Nonconservative collision - Work

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An 85kg baseball player collides with a 95kg catcher, both sliding together after the impact. The initial velocity of the player is 8.0 m/s, and the coefficient of kinetic friction is 0.70. The final velocity was calculated using conservation of momentum, leading to a kinetic energy equation that accounts for work done by friction. The correct distance they slide after the collision is approximately 1.04 meters, confirming the calculations. The discussion emphasizes the importance of understanding nonconservative work in collision scenarios.
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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
 
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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!
 
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