6. (15 points) Three material points with masses $m_{1}=100$ g, $m_{2}=200$ g, and $m_{3}=400$ g are arranged sequentially on a straight line. The distance between points 1 and 2 is 50 cm. The distance between points 2 and 3 is 200 cm. Determine the distance from point 1 to the center of mass of this system of three points.
Answer: $1.57 \mathrm{~m}$.

Solution. By the definition of the center of mass $\boldsymbol{x}_{\text {center }}=\frac{m_{1} \cdot 0+m_{2} \cdot 0.5+m_{3} \cdot 2.5}{m_{1}+m_{2}+m_{3}}=1.57 \mathrm{M}$.

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