Question

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?

Answer 1

Answer:

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.

Answer 2

The velocity of the mixed-pair skaters after their bodies meet is;

v = 0 m/s

We are told that there is no friction between the blades of their skates and the ice in the collisions. This therefore means that the momentum is conserved. Thus;

Initial momentum(p_i) = Final momentum(p_f)  

Let the ice skaters that collide be represented by 1 and 2.

Thus;

Mass of skaters 1 = m₁

Mass of skaters 2 = m₂

Initial speed of skaters 1 = u₁

Initial speed of skaters 2 = u₂

Now, the skaters have a common velocity after collision and as such using conservation of momentum, we have;

m₁u₁ + m₂u₂ = (m₁ + m₂)v

where v is their velocity after the collision

Now, we were told that the skaters were initially motionless before movement. This means that their initial velocities are both 0 m/s. Thus;

u₁ = u₂ = 0 m/s

Thus;

m₁(0) + m₂(0) = (m₁ + m₂)v  

(m₁ + m₂)v = 0

Since both masses can't be zero, the v = 0 m/s

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