Question

An electron has a kinetic energy K of 1 MeV and is incident on a proton at rest in the laboratory. Calculate the speed of the CMS frame (The centre of mass, or centre of momentum, (CMS) frame is that in which the sum of the momenta (i.e., the total momentum) of all particles is zero) moving relative to the laboratory.

(a) Express the initial energies Ee, Ep and initial momenta pe, pp of the electron and proton respectively (with c, me, mp, K).

(b) What are the total energy Etot and the total momentum ptot before collision?

(c) Express the transformed momentum p 0 tot in the CMS frame and obtain the speed v of the CMS frame with respect to the laboratory frame. The rest mass of the electron is 0.511 MeV/c2 . The rest mass of a proton is 938 MeV/c2

Answer #1

(a) Initial energy of electron, E_{e} = 1 MeV, Energy of
proton, E_{p} = 0,

Initial momentum of electron, p_{e} can be expressed
as

The momentum of the proton will be
p_{p} = 0 as it is at rest before collision.

(b) Total energy before collision will be

Total momentum before collision will be

(c) The velocity of electron is
v_{e}

the velocity of center of mass V can be expressed as

The total momentum
**P** of the system in the CMS frame:

What is the momentum of a proton with 950-MeV total energy?
(Rest energy of a proton is 938 MeV)

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