)mv
    c. 2/3(L/m)mv
    d. zero
    e. mL/M (mv)

    What is the angular momentum of the system after collision?

    a. Zero
    b. m(m/M)vL/2
    c. 2M/m(mvL)
    d. mvL
    e. mvL/2

    2. Relevant equations

    I'm unsure. I know from my textbook that L = r x p, though we really haven't done cross products so I'm unsure of what that means (it says it's mrvsin(θ), right hand rule direction)
    I also know about conservation of momentum, which I assume is mv=(m+M)vf. Also I know of kinetic energy conservation and torque, which causes angular momentum to change.. but I've read my textbook 5 times and I can't understand the steps to solve this very simple problem.

    3. The attempt at a solution

    At first I thought the initial answer to question 1 was zero, because if a particle moving in x collides at with an object in z, it would just rotate about the xz plane and not have any y momentum. But then I remembered that there's a right hand rule (which I don't understand much either) and I doubt the professor would put a question that didn't require calculations..

    In regards to question 2, since we're given the equation L = mrvsinx, and I'm assuming that it's just mrv since the angle is 90 (plus the answers don't have any sin's). But I don't understand how that would work in an inelastic collision (why would it be a ratio of M/m or m/M?)

    So my guesses are 1. D and 2. B. (because there's an M and a L/2 involved)

    I'm not just interested in just the answer, if someone could try to explain to me the general concept of how it works in a simple language I would really appreciate that. Thanks for your time!
     
  2. jcsd
  3. Oct 20, 2012 #2

    SammyS

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    Hello pwkr. Welcome to PF !

    What is the x-component of the momentum of the system before the collision ?

    What is the y-component of the momentum of the system before the collision ?

    What is the z-component of the momentum of the system before the collision ?

    What does conservation of momentum tell you regarding the momentum of the system before the collision compared to the momentum of the system after the collision ?


    To use the formula, mrvsin(θ), for angular momentum, you need to know how θ is determined. You are correct in thinking the θ approaches 90° as the piece of clay approaches the point of collision.

    What is the magnitude of the angular momentum of the system immediately before the collision ? Use conservation of angular momentum to answer the 2nd question.


    The right-hand rule is used to give the direction of the angular momentum vector, since angular momentum is a vector quantity, but you don't need this to answer these questions.
     
  4. Oct 21, 2012 #3
    Hello! thank you for your reply.

    Let's see:

    The x-component is just linear momentum (mv)
    y and z components are zero before the collision.

    Conservation of momentum tells me that the momentum before the collision should be the same after collision, so overall the whole momentum should be (mv)?

    I'm assuming that mv would be converted into (m+M)*r*v_f since the collision is inelastic. so.. p = mv = (m+M)*(L/2)*v_f ? but it's asking for the angular momentum and I just have a collection of formulas that represent it so in my head I find it hard to isolate. What I mean is that I only have one equation, mv=(m+M)(L/2)v_f and isolating vf and replugging it in wouldn't help me, so I wouldn't know what steps to do to find angular momentum.

    I think there's a concept I'm missing that is preventing me from solving this

    I just had a quick thought:

    the total momentum at the end would be mv(L/2), which would mean that v (relative to the origin) would decrease based on how much length the stick has as its radius. So my I guess my final answer would just be mv(L/2) and the y component is zero. Is that right?
     
    Last edited: Oct 21, 2012
  5. Oct 22, 2012 #4

    SammyS

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    Momentum is a vector quantity, so if it is unchanged by the collision, each of its components is unchanged.
     
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