Question

What is the momentum of a proton (in GeV/c) with KE=800MeV?

Answer 1

\[K =.8GeV\]

\[ K=\frac{p^2}{2m_p}\]
\[p=\sqrt{2Km_p}\]

\[m_p=.938GeV/c^2\]

Answer 2

Does that make sense?

Answer 3

It's relativistic, so classical equations don't apply.

Answer 4

I've been banging my head against this for an hour, I'd be very grateful if you actually helped me solve it.

Answer 5

I'm pretty sure this equation is what's used to solve it: E^2=p^2c^2+m^2c^4

Answer 6

E = total energy, which is kinetic energy + rest energy, which is 1739.57 MeV. But when I plug that into the equation and solve, I end up with something that's off by many orders of magnitude.

Answer 7

The answer should be in GeV, and I'm ending with something that's closer to a single eV. I'm beginning to hate physics.

Answer 8

hmmm, lemme see. 1 sec.... And physics LOVES you! It's just a coy mistress ^_^

Answer 9

I don't like coy mistresses... >.>

Answer 10

I like my mistresses straightforward and honest!

Answer 11

^_^

Answer 12

does 1.46 GeV/c sound right?

Answer 13

That was right! How did you get it?

Answer 14

She just needs to know that you understand her quirks ~_^ You had everything right, you just divided by the numerical value of c

\[E^2 = p^2c^2+m^2c^4\]
\[(1.739 GeV)^2=p^2c^2+(.938GeV/c^2)^2c^4\]
\[(1.739 GeV)^2=p^2c^2+(.938GeV/c^2)^2c^4\]
\[(1.739 GeV)^2=p^2c^2+(.938GeV)^2\]
\[(1.739 GeV)^2-(.938GeV)^2=p^2c^2\]
\[p=\sqrt{\frac{(1.739 GeV)^2-(.938GeV)^2}{c^2}}=\sqrt{(1.739 )^2-(.938)^2}\frac{GeV}{c}\]

The c is actually a part of the units!

Answer 15

(for that last step I just pulled the units out of the square root)

Answer 16

Where did the c^4 go in the third step?

Answer 17

Canceled with the c^2^2 I presume, but why didn't the exponent distribute to the .938 as well?

Answer 18

Yeah, the c^4's canceled out. What do you mean?

Answer 19

\[(.938 \frac{GeV}{c^2})^2c^4=(.938^2 \frac{GeV^2}{c^4})c^4 =(.938^2 \frac{GeV^2c^4}{c^4})=(.938^2 GeV^2) = (.938GeV)^2\]

Answer 20

~last term
\[=(.938 GeV)^2\]

Answer 21

So why doesn't that become .938^2GeV?

Answer 22

The GeV is also squared - there's nothing to cancel it out

Answer 23

i.e. .879844GeV?

Answer 24

if you wrote it like that, it would be
\[ .8798 \ GeV^2\]

Answer 25

just like if you had
\[(10 m)^2 = (100m^2)\]

Answer 26

Okay, I see, thanks a lot.

Answer 27

^_^ welcome! Always glad to act as a relationship counselor ^_^

Answer 28

Could you have solved this by leaving the mass in kg?

Answer 29

You could go through all the steps, but to get units of GeV would would have at some point had to convert kg into GeV/c^2

So to do it that way, just leave all of the c's as the letter c, then your final expression would look like
\[p=\sqrt{\frac{(1.739GeV)^2-(m_pc^2)^2}{c^2}}\]
then
\[1.782662×10^{−36} kg = 1 eV/c^2\]


Conversely, if you do everything in terms of meters, seconds, and kilograms you have to convert MeV into Joules in the beginning (then your numbers are crazy tiny though)

Answer 30

Are you still here?

Answer 31

hi, yah. The site kicked me off for a minute.

Answer 32

Me too, I guess they just did an update. Thanks for your help. I'm gonna close this one and ask another question.

Answer 33

Cool cool, and again, welcome ^_^