Let's examine what happens when the balls $A$ and $B$ collide. We assume that the balls have equal mass, so in the case of an elastic collision, $A$ and $B$ exchange velocities. Thus, ball $A$ travels along the same path that ball $B$ would have taken if the two balls had "passed through" each other.

![](https://cdn.mathpix.com/cropped/2024_05_02_8681b448b14159e69fdeg-1.jpg?height=550&width=469&top_left_y=300&top_left_x=817)

Now, imagine that the balls do not collide but pass through each other. The number of collisions, of course, is the same as the number of "pass-throughs." In this case, in 0.2 seconds, every ball returns to its starting point, while the tube bounces back from both ends. During this time, each ball passes through the other 99 balls, and does so twice. Indeed, for two balls, the bounces from the ends of the tube divide the 0.2-second time interval into five parts. The first and fifth parts are continuations of each other in terms of motion, so we can essentially divide the motion into four parts. Of these four parts, in two the balls move in opposite directions (and pass through each other), and in two they move in the same direction.

Thus, the number of pass-throughs in 0.2 seconds is

$$
2 \cdot \frac{100 \cdot 99}{2}=9900
$$

To this, we must add the collisions with the ends of the tube, which is 200. Therefore, in 0.2 seconds, 10100 collisions occur, and in 10 seconds, 505000 collisions.
A $1 \mathrm{~m}$ long, closed at both ends, horizontal pipe contains 100 small balls. Each ball has a speed of $10 \mathrm{~m} / \mathrm{s}$, and the balls collide with each other and the ends of the pipe completely elastically. How many collisions occur in 10 seconds?