How much total kinetic energy is needed in this case?

In summary, einstein's equation tells us that the mass of a particle is proportional to its velocity.
  • #1
eri3an
4
0
I have a problem with an exercise from school. could anyone help me?

Let say that you are searching for a new particel X by making collison between positrones and elektrones in an particelaccelerator. you guess that the new particles resting mass is 210 MeV/c^2 and you hope to produce it by:

(e+) + (e–) ---> X + photons

a) Let us say that the electron (e-) is in resting state in the laboratory system. How much energy is needed to accerate positrons (e+) to such a velocity for the reaction to take place?

b)On less energy demanding way of doing it is to let e+ och e– collide with each other from opposite directions with the same velocity How much total kinetic energy is needed in this case?

resting mass for an electrone or positron is 0.511 MeV/c^2

please I'm not interested in just getting an answer how do i solve it? I've tried by deriving from the E(a) + E(b) = E(c) + E(d) and tried to go using conservation of momentum but it doesn't seem to work :( Maybe I'm out on bad ice?
 
Last edited:
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  • #2
Hi eri3an,
you must remember Einstein's equation : E = mc^2, where m is the mass of the particle when it is moving.
 
  • #3
eri3an said:
I have a problem with an exercise from school. could anyone help me?

Let say that you are searching for a new particel X by making collison between positrones and elektrones in an particelaccelerator. you guess that the new particles resting mass is 210 MeV/c^2 and you hope to produce it by:

(e+) + (e–) ---> X + photons

a) Let us say that the electron (e-) is in resting state in the laboratory system. How much energy is needed to accerate positrons (e+) to such a velocity for the reaction to take place?

b)On less energy demanding way of doing it is to let e+ och e– collide with each other from opposite directions with the same velocity How much total kinetic energy is needed in this case?

resting mass for an electrone or positron is 0.511 MeV/c^2

please I'm not interested in just getting an answer how do i solve it? I've tried by deriving from the E(a) + E(b) = E(c) + E(d) and tried to go using conservation of momentum but it doesn't seem to work :( Maybe I'm out on bad ice?

I suggest you to study the problem in the CM (centre of mass) system.
Then you can pass to the system of the laboratory by Lorentz transformations.
 
  • #4
borisleprof said:
Hi eri3an,
you must remember Einstein's equation : E = mc^2, where m is the mass of the particle when it is moving.

i just though this statement could use some elaboration:
[tex]E=\gamma m_0c^2=\sqrt{m_0^2c^4+p^2c^2}[/tex]

i think using the real mass, i mean the thing you divide the force with to get the accelaration is better (and that's what borisleprof did).
but in most textbooks, they talk about the mass you measure at the rest frame...
 
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1. What is kinetic energy?

Kinetic energy is the energy an object possesses due to its motion. It is a scalar quantity and is dependent on the mass and velocity of the object.

2. How is kinetic energy calculated?

Kinetic energy is calculated using the equation KE = 1/2 * m * v^2, where m is the mass of the object in kilograms and v is the velocity in meters per second.

3. What is the unit of measurement for kinetic energy?

The unit of measurement for kinetic energy is joules (J). Other commonly used units are kilojoules (kJ) and calories (cal).

4. What factors affect the amount of kinetic energy needed?

The amount of kinetic energy needed is affected by the mass and velocity of the object. The greater the mass and velocity, the more kinetic energy is required.

5. How does kinetic energy relate to potential energy?

Kinetic energy and potential energy are both forms of energy. Kinetic energy is the energy of motion, while potential energy is the energy an object possesses due to its position or state. In certain cases, potential energy can be converted into kinetic energy and vice versa.

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