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!



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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, David Z
If this question can be reworded to fit the rules in the help center, please edit the question.


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


Not the answer you're looking for? Browse other questions tagged or ask your own question.