Question

Two people with different masses but equal speeds slide toward each other with little friction on ice with their arms extended straight out to the slide (so each has the shape of a “$\mathit{I}$”). Her right hand meets his right hand, they hold hands and spin 90°, then release their holds and slide away. Make a rough sketch of the path of the center of mass of the system consisting of the two people, and explain briefly. (It helps to mark equal time intervals along the paths of the two people and of their center of mass.)

Short answer

The velocity of the center of mass of the system is$\text{'}v\text{'}m/s$

Step-by-step solution

Identification of given data

• The mass of the first person is$m_1$
• The mass of the second person is $m_2$
• The speed of both persons is $v$

Concept of the center of mass of the system

The center of mass of the system is calculated by considering the average positions of all the objects acting in the system.

Calculation of the velocity of the center of mass of the system

The velocity of the center of mass of the system

$$\begin{gathered} V_{CM=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}} \\ =\frac{m_{1}v+m_{2}v}{m_1+m_2} \\ =\frac{v.\left(m_1+m_2\right)}{m_1+m_2} \\ =v \end{gathered}$$

Hence, the velocity of the center of mass of the system is $\text{'}v\text{'}m/s$

Drawing the rough sketch of the path of the center of mass of the system

This is the rough sketch of the path of the center of mass of the system consisting of the two people.