- #1
CaptainOfSmug
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Homework Statement
A 1200-kg car is backing out of a parking space at5.5m/s . The unobservant driver of a 1900-kgpickup truck is coasting through the parking lot at a speed of 4.0m/s and runs straight into the rear bumper of the car.
What is the change in internal energy of the two-vehicle system if the velocity of the pickup is 1.5 m/s backward after they collide?
Calculate the coefficient of restitution
Homework Equations
momentum
kinetic energy
internal energy
good ol fashion addition :P[/B]
The Attempt at a Solution
Okay so here's what I did first, I've been stuck on this for about an hour now and can't quite figure out where I went wrong.
I started by finding the final velocity of the car by using the momentum equation:
1900(4.0)+1200(-5.5)=1900(-1.5)+1200(vf)
vf=3.2083333~3.21m/s
I then went ahead and calculated all the kinetic energies (I'm calling the pickup 1 and the car 2)
K1i=.5(1900)(4.02
=15200J
K2i=.5(1200)(-5.5)2
=18150J[/B]
K1f=.5(1900)(-1.5)2
=2137.5J
K2f=.5(1200)(3.208333)2
=6176.028833J
Then I found the combined initial kinetic energy (now that I'm writing this I'm not sure that it matters)
K12i=.5(1900+1200)(-5.5-4.0)2
=139887.5J
Then the final kinetic energy:
K12f=(1900+1200)(2.208+1.5)2
=34361.01944J
I then found the change in kinetic energy:
ΔK=K12f-K12i
=-105526.4806J
Then from my book I read ΔE=-ΔK
So
ΔE= 105526.48J
Now from before had where I did each kinetic equation individually and added them up then subtracted the final from the initial I got ΔK=-25036.47J
I don't understand when I tried to combine them it didn't work they don't equal the same thing?? I hope one of the answers is correct... or close?
And as for calculating the coefficient of restitution
I just took the the combined final velocities by the initial and got 0.50 which means its inelastic.
Anyways, if someone could check my work and maybe help figure out what I'm doing wrong it would be greatly appreciated!