Question

Two skaters, one with mass 65 kg and the other with mass 40 kg, stand on an ice rink holding a pole of length 10 m and negligible mass. Starting from the ends of the pole, the skaters pull themselves along the pole until they meet. How far does the 40 kg skater move

Answer 1

The distance traveled by the skater is
`\bf{6.19~m}`.

Step-by-step explanation:

Given:

The mass of the first skater is,
`M_(1) = 40~kg`.

The mass of the second skater is,
`M_(2) = 65~kg`

The distance between the skaters is,
`d = 10~m`

As the skaters are moving towards each other, their center of mass will always be at rest.

Consider that the center of mass is at point C, as shown in the figure. Let's assume that the distance of center of mass from Skater A is
`x`.

From the figure, we can write

`x_(c) = (1)/((M_(1) + M_(2)))(-M_(1)x + M_(2)(10 - x))~~~~~~~~~~~~~~~~~~~~~~~(1)`

where
`x_(c)` is the position of center of mass along the pole.

As the center of mass is at rest, substituting
`0` for
`x_(c)` in equation (1), we have

`&& 0 = (1)/((M_(1) + M_(2)))[-M_(1)x + (10 - x)M_(2)]\\&or,& x = (10M_(2))/((M_(2) + M_(1)))~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)`

Substituting
`65~kg` for
`M_(2)` and
`40~kg` for
`M_(1)` in equation (2), we have

`x &=& (10(65~kg))/((65+40)~kg)~m\\~~~&=& 6.19~m`