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Relative to What? An Energy Conundrum

  1. Aug 15, 2013 #1
    Firstly, I am new to this forum and this is my first post, so if I've posted to the wrong place or somehow violated a forum rule, I apologize in advance. This is the most polite forum I have ever found, and I stand in awe of all of you ladies and gentlemen.

    Will one of you--or all of you--please point out the error of my thinking? This has been on my mind for at least a year, and still I cannot find an answer. I've tried to summarize the problem in the attached PDF.

    Attached Files:

  2. jcsd
  3. Aug 15, 2013 #2


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    Hello there! Welcome to the forum. I have taken some screenshots of example 13.2 from Kleppner and Kolenkow "An Introduction to Mechanics". The diagram drawn up by the authors, which should answer your conundrum, is contained in the *first* screenshot. The rest are just there for context. Here they are: http://postimg.org/gallery/1hz05sis/504dc6f9/

    EDIT: To clarify, your initial inertial frame in the second scenario is not attached to the second particle. What it is doing is comoving with the particle before the collision; if it were attached to the second particle then it wouldn't be an inertial frame because the particle's velocity does not remain constant. Immediately after the collision, the particle that the inertial frame was comoving with changes velocity after sticking onto the first particle and hence is now moving with respect to this inertial frame i.e. the inertial frame is no longer comoving with the second particle.
    Last edited: Aug 15, 2013
  4. Aug 15, 2013 #3


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    Welcome to our forums.

    There is the error. In the second case, you have to keep the reference frame (a change of reference frames does not preserve energy values), so the two masses will move after the collision in your reference frame - and this accounts for the difference of mv^2.

    This looks more like classical mechanics, so I moved the thread.
  5. Aug 15, 2013 #4


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    In the first case, the total momentum of the system is zero. So after the collision the total mass is at rest, and all of the kinetic energy of motion (m v^2, assuming each ball has mass m and |velocity| v) is converted to heat, as you said. In the frame where you are sitting on one of the balls of bubble gum, the total momentum of the system is not zero. Since momentum is conserved, the total mass is not at rest after the collision, but will be moving with velocity v. So the total mass has a kinetic energy after the collision of m v^2 ( (1/2) * 2m * v^2). Since the initial KE of the incoming mass was 2m v^2, as you said, m v^2 of it was converted to heat, the same as in the first case.
  6. Aug 15, 2013 #5
    I Think I Have It

    Thank all. I think I see the answer in momentum. Phyzguy, you did it. Ich verstehe! Thanks to all. I will publish my perception of your explanations as an addendum to my previous summary, hoping you will check out my reasoning.

    Y'all have preserved my sanity.
  7. Aug 17, 2013 #6
    I think I've got it now. As promised, I've attached the original summary (notation slightly changed for clarity) with an addendum reflecting my new thinking.

    I've found that when I'm trying to understand something it is most helpful to explain it to someone else, but the problem is finding someone willing to listen. Thus, I often resort to explaining to myself by writing a summary. If it seems I've gone all the way around the barn to get to the door, please understand that I was explaining it to a dummy, namely, me.

    Please review what I've added and comment, if you like.

    Attached Files:

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