The neutral pion is an elementary particle of the meson family that has a rest mass energy of 135 MeV. This particle is unstable and decays into two photons ("light particles" of no rest mass and energy E=pc). Consider now the following situation: a neutral pion has a kinetic energy of 270 MeV as measured in a given frame. Find the momentum, the direction of propagation and the energy of each of the two emitted photons, given that one of the photons is emitted in a direction perpendicular to the initial velocity of the neutral pion. My solution: by conservation of energy Kpion+Mpion*c^2=(p1+p2)c^2 p1+p2=(270+135)/c=405/c by conservation of momentum: x-direction: Ppion=p1cos(theta1) + p2cos(theta2)=p2cos(theta2) y-direction: 0=p1-p2sin(theta2) For the pion we have Ppion*c^2=(E^2-mc^2)^0.5=4.498E-15*c^2 My problem now is solving for theta2. theta2=arctan(P2y/P2x) I already know P2x=Ppion, but I can't get P2y. Thanks.