Question

A proton is accelerated to 0.999c.

a. What is the proton’s momentum?

b. By what factor does the proton’s momentum exceed its Newtonian momentum?

Short answer

The relativistic momentum is $1.12\times {10}^{-17}kgm/s$and the ratio of relativistic momentum and Newtonian momentum is 22.4.

Step-by-step solution

Given information 

We have given that, a proton is accelerated to 0.999c.

Part (a) Step 1.

We will use relativistic momentum formula to calculate photon's momentum.

Formula of relativistic momentum is :

$\rho =\frac{mv}{\sqrt{1-{\displaystyle \raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}$

Here, = density

m = mass

v = velocity

c = speed of light

Step 2.

Substituting the values in the formula :

$\rho =\frac{(1.67\times {10}^{-27}Kg)\times (0.999)\times (3\times {10}^{8}m/s)}{\sqrt{1-{(0.999)}^2}}$

$\rho =1.12\times {10}^{-17}kgm/s$

Part (b) Step 1.

The ratio of relativistic momentum and Newtonian momentum is:

$\frac{{\rho }_{relativistic}}{{\rho }_{newtonian}}=\frac{{\displaystyle \raisebox{1ex}{$mv$}\!\left/ \!\raisebox{-1ex}{$\sqrt{1-{\displaystyle \raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}$}\right.}}{mv}$

Step 2.

$\frac{{\rho }_{relativistic}}{{\rho }_{newtonian}}=\frac{1}{\sqrt{1-{\displaystyle \raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}}$

$\frac{{\rho }_{relativistic}}{{\rho }_{newtonian}}=22.4$