Two skaters A and B of mass 50 kg and 70 kg, respectively, stand facing each other, 6 metres apart on a horizontal smooth surface. They pull a rope stretched between them. How far has each moved when they meet?

Let m_A and m_B be the masses of skaters A and B and a_A and a_B their respective accelerations, when they pull at each other. From Newton's third law, action and reaction forces are equal in magnitude, i.e.

$$\begin{gathered} {\mathrm{m}}_{\mathrm{A}}{\mathrm{a}}_{\mathrm{A}}={\mathrm{m}}_{\mathrm{B}}{\mathrm{a}}_{\mathrm{B}} \\ \Rightarrow {\mathrm{m}}_{\mathrm{A}}\frac{{\mathrm{v}}_{\mathrm{A}}}{\mathrm{t}}={\mathrm{m}}_{\mathrm{B}}\frac{{\mathrm{v}}_{\mathrm{B}}}{\mathrm{t}} \\ \Rightarrow {\mathrm{m}}_{\mathrm{A}}{\mathrm{v}}_{\mathrm{A}}={\mathrm{m}}_{\mathrm{B}}{\mathrm{v}}_{\mathrm{B}} \\ \Rightarrow {\mathrm{m}}_{\mathrm{A}}^2{\mathrm{v}}_{\mathrm{A}}^2={\mathrm{m}}_{\mathrm{B}}^2{\mathrm{v}}_{\mathrm{B} \left(\mathrm{i}\right)}^2 \end{gathered}$$

Where v_A and v_B are their respective speeds and t is the time taken for them to meet. Let s_A and s_B be the distances travelled by them when they meet, we have:

$$\begin{gathered} 2{\mathrm{a}}_{\mathrm{A}}{\mathrm{s}}_{\mathrm{A}}={\mathrm{v}}_{\mathrm{A}}^2 \mathrm{and} 2{\mathrm{a}}_{\mathrm{B}}{\mathrm{s}}_{\mathrm{B}}={\mathrm{v}}_{\mathrm{B}}^2 \\ \mathrm{Using} \mathrm{these} \mathrm{equations} \mathrm{in} \mathrm{Eq}. \left(\mathrm{i}\right) \mathrm{noting} \mathrm{that} \\ {\mathrm{m}}_{\mathrm{A}}{\mathrm{a}}_{\mathrm{A}}={\mathrm{m}}_{\mathrm{B}}{\mathrm{a}}_{\mathrm{B}}, \mathrm{we} \mathrm{get} \frac{{\mathrm{s}}_{\mathrm{A}}}{{\mathrm{s}}_{\mathrm{B}}}=\frac{{\mathrm{m}}_{\mathrm{B}}}{{\mathrm{m}}_{\mathrm{A}}}=\frac{70}{50}=\frac{7}{5}. \\ \mathrm{Since} {\mathrm{s}}_{\mathrm{A}}+{\mathrm{s}}_{\mathrm{B}}=6\mathrm{m}; \\ \Rightarrow {\mathrm{s}}_{\mathrm{A}}=3.5 \mathrm{m} \mathrm{and} {\mathrm{s}}_{\mathrm{B}}=2.5 \mathrm{m}. \end{gathered}$$