a. Derive a velocity transformation equation for ${\mathrm{u}}_{\mathrm{y}}$and ${\mathrm{u}}_{\mathrm{y}}^{\text{'}}$. Assume that the reference frames are in the standard orientation with motion parallel to the x- and x′-axes.

b. A rocket passes the earth at $0.80\mathrm{c}$. As it goes by, it launches a projectile at $0.60\mathrm{c}$ perpendicular to the direction of motion. What is the particle’s speed, as a fraction of c, in the earth’s reference frame?

We have given,

Frame is moving in the along of x- axis and x' axis.

We have to find the their transformation velocity equation in y- axis and y'- axis.

Let assume an object A which is moving at constant velocity$({\mathrm{u}}_{\mathrm{x}}^{\text{'}},{\mathrm{u}}_{\mathrm{y}}^{\text{'}},{\mathrm{u}}_{\mathrm{z}}^{\text{'}})$ in the S'-frame. Where s′ frame is moving along the x'-axis with constant velocity v.

Let for an small increment in time ${\mathrm{dt}}^{\text{'}}$ and in the position along the x'-axis is $\mathrm{dx}\text{'}$.

Using Lorentz transformation, we can write,

$$\begin{gathered} \mathrm{dt}=\mathrm{\gamma }(\mathrm{dt}\text{'}+\frac{\mathrm{v}\mathrm{dx}\text{'}}{{\mathrm{c}}^2}) \\ \mathrm{dx}=\mathrm{\gamma }(\mathrm{dx}\text{'}+\mathrm{v}\mathrm{dt}\text{'}) \\ \mathrm{dy}=\mathrm{dy}\text{'} \\ \mathrm{dz}=\mathrm{dz}\text{'} \end{gathered}$$

Then its velocity will be derivative of the position components,

${\mathrm{u}}_{\mathrm{y}}=\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{dy}\text{'}}{\mathrm{\gamma }(\mathrm{dt}\text{'}+{\displaystyle \frac{\mathrm{v}\mathrm{dx}\text{'}}{{\mathrm{c}}^2}})}$

divide by $\mathrm{dt}\text{'}$above and below,

$$\begin{gathered} {\mathrm{u}}_{\mathrm{y}}=\frac{{\displaystyle \frac{\mathrm{dy}\text{'}}{\mathrm{dt}\text{'}}}}{\mathrm{\gamma }\left({\displaystyle \frac{\mathrm{dt}\text{'}+{\displaystyle \frac{\mathrm{v}\mathrm{dx}\text{'}}{{\mathrm{c}}^2}}}{\mathrm{dt}\text{'}}}\right)} \\ {\mathrm{u}}_{\mathrm{y}}=\frac{{\mathrm{u}}_{\mathrm{y}}^{\text{'}}}{\mathrm{\gamma }(1+{\displaystyle \frac{{\mathrm{vu}}_{\mathrm{x}}^{\text{'}}}{{\mathrm{c}}^2}})} \end{gathered}$$

Then,

$$\begin{gathered} {\mathrm{u}}_{\mathrm{y}}^{\text{'}}=\frac{\mathrm{dy}\text{'}}{\mathrm{dt}\text{'}}=\frac{\mathrm{dy}}{\mathrm{dt}\mathrm{\gamma }-\mathrm{\gamma }{\displaystyle \frac{\mathrm{v}\mathrm{dx}\text{'}}{{\mathrm{c}}^2}}} \\ {\mathrm{u}}_{\mathrm{y}}^{\text{'}}=\frac{{\mathrm{u}}_{\mathrm{y}}}{\mathrm{\gamma }(1-{\displaystyle \frac{{\mathrm{u}}_{\mathrm{x}}\mathrm{v}}{{\mathrm{c}}^2}})} \end{gathered}$$

We have given,

Rocket speed= $0.8\mathrm{c}$

Projectile speed =$0.6\mathrm{c}$

What is the speed of projectile with respect of rest frame.

From Lorentz transformation we can wrote,

${\mathrm{u}}_{\mathrm{y}}=\frac{{\mathrm{u}}_{\mathrm{y}}^{\text{'}}}{\mathrm{\gamma }(1+{\displaystyle \frac{{\mathrm{u}}_{\mathrm{x}}^{\text{'}}\mathrm{v}}{{\mathrm{c}}^2}})}$

Substitute the values

$$\begin{gathered} {\mathrm{u}}_{\mathrm{y}}=\frac{0.6\mathrm{c}}{\mathrm{\gamma }(1+{\displaystyle \frac{0\mathrm{v}}{{\mathrm{c}}^2}})} \\ {\mathrm{u}}_{\mathrm{y}}=0.6\mathrm{c}\times \sqrt{1-\frac{{(0.8\mathrm{c})}^2}{{\mathrm{c}}^2}} \\ {\mathrm{u}}_{\mathrm{y}}=0.36\mathrm{c} \end{gathered}$$

and then along the x-axis,

$$\begin{gathered} {\mathrm{u}}_{\mathrm{x}}=\frac{{\mathrm{u}}_{\mathrm{x}}^{\text{'}}+\mathrm{v}}{(1+{\displaystyle \frac{{\mathrm{u}}_{\mathrm{x}}^{\text{'}}\mathrm{v}}{{\mathrm{c}}^2}})} \\ {\mathrm{u}}_{\mathrm{x}}=\frac{0+\mathrm{v}}{1+0} \\ {\mathrm{u}}_{\mathrm{x}}=0.8\mathrm{c} \end{gathered}$$

Then the total speed observe is,

$$\begin{gathered} \mathrm{u}=\sqrt{{\mathrm{u}}_{\mathrm{x}}^{\text{'}2}+{\mathrm{u}}_{\mathrm{y}}^{\text{'}2}} \\ \mathrm{u}=\sqrt{{(0.36\mathrm{c})}^2+{(0.8\mathrm{c})}^2} \\ \mathrm{u}=0.88\mathrm{c} \end{gathered}$$