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Homework Statement
A football player weighing 71 kg running at 2.1 m/s sacks a stationary quarterback who weighs 52 kg. How far will the two move after the collision if the coefficient of friction between the cleats and ground is 0.052.
Homework Equations
m1v1' + m2v2' = mv'1
uk = Fk / Fn
The Attempt at a Solution
m1v1' + m2v2' = mv'1
(m1v1' + m2v2') / m = v'1
v'1 = 71 * 2.1 / 123
v'1 = 1.212 m/s
uk = Fk/Fn
Fk = Fnuk
Fk = 123 * 9.8 * 0.052
Fk = 62.68 N
Providing I did this right, I now have the speed they move after they are together and the friction in N. From here I am not sure how to calculate the distance they will move. Could somebody give me a hand?