Noncentral Elastic Collision: Return to Moving Particles

[Gavroy]

hi

i am describing a noncentral elastic collision between two particles. therefore i assumed that one of them is at rest. now i want to return back to the system where both particles are moving. do i only need to add the velocity of the particle at rest to both of them and leave all angles unchanged or is there something else i have to keep in mind? i am a little bit uncertain about this whole thing.

thank you for any kind of help!

 

[majon]

This link might help: www . phi . duke . edu/courses/217/MottScatteringReport/node8.html

 

[Gavroy]

sorry, link does not work.

 

[haruspex]

hi

i am describing a noncentral elastic collision between two particles. therefore i assumed that one of them is at rest. now i want to return back to the system where both particles are moving. do i only need to add the velocity of the particle at rest to both of them and leave all angles unchanged or is there something else i have to keep in mind? i am a little bit uncertain about this whole thing.

thank you for any kind of help!

Just take your reference frame as one in which one of the particles is initially at rest. I.e. subtract that velocity everywhere from the initial conditions and add it back in everywhere to the answer.

 

[PJC01]

I can provide some guidance on how to return to the system where both particles are moving after assuming one was at rest. In an elastic collision, the total kinetic energy and momentum of the system are conserved. Therefore, to return to the moving particles system, you should add the velocity of the stationary particle to both particles and leave all angles unchanged. This will ensure that the total momentum and kinetic energy of the system remain unchanged. However, it is always important to double check your calculations and make sure they are consistent with the laws of conservation of energy and momentum. If you are still uncertain, consulting with a colleague or conducting further research may help clarify any doubts.