# Conservation of relativistic momentum and energy

1. Feb 13, 2013

### FllOnBlckDys57

1. The problem statement, all variables and given/known data

The K0 meson is an uncharged member of the particle "zoo" that decays into two charged pions according to K0 ---> π+ + π-. The pions have opposite charges as indicated, and the same mass, mπ=140MeV/c^2. Suppose that a K0 at rest decays into two pions in a bubble chamber in which a magnetic field of 2.0T is present. If the radius of curvature of the pions is 34.4cm, find (a) the momenta and speeds of the pions and (b) the mass of the K0 meson.

mπ=140MeV/c^2

r=0.344m

B=2.0T

Problem figure:

http://img208.imageshack.us/img208/171/physpr.jpg [Broken]

2. Relevant equations

F=qv x B

F=ma

F=m(v^2/r)

E^2=p^2c^2+(mc^2)^2

p=mv/√(1-v^2/c^2)

E(K0)=E∏+ + E∏-

P(K0)=P∏+ + P∏-

3. The attempt at a solution

I feel like this problem should be started by first solving for the velocity of the particles. This could be done by combining F=qv x B and F=m(v^2/r) into m(v^2/r)=qv x B.

However, the charge (q) would still be unknown. Do pions have a standard charge? I can't seem to find it in my book. Or am I approaching this problem the wrong way?

Thanks!

Update:

I am pretty sure a positive pion has a charge of 1.602E-19 C and a negative pion has an opposite charge.

I solved for the mass of the pions in kg: (140MeV/c^2)(1.783x10^-30kg/1MeV)=2.4962x10^-28 kg

I then plugged these values into my equation:

m(v^2/r)=qvB

(2.4962x10^-28kg)(v^2/(0.344m))=(1.602x10^-19 C)(V)(2.0T)

Solving for v, I get: v=4.42x10^8 m/s

However, this can't be right since it is faster than the speed of light. Is there some sort of relativistic equation I should be using to solve for the velocity?

Last edited by a moderator: May 6, 2017
2. Feb 14, 2013

### tms

Correct. Charge comes in a standard size.
It is always good to apply a sanity check to your answers, as you did.

Since this is relativistic, you should use momentum in your calculations, and then get the velocity from that. Start with $F = qvB$ and remember the definition of force: $F = dp/dt \approx \Delta p/\Delta t$.