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  Body decay on the axis of an infinite wedge

+ 0 like - 0 dislike

Please help me to solve the following problem: on the axis of an infinite wedge that moves with velocity $\vec{V}$, the body decays with the formation of a lot of splinters that fly away uniformly in all directions with velocity $\vec{u}$. What should be the angle of the wedge that half of the splinters fall on its side surface?

The right answer is $\operatorname{tg}\frac{\alpha}{2} = \frac{u}{V}\sqrt{1-\frac{V^2}{c^2}}$.

As I understand, the figure to this problem looks like this 

If $\varphi$ is the angle between $\vec{u}$ and $Ox$, then $\vec{u}_{spl} = (u \cos \varphi, u \sin \varphi )$. By making the Lorentz transformations, we obtain that $$\vec{u^{'}}_{spl} = \left(\frac{u \cos \varphi - V}{1 - \frac{Vu \cos \varphi}{c^2}}, \frac{1}{\gamma} \frac{u \sin \varphi}{1 - \frac{Vu \cos \varphi}{c^2}} \right).$$

How can we take into account that the half should fall to the surface?

asked Dec 1, 2017 in Theoretical Physics by avenior (0 points) [ revision history ]

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