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Collision/Decay Problem

  1. Apr 3, 2015 #1
    1. The problem statement, all variables and given/known data
    An unstable particle of mass M = m1 + m2 decays into two particles of masses m1 and m2
    releasing an amount of energy Q. Determine the kinetic energies of the two particles in the
    ZMF. Given that m1/m2 = 4, Q = 1 MeV, and that the unstable particle is moving in the lab frame with kinetic energy 2.25MeV, find the maximum and minimum kinetic energies of the particle of mass m1 in the lab frame.

    2. Relevant equations
    See below

    3. The attempt at a solution

    Let the particle of mass m1 be A and of mass m2 be B.

    For the first part, pA* + pB* = 0 in the zero-momentum frame and so mAEA* = mBEB* where E is the kinetic energy of the particle in the ZMF.

    I think the other equation to use is EA* + EB* = Q but I'm not sure how to justify this.

    1) Why should the energy released in the decay (in the lab frame) be equal to the sum of energies of the resultant particles in the ZMF?

    For the second part, not sure how to start or even why there are a range of possible answers. Seems to me that the kinetic energy of the unstable particle in the lab frame gives directly its velocity in the lab frame, which is thus the velocity of the ZMF; meanwhile, the ZMF kinetic energies of the two particles after the decay can be converted into ZMF velocities. Then we just do vlab = v* + vZMF to find the lab velocities of the two particles. These lab velocities can be converted to lab KEs, done.

    2) How is there a range of kinetic energies of A possible, for the second part? How should I do it?
     
  2. jcsd
  3. Apr 3, 2015 #2

    vela

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    Does the problem statement really say ##M = m_1 + m_2##? I ask because it's inconsistent with the rest of the problem statement.

    Conservation of energy says that the total energy before the decay is equal to the total energy after the decay. Before the decay in the ZMF, the total energy is just ##Mc^2## because the particle is at rest. What makes up the total energy after the decay? Figure that out and set it equal to ##Mc^2##.

    As far as why you get a range of kinetic energies, you have to consider the direction of the particles in the ZMF and how that affects the results in the lab frame. Also, velocities don't add the way you think they do. You're using Galilean relativity; it works differently in special relativity when speeds are comparable to the speed of light.
     
    Last edited: Apr 3, 2015
  4. Apr 3, 2015 #3
    I'm fairly certain this problem is not one in which we are expected to use special relativity. The problem does say ##M = m_1 + m_2## and I would really think it is a standard Galilean-relativity momentum problem. Of course I would be happy to learn how to generalize into special-relativity the solution of this but first I'd like to do it the "normal" way (Galilean relativity).

    Q energy is released, in addition to which you have ##(\gamma_1-1)m_1c^2 + (\gamma_2-1)m_2c^2## from the two particles A and B. That's in special relativity anyway. Here I'd think we just have

    Q + (1/2)m1vA2 + (1/2)m2vB2 = (1/2)(m1+m2)u2

    where u is the initial velocity of M in the lab frame. We need to use the lab-frame for this equation because Q is presumably measured in the lab frame. But this is not at all the same equation as (1/2)m1vA2* + (1/2)m2vB2* = Q, where * represents ZMF velocities, which is what we need to use for the first part (according to the solutions).

    Can you explain how I should do this, and what is wrong with my reasoning in the OP regarding the second part?
     
  5. Apr 4, 2015 #4

    vela

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    The way I would interpret it is that Q is the energy released as measured in the center-of-mass frame.

    As I said earlier, you don't seem to be considering the direction that particle A moves. ##\lvert \vec{v}_\text{A} + \vec{v}_\text{zmf} \rvert^2## depends on the angle between ##\vec{v}_\text{A}## and ##\vec{v}_\text{zmf}##.
     
  6. Apr 5, 2015 #5
    I see. Then, the first part makes sense (as initial kinetic energy is 0 in the zero-momentum frame, as it moves along with the particle).

    Ah I see. So the problem is in the line: "the ZMF kinetic energies of the two particles after the decay can be converted into ZMF velocities". The kinetic energies can only be converted into magnitudes of velocity and thus do not consider the directions the velocities could take.

    Can you outline how I might go about solving this then?
     
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