Show that the total relativistic energy of a proton

AI Thread Summary
The discussion focuses on calculating the total relativistic energy of a proton in a moving bunch, demonstrating that it equals 5mc². Participants clarify the use of the equations for energy and relativistic factors, particularly emphasizing the importance of the Lorentz factor, γ. They confirm that the speed of the proton can be derived from the given ratio of v²/c² = 24/25, leading to the expression v = c√(24/25). The conversation highlights the simplicity of substituting values into the energy equation without needing exact mass or speed of light figures. Ultimately, the correct approach and understanding of the relativistic energy concept are achieved.
duggielanger
Messages
17
Reaction score
0

Homework Statement



The mass of a proton when at rest is m. According to an observer using the
detector frame, the speed of the anticlockwise moving bunch, A, is such that
va^2/c^2=24/25
Show that the total relativistic energy of a proton in bunch A, as observed in
the detector frame, is exactly 5mc^2, and work out the speed of the proton,
expressed as a decimal multiple of c, (to 5 significant figures).

Homework Equations


Right I think its these
E=mc^2
Etot=mc^2/√1-v^2/c^2
Etrans=mc^2/1-v^2/c^2
and maybe
E^2tot=p^2c^2+m^2c^4
p=mv/1-v^2/c^2

The Attempt at a Solution


Now I know that to get Etot you need e=mc^2 and Etrans and so maybe combing these equations will give the answer but I think this IS not the right to get 5mc^2.
So maybe its E^2tot equation I have to use.
Im just not sure.
 
Physics news on Phys.org
well if \frac{v^{2}}{c^{2}}=\frac{24}{25}

and the mass of the proton is m, and Etot = mc2\gamma

where \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

then why don't you just try plugging the values in?

I'm not sure what Etrans is supposed to beand to solve for the velocity of the proton all you need is \frac{v^{2}}{c^{2}}=\frac{24}{25}
 
Yeah getting that wrong about Etrans, just looked at my textbooks and was reading it wrong . I have an asnwwer for the last bit not sure if rights but this is what I have for that.
24/25 x 3.00 x10^8 =2.88000 x 10^8 ?

So are you saying just add the values for a proton = 1.67 x 10^-27
the speed of light 3.00 x 10^8
and then add 24=v and 25=c into the gamma part
 
if \frac{v^{2}}{c^{2}}=\frac{24}{25}

then v=c\sqrt{\frac{24}{25}}

and just leave it as a multiple of c


for the first part all you need to write is

E_{tot}=\frac{mc^{2}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

and then substitute \frac{v^{2}}{c^{2}} for \frac{24}{25}

you should be able to do it in your head, no need to put in the mass of the proton or the exact speed of light, since the answer you want is just 5mc2 where m is the mass of the proton and c is the speed of light
 
Thank you very much I have now , its taken two days for that to sink in !
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Replies
54
Views
10K
Replies
2
Views
1K
Replies
3
Views
205
Replies
1
Views
2K
Replies
2
Views
1K
Replies
8
Views
3K
Replies
19
Views
2K
Back
Top