- #1

- 110

- 0

A particle called a Kaon is moving at 0.8c through a detector when it decays into two pions.

Kaon particle: mass = 493.7 MeV/c^2

Pion+: mass = 139.6 MeV/c^2

Pion0: mass = 135.0 MeV/c^2

1) Apply conservation of momentum/energy to determine the momentum of each pion in the kaon frame of reference (where K is at rest).Use the fact that the momentum of the original kaon in its own frame is zero and don't forget to use the relativistic momentum energy conservation equation. Leave the momentum in MeV/c units.

Since the momentum of the kaon is 0 because it is at rest in its own frame, the magnitudes of the momentum of the pions should be equal.

##E_k = m_k c^2 = 493.7 \frac{MeV}{c^2} = (m_{\pi +} + m_{\pi 0}) c^2 + (p_{\pi +} + p_{\pi 0})c##

##493.7 \frac{MeV}{c} = (139.6 + 135.0) \frac{MeV}{c} + 2|p|##

##219.1 \frac{MeV}{c} = 2|p|##

##p = 109.55 \frac{MeV}{c}##

Solving for velocities:

##p{\pi +} = 109.55 \frac{MeV}{c} = \gamma_{\pi +} m_{\pi +} v_{\pi +} = \gamma_{\pi +} 139.6 \frac{MeV}{c^2} v_{\pi +}##

From this:

##v_{\pi +} = .61733 (c), and thus : \gamma_{\pi +} = 1.27113##

And:

##v_{\pi 0} = .63012 (c), and thus: \gamma_{\pi 0} = 1.28783##

From this there is no Conservation of energy:

##E_k = E_{\pi +} + E_{\pi 0} = 493.7 \frac{MeV}{c^2} = \gamma_{\pi +} m_{\pi +} + \gamma_{\pi 0} m_{\pi 0} = doesn't = 493.7!##

(I left units off)

2)Now transform the momentum-energy vectors for each pion and the kaon to the detector frame. Assume that the pion+ is moving in the direction the Kaon was moving in the detector frame.

3) Check to see that the momentum-energy conservation law is followed in the detector frame.

4) If the pions are moving off perpendicular to the Kaon's motion in the detector frame, calculate the momentum energy vectors for the pions in the detector frame and again check to see that momentum-energy is conserved.