Ten very small rubber balls begin equally spaced inside a $55 \mathrm{~m}$ long tube. They instantly begin to roll inside the tube at a constant velocity of $1 \mathrm{~m} / \mathrm{s}$. When a ball reaches an end of the tube, it falls out of the tube. When two balls bump into each other, they both instantly reverse directions but continue to roll at $1 \mathrm{~m} / \mathrm{s}$. Five configurations giving the initial direction of movement of each ball are shown. All gaps indicated in the diagram are the same length and are equal in length to the distance from the ends of the tube to the nearest ball. For which configuration will it take the least time for more than half of the balls to fall out of the tube?

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-047.jpg?height=434&width=1369&top_left_y=1607&top_left_x=318)
Since the rubber balls are very small and the tube is very long $(55 \mathrm{~m})$, we treat the balls as points with negligible width.

Since the 10 balls begin equally spaced along the tube with equal spaces before the first ball and after the last ball, then the 10 balls form 11 spaces in the tube, each of which is $\frac{55}{11}=5 \mathrm{~m}$ long.

When two balls meet and collide, they instantly reverse directions. Before a collision, suppose that ball $A$ is travelling to the right and ball $B$ is travelling to the left.

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-082.jpg?height=92&width=1084&top_left_y=518&top_left_x=583)

After this collision, ball $A$ is travelling to the left and ball $B$ is travelling to the right.

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-082.jpg?height=92&width=1084&top_left_y=694&top_left_x=583)

Because the balls have negligible size we can instead pretend that balls $A$ and $B$ have passed through each other and that now ball $A$ is still travelling to the right and ball $B$ is travelling to the left. The negligible size of the balls is important here as it means that we can ignore the fact that the balls will travel slightly further by passing through each other than they would by colliding.

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-082.jpg?height=90&width=1070&top_left_y=1077&top_left_x=587)

In other words, since one ball is travelling to the left and one is travelling to the right, it actually does not matter how we label them.

This means that we can effectively treat each of the 10 balls as travelling in separate tubes and determine the amount of time each ball would take to fall out of the tube if it travelled in its original direction.

In $(\mathrm{A})$,

- the first ball is $50 \mathrm{~m}$ from the right end of the tube, so will take $50 \mathrm{~s}$ to fall out
- the second ball is $45 \mathrm{~m}$ from the right end of the tube, so will take $45 \mathrm{~s}$ to fall out
- the third ball is $40 \mathrm{~m}$ from the right end of the tube, so will take $40 \mathrm{~s}$ to fall out
- the fourth ball is $20 \mathrm{~m}$ from the left end of the tube, so will take $20 \mathrm{~s}$ to fall out (note that this ball is travelling to the left)

and so on.

For configuration (A), we can follow the method above and label the amount of time each ball would take to fall out:

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-082.jpg?height=101&width=1084&top_left_y=2012&top_left_x=583)

We can then make a table that lists, for each of the five configuration, the amount of time, in seconds that each ball, counted from left to right, will take to fall out:

|  |  |  |  |  |  |  |  |  |  |  |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Configuration | $\# 1$ | $\# 2$ | $\# 3$ | $\# 4$ | $\# 5$ | $\# 6$ | $\# 7$ | $\# 8$ | $\# 9$ | $\# 10$ |
| (A) | 50 | 45 | 40 | 20 | 30 | 30 | 35 | 15 | 45 | 5 |
| (B) | 5 | 45 | 15 | 35 | 30 | 25 | 20 | 40 | 45 | 50 |
| (C) | 50 | 10 | 15 | 35 | 30 | 30 | 35 | 15 | 45 | 5 |
| (D) | 5 | 45 | 40 | 20 | 30 | 30 | 35 | 40 | 45 | 5 |
| (E) | 50 | 10 | 40 | 20 | 30 | 30 | 35 | 15 | 45 | 50 |

Since there are 10 balls, then more than half of the balls will have fallen out when 6 balls have fallen out.

In (A), the balls fall out after 5, 15, 20, 30,30,35,40,45, 45, and 50 seconds, so 6 balls have fallen out after 35 seconds.

The corresponding times for (B), (C), (D), and (E) are 35, 30, 35, and 35 seconds.

Therefore, the configuration for which it takes the least time for more than half of the balls to fall out is $(\mathrm{C})$.

ANSWER: $(\mathrm{C})$