Question

In a high-energy collision between a cosmic-ray particle and a particle near the top of the Earth's atmosphere, 120 km above sea level, a pion is created. The pion has a total energy E of 1.35 × 105 MeV and is travelling vertically downward. In the pion's rest frame, the pion decays 35.0 ns after its creation. The rest energy of a pion is 139.6 MeV (a) Calculate the speed of the pion and the Lorentz factor. (10 points) (b) At what altitude above sea level, as measured from Earth's reference frame, does the decay occur? Leave your result in kilometers. (10 points)

Answer 1

a) Lorentz factor = Total Energy/Rest Energy gamma = 1.35 times 10^5/139.6 = 967.05 gamma = 1/squareroot 1 - v^2/c^2 = 967.05 1 - v^2/c^2 = 1/(967.05)^2 v = squareroot 1 - 1/(967.05)^2 times C = 299792298 m/s first 160 m/s less that of light b) For it is decayed 35 ns later Therefore distance travelled in pi frame L = 299792298 times 35 times 10^-9 = 10.493 m This distance for the Earth frame = L gamma = 10, 147 m Therefore 10.147 km It decays 120-10.145 = 109.85 km