# Nonconservative collision - Work

• en bloc
In summary, a 85kg baseball player running at 8.0 m/s crashes into a 95kg catcher at rest and the two players slide together along the base path. Using the equations Wnc = ΔKE + ΔPE and mv + mv = (m+m)v, the final velocity is found and used to find the distance the players will slide, which is approximately 1.04m.

## Homework Statement

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?

## Homework Equations

1) Wnc = ΔKE + ΔPE

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

## 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:
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?

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

QuarkCharmer said:
Well, after they are done sliding their final KE is 0 right?

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!

Thats it!