# Homework Help: Collision Problem

1. Apr 14, 2007

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

m1v1' + m2v2' = mv'1
uk = Fk / Fn

3. 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?

2. Apr 14, 2007

### 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.

3. Apr 14, 2007

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?

Last edited: Apr 14, 2007
4. Apr 14, 2007

### 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.