oh wait the gamma in that equation is the gamma of the protons so first i have to solve for u'
ok so using this i got that the initial speeds of the protons in the COM frame was 0.36c
then knowing that the initial speed of the one proton in the lab frame of reference was zero i found that the...
would i just use the energy equation because if the pion is also at rest in the COM frame, then the kinetic energy of the two protons is just equal to the rest mass of the pions so i have to solve for v in the gamma of the kinetic energy of the protons?
2(gamma-1)*m(proton)*c^2 = m(pion)*c^2
ok you a lot easier
so in the centre of mass frame all the particles have a final velocity of zero then?
but how do i find the speed between the lab frame and the centre of mass frame?
ok well in the centre of mass frame I am getting that for energy:
2K(proton) = m(pion)c^2 + K(pion)
because the final kinetic energy of the protons is zero and the rest mass energy of the protons just cancel out on each side
for momentum:
gamma*m(proton)*u1 + gamma*m(proton)*(-u1) =...
well because in the question, one proton is at rest and the other collides with it
so I am trying to find the frame that sees the two protons having equal but opposite speeds, this other frame is S'
but in the frame that the questions speaks of, one proton has speed u1 and the other has a...
ok so i got a quadratic equation for v
(u1/c^2)v^2 - 2v + u1 = 0
i got this by setting u2=0 because its at rest in the initial frame of the question
so i guess i have to use the quadratic formula to find v?
im not sure what to do know even with the relative speed of the frames, all i know is...
oh I am sorry, I've never heard of it being referred to as that
so what should i do?
should i find the speed of the frames relative to each other using the lorentz transformations by setting:
u'1=u'2
what do you mean the centre of mass frame? of which particle?
and I am not sure if this will help cause even in the other frames i won't have anymore information that what i have now right?
Homework Statement
proton hits a stationary proton which then produces a pion with the following reaction
p + p = p + p +pion
if the initial proton has just enough energy to produce the pion what are the speeds of the final protons and pion in the laboratory frame?
m(pion)=0.144m(proton)...
actually i just checked, it does make a significant difference with the electrons
ill probably just ask my professor about that, cause he didnt specify in the question that we consider the mass of the electrons too
lmao, I am sorry i wasnt more excited, but i really do appreciate the help
and i was thinking the same thing about the electrons, but i think that since electrons are so much smaller than protons, its affect will be insignificant
once again thanks so much!
ya the masses are given
proton being 1.00728u and He being 4.00150u, and 4 protons are used, so its not exactly the same, would the difference in the mass of the 4 protons and the He nucleus be the mass equivalence of the energy given off?