PLEASE HELP! THANKS IN ADVANCE Step 1 (a) Over the short time interval of the co

Question

PLEASE HELP! THANKS IN ADVANCE

Step 1 (a) Over the short time interval of the collision, external forces do not impart significant impulse to the players. The two players move together as a unit after the tackle, so the collision is completely inelastic (b) We choose east to be the positive x-direction as shown in the diagram. The fullback running east has a mass of m1-90.0 kg and an initial speed of v1/-5.90 m/s. His opponent has a mass of m2 = 96.0 kg and is running north at an initial speed of v2j 3.00 m/s. The velocity of their combined mass immediately after the tackle is denoted by vf at an angle 6 from the positive x-direction ty (north) Dii +x (east) We apply conservation of momentum to both the x- and y-components of momentum. For conservation of momentum in the eastward x-direction, PxfPxi. We have (m1 + m2) v cos 6m1V1i 0 or, for the x-component of the final velocity Vxf resulting from the tackle, we have m1V1i (m1 + m2) k9 m/s m/s kg

Explanation / Answer

(theta = A)

Vxf = Vf*cos A = m1*V1i/(m1 + m2)

m1 +m2 = 90 + 96 = 186 kg

Vf*cos A = 90*5 kg.90 m/sec/(186 kg) = 2.85 m/sec

Bold part is your answer

Similarly

Vyf = Vf*sin A = m2*V2i/(m1 + m2)

Vf*sin A = 96*3.00/(90 + 96) = 1.55 m/sec

Now

Vxf = 2.85 m/sec

Vyf = 1.55 m/sec

Vf = sqrt (Vxf^2 + Vyf^2)

Vf = sqrt (2.85^2 + 1.55^2)

Vf = 3.24 m/sec

Angle theta will be

A = arctan (Vyf/Vxf)

A = arctan (1.55/2.85) = 28.54 deg

Please Upvote and let me know if you have any doubt.

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