|Apr12-12, 12:05 PM||#1|
Special Relativity 2-D Collision Problem
1. The problem statement, all variables and given/known data
This is my first time posting here, so please let me know if I do not have the proper format.
A particle of mass M is at rest in the laboratory when it decays into three identical particles, each of mass m. Two of the particles have velocities and directions as shown (one travels in the -x direction with a speed of 4c/5 and another travels in the -y direction with a speed of 3c/5. Calculate the direction and speed of the third particle (shown on the diagram moving with +x and +y components at some angle above the x axis) and find the ratio M/m.
I'm not sure (1) if the way I solved for v and the ratio is correct and (2) how to find the direction of the third particle. I originally solved it by saying the components of momentum are equal, but I don't think that would be correct because it is relativistic. Any help would be greatly appreciated!
2. Relevant equations
ρ = mv[itex]\gamma[/itex]
E = mc2[itex]\gamma[/itex]
3. The attempt at a solution
By conservation of momentum, ρi = ρf, so:
0 = (3/4)mc + (4/3)mc + v3[itex]\gamma[/itex]3
v3[itex]\gamma[/itex]3 = -(3/4)c - (4/3)c
After plugging in gamma to solve for v3:
v/sqrt(1-v2/c2) = -(25/12)c
v = +/- sqrt(c2/((144/625) + 1))
v = 0.9c
By conservation of energy, Ei = Ef, so:
Mc2 = (5/4)mc2 + (5/3)mc2 + 2.294mc2 where 2.294 is [itex]\gamma[/itex]3 (found using v solved for above)
so M = 5.21m
M/m = 5.21
|Apr12-12, 02:20 PM||#2|
Never mind, I think I got the answer. The components of relativistic momentum are just equal.
|relativity, special relativity|
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