Respuesta :
Explanation:
Given that,
Velocity of the proton in lab frame [tex]u_{x} = 0.6c[/tex]
Velocity of the observer v= 0.8c
We need to calculate the velocity of the proton with respect to the observer
Using formula of velocity
[tex]u'=\dfrac{v-u_{x}}{1-\dfrac{u_{x}v}{c^2}}[/tex]
[tex]u'=\dfrac{0.8c-0.6c}{1-\dfrac{0.6c\times0.8c}{c^2}}[/tex]
[tex]u'=c(\dfrac{0.8-0.6}{1-(0.8\times0.6)})[/tex]
[tex]u'=0.385c[/tex]
(a). We need to calculate the total energy of the proton in the lab frame
Using formula of kinetic energy
[tex]K.E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{u_{x}^2}{c^2}}}-1)[/tex]
Where, Proton mass energy = m₀c²
Put the value into the formula
[tex]K.E=938.28\times(\dfrac{1}{\sqrt{1-\dfrac{(0.6c)^2}{c^2}}}-1)[/tex]
[tex]K.E=938.28\times(\dfrac{1}{\sqrt{1-(0.6)^2}}-1)[/tex]
[tex]K.E=234.57\ MeV[/tex]
(b). We need to calculate the kinetic energy of the proton in the observer
Using formula of kinetic energy
[tex]K.E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{u'^2}{c^2}}}-1)[/tex]
[tex]K.E=938.28\times(\dfrac{1}{\sqrt{1-\dfrac{(0.385)^2}{c^2}}}-1)[/tex]
[tex]K.E=938.28\times(\dfrac{1}{\sqrt{1-(0.385)^2}}-1)[/tex]
[tex]K.E=78.366\ MeV[/tex]
(c). We need to calculate the momentum of the proton with respect to observer
Using formula of momentum
[tex]P_{obs}=\dfrac{m_{0}u'^2}{\sqrt{1-\dfrac{u'^2}{c^2}}}[/tex]
We know that,
Proton mass energy = m₀c²
[tex]m_{0}=\dfrac{938.28}{c^2}[/tex]
[tex]P_{obs}=\dfrac{\dfrac{938.28}{c^2}\times(0.385c)^2}{\sqrt{1-\dfrac{(0.385c)^2}{c^2}}}[/tex]
[tex]P_{obs}=\dfrac{938.28\times(0.385)^2}{\sqrt{1-0.385^2}}[/tex]
[tex]P_{obs}=150.69\ MeV/c[/tex]
Hence, This is required solution.
