Answer to Question #242907 in Physics for rshiii

Question

10. 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 Gretzkey 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

Momentum conservation:

$$mv_{1}-Mu_1=mv_2-Mu_2,\\\space\\ u_2=\frac{mv_2-mv_1+Mu_1}{M}=-4.66\text{ m/s}.$$

The defender moves in the opposite direction.

The change in kinetic energy:

$$\Delta E=\frac12(mv_1^2+Mu_1^2-mv_2^2-Mu_2^2)=6580\text{ J}.$$