# Relativistic speed involving particle decay

1. Mar 3, 2009

### kpou

1. The problem statement, all variables and given/known data
A particle of rest mass M0 is at rest in the laboratory when it decays into three identical particles, each of rest mass m. Two of the particles have velocity
u1=-4/5c i(vector)
u2=-3/5c j(vector)
Calculate the direction and speed of particle 3

2. Relevant equations
pf-pi=0
p=(gamma)mu

3. The attempt at a solution

The starting momentum = 0 since it is at rest, so the momentum of the three particles should add up to 0.
p1=m/(sqrt(1-(-4/5)^2)) -i(vector)
p2=m/(sqrt(1-(-3/5)^2)) -j(vector)
p3=m/(sqrt(1-(v/c)^2)) ij(vector)

the total momentum is p1+p2+p3
I calculated m/.6 -i + m/.8 -j + m/(sqrt(1-v/c)^2)) ij
I'm not sure how to break this up to find v.
I had tried to use u=(sqrt(ux^2+uy^2+uz^2)) but that = 1 for this.
I think to use that equation I need to break down the ij vector of particle 3
To do this it should be atanx=(3/4), but the answer is actually atanx=(9/16) I'm not sure why it's squared. Likely I'm forgetting something very basic that messed up the whole problem.

Main points of interest:
find out i and j vectors of particle 3
find out why the units are squared as in atanx=(9/16) and is that 9/25 (3/5)^2? if so that doesn't compute out using sin 29 = (9/25)/h

M/m should be (gamma)1*m+(gamma)2*m+(gamma)3*m since they all have different gammas.

Last edited: Mar 3, 2009
2. Mar 3, 2009

### kpou

My momentum equations were missing velocity. i only had mass in the numerator. p1=4/3 i(vector) p2=3/4 j(vector)
p3=m (sqrt ((4/3)^2+(3/4)^2))

much thanks to Bekki from yahoo answers :)