# Special relativity momentum and energy conservation

1. Nov 3, 2016

### Taylor_1989

1. The problem statement, all variables and given/known data
Two identical particles of mass m travel towards each other at speed v; they combine and form a single new particle. By employing conservation of momentum and conservation of energy, what is the mass of this new particle in

2. Relevant equations
Relativistic momentum and total energy , possibly lorentz velocity transformation

3. The attempt at a solution

I am completely lost to where to start with this, because of two reasons. I can see if I need to use a frame of reference, I not sure if the v<<c and do I take the velocity as a vector?

I know the momentum and total energy are consevered in relativity but in the question I see that if the final momentum is equal to the sum of the initial surely this 0 because they are moving at the same speed and they are the same mass, so how can they even move at a speed after the collision. My lecture has not explained this very well in his notes and I am very lost could someone please help me.

Also tell me what the latex tags are for the physics forum. Big thanks in advance.

2. Nov 3, 2016

### PeroK

I think it's safe to assume that "towards each other" means a head-on collision. So, momentum before the collision is 0, as you suspected.

Latex in line $E = mc^2$ or isolated: $$E = mc^2$$

3. Nov 3, 2016

### phinds

Something to think about to clarify your thinking on the velocity and Lorentz transform: would the answer be any different if you were to use a reference frame in which one of the particles is at rest?

4. Nov 3, 2016

### PeroK

Personally, I would prefer the centre-of-momentum frame for his one!

5. Nov 3, 2016

### Taylor_1989

Thank for the response guy, I just had another look and it dawned on me. I mean momentum would be 0 and then I could total energy of the system is equal to the rest energy hence [math]E=mc^2[\math]

6. Nov 3, 2016

### Buffu

You have to enclose the math in side two # or two $eg: ## e = mc^2 ## or$$e = mc^2$\$

just don't put ` in between.It is to stop the rendering of math.

7. Nov 3, 2016

### Ray Vickson

No. If the collision is perfectly inelastic (all kinetic energy converted to mass) then the mass afterwards would NOT be $2m$ in the relativistic case. (In the classical case we would not have conservation of energy because mass end energy are not equivalent in that regime.)