Two ice skaters have masses m 1 and m 2 and are initially stationary. Their skat

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 2 times as far as skater 2. What is the ratio m1 / m2 of their masses?

Explanation / Answer

The formula we will start with is vf2 = vo2 + 2ad

They started from rest so initial velocity is zero. The accelerations are given the be the same

Solve for a

a = vf2/2d

Thus

v1f2/2d1 = v2f2/2d2 We are told that d1 = 2d2, so by substitution

v1f2/4d2 = v2f2/2d2

Simplify

v1f2/2 = v2f2 which is 2v2f2 = v1f2

or 2(v2f) = v1f

Then by conservation of momentum

m1v1 = m2v2

m1/m2 = v2/v1

m1/m2 = (2(v2f)/(v2f)

Simplify

m1/m2 = 2

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