Question

Mixed-pair ice skaters performing in a show are standing motionless at arms 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?

Short answer

In all type of collision, momentum of the system remains conserved.

Step-by-step solution

 Step 1: Definition of Momentum

The product of a particle's mass and velocity is called momentum. The term "momentum" refers to a quantity that has both a magnitude and a direction. The temporal rate of change in momentum is equal to the force applied on the particle, according to Isaac Newton's second law of motion.

Let’s say mass of first ice skater\( = {m_1}\)

Mass of second ice skater\( = {m_2}\)

Velocity of first ice skater\( = {v_1}\)

Velocity of second ice skater \( = {v_2}\)

So, the initial momentum of the ice skaters is

\(\begin{aligned}{P_{initial}} &= {P_1} + {P_2}\\{P_{initial}} &= {m_1}{v_1} + {m_2}{v_2}\end{aligned}\)

Calculating the velocity of the ice skaters when their bodies meet 

There is no external force between them when their bodies meet. So, in the absence of net external force momentum of the system remains conserved. After meeting of their bodies total mass becomes\(({m_1} + {m_2})\)and they have common velocity \(v\).Therefore, from he conservation of linear momemtum

\(\begin{aligned}{P_{initial}} &= {P_{final}}\\{m_1}{v_1} + {m_2}{v_2} &= ({m_1} + {m_2})v\\v &= \dfrac{{{m_1}{v_1} + {m_2}{v_2}}}{{{m_1} + {m_2}}}\end{aligned}\)

Hence,their velocity when their bodies meet, is \(\dfrac{{{m_1}{v_1} + {m_2}{v_2}}}{{{m_1} + {m_2}}}\).

If they are motionless then

\(\begin{aligned}{v_1} &= 0\\{v_2} &= 0\\v &= \dfrac{{{m_1} \times 0 + {m_2} \times 0}}{{{m_1} + {m_2}}} = 0\end{aligned}\)

In this condition their net velocity will also be zero when their bodies meet.