## A particle decays into two daughter particles.Find velocity pf daughter particles?

1. The problem statement, all variables and given/known data

A particle of rest mass m moving with speed c/2 decays into two particles of rest mass 2m/5 each.The daughter particles move in the same line as the direction of motion of the original particle.Then what are the velocities of daughter particle?

2. Relevant equations

Momentum conservation:

p = p1+p2

Energy conservation:
(p2c2+m2c4)1/2=(p12c2+m12c4)1/2+(p22c2+m22c4)1/2

3. The attempt at a solution

Can i assume that both daughter particle has same momentum(velocity)?
otherwise the second equation becomes really complicated.:(
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 Recognitions: Gold Member Homework Help Science Advisor Staff Emeritus Try solving the problem in the rest frame of the original particle, and then transform back to the lab frame.
 In the frame of original particle, the daughter particles will have equal and opposite momentum. Then?how to proceed?

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## A particle decays into two daughter particles.Find velocity pf daughter particles?

What are the energy and the magnitude of the momentum of each particle?
 velocity of the particles in the frame of original particle,$$v=\frac{\frac{c% }{2}+v^{\prime }}{1+\frac{v^{\prime }c}{2c^{2}}}$$ magnitude of momentum,$p=\frac{2}{5}\frac{mv}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$ Energy =$$\left( p^{2}c^{2}+m^{2}c^{4}\right) ^{1/2}$$ am i correct? But things seems much more complicated.
 Recognitions: Gold Member Homework Help Science Advisor Staff Emeritus You're kind of going backwards. First, what is the energy of the particle in its rest frame? Then by conservation of energy, what can you say about the energy of the decay products? Then calculate their momenta.
 At rest frame, Energy = mc2 By conservation of energy, $$2(p^{2}c^{2}+\frac{4}{25}m^{2}c^{4})^{1/2}=mc^{2}$$ $$p=\frac{3}{10}mc$$ is that right?
 and velocity of the daughter particle in this frame, $$v=\frac{3}{5}c$$
 And on converting it to lab frame,i got, $$v=\frac{c}{7}$$ I think now i started to understand. Thanks a lot

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 Quote by humanist rho At rest frame, Energy = mc2 By conservation of energy, $$2(p^{2}c^{2}+\frac{4}{25}m^{2}c^{4})^{1/2}=mc^{2}$$ $$p=\frac{3}{10}mc$$ is that right?
Looks good. You'll find in general that when trying to solve problems in relativity like this one, it's best to stick with energy E and momentum p when you can and only resort to working with velocity v/c when absolutely necessary.

 Quote by humanist rho and velocity of the daughter particle in this frame, $$v=\frac{3}{5}c$$
To be a bit pedantic, this is the speed of each daughter particle. You still have to account for their directions.

 Quote by humanist rho And on converting it to lab frame,i got, $$v=\frac{c}{7}$$ I think now i started to understand. Thanks a lot
What's the speed of the other daughter particle in the lab frame?
 In the frame of original particle, Each new particle has velocity ,±5c/3 . In the lab frame one particle has velocity +11c/13 and other has velocity -c/7. Why aren't they equal? even when both of them have same mass and is produced from same particle?
 If they were equal in the lab frame that would violate conservation of momentum, since the original particle has non-zero momentum...
 yes i understood. thanks. :)