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Conservation of Energy and Momentum of Particles

  1. Apr 6, 2005 #1
    An atomic nucleus initially moving at 500 m/s emits an alpha particle in the direction of its velocity, and the new nucleus slows to 480 m/s. If the alpha particle has a mass of 4.0 u and the original nucleus has a mass of 226 u, what speed does the alpha particle have when it is emitted?

    Well this is what I came up with so far:
    m1=226 V1=500 V1'=?
    m2=4 V2=? V2'=480

    Use Conservation of Momentum and conservation of kinetic energy to get these equations, respectively:

    226(500) + 4(V2) = 226(V1') + 4(480) and...

    .5(226)(500)^2 + .5(4)(V2)^2 = .5(226)(V1')^2 + .5(4)(480)^2

    Hopefully this is correct so far....now I know I'm supposed to do substitution next, but I'm having some algebra problems.

    I got V2= -2.77e4 + 56.5(V1')
    I'm not sure if this is correct I'm ashamed to say it, but I'm having alot of trouble doing the algebra after I substitute it into the other equation.
  2. jcsd
  3. Apr 6, 2005 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Energy is not conserved. All you need is conservation of momentum. Realize that after the decay the old nucleus (226u) becomes two particles: the alpha particle (4u) and a new, smaller nucleus (222u).
  4. Apr 6, 2005 #3
    Thanks, my brain just isn't working anymore. I spend about 45minutes trying to do a ridiculous substitution method. You guys are saviors.
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