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

[CPA]

Actually, it was my daughter's homework problem. Two part - we got the first question correct, but were stumped on the second. BTW, this was submitted yesterday, so don't give me any lip about "do your own homework". I'm just curious as to how to get the answer.

In any case, the numbers may be off because, well, it was two days ago and I don't have a photographic memory, but I think these are close.

Given:

A locomotive at 45,000kg was moving in a direction at 2.03m/s when it collided and hooked into two attached locomotives at 90,000kg going in the same direction as the first at .97m/s.

1) What was the new velocity after the collision?

The correct answer was 1.32m/s (again, the numbers above may be slightly off, but I'm pretty sure from the memory this was the correct answer for this question)

2) What was the loss of kinetic energy after collision?

This is the question that had us stumped. All of the examples we could find online were for one object moving and the second not moving. I won't go through the various ways I attempted to calculate this, just know, each answer we submitted was wrong. So, I'm just curious how to figure out loss of kinetic energy from a collision of two objects going in the same direction at different speeds.

Thanks for the help, and a cookie to whoever sets us right.

 

[PokerGuy]

Thanks for the reminder of why I chose to go into a field that does not involve this kind of stuff. I'll leave it to those a lot smarter than I

 

[CPA]

Thanks for the reminder of why I chose to go into a field that does not involve this kind of stuff. I'll leave it to those a lot smarter than I

I know, right. Even though my job is heavy in math, I just couldn't grasp this one. I was so pissed with myself, I threw my hands up in disgust and walked off. LOL. Of course, after about 5 minutes, I started searching again, but after another 30 minutes or so, just couldn't figure it out. Unfortunately, she has to do this work online and only gets 7 attempts at an answer, each attempt reducing the points awarded for a correct answer. I think we go to the 6th attempt and realized it was a lost cause.

 

[disappoint]

Seems pretty straight forward. I'm surprised you couldn't get the answer. Ke=1/2mV² so just calculate the kinetic energy before the collision, after the collision and subtract right? Seems too easy maybe I'm missing something.

 

[Aikouka]

From what I found, you need the equation K = ½mv²

 

[Paratus]

Actually, it was my daughter's homework problem. Two part - we got the first question correct, but were stumped on the second. BTW, this was submitted yesterday, so don't give me any lip about "do your own homework". I'm just curious as to how to get the answer.

In any case, the numbers may be off because, well, it was two days ago and I don't have a photographic memory, but I think these are close.

Given:

A locomotive at 45,000kg was moving in a direction at 2.03m/s when it collided and hooked into two attached locomotives at 90,000kg going in the same direction as the first at .97m/s.

1) What was the new velocity after the collision?

The correct answer was 1.32m/s (again, the numbers above may be slightly off, but I'm pretty sure from the memory this was the correct answer for this question)

2) What was the loss of kinetic energy after collision?

This is the question that had us stumped. All of the examples we could find online were for one object moving and the second not moving. I won't go through the various ways I attempted to calculate this, just know, each answer we submitted was wrong. So, I'm just curious how to figure out loss of kinetic energy from a collision of two objects going in the same direction at different speeds.

Thanks for the help, and a cookie to whoever sets us right.

You can calculate the total KE of the system before the impact using:
.5(MassTrain1)(VelocityTrain1)^2 +.5(MassTrain2)(VeolcityTrain2)^2

And then compare it with the after impact:

.5(MT1+MT2)(combined velocity)^2

I'm not sure what method you used to solve the first part to get the combined velocity but in the absence of friction or deformation or something else the KE before the impact must equal the KE after the impact. So I would expect using the equations above they would come out equal. If it doesn't then we're missing something or you've made a mistake. (The numbers above don't seem to match)

Good luck!

 

[Jaepheth]

Since it doesn't specify any sort of height or the possibility of movement in the up/down direction then all the energy is kinetic.

Conservation of energy would dictate that there is no loss of energy in the system as a whole. So the answer would be 0 EDIT: Incorrect, see later posts

Seems like a trivial answer, and I learned in college to be very, very wary of interpreting a question in such a way that the answer is trivial; so they could be asking what is the loss of kinetic energy for the first car only (e.g. how much energy was transferred to the slower two cars in the collision)

E = 1/2 mv^2

Before collision:
45,000 / 2 * 2.03^2 = 92,720.25

After collision:
45,000 / 2 * 1.32^2 = 39,204

A difference of 53,516.25

 

[CPA]

Seems pretty straight forward. I'm surprised you couldn't get the answer. Ke=1/2mV² so just calculate the kinetic energy before the collision, after the collision and subtract right? Seems too easy maybe I'm missing something.

Yeah, we did that, which would give you 92,700+42,300-117,450 = 17,750
That answer didn't take. Maybe the system was incorrect or maybe we were fat-fingering something.

 

[CPA]

Since it doesn't specify any sort of height or the possibility of movement in the up/down direction then all the energy is kinetic.

Conservation of energy would dictate that there is no loss of energy in the system as a whole. So the answer would be 0

Seems like a trivial answer, and I learned in college to be very, very wary of interpreting a question in such a way that the answer is trivial; so they could be asking what is the loss of kinetic energy for the first car only (e.g. how much energy was transferred to the slower two cars in the collision)

E = 1/2 mv^2

Before collision:
45,000 / 2 * 2.03^2 = 92,720.25

After collision:
45,000 / 2 * 1.32^2 = 39,204

A difference of 53,516.25

Why do you not include the weight of the second set of locomotives (90,000kg) that are going at .97m/s. Isn't there kinetic energy involved with them?

 

[Exterous]

in the absence of friction or deformation or something else

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

 

[Paratus]

Since it doesn't specify any sort of height or the possibility of movement in the up/down direction then all the energy is kinetic.

Conservation of energy would dictate that there is no loss of energy in the system as a whole. So the answer would be 0

Seems like a trivial answer, and I learned in college to be very, very wary of interpreting a question in such a way that the answer is trivial; so they could be asking what is the loss of kinetic energy for the first car only (e.g. how much energy was transferred to the slower two cars in the collision)

E = 1/2 mv^2

Before collision:
45,000 / 2 * 2.03^2 = 92,720.25

After collision:
45,000 / 2 * 1.32^2 = 39,204

A difference of 53,516.25

That's a good point. You really can't solve the first part without knowing KE is conserved across the system. So the second part might actually mean how much KE was transferred from the 1st train to the 2nd.

Why do you not include the weight of the second set of locomotives (90,000kg) that are going at .97m/s. Isn't there kinetic energy involved with them?

You might consider answering the question the loss of KE for the system was 0
And the loss for train 1 was what Jaepheth shows using the corrected numbers.

 

[Aikouka]

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.)

 

[Paratus]

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

Good news for me. At work all the action takes place in a vacuum.

 

[disappoint]

Yeah, we did that, which would give you 92,700+42,300-117,450 = 17,750
That answer didn't take. Maybe the system was incorrect or maybe we were fat-fingering something.

Why do you not include the weight of the second set of locomotives (90,000kg) that are going at .97m/s. Isn't there kinetic energy involved with them?

I think you misunderstood the question. It's not asking for loss of Ke for both trains. Just the first one. The one that collided.

This is why we label things. Locomotive A collided into locomotives B and C...etc...

Not "A locomotive collided into another locomotive attached to another one filled with clowns shooting silly string at each other. Calculate how much silly string they would need to cover some guy sitting 3 seats over in a Spiderman costume with a web of silly string and lies. Lies and slander."

 

[CPA]

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.)

1.32 was provided and accepted. The system they use doesn't require exact precision. Usually only out to one or two decimals. So, using 1.32 I got 17,550 which was not correct. Maybe for purposes of the kinetic energy question, we needed to go out 3 decimals. If that's right, then the system is flawed. Why allow only two decimals for the first answer if three are required for the second.

 

[CPA]

I think you misunderstood the question. It's not asking for loss of Ke for both trains. Just the first one. The one that collided.

hmmm...I just assumed it was asking in total. It was not specific.

 

[BarkingGhostar]

I think you misunderstood the question. It's not asking for loss of Ke for both trains. Just the first one. The one that collided.

This is why we label things. Locomotive A collided into locomotives B and C...etc...

Not "A locomotive collided into another locomotive attached to another one filled with clowns shooting silly string at each other. Calculate how much silly string they would need to cover some guy sitting 3 seats over in a Spiderman costume with a web of silly string and lies. Lies and slander."

We have a winner. It is in the wording of the second problem in that it really is asking the loss of KE of the original loco.

 

[CPA]

I think you misunderstood the question. It's not asking for loss of Ke for both trains. Just the first one. The one that collided.

This is why we label things. Locomotive A collided into locomotives B and C...etc...

Not "A locomotive collided into another locomotive attached to another one filled with clowns shooting silly string at each other. Calculate how much silly string they would need to cover some guy sitting 3 seats over in a Spiderman costume with a web of silly string and lies. Lies and slander."

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

 

[Paratus]

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.

 

[OutHouse]

<<<< Runs out of thread like richard pryor running down the street on fire.

 

[Jaepheth]

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.)

and that's why you don't skip the math when you're a long time out of physics practice.

Formulae: http://hyperphysics.phy-astr.gsu.edu/hbase/inecol.html

Discussion: http://physics.stackexchange.com/qu...energy-be-conserved-in-an-inelastic-collision

 

[herm0016]

1.32 was provided and accepted. The system they use doesn't require exact precision. Usually only out to one or two decimals. So, using 1.32 I got 17,550 which was not correct. Maybe for purposes of the kinetic energy question, we needed to go out 3 decimals. If that's right, then the system is flawed. Why allow only two decimals for the first answer if three are required for the second.

Why? because the dolts with education degrees don't understand Significant Digits and are writing the questions from some book.

 

[Aikouka]

and that's why you don't skip the math when you're a long time out of physics practice.

Formulae: http://hyperphysics.phy-astr.gsu.edu/hbase/inecol.html

Discussion: http://physics.stackexchange.com/qu...energy-be-conserved-in-an-inelastic-collision

Your formula example is for an animate object colliding with an inanimate object?

(To be clear, I haven't done physics since high school.)

 

[Jaepheth]

Your formula example is for an animate object colliding with an inanimate object?

(To be clear, I haven't done physics since high school.)

Doesn't matter; there are reference frames in which either mass could be considered at rest.
If we have the observer moving at the same velocity as the slower cars then all the initial velocities just get reduced by .97

momentum: 45000*1.06 + 0 = 47700 then 47700/135000 = .353333

then if we slow the observer back down we'd add back the .97 and get .353333 + .97 = 1.32333 (same final velocity, different reference frame)

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

So the KE is lost (transferred to heat/sound/deformation/etc) because it takes work to manipulate the mass' velocities so that they match after the collision; otherwise you'd have an elastic collision where the trains just bounce off of each other.

 

[Paratus]

Ok last time:

Part 1) KEt1 + KEt2 = KEt(1+2)

T1---> T2-> = T1T2-->

KEt1 = .5(45000kg)(2.03m/s)^2 = 92720.25J
KEt2 = .5(90000kg)(.97m/s)^2 = 42340.5J

KEt1 +KEt2 = 135,060.75J

135,060.75J = .5(T1 +T2)(V)^2
135,060.75J = .5(45000+90000)(V)^2
V = 1.415m/s

Part 2

Kinetic energy of the system is conserved so kinetic energy loss of the entire system is 0

Kinetic energy lost by Train 1 = KEt1 before - KEt1after

KEt1 before impact = .5(45000kg)(2.03m/s)^2 = 92720.25J

KEt1 after impact = .5(45000kg)(1.415m/s)^2 = 45020.25J

Kinetic energy lost by Train1 = 92720.25J - 45020.25J = 47,700J


By providing both answers to part 2 you protect yourself from the poorly written question.