Solving Momentum & Energy Conservation: Find Daughter Particle Velocities

[humanist rho]

Homework Statement

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?

Homework Equations

Momentum conservation:

p = p₁+p₂

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

The Attempt at a Solution

Can i assume that both daughter particle has same momentum(velocity)?
otherwise the second equation becomes really complicated.:(

 

[vela]

Try solving the problem in the rest frame of the original particle, and then transform back to the lab frame.

 

[humanist rho]

In the frame of original particle, the daughter particles will have equal and opposite momentum.

Then?how to proceed?

 

[vela]

What are the energy and the magnitude of the momentum of each particle?

 

[humanist rho]

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.

 

[vela]

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.

 

[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?

 

[humanist rho]

and velocity of the daughter particle in this frame,
$$v=\frac{3}{5}c$$

 

[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

 

[vela]

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.

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.

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?

 

[humanist rho]

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?

 

[niklaus]

If they were equal in the lab frame that would violate conservation of momentum, since the original particle has non-zero momentum...

 

[humanist rho]

yes i understood. thanks. :)