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(I) What is the momentum of a proton traveling at $v = 0.68c$?

$P=4.6 \times 10^{-19} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}$

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Cornell University

Numerade Educator

Hope College

University of Winnipeg

So in this problem you have a proton moving really fast at sixty eight percent or point six eight times the speed of light. And you're asked to find the momentum of this pota. Okay, So from equation twenty six dash form, we know that the momentum of a relativistic Lee traveling park, part moving particle which is to say, moving at significant fractions of speed of light that is given by gamma times M times feet M is the mass of the of the proton. V is the velocity of the proton. Gamow is a factor relating that that tells you how fast relative to the speed of light. The particle is moving and that's given by one minus V squared over C squared. Ah, Now we know that V is point six eight times speed of light, which means that V oversea is point six eight. And so you square that and you get this factor in here. Okay, so let's plug in some numbers you have P equals the mass times. Velocity on top mass of a proton is given by one point six seven times ten to the negative twenty seven kilograms and velocity is point six, eight times the speed of light, which is very close to three times ten to the eighth meters per second. Okay. And you divide that by the we're gamma factor one minus the this kind of a C squared, which, if you recall here, is from here it point six eight eighty square That quite succeeds, squared. And then But those numbers and a calculator and you get that momentum is actually four point six times ten to the negative nineteen kilograms meters per second. Because, remember, this is in units of, uh, mass times mass times your velocity so they haven't.