We use the formula, relativistic momentum of proton
[tex]p =\frac{mv}{(1-\frac{v^2}{c^2})^1/2 }[/tex]
Here, m is mass of proton and its value [tex]1.672621898\times 10^{-27} \ kg[/tex] and c is speed of light and its value is [tex]3\times 10^{8} \ m/s[/tex].
Given, [tex]v= 0.742 c[/tex].
Substituting the values in above formula, we get
[tex]p = \frac{(1.672621898\times 10^{-27} \ kg) (0.742 \times 3 \times 10^8 m/s )}{(1-\frac{(0.742c)^2}{c^2})^1/2 } = 5.55 \times 10^{-19} \ kg \ m/s[/tex]
Thus, the momentum of proton is [tex]5.55 \times 10^{-19} \ kg \ m/s[/tex].
