# Potential and Thermal Energy

1. Aug 30, 2013

### PEZenfuego

1. The problem statement, all variables and given/known data
Starting from rest, a railroad car rolls down a hill 20 m high and hits another identical car at rest. The cars lock together after the collision. What fraction of the first car's change in potential energy is converted into thermal energy in the collision?

2. Relevant equations

ΔUth=mcΔT
Vg=mgh
K=(1/2)mv2

3. The attempt at a solution

My basic strategy here was that kinetic energy will not be conserved (due to the inelastic collision) so I need to find how much kinetic energy was lost and then the "lost" energy can be attributed to thermal energy. I can use conservation of energy equations to figure out how fast the car is going before the collision (v1i)

ΔVg+ΔK=0
mgΔh+(1/2)mv1i2=0
mghi=(1/2)mv1i2

Momentum is conserved, so
mv1i+mv2i=
2mvf.

That means that (1/2)v1i=vf

Then basically, Ki=Kf+Uth Note here, that this is kinetic energy for a different portion of the problem

Uth=Ki-Kf

Uth=(1/2)mv1i2-(1/2)(2m)(vf)2

Uth=(1/2)mv1i2-(1/2)(2m)((1/2)v1i)2

Uth=(1/2)mv1i2-(1/2)(2m)(1/4)v1i2

Uth=(1/4)mv1i2

Uth=(1/2)((1/2)mv1i2)

Uth=(1/2)(mghi)

Uth=(1/2)(Vg)

Anyway, I really feel as though I did something horrendously wrong here. Thanks for your time!

2. Aug 30, 2013

### barryj

Think about the problem this way. Regardless of high high the hill is, the change in PE is changed into KE at the time of collision. So ask yourself, if a car of mass m at some velocity v collides with an identical car of mass m, what is the final velocity of the car. Actually, you have already derived this relationship. So ask how much the KE is changed based on the conservation of momentum.

3. Aug 30, 2013

### PEZenfuego

So is it okay to consider the kinetic energy of the first car before and after the collision and the kinetic energy of the second car before and after the collision separately like I did?

4. Aug 30, 2013

### barryj

The KE of the first car will be (1/2)mv^2 before the collision. What will be the KE of both cars after the collision? The difference should be the loss of KE during the collision. So the PE of the first car will be converted into KE at the time of collision, then the KE after collision will, hopefully, less than the KE before the collision. Now you can work the percentages.

5. Aug 30, 2013

### PEZenfuego

That sounds like exactly what I did! So do you think my answer is correct?

6. Aug 30, 2013

### barryj

"What fraction of the first car's change in potential energy is converted into thermal energy in the collision?"

I don't see your answer. I would think it should be something like 1/2, or 3/5. or.....

7. Aug 30, 2013

### PEZenfuego

I got 1/2. Sorry, it is there at the bottom, just not in words.

8. Aug 30, 2013

### barryj

I don't think so. The initial KE is (1/2)mv^2. What is the final KE of the total mass (2m)? Now what is the ratio?
Remember that you have already calculated the velocity after the collision as it related to the velocity bvefore the colision.

9. Aug 30, 2013

### PEZenfuego

Now I'm confused again

Okay, step by step. Follow along and tell me where my mistake is. Momentum is conserved. Kinetic energy is not.

Step 1: The potential energy of the car an earth turns completely into kinetic energy of the car (ignoring kinetic energy of the earth).

Step 2:

mv1i+mv2i=mvf+mvf

Therefore, (1/2)v1i=vf

Step 3:

Kinetic energy before the collision: (1/2)(m)(v1i2)
Kinetic energy after the collision: (1/2)(2m)(vf)2

Now we plug in the equation we got from step 1.

Kinetic energy after the collision:
(1/2)(2m)((1/2)v1i)2
(1/2)(2m)(1/4)(v1i2)
(1/4)m(v1i2)

Kinetic energy before the collision: (1/2)(m)(v1i2)
Kinetic energy after the collision: (1/4)(m)(v1i2)

Step 4:

So, half of the kinetic energy is lost. Therefore, half of the kinetic is turned into thermal energy. Since the kinetic energy before the collision is equal to the potential energy, that implies that half of the potential energy is transformed into thermal energy.

I hope that makes sense. Where is my mistake?

10. Aug 30, 2013

### barryj

OOPS, my mistake. I think you are correct, 1/2

11. Aug 30, 2013

### PEZenfuego

Thanks for looking over it. One concern was that the height doesn't factor into this ratio. That seems a little strange to me, but I can accept it as true. This isn't necessarily a difficult problem, but it was difficult to keep track of the variables as there are essentially three stages to the problem rather than a simple before and after.

12. Aug 30, 2013

### barryj

Regardless of the heighth, the change in PE will be converted to some amount of KE. Then you have to figure out the ratio of the initial KE and the final KE as you did.