Energy Problems in Special Relativity
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Part 1
The speed of light is 2.998 * 10 ^8 m/s. Find the ratio of the total energy to the rest energy of a particle of rest mass m0 moving
with speed 0.26 c.
Part 2
The K^0 particle has a mass of 497.7 MeV/c^2 . It decays into aPi- and Pi+, each with mass 139.6 MeV/c2 . Following the decay of a K^ 0, one of the pions is at rest in the laboratory. Determine the kinetic energy of the K^ 0 prior to decay.
Answer in units of MeV.
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Solution Summary
The solution is given in the context of a discussion of Kaon decay and gamma radiation. 484 word and calculations
Solution Preview
The energy of a particle is given by:
E(v) = gamma(v) m c^2
Here gamma(v) = 1/sqrt[1-v^2/c^2], m is the rest mass. Note that in theoretical physics when we say mass, we always mean "rest mass".
If we divide the energy when the particle moves at some velocity v with the energy at rest we get:
E(v)/E(0) = gamma(v)/gamma(0) = gamma(v).
So, the ratio is just the gamma factor which for v = 0.26 c is 1.036
The Kaon decay problem can be solved by first considering the center of mass frame (here the total momentum is zero, this frame is also called the zero momentum frame). In fact, almost all problems in ...
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