Question

A 75.0 kg fullback running east with a speed of 6.00 m/s is tackled by a 82.0 kg opponent running north with a speed of 5.00 m/s. (a) Calculate the velocity of the players immediately after the tackle. (b)Calculate the direction of the players immediately after the tackle.(c)Determine the mechanical energy that is lost as a result of the collision. (d)Where did the lost energy go?

Answer 1

x: m•v1=(m1+m2) •v(x),
y: m•v2 =(m1+m2) •v(y),
v(x)= m•v1/(m1+m2),
v(y) =m•v2 /(m1+m2),

v=sqrt[v(x)²+v(y)²],
φ =arctan {v(y)/v(x)}.

∆K = Kf −Ki =
=(m1•v²/2 +m2•v²/2) - (m1 +m2) •v²/2.
mechanical energy -> thermal energy, sound, etc

Answer 2

Can you give me the values so I can compare, I don't think I am getting the right values!

Answer 3

x: m1•v1=(m1+m2) •v(x)
y: m2•v2 =(m1+m2) •v(y)
v(x)= m1•v1/(m1+m2) =75•6/(75+82)=2.87 m/s
v(y) =m2•v2 /(m1+m2)= 82•5/(75+82)=2.66 m/s

v=sqrt[v(x)²+v(y)²] =3.91 m/s
φ =arctan {v(y)/v(x)} =42.8⁰

∆K = Kf −Ki =(m1•v²/2 +m2•v²/2) - (m1 +m2) •v²/2 =
=(75•36/2 +82•25/2) –(75+82) •3.91²/2=
=1350 [phone] =1175 J

Answer 4

Thank you!