Two skaters, one with mass 65 kg and the other with mass 35 kg, stand on an ice rink holding a rope of length 15 m and negligible mass. Starting from the ends of the rope, the skaters pull themselves along the rope until they meet. How far does the 35 kg skater move? Ignore friction.

Respuesta :

Answer:

the two skaters will meet at a distance of 9.75 m from the initial position of 35 kg skater

Explanation:

Since there is no external force on the system of two skaters

So the two skaters will meet at the position of their center of mass

so let the position of 35 kg skater is origin

then the position of other skater is 15 m

now by the formula of center of mass we have

[tex]r_{cm} = \frac{m_1r_1 + m_2r_2}{m_1 + m_2}[/tex]

[tex]r_{cm} = \frac{35 (0) + 65(15)}{35 + 65}[/tex]

[tex]r_{cm} = 9.75 m[/tex]

so the two skaters will meet at a distance of 9.75 m from the initial position of 35 kg skater