# How much total kinetic energy is needed in this case?

1. Aug 8, 2006

### eri3an

I have a problem with an exercise from school. could anyone help me?

Let say that you are searching for a new particel X by making collison between positrones and elektrones in an particelaccelerator. you guess that the new particles resting mass is 210 MeV/c^2 and you hope to produce it by:

(e+) + (e–) ---> X + photons

a) Let us say that the electron (e-) is in resting state in the laboratory system. How much energy is needed to accerate positrons (e+) to such a velocity for the reaction to take place?

b)On less energy demanding way of doing it is to let e+ och e– collide with eachother from opposite directions with the same velocity How much total kinetic energy is needed in this case?

resting mass for an electrone or positron is 0.511 MeV/c^2

plz I'm not intrested in just getting an answer how do i solve it? I've tried by deriving from the E(a) + E(b) = E(c) + E(d) and tried to go using conservation of momentum but it doesn't seem to work :( Maybe i'm out on bad ice?

Last edited: Aug 8, 2006
2. Aug 8, 2006

### borisleprof

Hi eri3an,
you must remember Einstein's equation : E = mc^2, where m is the mass of the particle when it is moving.

3. Aug 10, 2006

### WigneRacah

I suggest you to study the problem in the CM (centre of mass) system.
Then you can pass to the system of the laboratory by Lorentz transformations.

4. Aug 10, 2006

### fargoth

i just though this statement could use some elaboration:
$$E=\gamma m_0c^2=\sqrt{m_0^2c^4+p^2c^2}$$

i think using the real mass, i mean the thing you divide the force with to get the accelaration is better (and thats what borisleprof did).
but in most text books, they talk about the mass you measure at the rest frame...

Last edited: Aug 10, 2006