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Homework Help: Momentum in elastic collision

  1. Jan 26, 2010 #1
    1. The problem statement, all variables and given/known data

    Uranium fission within a nuclear reactor produces high-speed neutrons. These must be slowed down for them to trigger a chain reaction within the uranium. In a heavy-water reactor the neutrons of mass m make multiple elastic collisions with deuterium nuclei of 2m, losing speed at each collision.

    (a) Suppose a fast-moving neutron of speed v makes a head-on elastic collision with a stationary deuterium nucleus. Show that its speed is reduced to v/3.

    2. Relevant equations

    p=mv - We probably have to use this
    Ek=0.5mv2 - Not sure if we have to use this, but the question IS in a section about kinetic energy. It just seems logical to use p=mv because using kinetic energy makes no sense.

    3. The attempt at a solution

    Ok it took me absolutely AGES to do this, but I don't think it's right:

    PD1 = mv = 2m x 0 = 0 N s
    PN1 = mv = 1m x v = mv

    PD1 + PN1 = PD2 + PN2 if no external forces are applied

    0 + mv = PD2 + PN2
    mv = PD2 + PN2
    PD2 = mv - PN2

    PD2 = PN2 - not sure

    Substitute above two equations together to show PN2:
    PN2 = mv - PN2
    2PN2 = mv
    PN2 = mv/2
    vN2 = PN2/2m
    vN2 = (mv/2)/2m
    vN2 = mv/4m - my rearranging may be wrong
    vN2 = v/3m

    But the answer is just v/3 not v/3m. Is this still right? Or is it just coincidence I got an answer so similar?
     
  2. jcsd
  3. Jan 26, 2010 #2
    v/3m cannot be right. it has units of m.kg/s
     
  4. Jan 26, 2010 #3
    The answer i got was -v/3. So the neutron bounces backwards with speed v/3 and the deuterium goes forward with 2v/3. I didn't read through your answer very thoroughly, but anywhere with mass on the bottom will be wrong. All the masses cancel out in my calculations
     
  5. Jan 26, 2010 #4
    Could you post your calculation please? You don't even have to explain it, but I've been trying to do this all night and I don't have a clue. Got an exam on thursday l0L!
     
  6. Jan 26, 2010 #5
    bah fine. I'm a lazy individual.
    v is initial velocity
    v1 is final velocity of the neutron
    v2 is the final velocity of the deuterium

    mv = m . v1 + 2m . v2 ------------- (momentum)
    mv² = m . v1² + 2m . v2² ------------- (energy)

    All the masses cancel:

    v = v1 + 2 . v2 ------------- (momentum)
    v² = v1² + 2 . v2² ------------- (energy)

    isolate the momentum equation for v2
    v2 = (v - v1)/2
    v2² = (v² - 2*v1*v + v1²)/4

    sub that into your energy equation and isolate for v1.
     
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