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Energy and Momentum Thought Experiment

  1. Jul 2, 2015 #1
    Suppose two balls A and B are connected by a compressed massless spring initially, and they are not under any other forces. The spring is then released and the result is observed.

    Scenario 1: Ball A has the same mass as Ball B.
    Scenario 2: Ball A has greater mass than Ball B.

    Define the reference point to be the center of the spring in the initial state. (Thus the reference frame is inertial, as there are no other forces on the system initially.)

    Does Ball B has the same final velocity with respect to the reference point in both scenarios?
    Is Ball B's final relative velocity with respect to Ball A the same in both scenarios?
    Last edited: Jul 2, 2015
  2. jcsd
  3. Jul 2, 2015 #2
    In scenario 1, if the balls are symmetric relative to the center of the string (ie: same distance from it), and release the at the same time, then they'll have the same velocity, because they accelerated the same way, whereas in scenario 2, it's not the case, one of the ball is more massive then the other and by F = ma, It'll be less accelerated, thus will have less velocity than ball B
    The energy is conserved, no matter what , so Mv^2/2 + mV^2/2, M donates A's mass and V B's velocity, so V^2 = 2PE/m - (M/m)*v^2 so in B's frame of reference, A's velocity is √(2PE/m - (M/m)*v^2) + v, this quantity seems to be not constant when m varies, to be sure let's take a derivative, ((Mv^2)/m^2 - 2PE/m^2))/√...., that monster is constant mean that (Mv^2 - 2 PE)/m^2 = 0, Mv^2 - 2 PE = 0 which we know to be true in that frame of reference, sl altering one of the ball's mass won't change the final relative velocity, Hope I didn't make math mistakes, Cheers
  4. Jul 2, 2015 #3
    Thanks Noctisdark. A couple questions:
    - PE means potential energy? ie energy in the spring initially?
    - Your conclusion is that: 1. In Scenario 2 Ball B will have greater final velocity, and 2. Ball B's relative velocity to Ball A will be the same in both scenarios, correct?
  5. Jul 2, 2015 #4
    I think there will be a change in relative velocity. Though the total energy is conserved and the energy conversion in both the cases are same, the energy has a mass component and velocity component. However, if we say the mass by which B is decreased is equal to the mass by which A is increased, then we might have some interesting question...
  6. Jul 2, 2015 #5
    @ Noctisdark, could u help me please, I could not get the meaning of Mv^2 - 2 PE = 0 conclusion. Isnt it like kinetic energy of ball B is equal to the potential energy of the spring. Isnt it supposed to be shared by ball A & B?
  7. Jul 2, 2015 #6
    PE is the elastic potential energy (due to the strech of the spring), between who it's supposed to be shared depends on the frame of refrence, It you read what I've wrote again, you'll see that we took B frame of reference, in that frame B isn't moving so no kinetic energy, so one could.conjecture that all of the energy has gone for A's kinetic,
  8. Jul 2, 2015 #7


    Staff: Mentor

    You can write two equations in two unknowns in the center of momentum frame.

    Conservation of energy: ##MV^2/2+mv^2/2=PE##
    Conservation of momentum: ##MV+mv=0##

    Then solve for ##V## and ##v##, where the capital letters indicate one object and the lower case letters indicate the other.
  9. Jul 2, 2015 #8
    Ya. I did not note that. If B is the frame of reference, whatever the motion that we would see will be in the way of A. (Even if A is stationary and we are moving). Thank you. If you can bear me one more question. The resultant velocity then depends on the mass of the objects is it not? consider that the two balls are made much smaller, the conversion of energy to velocity will be higher is it not? I mean, if the balls connected to the springs are smaller, they will move faster is it not ?(Since the energy supplied by spring is same) So how we say that the relative velocity is not dependent on mass? Sorry if you had already answered it... The question might even be stupid but I wanted to ask anyway :)
  10. Jul 2, 2015 #9

    Doc Al

    User Avatar

    Staff: Mentor

    You can't. Use the equations that DaleSpam provided to prove it. (You gave a good reason why the relative velocity does depend on the mass.)
  11. Jul 2, 2015 #10
    I've done a great mistake, It does depend on mass, when I took the derivative with resprct to the mass, I've found that in B's frame Mv^2 - 2 PE = 0, this seems to be true but remember that B's frame isn't inertial so the relative velocity does depend on mass, sorry for the huge mistake, I've been out all the day and I only had access to my phone for few minutes, Sorry
    Last edited: Jul 2, 2015
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