8. (10 points) Person A and Person B run back and forth on a 120-meter straight path. Person A's speed is 5 meters/second, and Person B's speed is 3 meters/second. If they start from the same end at the same time and run for 15 minutes, how many times do they meet head-on during this period (excluding the endpoints)?
【Analysis】According to the problem, it is necessary to understand that when they meet face-to-face, the total distance covered by the two people is 2 one-way trips, $120 \times 2 = 240$ (meters), and the time taken is $240 \div (3+5) = 30$ (seconds), meaning they meet every 30 seconds (including at the endpoints). The time for meeting at the endpoints is the least common multiple (LCM) of the time it takes for each person to complete a one-way trip $(120 \div 3 = 40, 120 \div 5 = 24)$. The LCM is the first time they meet at the endpoint. Using $120 \div 30 = 4$, we know that out of 4 meetings, 1 is at the endpoint. Therefore, the total number of meetings within 15 minutes is: $15 \times 60 \div 30 = 30$, and the number of times they meet at the endpoint is the integer part of $30 \div 4$, which is 7. Thus, the number of times they meet face-to-face (excluding the endpoints) during this period is: 30 - 7 = 23.

【Solution】Solution: $240 \div (3+5) = 30$ (seconds)
$$
\begin{array}{l}
120 \div 3 = 40 \text { (seconds) } \\
120 \div 5 = 24 \text { (seconds) }
\end{array}
$$
The least common multiple of 40 and 24 is 120 (the first time they meet at the endpoint)
$$
\begin{array}{l}
120 \div 30 = 4 \\
15 \times 60 \div 30 = 30 \text { (times) } \\
30 \div 4 = 7 \cdots 2 \\
30 - 7 = 23 \text { (times) }
\end{array}
$$

Answer: They meet face-to-face 23 times (excluding the endpoints) during this period.