91Ó°ÊÓ

Let's consider a reference frame $S$with respect to which this object moves along its positive x-direction at a velocity of $0.8c$.

The total relativistic energy of this object is,

$E=y_{0.8.}m{c}^2$

$=\frac{5}{3}\left(3m_a\right)c^2$

$E=5m_o{c}^2$

The relativistic momentum of the object is,

$p_x={\mathrm{γ}}_{z_i}m{u}_x$

$={\mathrm{γ}}_{0.8c}\left(3m_o\right)(0.8c)$

$=\frac{5}{3}\left(2.4m_o{c}^2\right)$

$p_x=4m_{o}c$

The relativistic velocity transformation is expressed as,

$u_x^{\text{'}}=\frac{u_x-v}{1-\frac{u_{x}v}{c^2}}$

$=\frac{0.8c-0.5c}{1-(0.8×0.5)}$

$=\frac{0.3c}{0.6}$

$u_s^{\text{'}}=0.5c$

The total energy and momentum in this new frame moving at $0.5c$to the right is,

Energy:

$E^{\text{'}}=7^{\text{'}}m{c}^{\text{'}}$

localid="1657557711016" $=7_{1.5c}\left(3m_{\mathrm{ν}}\right)c^2$

$=1.16\left(3\right)m_o{c}^2$

$E^{\text{'}}=3.48m_v{c}^2$

Momentum:

$p_x^{\text{'}}={\mathrm{γ}}_{i{i}_z^{\text{'}}}m{u}_x^{\text{'}}$

$={\mathrm{γ}}_{0.,c}\left(3m_o\right)(0.5c)$

localid="1659092783461" $\begin{array}{rcl}& =& (1.16×3×0.5)m_{o}c\\ {p\text{'}}_x& =& 1.74m_{o}c\end{array}$

Momentum:

$p_x^{\text{'}}={\mathrm{γ}}_v\left[p_x-\frac{v}{c}\left(\frac{E}{c}\right)\right]$

$={\mathrm{γ}}_{osc}\left[4m_{o}c-\left(\frac{0.5c}{c}\right)\left(\frac{5m_e{c}^2}{c}\right)\right]$

$=1.16\left[4m_{e}c-2.5m_{e}c\right]$

$p_x^{\text{'}}=1.74m_{v}c$

Energy:

$E^{\text{'}}={\mathrm{γ}}_r\left[E-v{p}_s\right]$

$={\mathrm{γ}}_{0.5c}\left[5m_v{c}^2-(0.5c)\left(4m_{t}c\right)\right]$

$=1.16\left(3m_o{c}^2\right)$

$E^{\text{'}}=3.48m_o{c}^2$

Thus, the $(2-38)$equations are verified.