Two ice skaters have masses m1 and m2 and are initially stationary. Their skates

Question

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?

Explanation / Answer

Determine speed of each skater before they start slowing down. For m1: vf = 0 a1 = a d1 = 2d, where d is the distance m2 glides vi = ? Applying time-independent kinematics equation: vi = sqrt(2a1d1), since vf = 0 vi = sqrt(4ad) v1 = sqrt(4ad) For m2: vf = 0 a2 = a d2 = d vi = ? v2 = sqrt(2ad) v1/v2 = sqrt[(4ad)/(2ad)] v1 = (sqrt2) v2 Before the push and after the push, the momentum of the system is 0: pinitial = pfinal 0 = m1v1 + m2v2 -m2v2 = m1v1 -v2/v1 = m1/m2 m1/m2 = 1/sqrt2

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