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US Physics Test Problem

  1. Feb 13, 2008 #1
    1. The problem statement, all variables and given/known data This problem was given in the 2008 United States Physics Team qualifying test, called the F=ma Contest:
    21. Consider a particle at rest which may decay into two (daughter) particles or into three (daughter) particles. Which of the following is true in the two-body case but false in the three-body case? (There are no external forces.)
    A. The velocity vectors of the daughter particles must lie in a single plane.
    B. Given the total kinetic energy of the system and the mass of each daughter particle, it is possible to determine the speed of each daughter particle.
    C. Given the speed(s) of all but one daughter particle, it is possible to determine the speed of each daughter particle.
    D. The total momentum of the daughter particles is zero.
    E. None of the above.
    2. Relevant equations[tex]\sum \vec{p}=0[/tex]
    3. The attempt at a solution In the two-body case, A is definitely true because the velocities of the daughter particles must lie on the same line. In the three body case, the velocity of any of the three daughter particles can be written as a linear combination of the other two. Since the set of all linear combinations of a set of two vectors is a plane, the velocity vectors must all lie on the same plane. Since A holds for both cases, it is not the right answer.
    Since D obviously holds for both cases, it is also not the right answer. That leaves B, C, and E. I am not sure where to proceed.
     
  2. jcsd
  3. Feb 14, 2008 #2
    First of all, you know VCM=0, so i'm working on CM referencial. that means the relevant equation is true as long as m1v1(particle1)=-m2v2(particle2) (v is vector)

    A right for both
    B wrong for both: for 2 particles there are 3 equations(momentum in x and y and energy) for 4 variables(vx1,vy1,vx2,vy2)

    for a 2 particle example:
    as there's no U, total energy is just
    [tex]E=0.5*m_1 (v_{x1}^{2}+v_{y1}^{2})+0.5*m_2(v_{x2}^{2}+v_{x2}^{2})[/tex]
    you know that:
    [tex]m_1*v_{x1}=m_2*v_{x2}
    m_1*v_{y1}=m_2*v_{y2}[/tex]
    now, you don't have enough equations for solving in order to the four v's.

    C wrong for both: you don't know their masses, so conservation of momentum don't help.(for 2 particles)
    you have 2 equations(momentum) for x and y, you don't know total kinectic energy. So, you have 2 equations for 4 vars(2 masses and 2 velocities(the unknown vx and vy))

    D trivial true for both
    E if nothing excapted, i guess i would write this one right!
     
  4. Feb 14, 2008 #3
    It is indeed impossible to determine all of vx1,vy1,vx2 and vy2, but it's still possible to determine the magnitude of both velocities, but not the direction.
    The masses must fly off in opposite directions, because the sum of their momenta is 0. We can just do the calculation in a coordinate system that has its x-axis parallel to the direction of the momenta of the masses, and you get vy1=vy2 = 0. You then have 2 equations left for vx1 and vx2
     
  5. Feb 15, 2008 #4
    I don't think conservation of energy is applicable here. Let's assume that there is no potential energy before or after the particle disintegrates. Therefore, conservation of energy is equivalent to the conservation of kinetic energy. Clearly, the conservation of kinetic energy does not hold; before the disintegration, the system had no kinetic energy, but after the disintegration the system had a nonzero kinetic energy.

    The only explanation that I can think of is that the particle had some internal energy before it disintegrated. If this is the case, then in order to write a conservation of energy equation we would have to calculated the minimum internal energy of the particle before the collision.

    Any further help would be greatly appreciated.
    Thank You in Advance.
     
  6. Feb 15, 2008 #5
    1 indicates correct, 0 indicates not correct
    ...2 3
    A 1 0 cannot have a x & y velocity because conservation of momentum wouldnt be right
    B 1 0 shown below (because of the above)
    C 0 0 need to know their masses and velocity directions
    D 1 1 trivial
    E n/a but wrong

    B is right for 2 particle but not enough info for 3 particle (they are in two planes). For 1 plane however there is no vY only vX...

    Example

    After decay:

    <- o o--->

    If mass A is 4 and mass B is 2 & if KE is 32

    32= 0.5mAvA^2 - 0.5mBvB^2 neg sign as v2 is opposite direction to v1
    32- 0.5mAvA^2 = -0.5mBvB^2
    32- 2vA^2 = vB^2

    mAvA = -mBvB
    4vA = -2vB

    (1) 32 - 2vA^2 = vB^2
    (2) 4vA = -2vB

    (2) 2vA = -vB
    sub (2) into (1)

    32-vB^2 = vB^2
    2vB^2 = 32
    vB^2 = 16
    vB = 4 (not - since we have already considered the directions)

    sub vB into (2)
    2vA = -vB
    vA = -4/2 = -2

    However for a three particle decay there would necessarily be two planes in order for momentum to be conserved so LittlePigs explanation suffices for why it wont work for 3 particles.
     
    Last edited: Feb 15, 2008
  7. Feb 16, 2008 #6
    I disagree. Statement A holds for both the two body case and the tree body case. For the two body case, it is trivially true because not only is the motion limited to one plane, but it is also limited to only one line in that plane.

    The three body case is not as obvious. But consider the conservation of momentum:
    [tex]m_{1}\vec{v}_{1}+m_{2}\vec{v}_{2}+m_{3}\vec{v}_{3}=0.[/tex] Solving this equation for the velocity of the third particle, we get:
    [tex]\vec{v}_{3}=\frac{-m_{1}\vec{v}_{1}-m_{2}\vec{v}_{2}}{m_{3}}=\frac{-m_{1}}{m_{3}}\vec{v}_{1}+\frac{-m_{2}}{m_{3}}\vec{v}_{2}.[/tex]
    In other words, the velocity of the third particle is a linear combination of the velocities of the other two particles. Therefore, the velocity vector of the third particle must lie in the same plane as the velocities of the other two particles.
     
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