Chapter 9: 875865-9-7 Q (page 377)

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 “ $I$ ”). 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.)

Expert verified

The velocity of the center of mass of the system is $' v' m / s$

Step by step solution

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

The velocity of the center of mass of the system

$V_{C M = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}} = \frac{m_{1} v + m_{2} v}{m_{1} + m_{2}} = \frac{v . (m_{1} + m_{2})}{m_{1} + m_{2}} = v$

Hence, the velocity of the center of mass of the system is $' v' m / s$

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

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