elastic collision and kinetic energy
Physics and electronics discussion, revision, exam and homework help.
-
- Reputation:
- Thread Starter
- New Member
- Posts: 7
elastic collision and kinetic energyThis 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 -
- Reputation:
- Full Member
- Posts: 121
Re: elastic collision and kinetic energyI 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.(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
Using conservation of momentum
This simplifies to
Conservation of energy
Simplifies to
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
Substituting this value of y into conservation of momentum gives
Complete the square or use quadratic formula to solve for x and you end up with
Obviously we are not interested in the case when x=v so we shall use
Kinetic energy of neutron to begin with is
Kinetic energy of neutron after collision is
Final kinetic energy divided by original kinetic energy times 100 will give us the percentage of the neutrons k.e kept.
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.Last edited by Augmented hippo; 17-06-2012 at 09:31.
