Question

Relativistic momentum question

Answer 1

A proton has momentum with magnitude p0 when its speed is 0.400c. In terms of p0, what is the magnitude of the proton’s momentum when its speed is doubled to 0.800c?

I'm not entirely sure how to do this problem?

Answer 2

So, I'd use

\[\large p = \frac{ mv }{ \sqrt{1-\frac{ v^2 }{ c^2 }} }\] which is the relativistic momentum, so mv = p0?

Answer 3

So \[v^2 = (0.8c)^2?\] that's the part that troubles me the most, I guess. I'm just starting relativity so I'm trying to understand the process/ concept.

Answer 4

\[\huge p = \frac{ p_0 }{ \sqrt{1-(\frac{ 0.8c }{ c }})^2 }\] this way we could cancel out the c's. Not entirely sure though.

Answer 5

@Isaiah.Feynman @jim_thompson5910

Hey, do you guys have any idea how to approach this problem?

Answer 6

Not yet.

Answer 7

Alright, no worries, thanks anyways :P

Answer 8

Feynman was instrumental in the Manhattan Project, so I am sure this is like playing tic-tac-toe to him

Answer 9

I'm just trying to confirm if I did it correctly. So,

\[p = \frac{ p_0 }{ \sqrt{0.36} }\]

so solving this, \[p = 0.6 p_0\]

Answer 10

the equation looks legit

Answer 11

Lol, but I'm not sure if I'm putting all my variables correctly, I think there's something I'm missing, not sure.

Answer 12

are you asking me if \[p_0 = 0.400c = mv? \]

Answer 13

Yes, that's part of it

Answer 14

Well I'd think \[p_0 = m(0.400c)\]

Answer 15

so what is your relativistic mass?

Answer 16

But the question would want p = in terms of p0...and I doubt the mass would be there and we'd have to leave p0 there.

Answer 17

shouldn't you use
\[m_{rel} = \frac{m}{\sqrt{1-\frac{v^2}{c^2}}}\]

Answer 18

Well from what I think, could be wrong, the mass at 0 velocity is m, so I don't think that would work.

Answer 19

relativistic momentum
\[p = \gamma mv\]

Answer 20

HAHA you know the question is a trick one
it is just asking when v is doubled

Answer 21

SMH!

Answer 22

Yeah, but still getting the wrong answer >.<

Answer 23

you doubled v right?

Answer 24

Yeah, I replaced v UNDER the squareroot as 0.800c

Answer 25

\[\large p = \frac{ mv }{ \sqrt{1-\frac{ v^2 }{ c^2 }} } \implies \frac{ p_0 }{ \sqrt{1-\left( \frac{ 0.800c }{ c } \right)^2} }\]

Answer 26

IDK

Answer 27

it looks legit

Answer 28

Right?

Answer 29

that is what I would have done

Answer 30

maybe you're just punching the numbers wrong in the calculator

Answer 31

That could be it, one sec, let me try again.

Answer 32

Oh wait, I'll use wolframalpha :P

Answer 33

make sure you use proper parentheses to break things down properly

Answer 34

Yeah, wolfram is getting the same thing as me, but it's wrong, mhm.

Answer 35

let me see
put the link of the wolfram calculation

Answer 36

http://www.wolframalpha.com/input/?i=1%2Fsquareroot%281-0.80%5E2%29

Answer 37

Tricky question, it's probably right in front of our eyes, but we haven't noticed it.

Answer 38

kinda tripping me up now
I've recently taken this

Answer 39

I am going to read my textbook then get back with you

Answer 40

Alright thanks, I'll try working it out, and if I get it, I will definitely let you know.

Answer 41

http://finedrafts.com/files/CUNY/Physics/University%20Physics%20with%20Modern%20Physics,%2013th%20Edition.pdf
pp. 1243
let us both look at it

Answer 42

I think that's what it is
the relativistic mass has to be taken into account

Answer 43

Hey, I just scrolled down to the answers and it says 3.06p0, but I want to know how they got that?

Answer 44

try calculating rel p at 0.800 by itself

Answer 45

Not getting it

Answer 46

then I am going to wait for the heavy weight physics enthusiast here
@ganeshie

Answer 47

Haha, I think I got a few more ideas...I'll get back to you.

Answer 48

@nincompoop Hey I got it

Answer 49

\[P = \gamma m v~~~~P_0 = \gamma_0 m v_0\]

\[\gamma = (1-(v/c)^2)^{-1/2} ~~~~ \gamma_0 = (1-(v_0/c)^2)^{-1/2}\]

\[\gamma = 1.67~~~~\gamma_0 = 1.09\]

\[m = \frac{ p }{ \gamma v }~~~~m_0 = \frac{ p }{ \gamma_0 v_0 }\]


\[\frac{ p }{ \gamma v }= \frac{ p_0 }{ \gamma_0 v_0 }\]

now solving for p

\[p = \frac{ \gamma }{ \gamma_0 }*\frac{ v }{ v_0 } p_0 \implies \left( \frac{ 1.67 }{ 1.09 } \right)\left( \frac{ 0.8 }{ 0.4 } \right) = 3.06p_0\]

Answer 50

:) told ya laughing out loud

Answer 51

Haha, I was reading the book and was like eh lets give this a shot and it worked :P, it's one thing having the answer but knowing the process is wayyy more intriguing to find.

Answer 52

calculate it separately and compare just like when you're doing gravity vs elec force

Answer 53

good job

Answer 54

you can download the book and keep it
I used it