91Ó°ÊÓ

The mass of the one skater is $m_1=65kg.$

The mass of the second skater is$m_2=40kg.$

The length of the pole is$I=10m.$

We can use the concept of center of mass of the system. When skaters are pulling each other, the center of mass of the system will not move. Therefore, skaters will meet at the center of mass irrespective of pull from both the skaters.

Formula:

${\stackrel{→}{R}}_{cm}=\frac{m_1{\stackrel{→}{r}}_1+m_2{\stackrel{→}{r}}_2}{m_1+m_2}$

The center of mass of the system does not move. The skaters are meeting at the center of mass. Consider the $m_2$meets distance x from the center of mass.

Substitute the values in the equation (i)

$$\begin{gathered} 0=\frac{\left(65kg\right)(10m-x)+\left(40kg\right)(-x)}{65kg+40kg} \\ \left(65kg\right)(10m-x)-\left(40kg\right)\left(x\right)=0 \\ (65kg×10m)-(65kg×x)=\left(40kg\right)\left(x\right) \\ (65kg×10m)=\left(40kg\right)\left(x\right)+(65kg×x) \\ (65kg×10m)=(40kg+65kg)\left(x\right) \\ x=\frac{(65kg×10m)}{(40kg+65kg)} \\ =6.2m \end{gathered}$$
The distance covered by the skater $m_2$from the center of mass of the system is $x=6.2m$