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Thread: conservation of linear momentum

  1. #1
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    conservation of linear momentum

    Neutrons in a nuclear reactor have the Kinetic energy $\displaystyle 6.0\times10^{-13}$J . The kinetic energy of the neutron is reduced to $\displaystyle 6.0\times10^{-21}$ by a series of collision between the neutron and carbon nuclei in the moderator . On average , the fractional loss of the kinetic energy of the neutron after each collision is 0.14 . how many collisions does a neutron experience in the process ?

    My working :

    Let the initial Kinetic energy : $\displaystyle E_o$

    Kinetic energy after 1 collision = $\displaystyle (1-0.14)E_o$

    After n collisions , the kinetic energy will be

    (1) $\displaystyle n(0.86)E_o$ or

    (2) $\displaystyle (0.86)^nE_o$

    Which one is the correct one and why is it so ?? Thanks for your time .
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by thereddevils View Post
    Neutrons in a nuclear reactor have the Kinetic energy $\displaystyle 6.0\times10^{-13}$J . The kinetic energy of the neutron is reduced to $\displaystyle 6.0\times10^{-21}$ by a series of collision between the neutron and carbon nuclei in the moderator . On average , the fractional loss of the kinetic energy of the neutron after each collision is 0.14 . how many collisions does a neutron experience in the process ?

    My working :

    Let the initial Kinetic energy : $\displaystyle E_o$

    Kinetic energy after 1 collision = $\displaystyle (1-0.14)E_o$

    After n collisions , the kinetic energy will be

    (1) $\displaystyle n(0.86)E_o$ or

    (2) $\displaystyle (0.86)^nE_o$

    Which one is the correct one and why is it so ?? Thanks for your time .

    Let $\displaystyle E_n$ be the energy after n collisions.

    then, $\displaystyle E_1=.86E_0$
    $\displaystyle E_2=.86E_1$
    ..................................
    $\displaystyle E_n=.86E_{n-1}$
    ..................................

    which gives $\displaystyle E_n=(.86)^nE_0$
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  3. #3
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    Quote Originally Posted by malaygoel View Post
    Let $\displaystyle E_n$ be the energy after n collisions.

    then, $\displaystyle E_1=.86E_0$
    $\displaystyle E_2=.86E_1$
    ..................................
    $\displaystyle E_n=.86E_{n-1}$
    ..................................

    which gives $\displaystyle E_n=(.86)^nE_0$

    oh ok , i see now . Thanks a lot !!

    so it actually forms a geometric progression ,

    0.86Eo , (0.86)^2Eo , ... , (0.86)^nEo

    Thanks again,
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