# Homework Help: Relativistic Centre of Momentum Frame

1. Jan 4, 2016

### xoxomae

1. The problem statement, all variables and given/known data
A partical of mass m ha speed 0.546c relative to inertial frame S. The particle collides with an identical particle at rest in the inertial frame S. Relative to S and in terms of c, what is the speed of S' in which the total momentum of these particles is 0.
2. Relevant equations

3. The attempt at a solution
I can see that in the S' frame, Pa = - Pb so that means that Va = - Vb since both masses are equal. But I just can't get my head around why i need to use special relativity because everything is in reference to the S frame anyway.

2. Jan 4, 2016

### andrewkirk

That's correct, but it doesn't answer the question. They want you to tell them the value of Va. THe naïve non-relativistic answer is that it's half of 0.546c, ie 0.273c but that answer is wrong because it doesn't allow for the fact that, if you choose a frame that has velocity 0.273c relative to the first one, the velocities of the two particles will NOT be 0.273c and -0.273c respectively, because of the relativistic corrections.

3. Jan 4, 2016

### Staff: Mentor

Pa=-Pb and Va=-Vb are not in S.

4. Jan 4, 2016

### xoxomae

S' frame is moving at a velocity v w.r.t S frame.
Let velocity of Particle 1 = U1 in S frame and particle 2 = U2 in S frame therefore Particle 1 = U1' and particle 2 = U2' in S' Frame
Since particle 2 is at rest in S frame, its velocity in S' frame is equal to the velocity of the S' frame relative to S i.e. v=u2'
So in a centre of momentum frame u2'= - u1' = v

Using Einstein velocity addition formulas and setting c to 1

u2' = ( u2 - v ) / (1-v/c) = v

When solved gives v = 0.297c

Does this seem right?

5. Jan 4, 2016

Right.