Question

A proton in an accelerator attains a speed of 0.742c. What is the magnitude of the momentum of the proton?

Answer 1

We use the formula, relativistic momentum of proton

$p =\frac{mv}{(1-\frac{v^2}{c^2})^1/2 }$

Here, m is mass of proton and its value $1.672621898\times 10^{-27} \ kg$ and c is speed of light and its value is $3\times 10^{8} \ m/s$.

Given,  $v= 0.742 c$.

Substituting the values in above formula, we get

$p = \frac{(1.672621898\times 10^{-27} \ kg) (0.742 \times 3 \times 10^8 m/s )}{(1-\frac{(0.742c)^2}{c^2})^1/2 } = 5.55 \times 10^{-19} \ kg \ m/s$

Thus, the momentum of proton is $5.55 \times 10^{-19} \ kg \ m/s$.