good ol' relativity....again

[Sarahisme]

Hey just a small easy (i think/hope) question... have never done one of these types of questions before, so just thought i'd see how i'm doing

 

ok here goes...

 

[math]

T = m_0c^2[\gamma - 1]

[/math]

[math]

u_0 = rest \ energy = m_0c^2

[/math]

 

 

now

[math]

u_0 = 938 MeV = 1.502676\times10^{-10} J

[/math]

[math]

E = 1400 MeV = 2.2428\times10^{-10} J

[/math]

 

 

[math]

m_0 = \frac{u_0}{c^2} = 1.66964\times10^{-27} kg

[/math]

 

[math]

E = T + u_0

[/math]

so

[math]

T = E - u_0 = 462 MeV = 7.40124\times10^{-11} J

[/math]

 

Now solving

[math]

T = m_0c^2[\gamma - 1]

[/math]

 

we get

[math]

1-\frac{v^2}{c^2} = 0.67

[/math]

 

[math]

\therefore v = 1.723\times10^{8} (m/s)

[/math]

 

and for part (b)

[math]

E = mc^2

[/math]

so

[math]

m = kinetic mass = \frac{E}{c^2} = 8.224\times10^{-28} kg

[/math]

 

[math]

p = mv

[/math]

 

[math]

\therefore p = 1.417\times10^{-19} (kg.m^{-1}.s^{-1})

[/math]

 

phew, there we , how was that?

[Sarahisme]

oops, the question, forgot to add the question

 

well, here it is

[mezarashi]

I believe your answers are inconsistent with the relativistic energy equation which is:

 

[math]

E = \frac{mc^2}{\sqrt{1-\frac{u^2}{c^2}}}

[/math]

 

where the mc^2 component is the rest mass.

[Sarahisme]

ok, hang on i'll give it another go in a min

[swansont]

I believe your answers are inconsistent with the relativistic energy equation which is:

 

[math]

E = \frac{mc^2}{\sqrt{1-\frac{u^2}{c^2}}}

[/math]

 

where the mc^2 component is the rest mass.

 

Where do you see the inconsistency? I don't see any problems.

[Sarahisme]

wait...so are you saying that i am completely correct then swansont? or ....??

[mezarashi]

Where do you see the inconsistency? I don't see any problems.

 

Maybe my arithmetic has gone to hell, but pluging in the numbers, I didn't get the proper total energy back. Could be a miscalculation on my part, but its better if someone checks.

[Sarahisme]

Maybe my arithmetic has gone to hell, but pluging in the numbers, I didn't get the proper total energy back. Could be a miscalculation on my part, but its better if someone checks.

 

 

which equation are you talking about plugging stuff back into exactly?

[swansont]

I was just looking at the equations used, not the math. OK - I see the problem. The "relativistic mass" calculated is less than the rest mass, which is impossible.

 

p = [math]\gamma m_0 v[/math]

 

If you use relativistic mass, you need to use m = [math]\gamma m_0 [/math] which is not T/c², it's E/c² (total energy, not KE)

[Sarahisme]

hang on, i'll redo my answer, i think i see where i went wrong (hopefully anyway )

 

give me a min or two to type it up, .... hang on ...

[Sarahisme]

ok here is my updated answer

 

##############UPDATED ANSWER############

[math]

u_0 = 938 MeV = 1.502676 \times 10^{-10} J

[/math]

[math]

E = 1400 MeV = 2.2428 \times 10^{-10} J

[/math]

 

[math]

u_0 = m+0c^{2} = rest \ energy

[/math]

(where [math] m_0 = [/math] rest mass of the the proton)

 

[math]

T = m_0c^{2}[\gamma - 1] = kinetic \ energy

[/math]

 

[math]

E = mc^{2}

[/math]

(where m = kinetic mass of the proton)

 

[math]

m = \frac{E}{c^{2}} = \frac{2.2428 \times 10^{-10} J}{3 \times 10^{8} m/s} = 2.492 \times 10^{-27} kg

[/math]

 

 

[math]

u_0 = m_oc^{2}

[/math]

[math]

\gamma u_0 = \gamma m_oc^{2}

[/math]

 

but

[math]

\gamma m_0 = m

[/math]

 

so

[math]

\gamma u_0 = mc^{2}

[/math]

so

[math]

\sqrt(1- \frac{v^{2}}{c{2}}) = \frac{u_0}{mc^{2}}

[/math]

 

[math]

\therefore v = 2.227 \times 10^{8} \ m/s

[/math]

 

phew! , hows that for part (a) now? :S

 

ok and part (b)'s ######updated answer######:

 

[math]

p = mv

[/math]

(where m = the kinetic mass of the proton = [math] 2.492 \times 10^{-27} kg [/math]

 

from part (a), we know v = [math] 2.227 \times 10^{8} [/math] m/s

 

so

[math]

p = mv = (2.492 \times 10^{-27} kg)(2.227 \times 10^{8} \ m/s) = 5.550 \times 10^{-19} kg.m.s^{-1}

[/math]

[Sarahisme]

i think it was this part of my orginal answer that was wrong:

we get

[math]

1-\frac{v^2}{c^2} = 0.67

[/math]

 

oh and this bit of it too

so

[math]

m = kinetic mass = \frac{E}{c^2} = 8.224\times10^{-28} kg

[/math]

 

i guess i just must have been pressing the wrong buttons on my calculator!

[swansont]

i think it was this part of my orginal answer that was wrong:

we get

[math]

1-\frac{v^2}{c^2} = 0.67

[/math]

 

oh and this bit of it too

so

[math]

m = kinetic mass = \frac{E}{c^2} = 8.224\times10^{-28} kg

[/math]

 

i guess i just must have been pressing the wrong buttons on my calculator!

 

Yes, there should be a square root still on the left-hand side of the first equation, and you appear to have used kinetic energy rather than total energy in the second equation.

 

Updated answer looks better (I didn't check all the math, though) aside from a few LaTEX things, like remembering to square c in one place.

[Sarahisme]

ok, cool! , yep i think i got it right this time too (i've checked though it a few times, and it makes sense, and still gives me the same answer each time )