# Nonconservative collision - Work

1. Nov 7, 2011

### en bloc

1. The problem statement, all variables and given/known data
An 85kg baseball player is running towards home base at 8.0 m/s when he crashes into the catcher who is initially at rest. the two players slide together along the base path toward home plate. If the mass of the catcher is 95 kg and the coefficient of kinetic friction between the players and the ground is .70, how far will the players slide?

2. Relevant equations
1) Wnc = ΔKE + ΔPE

2) mv + mv = (m+m)v

3. The attempt at a solution
I found the final velocity using equation 2, and used it to find Δx in equation 1. I'm not sure if this is right.

.70*1764N*Δx*cos180=(1/2)*180kg*(3.78 m/s)^(2) - (1/2)85kg*(8 m/s)^(2)
-1235N*Δx = -1434N
Δx=1.16m

Last edited: Nov 7, 2011
2. Nov 7, 2011

### QuarkCharmer

The initial velocity seems correct.

For the next part try this:

$$\frac{1}{2}(M+m)v^{2} - W_{f} = \frac{1}{2}(M+m)v^{2}$$

That is, their initial kinetic energy, minus the work done by friction, equals their final kinetic energy. Well, after they are done sliding their final KE is 0 right?

$$\frac{1}{2}(M+m)v^{2} - u(M+m)gx = 0$$

3. Nov 7, 2011

### en bloc

Oooooh. That's right. KEf is 0, so equation 1 should've been like this:

Wnc= ΔKE + 0

.7 * 1764N * Δx * cos180 = 0 - 1/2 * 180kg * (3.78 m/s)^2
Δx= 1.04m

Thanks!

4. Nov 7, 2011

Thats it!