# conservation of linear momentum

• Jun 29th 2009, 01:17 AM
thereddevils
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 .
• Jun 29th 2009, 01:29 AM
malaygoel
Quote:

Originally Posted by thereddevils
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 $\displaystyle E_n=(.86)^nE_0$
• Jun 29th 2009, 05:03 AM
thereddevils
Quote:

Originally Posted by malaygoel
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,