aAccording to observer O a certain particle has a momentum o

(a)According to observer O, a certain particle has a momentum of p=784 MeV/c and a total relativistic energy of E=1175 MeV. What is the rest energy of this particle?
MeV/c**2

(b)An observer O\' in a different frame of reference measures the momentum of this particle to be p\'=867 MeV/c. What does O\' measure for the total relativistic energy of the particle?
Mev

Solution

E = sqrt[ p^2 c^2 + m0^2 c^4]

and rest energy = mo c^4

So m0^2 c^4 = (rest energy)^2

1175 MeV = sqrt [ (784 MeV /c)^2 (c^2) + (rest energy)^2]

1380625 = 614656 + (rest energy)^2

rest energy = 875.20 MeV ...........Ans

(B) Rest energy = m0 c^2 = 875.20

m0 = 875.20 MeV / c^2 ........rest mass

Momentum. p = Y m0 v

where Y = 1 / sqrt[ 1 - (v/c)^2]

867 MeV / c = (875.20 MeV/c^2) (v) / sqrt[ 1 - (v/c)^2]

751689 (1 - (v/c)^2) = 765975 v^2/c^2

(v/c)^2 = 0.495

v = 0.704 c

Y = 1 / sqrt [ 1- 0.704^2] = 1.408

total relativistic energy = Y m0 c^2

= 1.408 x 875.20 MeV/c^2 x c^2

= 1232.33 MeV .............Ans