A particle of mass m whose total energy is twice its rest energy collides with an identical particle at rest. If they stick together, what is the mass of the resulting composite particle? What is its velocity?

Write the expression for the relativistic mass $\mathit{m}\mathbf{=}\frac{{\mathbf{m}}_{\mathbf{0}}}{\sqrt{\mathbf{1}\mathbf{‐}{\displaystyle \frac{{\mathbf{u}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}}\mathbf{}\mathbf{}\mathbf{}}}$ …… (1)

Here,$m_0$is the rest mass of the particle, u is the velocity of the particle, and c is the speed of light.

As it is given that the total energy is twice its rest energy, write the required equation.

$E=2m_0{c}^2$

Substitute $E=m_0{c}^2$in the above equation.

$$\begin{gathered} m{c}^2=2m_0{c}^2 \\ m=2m_0 \end{gathered}$$

Substitute $2m_0$for in equation (1).

$$\begin{gathered} 2m_0=\frac{m_0}{\sqrt{1-{\displaystyle \frac{u^2}{c^2}}}} \\ 2\sqrt{1-\frac{u^2}{c^2}}=1 \\ 4\left(1-\frac{u^2}{c^2}\right)=1 \\ -\frac{u^2}{c^2}=\frac{1}{4}-1 \end{gathered}$$

On further solving,

$$\begin{gathered} -\frac{u^2}{c^2}=-\frac{3}{4} \\ u=\sqrt{\frac{3}{4}}c \end{gathered}$$

Using the conservation of total energy during a collision, write the equation for the relativistic mass of the composite particle.

$m{c}^2+m_0{c}^2=M{c}^2$

Here, M is the mass of the composite particle after a collision.

Substitute $2m_0$for in the above equation.

$$\begin{gathered} \left(2m_0\right)c^2+m_0{c}^2=M{c}^2 \\ 3m_0=M \end{gathered}$$

Using the special theory of relativity, write the expression for the relativistic mass of the composite particle.

$M=\frac{M_0}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}}......\left(2\right)$

Similarly , using the conservation of linear momentum during the collision, write the equation for the relativistic velocity of the composite particle.

$mu=Mv$

Substitute $2m_0$for $m,\sqrt{\frac{3}{4}c}$, for $u$and $3m_0$for$M$in the above equation to calculate the velocity of the composite particle.

$$\begin{gathered} \left(2m_0\right)\left(\sqrt{\frac{3}{4}c}\right)=\left(3m_0\right)v \\ 2\left(\frac{\sqrt{3}}{2}c\right)=3v \\ v=\left(\frac{\sqrt{3}}{3}\times \frac{\sqrt{3}}{\sqrt{3}}\right)c \\ v=\frac{c}{\sqrt{3}} \end{gathered}$$

Substitute $M$for $3m_0$and $\frac{c}{\sqrt{3}}$for $v$in equation (2) to calculate the mass of the composite particle.

$$\begin{gathered} 3m_0={\frac{M_0}{\sqrt{1-{\displaystyle \frac{{\left({\displaystyle \frac{c}{\sqrt{3}}}\right)}^2}{c^2}}}}}^{} \\ 3m_0=\frac{M_0}{\sqrt{1-{\displaystyle \frac{1}{3}}}} \\ 3m_0=\frac{M_0}{\sqrt{{\displaystyle \frac{2}{3}}}} \end{gathered}$$

On further solving,

$$\begin{gathered} \left(3m_0\right)\left(\sqrt{\frac{2}{3}}\right)=M_0 \\ \left(3m_0\right)\left(\sqrt{\frac{2}{3}}\times \sqrt{\frac{3}{3}}\right)=M_0 \\ \left(3m_0\right)\left(\sqrt{\frac{6}{3}}\right)=M_0 \\ {M}_0=\sqrt{6m_0} \end{gathered}$$

Therefore, the mass of the resulting composite particle is$M_0=\sqrt{6m_0}$, and its velocity is $v=\frac{c}{\sqrt{3}}$.