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elastic collision and kinetic energy

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    This should be simple but i just cant do it! Prove that an elastic head on collision between a neutron and a carbon atom which is initially stationary would result in the neutron losing about 72% of its kinetic energy to the carbon atom. I know that both KE and momentum are conserved but i can't seem to eliminate the unknowns in the equations!

    If any of you physicists out there can explain how it can be proved i'd be sooo grateful
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    (Original post by lilangela)
    This should be simple but i just cant do it! Prove that an elastic head on collision between a neutron and a carbon atom which is initially stationary would result in the neutron losing about 72% of its kinetic energy to the carbon atom. I know that both KE and momentum are conserved but i can't seem to eliminate the unknowns in the equations!

    If any of you physicists out there can explain how it can be proved i'd be sooo grateful
    I started by saying the neutron has mass m and velocity v. Carbon atom has mass 12m and is stationary. After collision I said neutron has velocity x and carbon atom has velocity y in the same direction of travel as the neutron was originally travelling in.
    Using conservation of momentum
    mv=mx+12my
    This simplifies to v=x+12y

    Conservation of energy
    \frac{mv^2}{2}=\frac{mx^2}{2}+6m  y^2
    Simplifies to v^2=x^2+12y^2

    It sounds like you got this far but didn't know what to do next. The key is you want to know x in terms of v. You don't need to eliminate all the variables, just y.

    Rearranging conservation of momentum for y gives y=\frac{v-x}{12}
    Substituting this value of y into conservation of momentum gives v^2=x^2+\frac{(v-x)^2}{12}
    12v^2=12x^2+(v-x)^2
    12v^2=12x^2+v^2-2vx+x^2
    13x^2-2vx-11v^2=0
    Complete the square or use quadratic formula to solve for x and you end up with x=\frac{v}{13}\pm\frac{12v}{13}
    Obviously we are not interested in the case when x=v so we shall use x= \frac{-11v}{13}
    Kinetic energy of neutron to begin with is \frac{mv^2}{2}
    Kinetic energy of neutron after collision is \frac{mx^2}{2}=\frac{m}{2}\times (\frac{-11v}{13})^2 = \frac{121mv^2}{338}
    Final kinetic energy divided by original kinetic energy times 100 will give us the percentage of the neutrons k.e kept. (\frac{121mv^2}{338})\div(\frac{  mv^2}{2}) = 0.72 0.72\times 100 = 72
    This is slightly different from what you said in your post as the neutron has kept 72% of it's K.E not lost 72%. Anyway hope that helps.

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Updated: June 17, 2012
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