Mixed-pair ice skaters performing in a show are standing motionless at arm's length just before starting a routine. They reach out, clasp hands, and pull themselves together by only using their arms. Assuming there is no friction between the blades of their skates and the ice, what is their velocity after their bodies meet?

Question

Physics

Zachery Stanton

3 April, 09:57

Mixed-pair ice skaters performing in a show are standing motionless at arm's length just before starting a routine. They reach out, clasp hands, and pull themselves together by only using their arms. Assuming there is no friction between the blades of their skates and the ice, what is their velocity after their bodies meet?

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Answers (1)

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Maynard · Answer 1

Their velocity is 0.

Explanation:

For all collisions, the momentum is conserved, thus the initial momentum must be equal to the final momentum:

Qi = Qf

The momentum is the mass multiplied by the velocity. So, in the beginning, the total momentum is the sum of the momentum of the ice skaters (A and B). And after the collision, they are together, thus they have the same velocity, and the final momentum will be (mA + mB) * V, where m is the mass and V is the velocity, so:

mA*VA + mB*VB = (mA + mB) * V

If they are motionless, than VA = VB = 0

mA*0 + mB*0 = (mA + mB) * V

V = 0.