91影视

Write the expression for the relativistic mass $\mathit{m}\mathbf{=}\frac{{\mathbf{m}}_{\mathbf{0}}}{\sqrt{\mathbf{1}\mathbf{鈥�}{\displaystyle \frac{{\mathbf{u}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}}\mathbf{}\mathbf{}\mathbf{}}}$ 鈥︹�� (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.

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

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,

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

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