Ice Skating conservation of momentum (conceptual problem)

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Homework Statement


Two ice skaters have masses m1 and m2 and are initially stationary. Their skates are identical. They push against one another, as in Figure 7.11, 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?

Homework Equations


m1v1 + m2v2 = m1vo1 + m2vo2

v = x/t

The Attempt at a Solution


At the beginning, the skaters are stationary:
m1v1 + m2v2 = m1(0 m/s) + m2(0 m/s)

since distance has doubled for skater 2:

m1(x/t) + m2(2x/t) = 0 N(s)

Factoring out velocity

(x/t)(m1 + 2m2) = 0
Therefore, m1/m2 = 1/2

What did I miss?
 

Answers and Replies

  • #2
kuruman
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What did I miss?
Is momentum conserved when the external force of friction acts on the skaters? You are told that the magnitude of the skaters' acceleration is the same. Is that important?

For future reference: If you cite a figure, please post it.
 
  • #4
Is momentum conserved when the external force of friction acts on the skaters? You are told that the magnitude of the skaters' acceleration is the same. Is that important?
Yes, momentum is always conserved. I don't know how acceleration is useful though, other than proving that the force of friction on each person is different.

Same t?
They have different times to come to a stop, as haruspex helped me realize. I just don't know the relationship between both times.

Would it be a 2:1 relationship based on the 2x displacement?
That would mean that m1/m2 = 1
 
  • #5
kuruman
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Yes, momentum is always conserved.
Momentum is not conserved when there are external forces acting on the center of mass. Here, friction is an external force acting on the CM, therefore momentum is not conserved.
Would it be a 2:1 relationship based on the 2x displacement?
That would mean that m1/m2 = 1
Why? Write some equations down. First you need to figure out the ratio of the speeds of the two skaters. You are supposed to neglect friction as the two push against each other, therefore the net force acting on one skater is the force exerted by the other skater. How can you use this fact to find the skaters' ratio of speeds at the time they separate?
 
  • #6
haruspex
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I just don't know the relationship between both times.
It's a matter of using the equation that has the right combination of variables.
Of distance, acceleration, time, initial velocity and final velocity, which are you trying to find and which three do you know? Pick the SUVAT equation with those four.
 

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