Question 39.7: Relativistic Leaders of the Pack Two motorcycle pack leaders......
Relativistic Leaders of the Pack
Two motorcycle pack leaders named David and Emily are racing at relativistic speeds along perpendicular paths as shown in Figure 39.15. How fast does Emily recede as seen by David over his right shoulder?
Conceptualize The two observers are David and the police officer in Figure 39.15. The event is the motion of Emily. Figure 39.15 represents the situation as seen by the police officer at rest in frame S. Frame S^{\prime} moves along with David.
Categorize Because the problem asks to find an observed velocity, we categorize this problem as one requiring the Lorentz velocity transformation. The motion takes place in two dimensions.
Analyze Identify the velocity components for David and Emily according to the police officer:
David: v_{x}=v=0.75 c \quad v_{y}=0
Emily: u_{x}=0 \quad u_{y}=-0.90 c
Using Equations 39.16 and 39.17,
u_{x}^{\prime}=\frac{u_{x}-v}{1-\frac{u_{x} v}{c^{2}}} (39.16)
u_y^{\prime}=\frac{u_y}{\gamma\left(1-\frac{u_x v}{c^2}\right)} \quad \text { and } \quad u_z^{\prime}=\frac{u_z}{\gamma\left(1-\frac{u_x v}{c^2}\right)} (39.17)
calculate u_{x}^{\prime} and u_{y}^{\prime} for Emily as measured by David:
\begin{aligned} & u_{x}^{\prime}=\frac{u_{x}-v}{1-\frac{u_{x} v}{c^{2}}}=\frac{0-0.75 c}{1-\frac{(0)(0.75 c)}{c^{2}}}=-0.75 c \\ & u_{y}^{\prime}=\frac{u_{y}}{\gamma\left(1-\frac{u_{x} v}{c^{2}}\right)}=\frac{\sqrt{1-\frac{(0.75 c)^{2}}{c^{2}}(-0.90 c)}}{1-\frac{(0)(0.75 c)}{c^{2}}}=-0.60 c \end{aligned}
Using the Pythagorean theorem, find the speed of Emily as measured by David:
u^{\prime}=\sqrt{\left(u_{x}^{\prime}\right)^{2}+\left(u_{y}^{\prime}\right)^{2}}=\sqrt{(-0.75 c)^{2}+(-0.60 c)^{2}}=0.96 c
Finalize This speed is less than c, as required by the special theory of relativity.