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Math Help - conservation of linear momentum

  1. #1
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    conservation of linear momentum

    Neutrons in a nuclear reactor have the Kinetic energy 6.0\times10^{-13}J . The kinetic energy of the neutron is reduced to 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 : E_o

    Kinetic energy after 1 collision = (1-0.14)E_o

    After n collisions , the kinetic energy will be

    (1) n(0.86)E_o or

    (2)  (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 6.0\times10^{-13}J . The kinetic energy of the neutron is reduced to 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 : E_o

    Kinetic energy after 1 collision = (1-0.14)E_o

    After n collisions , the kinetic energy will be

    (1) n(0.86)E_o or

    (2)  (0.86)^nE_o

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

    Let E_n be the energy after n collisions.

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

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

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

    which gives 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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