Body decay on the axis of an infinite wedge

[avenior]

Homework Statement

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?

Homework Equations

Lorentz transformations

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

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?

 

[haruspex]

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

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.

 

[avenior]

What aspect of the velocity vector determines whether the splinter will eventually collide with the moving surface?

The splinter will collide with the moving surface if $$-\operatorname{tg}\frac{\alpha}{2} < \frac{u \sin\varphi}{\gamma (u \cos\varphi - V)} < \operatorname{tg}\frac{\alpha}{2}$$

 

[haruspex]

The splinter will collide with the moving surface if $$-\operatorname{tg}\frac{\alpha}{2} < \frac{u \sin\varphi}{\gamma (u \cos\varphi - V)} < \operatorname{tg}\frac{\alpha}{2}$$

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.

 

[avenior]

Ok, thanks. But how does it relate to the fact that half should collide with the surface?

 

[haruspex]

Ok, thanks. But how does it relate to the fact that half should collide with the surface?

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

Did you find the sign correction?

 

[avenior]

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

$\cos \varphi < \frac{V}{u}$ and $\varphi$ should depend on $\frac{\alpha}{2}$, but I don't understand how to take into account half of the splinters.

 

[haruspex]

$\cos \varphi < \frac{V}{u}$ and $\varphi$ should depend on $\frac{\alpha}{2}$, but I don't understand how to take into account half of the splinters.

No, forget the wedge for a moment, just look at the source. What range of φ would constitute half?

 

[avenior]

No, forget the wedge for a moment, just look at the source. What range of φ would constitute half?

$\varphi \in \left[0,\pi\right]$ or $\varphi \in \left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]$?

 

[haruspex]

$\varphi \in \left[0,\pi\right]$ or $\varphi \in \left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]$?

Yes, one of those. You are measuring φ from the positive x axis, right? So which of those ranges do you think?

 

[avenior]

You are measuring φ from the positive x axis, right? So which of those ranges do you think?

Yes. Given the condition of the task, I think $\varphi \in \left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]$.

 

[haruspex]

Yes. Given the condition of the task, I think $\varphi \in \left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]$.

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?

 

[avenior]

$\operatorname{tg}\varphi^{'} = \frac{u \sin\varphi}{\gamma (u \cos\varphi - V)}$ 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!