A relativistic proton has a momentum of `1.0 * 10^(-17) kg * m/s` and a rest energy of `0.15` nanojoules. What is the kinetic energy of this proton? (`c = 3.00 * 10^8 m/s,` mass of proton = `1.67 * 10^(-27) kg` ).

Expert Answers

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As we know, a mass of a body changes for an observer when an observer moves relative to a body. Denote the speed as `v.`

For a body with a rest mass `m_0` its rest energy is `E_r=m_0 c^2.` Its full energy is `E=m c^2,` where `m` is its mass (not a rest mass), and its kinetic energy is the difference between full and rest energies, `E_k=m c^2-m_0 c^2.`

Recall the well-known formula for a relativistic mass, `m=m_0/sqrt(1-v^2/c^2).` Therefore a momentum is `p=mv=(m_0 v)/sqrt(1-v^2/c^2).`

Square this equality and obtain `p^2=(m_0^2 v^2)/(1-v^2/c^2),` or `p^2-p^2v^2/c^2=m_0^2v^2.` Solving this for `v^2` we obtain `v^2=p^2/(m_0^2+p^2/c^2)=(p^2c^2)/(p^2+E_r^2/c^2)` and therefore `1-v^2/c^2=1-p^2/(p^2+E_r^2/c^2)=E_r^2/(c^2p^2+E_r^2).`


Now we can compute a kinetic energy:

`E_k = mc^2-m_0c^2 = (m_0c^2)/sqrt(1-v^2/c^2)-E_r =E_r/(E_r/sqrt(c^2p^2+E_r^2))-E_r=`



`c` , `E_r` and `p` are given and we can find the numeric result:

 `E_k=sqrt(9*10^16*10^(-34)+2.25*10^(-20))-1.5*10^(-10) =`

 `= 10^(-10)*(sqrt(9*100+2.25)-1.5) approx 28.54*10^(-10) (J),`

or 2.854 nanojoules.

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