91Ó°ÊÓ

We have given,

Mass of A=$3\mathrm{m}$

Mass of B =$\mathrm{m}$

Mass of C =$2\mathrm{m}$

speed of B =$\frac{\mathrm{c}}{2}$

Speed of C =$\frac{\mathrm{c}}{4}$

We have to find the velocity and momentum of A in s' frame.

Using Lorentz's transformation, We can write

$$\begin{gathered} {\mathrm{u}}_{\mathrm{x}}\text{'}=\frac{{\mathrm{u}}_{\mathrm{x}}-\mathrm{v}}{1-{\displaystyle \frac{{\mathrm{u}}_{\mathrm{x}}\mathrm{v}}{{\mathrm{c}}^2}}} \\ {\mathrm{u}}_{\mathrm{x}}^{\text{'}}=\frac{0-{\displaystyle \frac{\mathrm{c}}{2}}}{1-0} \\ {\mathrm{u}}_{\mathrm{x}}^{\text{'}}=-\frac{\mathrm{c}}{2} \end{gathered}$$

and momentum will be,

$$\begin{gathered} \mathrm{P}\text{'}={\mathrm{mu}}_{\mathrm{x}}^{\text{'}} \\ \mathrm{P}\text{'}=3\mathrm{m}(-\frac{\mathrm{c}}{2}) \\ \mathrm{P}\text{'}=-\frac{3}{2}\mathrm{mc} \end{gathered}$$

We have given,

Mass of B =$\mathrm{m}$

Speed of B= $\frac{\mathrm{c}}{2}$

Mass of C =$2\mathrm{m}$

Speed of C= $\frac{\mathrm{c}}{4}$

We have to find the speed and momentum of B and C in s' frame.

Using Lorentz's transformation, We can write velocity of B is

$$\begin{gathered} {\mathrm{u}}_{\mathrm{x}}\text{'}=\frac{{\mathrm{u}}_{\mathrm{x}}-\mathrm{v}}{1-{\displaystyle \frac{{\mathrm{u}}_{\mathrm{x}}\mathrm{v}}{{\mathrm{c}}^2}}} \\ {\mathrm{u}}_{\mathrm{x}}^{\text{'}}=\frac{-{\displaystyle \frac{\mathrm{c}}{2}}-{\displaystyle \frac{\mathrm{c}}{2}}}{1-{\displaystyle \frac{1}{{\mathrm{c}}^2}}(-{\displaystyle \frac{\mathrm{c}}{2}})\left({\displaystyle \frac{\mathrm{c}}{2}}\right)} \\ {\mathrm{u}}_{\mathrm{x}}^{\text{'}}=-\frac{4\mathrm{c}}{5} \end{gathered}$$

and velocity of C is,

$$\begin{gathered} {\mathrm{u}}_{\mathrm{x}}\text{'}=\frac{{\mathrm{u}}_{\mathrm{x}}-\mathrm{v}}{1-{\displaystyle \frac{{\mathrm{u}}_{\mathrm{x}}\mathrm{v}}{{\mathrm{c}}^2}}} \\ {\mathrm{u}}_{\mathrm{x}}^{\text{'}}=\frac{{\displaystyle \frac{\mathrm{c}}{4}}-{\displaystyle \frac{\mathrm{c}}{2}}}{1-{\displaystyle \frac{1}{{\mathrm{c}}^2}}\left({\displaystyle \frac{\mathrm{c}}{4}}\right)\left({\displaystyle \frac{\mathrm{c}}{2}}\right)} \\ {\mathrm{u}}_{\mathrm{x}}^{\text{'}}=-\frac{2\mathrm{c}}{7} \end{gathered}$$

Now the momentum will be,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{B}}^{\text{'}}={\mathrm{m}}_{\mathrm{B}}{\mathrm{u}}_{\mathrm{B}} \\ {\mathrm{P}}_{\mathrm{B}}^{\text{'}}=\mathrm{m}(-\frac{4}{5}\mathrm{c}) \\ {\mathrm{P}}_{\mathrm{B}}^{\text{'}}=-\frac{4}{5}\mathrm{mc} \end{gathered}$$

and momentum of C is,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{C}}^{\text{'}}=\left(2\mathrm{m}\right)(-\frac{2\mathrm{c}}{7}) \\ {\mathrm{P}}_{\mathrm{C}}^{\text{'}}=-\frac{4\mathrm{mc}}{7} \end{gathered}$$

We have find the final momentum in s' frame.

since,

$$\begin{gathered} \mathrm{P}={\mathrm{P}}_{\mathrm{B}}^{\text{'}}+{\mathrm{P}}_{\mathrm{C}}^{\text{'}} \\ \mathrm{P}=-\frac{4}{5}\mathrm{mc}-\frac{4}{7}\mathrm{mc} \\ \mathrm{P}=-\frac{48}{35}\mathrm{mc} \end{gathered}$$

We have to find that is momentum is conserved or not in s' frame.

Since, in S frame the total momentum is zero after decomposition.

But in S' frame it is not zero. then we can say that the momentum is not conserved in the S' frame.