Nonconservative collision - Work

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Homework Help Overview

The problem involves a collision between a baseball player and a catcher, focusing on the subsequent sliding motion and the work done against friction. The context is rooted in concepts of momentum conservation and nonconservative work in physics.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculation of final velocity post-collision and the application of work-energy principles to find the distance slid. There are attempts to clarify the relationship between initial kinetic energy, work done by friction, and final kinetic energy.

Discussion Status

Several participants have engaged in exploring the correct application of equations related to kinetic energy and work done by friction. Guidance has been offered regarding the interpretation of final kinetic energy and its implications for the calculations. There appears to be a productive exchange of ideas, with some participants confirming the correctness of initial calculations.

Contextual Notes

There is an ongoing discussion about the assumptions regarding the final kinetic energy being zero after sliding, as well as the implications of the coefficient of friction in the calculations. Participants are also considering the effects of the combined mass of the players on the sliding distance.

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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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