# Special theory of relativity, a collision between two partic

1. Feb 18, 2015

### 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

2. Feb 18, 2015

### Orodruin

Staff Emeritus
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 $v/c = pc/E$.

3. Feb 18, 2015

### alizeid

Thanks for the tip but is it correct answer?

4. Feb 18, 2015

### Orodruin

Staff Emeritus