- #1
Amrator
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- Homework Statement
- A and B both start at the origin and simultaneously head off in opposite directions, each with speed ##\frac{3c}{5}## with respect to the ground. A moves to the right and B moves to the left. Consider a mark on the ground at ##x = L##. As viewed in the ground frame, A and B are a distance ##2L## apart when A passes this mark. As viewed by A, how far away is B when A coincides with the mark?
- Relevant Equations
- $$t_{observed} = \gamma t_{proper}$$
$$L_{observed} = \frac{L_{proper}} {\gamma}$$
$$t_R - t_L = \frac{Lv}{c^2}$$
$$V = \frac{u+v}{1+\frac{uv}{c^2}}$$
I don't know why I'm so puzzled by this problem; it's only one star. So first, I drew a picture of A and B in the ground frame. Then I drew B and the ground in A's frame. I then used the velocity addition formula to obtain the velocities of both B and the ground relative to A.
$$\frac{v_{B}^{A} + v_{gound}^{A}}{1 + \frac{v_{B}^{A} v_{gound}^{A}}{c^{2}}} = \frac{-\frac{3c}{5} - \frac{3c}{5}}{1 + \frac{9}{25}} = -\frac{15c}{17}$$
And I don't where to go from here.
$$\frac{v_{B}^{A} + v_{gound}^{A}}{1 + \frac{v_{B}^{A} v_{gound}^{A}}{c^{2}}} = \frac{-\frac{3c}{5} - \frac{3c}{5}}{1 + \frac{9}{25}} = -\frac{15c}{17}$$
And I don't where to go from here.
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