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