Two motorcycle pack leaders named David and Emily are racing at relativistic speeds along perpendicular paths, as shown in Figure 39.17. How fast does Emily recede as seen by David over his right shoulder?

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Accepted Answer

Figure 39.17 represents the situation as seen by a police officer at rest in frame S, who observes the following:

David: [latex] u_{x}=0.75 c \quad u_{y}=0 [/latex].

Emily: [latex] u_{x}=0 \quad u_{y}=-0.90 c [/latex].

To calculate Emily’s speed of recession as seen by David, we take [latex] S ^{\prime} ext { to move along with David and then calculate } u_{x}^{\prime} ext { and } [/latex][latex] u_{y}^{\prime} ext { for Emily using Equations } 39.16 ext { and } 39.17 ext { : } [/latex]

[latex] u_{x}^{\prime}=\frac{u_{x}-v}{1-\frac{u_{x} v}{c^{2}}} [/latex] (39.16).

[latex] u_{y}^{\prime}=\frac{u_{y}}{\gamma\left(1-\frac{u_{x} v}{c^{2}} ight)} \quad ext { and } \quad u_{z}^{\prime}=\frac{u_{z}}{\gamma\left(1-\frac{u_{x} v}{c^{2}} ight)} [/latex] (39.17).

[latex] 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 [/latex].

[latex] u_{y}^{\prime}=\frac{u_{y}}{\gamma\left(1-\frac{u_{x} v}{c^{2}} ight)}=\frac{\sqrt{1-\frac{(0.75 c)^{2}}{c^{2}}}(-0.90 c)}{\left(1-\frac{(0)(0.75 c)}{c^{2}} ight)} [/latex].

[latex] =-0.60 c [/latex].

Thus, the speed of Emily as observed by David is

[latex]u^{\prime}=\sqrt{\left(u_{x}^{\prime} ight)^{2}+\left(u_{y}^{\prime} ight)^{2}}=\sqrt{(-0.75 c)^{2}+(-0.60 c)^{2}} [/latex].

=0.96c.

Note that this speed is less than c, as required by the special theory of relativity.

Question 39.9: Two motorcycle pack leaders named David and Emily are racing...

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