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Elastic collision between neutrons and deuterons

  1. Sep 18, 2007 #1
    1. The problem statement, all variables and given/known data
    Canadian nuclear reactors use heavy water moderators in which elastic collisions occur between neutrons and deuterons of mass 2.0u.
    a.) What is the speed of a neutron expressed as a fraction of its original speed, after a head-on elastic collision with a deuteron which is initially at rest?
    b.) What is its kinetic energy, expressed as a fraction of its original kinetic energy?
    c.) How many such successive collisions will reduce the speed of a neutron to 1/6600 of its original value?

    2. Relevant equations
    If the 2nd particle is at rest
    [tex]V_{A} = \frac{m_{A} - m_{B}}{m_{A}+m_{B}}[/tex]* V2
    [tex]V_{B} = \frac{2m_{A}}{m_{A} + m_{B}}[/tex]* V2

    3. The attempt at a solution
    let n = neutrons , dn = deuterons
    VnMn = MnVn2 + MdnVdn2

    using equation above:
    [tex]V_{n2} = \frac{2.0u - 2.0u}{2.0u+2.0u} V_{n}[/tex]
    Vn2 = 0

    b.) K2 = ?
    since Vn2 =0
    VnMn = Vdn2
    K2 = 1/2 (2.0u)(Vdn2)2
    K2 = u(Vdn2)^2 <<<< K equals velocity of deuterons squared times u

    c.) (2.0)(1/6600Vn) =

    3300 collisions
    I honestly dont know about this
    Last edited: Sep 18, 2007
  2. jcsd
  3. Sep 18, 2007 #2


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    Staff: Mentor

    For an elastic collision one applies conservation of momentum and energy.

    Intially, the deuteron at rest, Vd, is zero, so it has not momentum or KE.

    The neutron will not lose all its energy or momentum, but will recoil 180°.

    So write the momentum and energy equations and one should end up with

  4. Sep 18, 2007 #3
    but i thought masses are the same:
    which makes
    v{initial speed of neutron} = (m1 - m2) / (m1 + m2) * V{final speed of deuteron}

    equal to zero? and the deuteron goes to move which has a speed equal to the initial speed of neutron?...
  5. Sep 18, 2007 #4


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    Homework Helper

    deuteron has a mass of 2.0u. Neutro has a mass of 1.0u
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