Posted by Megan on Wednesday, October 17, 2012 at 11:03am.
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?

Physics URGENT!!!  Elena, Wednesday, October 17, 2012 at 11:16am
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

Physics URGENT!!!  Megan, Wednesday, October 17, 2012 at 11:26am
Can you give me the values so I can compare, I don't think I am getting the right values!

Physics URGENT!!!  Elena, Wednesday, October 17, 2012 at 11:36am
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 +1025 1200 =1175 J

Physics URGENT!!!  Megan, Wednesday, October 17, 2012 at 11:41am
Thank you!
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