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PhMichael
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The velocity of B w.r.t A is:
[tex]\vec{v}=0.6c\left ( \frac{\hat{x}-\hat{y}}{\sqrt{2}} \right )[/tex].
what is [tex]\alpha[/tex] (the 90 degrees of the triangle in A's frame) in B's frame?
2. The attempt at a solution
AC and BC will each be lengthened in B's frame, so that:
[tex] AC' = \frac{1}{\sqrt{1-(v_{x}/c)^{2}}} \cdot 12 = \frac{60 \sqrt{82}}{41} [/tex]
[tex] BC' = \frac{1}{\sqrt{1-(v_{y}/c)^{2}}} \cdot 12 = \frac{60 \sqrt{82}}{41} [/tex]
where,
[tex]v_{x}=v_{y}=\frac{0.6c}{\sqrt{2}} [/tex]
But here I'm stuck and can't proceed ... any guidance will be appreciated :)
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Homework Statement
The velocity of B w.r.t A is:
[tex]\vec{v}=0.6c\left ( \frac{\hat{x}-\hat{y}}{\sqrt{2}} \right )[/tex].
what is [tex]\alpha[/tex] (the 90 degrees of the triangle in A's frame) in B's frame?
2. The attempt at a solution
AC and BC will each be lengthened in B's frame, so that:
[tex] AC' = \frac{1}{\sqrt{1-(v_{x}/c)^{2}}} \cdot 12 = \frac{60 \sqrt{82}}{41} [/tex]
[tex] BC' = \frac{1}{\sqrt{1-(v_{y}/c)^{2}}} \cdot 12 = \frac{60 \sqrt{82}}{41} [/tex]
where,
[tex]v_{x}=v_{y}=\frac{0.6c}{\sqrt{2}} [/tex]
But here I'm stuck and can't proceed ... any guidance will be appreciated :)
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