Two ice skaters have masses m1 and m2 and are initially stationary. Their skates are identical. They push against one another, as in Figure 7.10, 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?

Answer 1

If they decelerate at the same rate, and skater 1 goes twice as far, she also glides for a period that is twice as long. She must have started with twice the velocity of the other skater. V1 = 2 V2

Skater 1 and skater 2 has equal momenta at pushoff.

Therefore m1*V1 = m2*V2, and

m1/m2 = V2/V1 = 1/2