The rest mass of a proton is 1.673 ｘ 10⁻²⁷ kg.  At what speed would the mass of the proton be tripled? (Answer: 0.9428c)



Given Data:

Rest mass of the Proton = m₀ = 1.673 ｘ 10⁻²⁷ kg

Relativistic mass of the Proton = m = 3m₀

To Find:

Speed of Proton = v =  ? 




Solution:


According to special theory of relativity, the observed (increase) mass m is given by

m = `\frac {m₀}{sqrt {1 - frac {v^2}{c^2}}}`   

putting the value of m

3m₀ = `\frac {m₀}{sqrt {1 - frac {v^2}{c^2}}}` 

or

3 = `\frac {1}{sqrt {1 - frac {v^2}{c^2}}}`

or (reversing the fraction on both sides



`\frac {1}{3}` = `\sqrt {1 - frac {v^2}{c^2}}`


Taking square both sides we get


`\frac {1}{9}` = 1 - `\frac {v^2}{c^2}`


or


`\frac {v^2}{c^2}` = 1 - `\frac {1}{9}`


`\frac {v^2}{c^2}` = `\frac {8}{9}`

v² = 0.889 c²

taking square root on both sides

`\sqrt {v²}` = `\sqrt {0.889 c²}`

v = 0.942 c --------------------Ans.

Thus, when the speed of the proton is 0.942c, its mass increases to triple.



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