# Relativistic addition of velocities in the y-direction? [closed]

I've been having some difficulty with the problem described below, please have a look:

Two particles with identical masses (m) undergo a collision. Before the collision they move with velocities (c/2,0,0) and (-c/2,0,0), and after the collision, they move with velocities (0,c/2,0) and (0,-c/2,0) in frame S.

How can I find the velocity of the particles AFTER the collision in frame S' where one of the particles is at rest prior to collision??

Working so far: Using the RELATIVISTIC SPEED ADDITION FORMULA for the normal x-axis case, I have obtained $u'x=4/5c$ for the particle not at rest BEFORE the collision. HOW DO I SOLVE THE PROBLEM FOR THE Y-AXIS CASE AFTER THE COLLISION IN THE S' FRAME?

Also, how can I show there is a conservation of momentum before and after collision using the formula p=(gamma)mv? The different direction of velocity after the collision is confusing me...since momentum is conserved linearly right?

A detailed description would be appreciated!

THANK YOU SO MUCH! :)

## closed as off-topic by John Rennie, David Z♦Mar 3 '15 at 12:47

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Let's take the two frames $S$ and $S'$ to coincide at the moment of the collision, so the collision occurs at $(t, x, y) = (0, 0, 0)$ in both frames. We'll take the velocity of $S'$ relative to $S$ to be $v$ (i.e. $v = c/2$), and well take this to be along the $x$ axis in S.

In $S$ the positions of the particles at a time $t$ after the collision are $(t, 0, +vt)$ and $(t, 0, -vt)$, and their relative velocity is just the separation between the two points divided by the time.

So just use the Lorentz transformations to see where these two points are in $S'$, then divide the separation in $S'$ by the time in $S'$ to get the relative velocity in $S'$. The Lorentz transformations are:

$$t' = \gamma \left( t - \frac{vx}{c^2} \right )$$

$$x' = \gamma \left( x - vt \right)$$

$$y' = y$$

There is a similar problem discussed in Light & Observer moving perpendicular to each other