# Chapter 8 - Rotational Motion and Equilibrium - Learning Path Questions and Exercises - Conceptual Questions - Page 303: 24

a). Linking of arms by the two skaters moving in opposite direction is similar to head-on collision and hence the angular momentum will remain same. So, $mv-mv=mv^{'}$ Hence, $v^{'}=0$ Hence, the velocity of their center of mass after they link arms will be zero. b). Their initial linear kinetic energies would be converted and the two skaters would start to rotate about the combined center of mass.

#### Work Step by Step

a). Linking of arms by the two skaters moving in opposite direction is similar to head-on collision and hence the angular momentum will remain same. So, $mv-mv=mv^{'}$ Hence, $v^{'}=0$ Hence, the velocity of their center of mass after they link arms will be zero. b). Their initial linear kinetic energies would be converted and the two skaters would start to rotate about the combined center of mass.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.