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# A proton has momentum with magnitude $p_{0}$ when its speed is 0.400$c .$ In terms of $p_{0},$ what is the magnitude of the proton's momentum when its speed is doubled to 0.800$c ?$

## $3.06 p_{0}$

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the speed of the proton is substantial fraction off the speed of light, so we must use the relativistic formula for moment. We know that the momentum of the Proton P is he put together M V, where m is its mass and be isn't speed, but its initial momentum will call peanuts. You're not just some key. Yeah, my note I m and be not. That was when its speed is 0.4 10 c. So we can also see from here that p divided by peanut. If you divide between creations is equal to gamma, The over gamma note be not with the masters canceling, and we also know that the over the not is equal to its speed is doubled. So what effect is that? Having a momentum? Well, let's calculators. So first, let's calculate gamma and gamma. Not so gamma. Note. Easy for two one divided by square hoods one minus B not squared. Oversea squid that's equal to one over this square to it off one minus zero point full. All squid sees. Cancel minions to get gamma. Note. The initial gamma is one point zero 911 Next, who calculated Emma when the speed is doubled. So Emma is equal to again. One over the square root off one minus B squared in this case of a sea squid and this is equal to one over the square root off one minus. The new speed is zero point eight time. See squid. He 2% the speed of light. And hence we get gamma to be one or in 66 soon. So since we have gamma and gamma not we can use our expression about to find the new momentum, Pete. So the momentum p is equal to he not and to and the ratio of the Gammas 1.667 divided by 1.0 91 against your momentum. He is a good three 0.6 Not what this means is that even though the speed doubles the momentum more than triples

University of Kwazulu-Natal

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