What will the final velocity of the particles be?

[Lotto]

Homework Statement

I have a particle with invariant mass $m_0$ moving with speed $v=0.999c$. It inelasticely collides with a particles at rest of the same mass $m_0$. They connect and move with final speed $u$. What will be the speed?

Relevant Equations

I would use the law of conservation of momentum, so $\frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{2m_0 u}{\sqrt{1-\frac{u^2}{c^2}}}$.

So I would get $u=\frac{vc}{\sqrt{4c^2-3v^2}}=0.996c$. But the right answer is $0.956c$. Where is my mistake?

 

[TSny]

$\frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{2m_0 u}{\sqrt{1-\frac{u^2}{c^2}}}$.

You have mistakenly assumed that the final mass is $2 m_0$.

 

[Lotto]

You have mistakenly assumed that the final mass is $2 m_0$.

And what is the final mass?

 

[topsquark]

Homework Statement: I have a particle with invariant mass $m_0$ moving with speed $v=0.999c$. It inelasticely collides with a particles at rest of the same mass $m_0$. They connect and move with final speed $u$. What will be the speed?
Relevant Equations: I would use the law of conservation of momentum, so $\frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{2m_0 u}{\sqrt{1-\frac{u^2}{c^2}}}$.

So I would get $u=\frac{vc}{\sqrt{4c^2-3v^2}}=0.996c$. But the right answer is $0.956c$. Where is my mistake?

You are conserving 4-momentum, not the 3-momentum. So the 1 component is
$\gamma m_0 v = \gamma^{\prime} M u$

and the time component is
$\dfrac{1}{c}\sqrt{(m_0c^2)^2+(m_0 v c)^2} + m_0 c = \dfrac{1}{c}\sqrt{(Mc^2)^2+(M u c)^2}$

-Dan

 

[TSny]

And what is the final mass?

There are two conserved quantities: the total energy of the system and the total momentum of the system. So, you can set up two equations for the two unknowns: $M$ and $u$, where $M$ is the mass of the final particle.

Or, note that the mass of the final particle must satisfy the relation $$M^2=\left (\frac{E_f}{c^2}\right)^2 -\left(\frac{p_f}{c}\right)^2, $$ where $E_f$ and $p_f$ are the energy and momentum of the final particle. Using conservation of energy and momentum, this may be written as $$M^2=\left (\frac{E_{0,sys}}{c^2}\right)^2 -\left(\frac{p_{0,sys}}{c}\right)^2. $$ Here, $E_{0,sys}$ and $p_{0,sys}$ are the total energy and total momentum of the initial system.

 

[Lotto]

OK, so I suppose that I can also write
$$m_0c^2+\frac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{Mc^2}{\sqrt{1-\frac{u^2}{c^2}}}.$$

So the new invariant mass is $M=6.855m_0$. But how is it possible that the mass of the system is bigger than at the beginning? Shouldn't be mass conserved? How to explain it?

 

[TSny]

OK, so I suppose that I can also write
$$m_0c^2+\frac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{Mc^2}{\sqrt{1-\frac{u^2}{c^2}}}.$$

So the new invariant mass is $M=6.855m_0$.

Yes.
Actually, I get $M = 6.84m_0$. But our difference is probably just "round-off error".

But how is it possible that the mass of the system is bigger than at the beginning? Shouldn't be mass conserved? How to explain it?

If you calculate the kinetic energy $T_0$ of the moving particle before the collision, you will find that $T_0 = 21.37 m_0c^2$.

Likewise, you can show that the kinetic energy $T_f$ of the final particle is $T_f = 16.53 m_0 c^2$.

Thus, there has been a decrease in kinetic energy during the collision of $(21.37 - 16.53)m_0c^2 = 4.84 m_0c^2$.

Note that the rest mass energy of the final particle is greater than the sum of the individual initial rest mass energies by the amount $6.84 m_0c^2 - 2m_0c^2 = 4.84m_0c^2$. This matches the loss in kinetic energy of the system.

So, we could interpret this as saying that some of the initial kinetic energy of the system has been converted into part of the rest mass of the final particle. Or we could say that there is an equivalence of the loss of kinetic energy and the gain in rest mass energy. This is an illustration of the "equivalence of mass and energy".

 

[PeroK]

Shouldn't be mass conserved? How to explain it?

Rest mass is not a conserved quantity. Invariant mass (of a closed system) is conserved.