Respuesta :
Answer:
(a). The work done is 7001 MeV.
(b). The momentum of this proton is [tex]4.20\times10^{8}\ kg-m/s[/tex].
Explanation:
Given that,
Speed = 0.993 c
We need to calculate the work done
Using work energy theorem
The work done is equal to the kinetic energy relative to the proton
[tex]W=K.E[/tex]
[tex]W=\dfrac{m_{p}c^2}{\sqrt{1-\dfrac{v^2}{c^2}}}-m_{p}c^2[/tex]
Put the value into the formula
[tex]W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(\dfrac{0.993c}{c})^2}}-1.67\times10^{-27}\times(3\times10^{8})^2[/tex]
[tex]W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(0.993)^2}}-1.67\times10^{-27}\times(3\times10^{8})^2[/tex]
[tex]W=1.122\times10^{-9}\ J[/tex]
[tex]W=7001\ MeV[/tex]
(b). We need to calculate the momentum of this proton
Using formula of momentum
[tex]p=\dfrac{m_{0}v}{\sqrt{1-\dfrac{v^2}{c^2}}}[/tex]
Put the value into the formula
[tex]p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(\dfrac{0.993c}{c})^2}}[/tex]
[tex]p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(0.993)^2}}[/tex]
[tex]p=1.404\times10^{-26}c[/tex]
[tex]p=4.20\times10^{8}\ kg-m/s[/tex]
Hence, (a). The work done is 7001 MeV.
(b). The momentum of this proton is [tex]4.20\times10^{8}\ kg-m/s[/tex].
