A 92-kg fullback moving south with a speed of 5.8 m/s is tackled by a 110-kg lineman running west with a speed of 3.6 m/s. Assuming momentum conservation, determine the speed and direction of the two players immediately after the tackle. Give the direction as an angle, in degrees, south of west.

Respuesta :

Answer:

The speed and direction of the two players immediately after the tackle are 3.3 m/s and 53.4° South of West

Explanation:

given information:

mass of fullback, [tex]m_{x}[/tex] = 92 kg

speed of full back, [tex]v_{x}[/tex] = 5.8 to south

mass of lineman, [tex]m_{y}[/tex] =110 kg

speed of lineman, [tex]v_{y}[/tex] = 3.6

according to conservation energy,

assume that the collision is perfectly inelastic, thus

initial momentum = final momentum

                            [tex]P_{ix}[/tex] = [tex]P_{x}[/tex]'

                          m₁v₁ = (m₁+m₂)[tex]v_{x}[/tex]'

                             [tex]v_{x}[/tex]' = m₁v₁/(m₁+m₂)

                                  = (92) (5.8)/(92+110)

                                  = 2.64 m/s

                            [tex]P_{iy}[/tex] = [tex]P_{y}[/tex]'

                         m₂v₂ = (m₁+m₂)[tex]v_{y}[/tex]'

                             [tex]v_{y}[/tex]' = m₁v₁/(m₁+m₂)

                                  = (110) (3.6)/(92+110)

                                  = 1.96 m/s

thus,

[tex]v[/tex]' = √[tex]v_{x}[/tex]'²+[tex]v_{y}[/tex]'²

  = 3.3 m/s

then, the direction of the two players is

θ = 90 - tan⁻¹([tex]v_{y}[/tex]'/[tex]v_{x}[/tex]')

  = 90 - tan⁻¹(1.96/2.64)

  = 53.4° South of West

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