Answer to Question #113264 in Mechanics | Relativity for Moel Tariburu

Question

A pion lives 1.50 x 10-16 s as measured in the laboratory, and 0.90 x 10-16 s when at rest
relative to an observer.
i) Determine the speed of the pion.
[2.5]
ii) Calculate the kinetic energy in MeV of the particle, given its rest mass energy is
139.6 MeV?

Answer 1

Solution.

$\tau =1.50\sdot10^{-16}s;$

$\tau_0 =0.90\sdot10^{-16}s;$

$c = 3\sdot10^8m/s;$

$\tau= \dfrac{\tau_0}{\sqrt{1-\dfrac{\upsilon^2}{c^2}}}; \implies$ $\upsilon = c\sqrt{1-\dfrac{\tau_0^2}{\tau^2}};$

$\upsilon = c\sqrt{1-\dfrac{(0.90\sdot10^{-16})^2}{(1.5\sdot10^{-16})^2}}=1.92\sdot10^8m/s;$

$E = \dfrac{mc^2}{\sqrt{1-\dfrac{\upsilon^2}{c^2}}}$ ;

$E =\dfrac{139.6MeV}{\sqrt{1-\dfrac{(1.92\sdot10^8m/s)^2}{(3\sdot10^8m/s)^2}}}=232.7MeV;$

Answer:$\upsilon = 1.92\sdot10^8m/s;$

$E = 232.7MeV.$