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 =(mv)/((1-(v^2)/(c^2))^1/2 )`

Here, m is mass of proton and its value
`1.672621898* 10^(-27) \ kg` and c is speed of light and its value is
`3* 10^(8) \ m/s`.

Given,
`v= 0.742 c`.

Substituting the values in above formula, we get

`p = ((1.672621898* 10^(-27) \ kg) (0.742 * 3 * 10^8 m/s  ))/((1-((0.742c)^2)/(c^2))^1/2 ) = 5.55 *  10^(-19) \ kg \ m/s`

Thus, the momentum of proton is
`5.55 *  10^(-19) \ kg \ m/s`.