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One Dimensional Collision dealing with Reference Frames

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

    A 20g ball of clay is shot to the right at 12m/s toward a 40g ball of clay at rest. The two balls of clay collide and stick together. Call this reference frame S.


    2. Relevant equations

    What is the velocity of a reference frame S' in which the total momentum is zero?

    3. The attempt at a solution

    How I interpreted this is I calculated what velocity the 40 g ball would need to move for total momentum to equal zero. I came up with (.02kg)(12m/s)+(.04kg)(-6m/s)=0

    Having found its velocity to be -6 m/s to satisfy a total momentum of zero I figured a reference frame moving 3 m/s would make the 20 g ball appear like it was traveling at 9 m/s and the 40 g ball at -9 m/s.

    But this still doesn't sit right with me. I don't know why I went through the trouble of making their velocities appear equal but opposite, it seems unnecessary.

    So my other though is maybe they mean the total momentum of the reference frame should be zero in which case the reference frame would simply have a velocity of 0 m/s.

    Do either of these sound right? The wording of the question seems to complicate things.

    Thanks in advance
     
  2. jcsd
  3. Mar 1, 2012 #2

    tiny-tim

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    Hi JimiJams! :smile:
    hmm …

    let's start again:

    describe in english what the collision would look like in in frame S'
     
  4. Mar 1, 2012 #3
    In my opinion, what the question is asking you is:

    To the reference frame that is watching the two balls collide, what must be its speed such that the total momentum of the collision is 0.

    So the speeds of the two balls of clay need to be subtracted by some speed (which would be the speed of S') so that their new speed is in the reference frame of the observer.

    My equation looks something like (m1)(12-x)+(m2)(0-x)=0, where x is the speed of the observer.

    By solving this you'll get a number which will satisfy the equation when you plug in the new speeds of the to objects relative to the observer.
     
  5. Mar 1, 2012 #4
    Tiny Tim, that's the best I can do. It's word for word from the book so...

    EricV, you completely cleared it up for me. I was heading in that direction except I was strictly thinking of velocity and not taking into account mass. So I kept trying to figure a way to make each ball appear like it was moving 0 m/s. obviously 12 m/s right only works for the 20g ball and 0 m/s only works for the 40g ball. MASS! DUH! Thanks a lot!
     
  6. Mar 1, 2012 #5

    tiny-tim

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    Yes, I know, but the book describes it in one frame, and it often helps to describe it in the other frame …

    what would the collision look like in in frame S' ? :wink:
     
  7. Mar 1, 2012 #6
    JimiJams,

    You're welcome. However, be sure to follow tiny-tim's advice as well next time. The very first thing I did when I read the problem was I visualized it from the perspective of the S' observer, and after that the answer came quickly.

    Eric
     
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