Physics homework help - that's right I admit it upfront

[Matthiasa]

This is why we don't ask physics questions here, although a few did get it correct followed by others saying it wasn't ignoring significant figures. The statement of "collided and hooked" is trying to indicate that it is an inelastic collision. In practice means the difference in KE which gets converted primarily into heat, sound, and or KE in other directions with the parts knocked off. Noting that the energy used in deformation of a material by x amount gets converted into heat.

Also note that is a rough approximation of what happens.

 

[CPA]

The correct answer for part 1 is 1.415m/s

The correct answer for part 2 is 0 KE energy loss for the system
And KE of train 1before impact minus KE of train1 after the impact.

the correct answer for 1 was 1.32. I am sure of that.

 

[disappoint]

The question did not have them labeled, so I didn't in my OP.

I'm not blaming you.

 

[CPA]

This is why we don't ask physics questions here, although a few did get it correct followed by others saying it wasn't ignoring significant figures. The statement of "collided and hooked" is trying to indicate that it is an inelastic collision. In practice means the difference in KE which gets converted primarily into heat, sound, and or KE in other directions with the parts knocked off. Noting that the energy used in deformation of a material by x amount gets converted into heat.

Also not that is a rough approximation of what happens.

uhhhhhhhh...droooool

 

[CPA]

I'm not blaming you.

I know. Just letting you know why we didn't associate the second question to be specific to the first locomotive only. In any case, thanks for the explanation. I'm going to see if my daughter ever got the correct answer from her teacher.

 

[sontakke]

This forum disappoints. Very very important question was never asked!

 

[DigDog]

K = ½mv²

everybody knows that, and i totally knew that before i read this thread and not copy-pasta from someone else.

 

[Paratus]

the correct answer for 1 was 1.32. I am sure of that.

Ok. Then ignore the numbers I calculated. One of the other numbers in the OP must be different. Anyway if your daughter just follows the process I did she'll get the right answer.

 

[Matthiasa]

I know. Just letting you know why we didn't associate the second question to be specific to the first locomotive only. In any case, thanks for the explanation. I'm going to see if my daughter ever got the correct answer from her teacher.

That's because the second question would almost never be just the first locomotive only... The value you used to 3 significant figures was wrong as was already pointed out.

 

[BoberFett]

One of my biggest take-aways from my college classes is that physics is so much easier if you ignore friction, air resistance, deformation and\or everything happens in a vacuum

Assume the cow is spherical and in a vacuum...

 

[KayGee]

Ok. Then ignore the numbers I calculated. One of the other numbers in the OP must be different. Anyway if your daughter just follows the process I did she'll get the right answer.

The numbers in the OP are correct; your process is wrong. In an inelastic collision ("collided and hooked"), linear momentum is conserved; kinetic energy is not.

 

[Paratus]

The numbers in the OP are correct; your process is wrong. In an inelastic collision ("collided and hooked"), linear momentum is conserved; kinetic energy is not.

Well now don't I feel like a dumbass.

You're correct. It's an inelastic collision.

Sorry, CPA 1.32 is correct.

 

[Aikouka]

Then KE lost = (45000/2 * 1.06^2 + 0) - (135000/2 * 0.35333^2) = 16854.02 (Same energy lost, different reference frame)

I'm kind of confused, because I thought you were saying that I'm wrong... but that's the same answer that I got.

 

[Jaepheth]

I'm kind of confused, because I thought you were saying that I'm wrong... but that's the same answer that I got.

No, I'm agreeing with you; your calculation was correct. I meant I shouldn't have glossed over the math.
My original post was wrong because while energy is conserved, kinetic energy is lost when the trains "hook" together.

It just doesn't matter how fast a third party (the question writer) is observing at, because it's the relative speed between the two trains where your answer is going to come from.

 

[gururu2]

I find everything about physics interesting except for the hypothetical questions that render any level of instruction inadequate.

 

[Exterous]

Assume the cow is spherical and in a vacuum...

 

[Daedalus685]

I find everything about physics interesting except for the hypothetical questions that render any level of instruction inadequate.

http://www.smbc-comics.com/?id=2783

Also, I'm glad my physics degree allowed me to solve the question correctly (using momentum conservation)... even if I'm way too late to the party to be helpful.

 

[DrPizza]

Just saw this thread. KayGee is correct - there's a conservation of energy, but not a conservation of kinetic energy. Really cool demonstration to show where the energy in a collision goes. Macroscopically, there's no such thing as a perfectly elastic collision. ALL collisions lose kinetic energy, usually in the form of heat. (I point out "macroscopically", because "heat" in matter is kinetic energy of the molecules in that matter.) If perfectly elastic macroscopic collisions were possible, then perpetual motion machines would be possible.
But, as a demonstration, get two 1 kilogram steel or iron spheres. Smash them together in your hands as hard as you can. Don't get a finger between them. (Been there, done that.) Now, have someone hold a piece of paper so that it's dangling in the air. Smash the spheres together, sandwiching the paper. Repeat a couple of times. The smell of burnt paper is unmistakable. If you look closely, you'll see burn marks on the paper.

Incidentally, if you get a finger between them, then subsequent smashes involve blood splatter. (Been there, done that, to half a dozen students' papers before I noticed. Yes, it hurt. But, I tried not to let anyone realize I smashed my finger.)

 

[Brian Stirling]

For #1, I got ... [ 45000 * 2.03 + 90000 * .97 ] / (45000 + 90000) = 1.323..

For #2, I got ... ([.5 * 45000 * 2.03^2] + [.5 * 90000 * .97^2]) - [.5 * 135000 * 1.323^2] = 16854

(Note, the answer won't be the exact same as what I have above, but that's because I inserted the equation from #1 in place of 1.323, which gave a more precise answer.)

Yep, that's what I got...


Brian

 

[Imp]

Got 54% in first year dynamics... probably shouldn't try to help.