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Body decay on the axis of an infinite wedge

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  1. Dec 5, 2017 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations
    Lorentz transformations

    3. The attempt at a solution
    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
    h_1512489612_4585045_d5ef432bbb.png

    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?
     
    Last edited: Dec 5, 2017
  2. jcsd
  3. Dec 5, 2017 #2

    haruspex

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    What aspect of the velocity vector determines whether the splinter will eventually collide with the moving surface? Hint: for a moving infinite plane surface, it might as well be moving perpendicularly to itself. Any sideways movement is irrelevant.
     
  4. Dec 5, 2017 #3
    The splinter will collide with the moving surface if $$-\operatorname{tg}\frac{\alpha}{2} < \frac{u \sin\varphi}{\gamma (u \cos\varphi - V)} < \operatorname{tg}\frac{\alpha}{2}$$
     
  5. Dec 5, 2017 #4

    haruspex

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    Doesn't look quite right.
    If u cos(φ) > V there should be no possibility of collision. For u cos(φ) < V, the probability should increase as V increases.
     
  6. Dec 6, 2017 #5
    Ok, thanks. But how does it relate to the fact that half should collide with the surface?
     
  7. Dec 6, 2017 #6

    haruspex

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    If half are colliding with the wedge, what range of values of φ would that be?

    Did you find the sign correction?
     
  8. Dec 6, 2017 #7
    ##\cos \varphi < \frac{V}{u}## and ##\varphi## should depend on ##\frac{\alpha}{2}##, but I don't understand how to take into account half of the splinters.
     
  9. Dec 6, 2017 #8

    haruspex

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    No, forget the wedge for a moment, just look at the source. What range of φ would constitute half?
     
  10. Dec 6, 2017 #9
    ##\varphi \in \left[0,\pi\right]## or ##\varphi \in \left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]##?
     
  11. Dec 6, 2017 #10

    haruspex

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    Yes, one of those. You are measuring φ from the positive x axis, right? So which of those ranges do you think?
     
  12. Dec 6, 2017 #11
    Yes. Given the condition of the task, I think ##\varphi \in \left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]##.
     
  13. Dec 6, 2017 #12

    haruspex

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    Ok, so if a splinter only just makes it to the wedge, what is φ for that splinter? According to your eqn in post #3 (but after correcting the sign) what value of α says that such a splinter would just make it to the wedge?
     
  14. Dec 7, 2017 #13
    ##\operatorname{tg}\varphi^{'} = \frac{u \sin\varphi}{\gamma (u \cos\varphi - V)}## and ##\varphi = \frac{\pi}{2}##. Hence tangent of the boundary angle in a moving coordinate system ##\operatorname{tg}\varphi^{'} = -\frac{1}{\gamma}\frac{u}{V}##. But ##\frac{\alpha^{'}}{2} = \pi - \varphi^{'}## and we got the answer. Thanks!
     
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