Two ice skaters have masses m1 and m2 and are initially stationary. Their skates are identical. They push against one another, as shown below, 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 3 times as far as skater 2. What is the ratio m1 / m2 of their masses?

Question

Ahmad alajmi · Accepted Answer

Two ice skaters stand at rest in the center of an ice rink. When they push off against each 45-kg skater acquires a speed of 0.62 m/s. If the speed of the other skater is 0.89 m/s, what is that skater's mass?

drwls · Answer

Their initial velocities are in a ratio V1/V2 = m2/m1

Their kinetic friction forces are in a ratio
F1/F2 = M1/M2

Gliding time t is proportional to
V/a = V m/F

Gliding distance D is proportional to
V*t = V^2*m/F

D1/D2 = 3 = [V1^2*m1/F1)][m2/(F2*V2^2)]
= (V1/V2)^2*(m1/m2)*F2/F1
= (m2/m1)^2

m2/m1 = 3^1/2 = 1.732
m1/m2 = 0.577