Answer to Question #248234 in Mechanics | Relativity 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 hisoriginal 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) Let's first find the masses of hockey star and defender, respectively:

$$m_1=\dfrac{W_1}{g}=\dfrac{756\ N}{9.8\ \dfrac{m}{s^2}}=77.1\ kg,$$ $$m_2=\dfrac{W_2}{g}=\dfrac{900\ N}{9.8\ \dfrac{m}{s^2}}=91.8\ kg.$$

We can find the velocity of the defender immediately after the collision from the law of conservation of momentum:

$$m_1v_{1i}-m_2v_{2i}=m_1v_{1f}-m_2v_{2f},$$ $$v_{2f}=\dfrac{m_1(v_{1f}-v_{1i})+m_2v_{2i}}{m_2},$$ $$v_{2f}=\dfrac{77.1\ kg\cdot(1.50\ \dfrac{m}{s}-13.0\ \dfrac{m}{s})+91.8\ kg\cdot5.0\ \dfrac{m}{s}}{91.8\ kg},$$ $$v_{2f}=-4.66\ \dfrac{m}{s}.$$

The sign minus means that the defender moves in the opposite direction after the collision.

(B) Let's first find the total kinetic energy of the players before and after the collision:

$$KE_{tot,i}=\dfrac{1}{2}m_1v_{1i}^2+\dfrac{1}{2}m_2v_{2i}^2,$$ $$KE_{tot,i}=\dfrac{1}{2}\cdot77.1\ kg\cdot(13.0\ \dfrac{m}{s})^2+\dfrac{1}{2}\cdot91.8\ kg\cdot(5.0\ \dfrac{m}{s})^2=7662.45\ J,$$ $$KE_{tot,f}=\dfrac{1}{2}m_1v_{1f}^2+\dfrac{1}{2}m_2v_{2f}^2,$$ $$KE_{tot,i}=\dfrac{1}{2}\cdot77.1\ kg\cdot(1.5\ \dfrac{m}{s})^2+\dfrac{1}{2}\cdot91.8\ kg\cdot(-4.66\ \dfrac{m}{s})^2=1083.5\ J.$$ $$\Delta KE=KE_{tot,i}-KE_{tot,f}=7662.45\ J-1083.5\ J=6579\ J.$$