Relativity and Momentum

1. Aug 7, 2008

UAPhys03

1. The problem statement, all variables and given/known data
What is the momentum of a proton with a kinetic energy of 750MeV?

2. Relevant equations
E = m0c2 + K
E2 = p2c2 + m0c2
Mp = 1.67 x 10-21

3. The attempt at a solution

I think whats confusing me is K = 750MeV, but I tried to solve for p

First, I converted 750MeV to 750x106 eV

Then solved for E in the first equation
E = 750x106 eV

Then used that to solve for p in the second equation
But i get a wrong answer.

The correct answer is 1404MeV/c

Last edited: Aug 7, 2008
2. Aug 7, 2008

hage567

In your second equation, $$m_oc^2$$ needs to be squared. So it should be $$E^2 = p^2c^2 + (m_oc^2)^2$$

You also need to express the proton mass in electronvolts. Are you doing that?

3. Aug 8, 2008

Redbelly98

Staff Emeritus
In other words, you get that E = K. But that's only true when m0=0, which is not the case for a proton.

You did mean to say proton, and not photon?

4. Aug 8, 2008

UAPhys03

Hey,

So I fixed the equation and figured it out

E = m0c2 + KE

m0c2 = 1.5x10-10j = 939.4MeV

E = 939 + 750 = 1689MeV

E2 = p2c2 + m02c2

16892 = p2c2 + 9392

p2c2 = 191000

pc = 1404MeV

p = 1404MeV/c

Thanks!

5. Aug 9, 2008

apunisheriii

is it okay to solve like this??

1/2(Mp)v2=K
1/2(p2)/Mp=K
p=square root of 2K(Mp)
p=6.34x10^-19 kg m/s

6. Aug 9, 2008

Andrew Mason

Your arithmetic is not correct. It should be: p2c2 = 1,970,000

AM

7. Aug 9, 2008

Andrew Mason

Not for relativistic speeds. A proton with 750 Mev of kinetic energy is moving at relativistic speed. At non-relativistic speed you can use:

$$\text{KE} = \frac{1}{2}mv^2 = \frac{p^2}{2m}$$

$$p = \sqrt{2m\text{KE}}$$

You had it right if you were using Mp for proton mass and p for momentum, but it is confusing.

AM

8. Aug 9, 2008

apunisheriii

How can i know whether it is in non-relativistic or relativistic speed???

9. Aug 9, 2008

Redbelly98

Staff Emeritus
For non-relativistic speeds, KE has to be a lot less than mc2. Equivalently, this means the speed is a lot less than c.

In this problem, KE is 750 MeV. That is comparable to mc2 = 940 MeV, so it is relativistic in this case.

10. Aug 9, 2008

UAPhys03

So for relativistic speed wouldn't you just substitute that equation,

K = 1/2((lorenz factor) - 1)mv2

substitute for p

K/((lorenz factor) - 1) = p2/2m

and you could figure out how fast the proton is moving given the kinetic energy?

11. Aug 9, 2008

Redbelly98

Staff Emeritus
No, the relation is

K = ( (Lorenz factor) - 1 ) m c2

Given K, solve for the Lorenz factor, from which you can get v.

12. Aug 10, 2008

apunisheriii

So can i use E=hf to calculate the momentum as it's relativistic??

13. Aug 10, 2008

Redbelly98

Staff Emeritus
Why would you think that? Momentum does not appear anywhere in your equation ???

14. Aug 11, 2008

apunisheriii

i thought the wavelength in the equation E=hc/lambda is the same as the de Broglie's wavelength......
is the de Broglie's wavelength of a electron with non-relativistic speed couldnt be sub into the E=hf equation??

15. Aug 11, 2008

Redbelly98

Staff Emeritus
Sometimes, not always. There actually 3 "speed regimes", not 2, to think about.

1. Non-relativistic: K is much less than mc2
2. Moderately relativistic: K is comparable to mc2
3. Extremely relativistic: K is much larger than mc2

That equation, E = hf = hc/lambda = pc , is valid in the extremely relativistic case. Another way of seeing this is from two equations that are true in all 3 cases:

E2 = (pc)2 + (mc2)2
and
E = K + mc2

In case 3 (and only then) we can neglect the mc2 terms because it is much smaller than the other quantities, and we get

E = K = pc (and also E = hc/lambda = hf)

Hope that helps.

16. Aug 12, 2008

apunisheriii

o,
finally i understood!!
thnx ya!!