Collision Problem: Football Player & QB, Coeff of Friction 0.052

[Faded Maximus]

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?

 

[hage567]

Consider the work done by the friction to bring the players to a stop. You know how much energy there is, since you know the velocity of the players after impact.

 

[Faded Maximus]

I understand what you mean:

w = fd
d = w / f

but I am unaware of how I know how much energy there is from the velocity.

edit: would you substitue w for Ek, which would be 1/2mf^2 - 1/2mvi^2?

 

[hage567]

If you write out the energy conservation for before and after impact and include the energy lost due to friction, it is a bit more obvious,

0.5m1v1^2 = 0.5mv^2 + fd

where fd is the energy lost due to friction.