Answer to Question #277631 in Physics for Kristine

Question

Ice hockey star Wayne Gretzky is skating at 13.0 𝑚/𝑠 toward a defender, who, in turn, is skating at 5.00 𝑚/𝑠 toward Gretzky, shown in Figure right. Gretzky’s weight is 756 𝑁; that of the defender is 900 𝑁. Immediately after the collision, Gretzky is moving at 1.50 𝑚/𝑠 in his original direction. Neglect external horizontal forces applied by the ice to the skaters during the collision. A) What is the velocity of the defender immediately after the collision? b) Calculate the change in the combined kinetic energy of the two players.

Answer 1

a) By momentum conservation:

$$m_Gv_G-m_Dv_d=m_Gu_G+m_du_d,\\\space\\ u_d=4.66\text{ m/s}$$

in the direction of Gretzky.

b) The change in kinetic energy is

$$\Delta K=\frac12(m_Gv_G^2+m_dv_d^2-m_Gu_G^2-m_du_d^2),\\ \Delta K=63468\text{ J}.$$