Special theory of relativity, a collision between two partic

[alizeid]

A particle of mass m comes with the speed 0.6c and collides with another particle of mass m which is at rest. In the collision melts the particles together and form a particle. What is the mass and velocity of the particle is formed?

solution:

The momentum and the total energy is conserved
E_{before}=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}+mc^2,v=0.6c \\ \Rightarrow E_{before}=2.25mc^2 \\
E_{after}=\frac{Xmc^2}{\sqrt{1-\frac{z^2}{c^2}}}\\ , where xm is the new mass and z is the velocity

Now, the energy is conserved
\frac{Xmc^2}{\sqrt{1-\frac{z^2}{c^2}}}=2.25mc^2\Leftrightarrow X^2=2.25^2(1-\frac{z^2}{c^2})\\

Now we do the same for momentum:
P_{befor}=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}, v=0.6c\Rightarrow P_{befor}=0,75mc
momentum after the collision:
P_{after}=\frac{mz}{\sqrt{1-\frac{z^2}{c^2}}}

the momentum is conserved:
\frac{mz}{\sqrt{1-\frac{z^2}{c^2}}}=0.75mc\Leftrightarrow xz=0.75c\cdot\sqrt{1-\frac{z^2}{c^2}}\\ \Leftrightarrow x^2z^2=0.75^2c^2\cdot(1-\frac{z^2}{c^2})
we have from the first equation that x^2=2.25^2\cdot(1-\frac{z^2}{c^2}):
x^2z^2=0.75^2c^2\cdot(1-\frac{z^2}{c^2})\\ \Leftrightarrow 2.25^2\cdot(1-\frac{z^2}{c^2})z^2=0.75^2c^2\cdot(1-\frac{z^2}{c^2})\\ z^2=\frac{0.75^2}{2.25^2}c^2\\ \Leftrightarrow z=0.33c

Now we have z, and then we can solve X. I get x = 2.12. Thus the mass to be 2.12m and the speed 0.33c. Is this true? Thanks for the help

 

[Orodruin]

Te momentum of the new particle after the collision must be based on the mass after collision. Otherwise you have more or less the idea. Just one comment, for any particle, you can use the relations $E^2 = p^2c^2 + m^2c^4$ and $v/c = pc/E$.

 

[alizeid]

Thanks for the tip but is it correct answer?

 

[Orodruin]

Yes. The answer looks reasonable.

 

[alizeid]

Thank you very much!