# Body decay on the axis of an infinite wedge

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1. Dec 5, 2017

### avenior

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

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. Dec 5, 2017

### haruspex

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.

3. Dec 5, 2017

### avenior

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

4. Dec 5, 2017

### haruspex

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.

5. Dec 6, 2017

### avenior

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

6. Dec 6, 2017

### haruspex

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

Did you find the sign correction?

7. Dec 6, 2017

### avenior

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

8. Dec 6, 2017

### haruspex

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

9. Dec 6, 2017

### avenior

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

10. Dec 6, 2017

### haruspex

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

11. Dec 6, 2017

### avenior

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

12. Dec 6, 2017

### haruspex

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?

13. Dec 7, 2017

### 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!