A75.0-kg fullback running east with a speed of 4.00 m/s is tackled by a 91.0-kg opponent running north with a speed of 3.00 m/s. what is the direction they are moving after the totally inelastic collision (a perfect tackle)?

515.35 J

Explanation:

x: m1 x v1=(m1+m2) x v(x)


y: m2 x v2 =(m1+m2) x v(y)


v(x)= m1 x v1/(m1+m2) =75x4/(75+91)= 1.81 m/s


v(y) =m2 x v2 /(m1+m2)= 91 x 3 /(75+91)= 1.64 m/s

v=sqrt[v(x)^2 +v(y)^2] = 2.44 m/s


φ =arctan {v(y)/v(x)} = 47.82 degrees 

∆K = Kf −Ki =(m1 x v^2/2 +m2 x v^2/2) - (m1 +m2) x v^2/2 =


=(75 x 16/2 +91 x 9/2) –(75+91) x 2.44^2/2

= (600 + 409.5 ) - (166 x 2.98) = 515.35 J

1= 26.17

2= 1.17

3=30.86

4=5.86

just took the assigment on edge