Two skaters on a frictionless icy surface push apart from one another. One skater has a mass M much greater than the mass m of the second skater. After some time the two skaters are at a distance d apart. How far has the lighter skater moved from her original position?

### 1 Answer | Add Yours

Two skaters on a frictionless surface of ice push each other. The mass of one of them is M and the mass of the other is m. After some time they are a distance D apart. Let the lighter skater with mass m have moved a distance x, the distance moved by the other skater with mass M is D - x.

If a force F was applied by each of them on the other when they pushed each other, the resulting acceleration of the lighter skater is F/m and the acceleration of the other skater is F/M.

The time taken by the two skaters to move apart by distance D is `sqrt((2*D)/(F/m + F/M))` . The distance moved by the skater in this duration of time is `(1/2)*(F/m)*(sqrt((2*D)/(F/m + F/M)))^2`

= `(1/2)*(F/m)*((2*D)/(F/m + F/M))`

= `(1/m)*(D/((m+M)/(m*M)))`

= `(D*M)/(m+M)`

The lighter skater moves a distance `(D*M)/(m+M)` from her original position.

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes