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Freight car-caboose collision finding mass without velocities

by hangten1039
Tags: carcaboose, collision, freight, mass, velocities
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hangten1039
#1
Nov1-09, 10:24 AM
P: 17
A 34000-kg 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!
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Doc Al
#2
Nov1-09, 10:30 AM
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Quote Quote by hangten1039 View Post
m1v1= (m1+m2)vf
OK.
KEbefore= .23KEcollision
Rethink this one more carefully.

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!
What's the definition of KE? Write vf in terms of v1 (using your first equation).
hangten1039
#3
Nov1-09, 10:35 AM
P: 17
so would i get for kinetic equation:
.5m1(v1^2)= .23(.5)(m1+m2)(m1v1/(m1+m2))

hangten1039
#4
Nov1-09, 10:38 AM
P: 17
Freight car-caboose collision finding mass without velocities

nevermind i would actually get: .5(m1)((m1+m2)(vf)/m1) = .23(.5)(m1+m2)vf
Doc Al
#5
Nov1-09, 10:39 AM
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Quote Quote by hangten1039 View Post
so would i get for kinetic equation:
.5m1(v1^2)= .23(.5)(m1+m2)(m1v1/(m1+m2))
Two problems:
(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?
hangten1039
#6
Nov1-09, 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))
hangten1039
#7
Nov1-09, 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
Doc Al
#8
Nov1-09, 10:55 AM
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Quote Quote by hangten1039 View Post
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))
You still have the square wrong on the RHS. The entire vf term must be squared.

Your equation (once you correct the above error) says that the initial KE = 23% of the final KE. Does that make sense?
Doc Al
#9
Nov1-09, 10:56 AM
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Quote Quote by hangten1039 View Post
I still have initial velocity as unknown or final velocity, so i'm not sure how to get rid of that
Once you get the KE equation correct, the velocity cancels.
hangten1039
#10
Nov1-09, 11:09 AM
P: 17
so actually I would have to take 23% of the initial kinetic energy
Doc Al
#11
Nov1-09, 11:10 AM
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Quote Quote by hangten1039 View Post
so actually I would have to take 23% of the initial kinetic energy
You start with one dollar, but then lose 23 cents. What's left?

So the final KE is what percentage of the initial?
hangten1039
#12
Nov1-09, 11:14 AM
P: 17
the final is 77% of the initial
Doc Al
#13
Nov1-09, 11:18 AM
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Quote Quote by hangten1039 View Post
the final is 77% of the initial
Good. Use that to write your KE equation.


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