How Does Collision Affect Internal Energy in a Two-Vehicle System?

[CaptainOfSmug]

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(v_f)
v_f=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)
K_1i=.5(1900)(4.0²
=15200J
K_2i=.5(1200)(-5.5)²
=18150J[/B]
K_1f=.5(1900)(-1.5)²
=2137.5J
K_2f=.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)
K_12i=.5(1900+1200)(-5.5-4.0)²
=139887.5J
Then the final kinetic energy:
K_12f=(1900+1200)(2.208+1.5)²
=34361.01944J

I then found the change in kinetic energy:
ΔK=K_12f-K_12i
=-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!

 

[vela]

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

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(v_f)
v_f=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)
K_1i=.5(1900)(4.0²
=15200J
K_2i=.5(1200)(-5.5)²
=18150J
K_1f=.5(1900)(-1.5)²
=2137.5J
K_2f=.5(1200)(3.208333)²
=6176.028833J

Up to here is fine.

Then I found the combined initial kinetic energy (now that I'm writing this I'm not sure that it matters)
K_12i=.5(1900+1200)(-5.5-4.0)²
=139887.5J
Then the final kinetic energy:
K_12f=(1900+1200)(2.208+1.5)²
=34361.01944J

This doesn't make sense. An object with mass $m$ moving with speed $v$ has kinetic energy $\frac 12 mv^2$. I don't know what $\frac 12 (m_1+m_2)(v_1+v_2)^2$ would represent in this problem. You don't have a body with combined mass $m_1+m_2$ moving at a speed $v_1+v_2$.

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

This was the correct way to calculate the change.

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!

 

[BvU]

The expressions for K12i and K12f are not correct. The combined mass does not move at the sum of the speeds.