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**1.**Consider the decay of particle A with rest mass [tex]M_{0A}[/tex] into two particles, labelled particle 1 and particle 2. The energy of particle 1 is denoted by [tex]E_1[/tex] and the rest mass by [tex]m_{01}[/tex], similarly for particle 2 the energy is [tex]E_2[/tex] and the rest mass [tex]m_{02} [/tex].

**). in the rest frame of particle A write down expressions describing the conservation of momentum and energy in this decay process.**

i

i

Since this is the rest frame of particle A, the particle is stationary. This implies it has no momentum so it's total energy is just [tex]E_{initial}=\gamma M_{0A} c^2[/tex].

The question is unclear about whether the particles produced have a velocity or not. I decided to say they did (since I could always say it was 0) and called them [tex]v_1[/tex] for particle 1 and [tex]v_2[/tex] for particle 2.

The total energy of particle 1 is therefore [tex]E_1=\gamma m_{01} c^2[/tex] and the total energy of particle 2 is therefore [tex]E_2=\gamma m_{02} c^2[/tex].

Since energy has to be conserved, [tex]E=E_1+E_2[/tex] or [tex]\gamma M_{0A}c^2=\gamma m_{01} c^2+\gamma m_{02}c^2 \ \Rightarrow \ M_{0A}=m_{01}+m_{02}[/tex].

This didn't look right (even though I can't see where it could be wrong) so I decided to use the equation [tex]E^2=\rho ^2 c^2 + m^2 c^4 [/tex]:

E remains the same since [tex]\rho=0[/tex].

[tex]E_1=\sqrt{\rho_1^2 c^2+m_{01}^2c^4}[/tex]

[tex]E_2=\sqrt{\rho_2^2c^2+m_{02}^2c^4[/tex]

Hence [tex]E=E_1+E_2 \Rightarrow \ M_{0A}c^2=\sqrt{\rho_1^2 c^2+m_{01}^2c^4}+\sqrt{\rho_2^2c^2+m_{02}^2c^4[/tex].

This looked a bit better except I can't get the second part of the question from this!

Any help would be appreciated.

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