Question

A proton moves along the x-axis in the laboratory with velocity uy = 0.6c. An observer moves with a velocity of v=0.8c along the direction of motion of the proton. a. What is the total energy of the proton in the lab frame? b. What kinetic energy does the observer measure for the proton? c. What momentum (in units of MeV/c) does the observer measure for the protons?

Answer 1

Explanation:

Given that,

Velocity of the proton in lab frame $u_{x} = 0.6c$

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

$u'=\dfrac{v-u_{x}}{1-\dfrac{u_{x}v}{c^2}}$

$u'=\dfrac{0.8c-0.6c}{1-\dfrac{0.6c\times0.8c}{c^2}}$

$u'=c(\dfrac{0.8-0.6}{1-(0.8\times0.6)})$

$u'=0.385c$

(a). We need to calculate the total energy of the proton in the lab frame

Using formula of kinetic energy

$K.E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{u_{x}^2}{c^2}}}-1)$

Where, Proton mass energy = m₀c²

Put the value into the formula

$K.E=938.28\times(\dfrac{1}{\sqrt{1-\dfrac{(0.6c)^2}{c^2}}}-1)$

$K.E=938.28\times(\dfrac{1}{\sqrt{1-(0.6)^2}}-1)$

$K.E=234.57\ MeV$

(b). We need to calculate the kinetic energy of the proton in the observer

Using formula of kinetic energy

$K.E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{u'^2}{c^2}}}-1)$

$K.E=938.28\times(\dfrac{1}{\sqrt{1-\dfrac{(0.385)^2}{c^2}}}-1)$

$K.E=938.28\times(\dfrac{1}{\sqrt{1-(0.385)^2}}-1)$

$K.E=78.366\ MeV$

(c). We need to calculate the momentum of the proton with respect to observer

Using formula of momentum

$P_{obs}=\dfrac{m_{0}u'^2}{\sqrt{1-\dfrac{u'^2}{c^2}}}$

We know that,

Proton mass energy = m₀c²

$m_{0}=\dfrac{938.28}{c^2}$

$P_{obs}=\dfrac{\dfrac{938.28}{c^2}\times(0.385c)^2}{\sqrt{1-\dfrac{(0.385c)^2}{c^2}}}$

$P_{obs}=\dfrac{938.28\times(0.385)^2}{\sqrt{1-0.385^2}}$

$P_{obs}=150.69\ MeV/c$

Hence, This is required solution.