I need step 5. Ill attach the previous steps. Thank you Step 1 (a) Over the shor
ID: 2033534 • Letter: I
Question
I need step 5. Ill attach the previous steps. Thank you
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 v1j5.90 m/s. His opponent has a mass of m296.0 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 ? from the positive x-direction y(north) tx (cast +0 We apply conservation of momentum to both the x- and y-components of momentum. For conservation of momentum in the eastward x-direction, PxfPx We have (m1 + m2) vf cos ? = m1v1i + O or, for the x-component of the final velocity vxf resulting from the tackle, we have miV1 90 k 5.9 m/s 2.852.85 m/ 186 kg Step 2 For conservation of momentum in the northward y-direction, Pf- Pyi We have (m1 m2) vf sin ? 0 + m2v2i or, for the y-component of the final velocity vvfof the tackle, we have m2V2i m1 m2 3 m/s 1.55 m/s. 186 kgExplanation / 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
Vf = 3.24 m/sec at 28.54 deg north of east.
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