Two ice skaters have masses m1 and m2 and are initially stationary

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Two ice skaters have masses m1 and m2 and are initially stationary. Their skates are identical. They push against one another and move in opposite directions with different speeds. While they are pushing against each other, any kinetic frictional forces acting on their skates can be ignored. However, once the skaters separate, kinetic frictional forces eventually bring them to a halt. As they glide to a halt, the magnitudes of their accelerations are equal, and skater 1 glides twice as far as skater 2. What is the ratio m1/m2 of their masses?

Let v1 and v2 be the initial velocities gained when they push against one another.
By Law of conservation of momentum, m1v1 = m2v2
Let s2 be the distance that the second skater glides before coming to rest.
Then, 2s2 would be the distance that the first skater glides before coming to rest.
If a is the acceleration in both the cases, then for the first skater:
02 = v12 - 2a(2s2), or v12 = 4as2
For the second skater: 02 = v22 - 2a(s2), or v22 = 2as2
Therefore,