Example 3 On a circle with a circumference of $300 \mathrm{~cm}$, there are two balls, A and B, moving in uniform circular motion at different speeds. Ball A starts from point A and moves in a clockwise direction, while ball B starts from point B at the same time and moves in a counterclockwise direction. The two balls meet at point C. After meeting, the two balls reverse their directions and continue moving in uniform circular motion, but ball A's speed is doubled, and ball B's speed is halved. They meet for the second time at point D (as shown in Figure 1). It is known that $\overparen{A M C}=40 \mathrm{~cm}$ and $\overparen{B N D}=20 \mathrm{~cm}$. Find the length of $\overparen{A C B}$.
Solution: Let the length of $\overparen{A C P}$ be $x \mathrm{~cm}$, and the original speeds of A and B be $v_{1}$ and $v_{2}\left(v_{1} \neq v_{2}\right)$, respectively. According to the problem, we have
$$
\left\{\begin{array}{l}
\frac{40}{v_{1}}=\frac{x-40}{v_{2}}, \\
\frac{300-20-(x-40)}{2 v_{1}}=\frac{x-40+20}{\frac{1}{2} v_{2}} .
\end{array}\right.
$$
(1) $\div$ (2) gives $\frac{2 \times 40}{320-x}=\frac{x-40}{2(x-20)}$.

Solving, we get $x_{1}=120, x_{2}=80$.
When $x=80$, substituting into (1) gives $v_{1}=v_{2}$, which contradicts the problem statement, so it is discarded. Therefore, the length of $\overparen{A C B}$ is $120 \mathrm{~cm}$.