During a hockey game, two hockey players (m1 = 82kg and m2 = 76kg) collide head on in a 1 dimensional perfectly instant collision. If the first hockey player is moving to the left at a velocity 4.2 m/s and the second hockey player is moving in the opposite direction at a velocity of 3.4 m/s, how fast are they both moving after they collide, assuming they stick together after they collide? How much kinetic energy is lost as a result of the collision?

Respuesta :

Given:

The mass of player 1 is m1 = 82 kg

The initial velocity of player 1 is

[tex]v_{i1}=\text{ 4.2 m/s}[/tex]

towards left.

The mass of player is m2 = 76 kg

The initial velocity of player 2 is

[tex]v_{i2}=\text{ -3.4 m/s}[/tex]

in opposite direction.

Required:

The final velocity after the collision assuming they stick together.

The loss of kinetic energy after collision.

Explanation:

According to the conservation of momentum, the final velocity will be

[tex]\begin{gathered} m1v_{i1}+m2v_{i2}=(m1+m2)v_f \\ v_f=\frac{m1v_{i1}+m2v_{i2}}{m1+m2} \\ =\text{ }\frac{(82\times4.2)-(76\times3.4)}{82+76} \\ =0.544\text{ m/s} \end{gathered}[/tex]

The loss in kinetic energy will be

[tex]\begin{gathered} \Delta KE=KE_{after}-KE_{before} \\ =\frac{1}{2}(m1+m2)(v_f)^2-\frac{1}{2}m1(v_{i1})^2-\frac{1}{2}m2(v_{i2})^2 \\ =\frac{1}{2}(82+76)(0.544)^2-\frac{1}{2}\times82\times(4.2)^2-\frac{1}{2}\times76\times(3.4)^2 \\ =23.38-723.24-439.28 \\ =-1139.14\text{ J} \end{gathered}[/tex]

Final Answer:

The final velocity after the collision assuming they stick together is 0.544 m/s.

The loss of kinetic energy after the collision is 1139.14 J.