freight carcaboose collision finding mass without velocitiesby hangten1039 Tags: carcaboose, collision, freight, mass, velocities 

#1
Nov109, 10:24 AM

P: 17

A 34000kg railroad freight car collides with a stationary caboose car. They couple together, and 23 percent of the initial kinetic energy is dissipated as heat, sound, vibrations, and so on. What is the mass of the caboose?
m1v1= (m1+m2)vf KEbefore= .23KEcollision I don't know how to set up the last equation so that velocities cancel. I am not sure how to set up the kinetice energy equation, please help! thank you! 



#2
Nov109, 10:30 AM

Mentor
P: 40,907





#3
Nov109, 10:35 AM

P: 17

so would i get for kinetic equation:
.5m1(v1^2)= .23(.5)(m1+m2)(m1v1/(m1+m2)) 



#4
Nov109, 10:38 AM

P: 17

freight carcaboose collision finding mass without velocities
nevermind i would actually get: .5(m1)((m1+m2)(vf)/m1) = .23(.5)(m1+m2)vf




#5
Nov109, 10:39 AM

Mentor
P: 40,907

(1) You forgot to square the velocity on the right hand side. (2) The final KE is less. If 23 % of the initial KE is "lost", what's left? 



#6
Nov109, 10:46 AM

P: 17

I'm not sure I understand what you are explaining in part 2..
so my equation would be .5m1(v1^2)= .23(.5)(m1+m2)(m1v1^2/(m1+m2)) 



#7
Nov109, 10:46 AM

P: 17

I still have initial velocity as unknown or final velocity, so i'm not sure how to get rid of that




#8
Nov109, 10:55 AM

Mentor
P: 40,907

Your equation (once you correct the above error) says that the initial KE = 23% of the final KE. Does that make sense? 



#10
Nov109, 11:09 AM

P: 17

so actually I would have to take 23% of the initial kinetic energy




#11
Nov109, 11:10 AM

Mentor
P: 40,907

So the final KE is what percentage of the initial? 



#12
Nov109, 11:14 AM

P: 17

the final is 77% of the initial



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