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Collisions homework help

  1. May 28, 2010 #1
    1. The problem statement, all variables and given/known data

    A nuclear reactor (see below) contains nuclei that fission when bombarded by slow neutrons. The fission reaction produces fast neutrons which need to be moderated (i.e. slowed down) in collisions before they can cause further fissions. In this problem we consider a reactor which uses graphite (i.e. carbon) as moderator. Assume the neutron and carbon nuclear masses are 1.67 × 10–27 kg and 20.0 × 10–27 kg respectively.

    Consider a head-on collision between a neutron with speed 10.0 Mm/s and a stationary carbon nucleus.

    (i) Calculate the speed of the neutron after the collision

    (ii) Calculate the speed of the carbon nucleus after the collision

    3. The attempt at a solution

    (i) I assume this is a perfectly elastic collision so, momentum before = momentum after

    mnvn,i = (mn+mc)V

    10 × (1.67 × 10–27) = [(1.67 × 10–27)+(20.0 × 10–27)]V

    V=4.5 Mm/s

    But the correct answer is 8.46 Mm/s. How did they get that?

    (ii) What do they mean by "the carbon nucleus"? And how do we obtain its speed? (by the way it has to be 1.54 Mm/s)
    Last edited: May 28, 2010
  2. jcsd
  3. May 28, 2010 #2


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    Re: Collisions

    The equation you gave above is for a perfectly inelastic collision, where the nulceus and the carbon atom stick together after the collision.

    To solve this problem you should assume a perfectly elastic collision. Youll need two equations for this. One for conservation of momentum, and other for conservation of energy. The neutron and the carbon nucleus will each have different velocities after the collision.
    Again, conservation of momentum, and conservation of energy. You'll have two equations and two unknowns. Since it states that the collision is a "head-on" collision, you can treat it as a 1-dimensional problem (i.e. you don't have to break it up into separate x, y, and z components). Hope that helps! :biggrin:
  4. May 29, 2010 #3
    Re: Collisions

    Sorry, I didn't realize that. Actually, I was meant to write "perfectly inelastic" because I thought the carbon and neutron stick together after the collision! Anyway,

    mnvn,i = mnvn,f+mcvc,f

    10 × (1.67 × 10–27) = (1.67 × 10–27)vn,f+(20.0 × 10–27)vc,f

    Now in order to use this equation to find the final speed of the neutron, I have to first find the final velocity of the carbon (which is what part (ii) is asking)! But how can I use the following equation when I don't have the height?


  5. May 29, 2010 #4


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    Re: Collisions

    Ugh.... :uhh: Gravity doesn't fit in this. That's not really what I meant.

    Conservation of momentum and conservation of (kinetic) energy means,

    mcvc + mnvn = mcvc' + mnvn'

    (1/2)mcvc2 + (1/2)mnvn2 = (1/2)mcvc'2 + (1/2)mnvn'2
  6. May 30, 2010 #5
    Re: Collisions

    But how can I use these? I mean, we have

    mnvn,i + 0 = mnvn,f + mcvc,f

    in this equation we have two unknowns, the final velocities of both the carbon and the neutron.


    [tex]\frac{1}{2}(1.67 \times 10^{-27})10^2 =\frac{1}{2}(1.67 \times 10^{-27}) v_{n,f}^2 + \frac{1}{2} (2 \times 10^{-27}) v_{c,f}^2[/tex]

    Again here we have the same two unknowns. So we can't really use one of these equations to solve the other one. :rolleyes:
  7. May 31, 2010 #6


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    Re: Collisions

    These are called independent simultaneous equations. And if you have 2 unknowns, you can solve for them as long as you have 2 independent simultaneous equations. There is more than one way to solve these equations. I'll give you two examples of simple ways to solve these.

    Suppose we have

    [tex] x + 3y = 11 [/tex]

    [tex] y - x = 1 [/tex]

    One way to solve for x ahd y is substitution. In the second equation, solve for y.

    [tex] y = 1 + x [/tex]

    Then take that result and plug it in to the first equation, substituting (1 + x) wherever there is a y.

    [tex] x + 3(1+x) = 11 [/tex]

    [tex] x + 3 + 3x = 11 [/tex]

    [tex] 4x = 8 [/tex]

    [tex] x = 2 [/tex]

    Now we go back and substitute 2 into x in one of our above equations, and solve for y.

    [tex] y = 1 + x = 1 + 2 = 3[/tex]

    Here is another way to solve for the variables x and y, where we carefully add both equations together. Starting with

    [tex] x + 3y = 11 [/tex]

    [tex] y - x = 1 [/tex]

    we multiply both sides of one of the equations by whatever it takes such that one of the variables cancels out when we add the equations together. As it happens here, there is an x in one equation and a -x in the other. So in this case they are already ready to be added.

    [tex] (x + 3y) + (y - x) = 11 + 1 [/tex]

    [tex] 4y = 12 [/tex]

    [tex] y = 3 [/tex]

    Substituting 3 into y in one of the above equations allows us to solve for x.

    [tex] 3 - x = 1 [/tex]

    [tex] x = 2 [/tex]

    There are other ways to solve systems of independent simultaneous equations too. Most of them are part of a topic of math called linear algebra. But I won't go into that here.
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