13.408 A material particle $m_{1}$ oscillates between points $A$ and $B$, which are 3.01 m apart. The particle's speed is constant, and it does not stop at the endpoints. After 11 s from the departure of particle $m_{1}$ from point $A$, another particle $m_{2}$ starts moving from point $B$ with a constant but lower speed. This particle, moving towards point $A$, meets particle $m_{1}$ twice, specifically 10 and 45 s after the departure of the second particle. Determine the speeds of the particles.
Solution. Let $x$ and $y$ be the speeds (cm/s) of particles $m_{1}$ and $m_{2}$. By the time particle $m_{2}$ exits, the distance between $m_{1}$ and $m_{2}$ is $301-11 x$ cm, hence $301-11 x=10(x+y)$. Let $C$ be the point of the first meeting of the particles. Then $B C=10 y$, and therefore, by the time of the second meeting (i.e., 35 s after the first), particle $m_{l}$ will have traveled a distance $C B+B C=20 y$ greater than particle $m_{2}$ (counting from the moment of the first meeting). Therefore, $\left\{\begin{array}{l}301-11 x=10(x+y), \\ 20 y=35(x-y)\end{array} \Rightarrow\left\{\begin{array}{l}x=11, \\ y=7 .\end{array}\right.\right.$

Answer: 11 cm/s, 7 cm/s.