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Centripetal Force and Circular Motion

The velocity ...

Question

The velocity components of a particle moving in the xy plane of the reference frame $K$ are equal to $v_{x}$ and $v_{y}$ . Find the velocity $v^{'}$ of this particle in the frame $K^{'}$ which moves with the velocity $V$ relative to the frame $K$ in the positive direction of its x axis.

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Solution

By the velocity addition formula:
$v_{x}^{'} = \frac{v_{x} - V}{1 - \frac{V v_{x}}{c^{2}}}$ , $v_{y}^{'} = \frac{v_{y} \sqrt{1 - \frac{V^{2}}{c^{2}}}}{1 - \frac{v_{x} V}{c^{2}}}$
And, $v^{'} = \sqrt{{v_{x}^{'}}^{2} + {v_{y}^{'}}^{2}} = \frac{\sqrt{{(v_{x} - V)}^{2} + v_{y}^{2} (1 - \frac{V^{2}}{c^{2}})}}{1 - \frac{v_{x} V}{c^{2}}}$

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