# Math Help - conservation of linear momentum

1. ## 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 .

2. Originally Posted by thereddevils
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 $E_n=(.86)^nE_0$

3. Originally Posted by malaygoel
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,