Particle moves in square, what is the expected distance before first return to edge?

There is a unit square with a particle moving in it. After the particle collides with an edge, the angle of reflection is random and is drawn from the uniform distribution on $[-\frac{\pi}{2}, \frac{\pi}{2}].$

The question is to find the average distance the particle covers before it returns to the same edge next time (I guess that after a large number of collusions the starting point is not important).

To me, this sounds like a question about the stationary distribution of a Markov chain with a continuum of states. However, the problem actually is taken from a physics olympiad for high school students. It is claimed that the answer is $2\sqrt{2}$.

If it is not a mistake, there probably is an intuitive non-rigorous argument why the answer is $2\sqrt{2}$.

Update 1: I was asked to post the original text of the problem here. It is somewhat different from what I wrote above, but I believe that this is basically the same question:

Problem: In a computer model, movement of a particle inside of a square is simulated. Square has sides of length L, the speed of the point is V. After a collusion with an edge point bounces at a random angle (equiprobable from -90 to 90 degrees) with the same speed. Estimate the number of collusions with one of the sides after a large period of time T.

Answer: $\frac{TV}{2\sqrt{2}L}$.

Update 2: There were attempts to do a simulation (see comments below), and the results tend to be somewhat smaller than $2\sqrt{2}$.
Also, in my simulation the distribution of collision points is not uniform (points close to angles are more frequent) and distribution of distance from bounce to bounce is asymmetric and bimodal.

One possible heuristic reasoning behind the answer (for the "official" problem) is the following:

1.The events of hitting each side are equiprobable (good estimate if T is big) ,so the number of hits for a fixed side is estimated as : $\frac{(total-no-of-hits)}{4}$

2.Total traveled distance in time $T$ is $TV$

3.The longest traveled distance between two hits (between any two sides) is $\sqrt{2}L$(on the diagonal). The shortest is 0 . A rough estimate for the average distance between two hits will be $\frac{\sqrt{2}L}{2}$ (this is the most problematic part).

4.Now we can estimate the total number of hits of any side as $\frac{2TV}{\sqrt{2}L}$

5.For a fixed side the total number of hits will be $\frac{2TV}{4\sqrt{2}L}=\frac{TV}{2\sqrt{2}L}$

Update: With these estimates the answer to the initial question is indeed $2\sqrt{2}L$ . If we introduce two random variables , X- the length of the path between two collisions and Y - the number of sides hit until the particle returns to the side where it started will be $Z=X + YX$ . Since X and Y are independent , using the estimates from above and observing that Y follows a geometric distribution with $p=\frac{1}{3}$ we conclude that the expectation of Z is : $E[Z] = \frac{\sqrt{2}L}{2} + 3\frac{\sqrt{2}L}{2}$ = $2\sqrt{2}L$