Special Relativity- momentum and energy

In summary, in this conversation, the questioner is attempting to solve a problem involving an electron accelerated in a linear accelerator and a collision between two particles. In the first part, the questioner is unsure about the assumptions made, the units of gamma and the fact that they keep getting 40 as an answer. In the second part, the questioner presents their attempt at solving the problem, but they are unsure if they have done it correctly. They then ask for clarification and receive guidance on how to correctly approach the problem. After making some corrections, the questioner is still getting the same answer and asks for further help.
  • #1
C.E
102
0
1. a. An electron of rest mass (0.511Mev) is accelerated in a linear accelerator so it has a total energy of 40 Gev.
i).What fraction of the total energy is kinetic energy?
ii).what is the reletivistic linear momentum of the electron?
iii).What is its relativistic mass?
b.A particle of rest mass m travels so that its total energy is twice its rest mass energy. It collides with a particle of rest mass m and forms a composite particle.
i). What is the momentum of each of the initial particles?
ii).What is the momentum of the new particle produced in the collision.
iii).What is the rest mass of the new particle

3. Here is my attempt at part a, there are a few things I am unsure on firstly, in answering this question I have assumed potential energy is negligeable, i.e. there is only kinetic and rest mass energy, is that ok? Secondly I am a bit worried by the fact I keep getting 40 as an answer and finally, I was unsure on units, what are the units of gamma?

a. (i). (40 x 10^9 - 0.511 x 10^6)/40 x 10^9 =0.99999.
ii) E^2=p^2c^2 +m^2c^4

so p=40Gev/c

iii).E=gmc^2 where g is gamma.
Therefore g= (40x 10^9)/ 0.511 x 10^6) =78278.
Hence relativistic mass= 78278 x 0.511 x 10^-3= 40 Gev/c^2.

Here is part b.
b. (i). stationary particle: 0.
moving particle:mc^2sqrt(3) using the same equation as in aii.
(ii) sqrt(3)mc^2 by conservation laws.
(iii)Total energy mc^2 + 2mc^2 =3mc^2
rest energy=sqrt(total energy^2-(momentumxc)^2)
=sqrt(6m^2c^4)=mc^2sqrt(6)

Have I done this question right?
 
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  • #2
Hi C.E! :smile:

(what happened to that γ I gave you before? :wink:)
C.E said:
1. a. An electron of rest mass (0.511Mev) is accelerated in a linear accelerator so it has a total energy of 40 Gev.
i).What fraction of the total energy is kinetic energy?

Here is my attempt at part a, there are a few things I am unsure on firstly, in answering this question I have assumed potential energy is negligeable, i.e. there is only kinetic and rest mass energy, is that ok? Secondly I am a bit worried by the fact I keep getting 40 as an answer and finally, I was unsure on units, what are the units of gamma?

a. (i). (40 x 10^9 - 0.511 x 10^6)/40 x 10^9 =0.99999.

Nope … start again …

if energy/rest-mass = 40G/.511M, what is γ? :smile:

(and yes, there is only kinetic and rest mass energy)
 
  • #3
Sorry the 0.511Mev is rest mass energy, (not rest mass), doesa that make a difference to your last post?
 
  • #4
C.E said:
Sorry the 0.511Mev is rest mass energy, (not rest mass), doesa that make a difference to your last post?

Nooo … that was obvious from the question.
 
  • #5
Ok, here is my new attempt (finding gamma first)
Energy= gamma x rest energy
Hence gamma=78278
Kinetic energy= (gamma -1) x rest mass energy = 78277
Giving: kinetic energy / total energy = 78277/78278 =0.99999.
I still get the same answer, what is wrong with the above?
 

Related to Special Relativity- momentum and energy

1. What is the principle of relativity in special relativity?

The principle of relativity states that the laws of physics should be the same for all observers in uniform motion, regardless of their relative velocity. This means that there is no preferred frame of reference in the universe.

2. How does special relativity change our understanding of momentum and energy?

In special relativity, momentum and energy are no longer separate quantities, but are instead components of a unified concept called four-momentum. This means that the total four-momentum of a system is conserved, rather than just the individual momentum and energy of each object.

3. What is the equation for relativistic momentum?

The equation for relativistic momentum is p = mγv, where p is momentum, m is rest mass, γ is the Lorentz factor (equal to 1/√(1-v^2/c^2)), and v is velocity. This equation takes into account the effects of relativistic speeds on an object's momentum.

4. How does special relativity affect the concept of kinetic energy?

In special relativity, the equation for kinetic energy is modified to include the effects of relativity. The equation is K = (γ-1)mc^2, where K is kinetic energy, γ is the Lorentz factor, m is rest mass, and c is the speed of light. This means that the kinetic energy of an object increases significantly as its velocity approaches the speed of light.

5. Can special relativity be applied to all situations?

No, special relativity only applies to objects moving at constant velocities in a straight line. It does not apply to accelerating objects or objects affected by gravitational fields. For these situations, general relativity must be used.

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