Physics problem: momentum and its conservation. Help with lingo!

Posts: 8761 - Karma: 139

Karma: 139

Posts: 8761

Crazy Fool

So I have to present a solution to a physics problem in class on wednesday and I am a little confused on some of the terminology on the problem. I'll put the problem then ask my questions. any help would be greatly appreciated. FYI, I can't ask my teacher because I haven't been to school in over a week.

A 975 kg two-stage rocket is traveling at a speed of 5.8x10^3 m/s (or 5800 m/s) with respect to Earth when a pre-designed explosion separates the rocket into two sections of equal mass that then move at a speed of 2.20x10^3 m/s (2200 m/s) relative to each other along the original line of motion.

a)What are the speed and direction of each section (relative to Earth) after the explosion?

b)How much energy was supplied by the explosion? [Hint: What is the change is KE (kinetic energy) as a result of the explosion]

***The first road block I am at is the phrase "relative to each other along the original line of motion." Does this mean that the combined speed of the two rockets after the explosion is 2200 m/s? Or does this mean that it is a vector and the velocity up added to the velocity down is equal to 2200m/s I can't start the problem till I figure this out.

Nov 19 2007 4:40PM

1 0

thegreyarea

Posts: 3265 - Karma: 34

Karma: 34

Posts: 3265

Addict

"relative to each other along the original line of motion."
This means that assuming the rocket is moving in a vertical line, perpendicular to the earth, when the explosion goes off, half the rocket shoots upwards and half of it shoots down, still on that same vertical line.

Relative to each other means that if you were sitting on one of those rockets, the other rocket is moving away from you at 2200 m/s.

Nov 19 2007 4:46PM

0 0

kepler.

Posts: 8761 - Karma: 139

Karma: 139

Posts: 8761

Crazy Fool

Thanks, but how am i supposed to use the conservation of momentum to solve for the two pieces' velocities after the explosion? Should I treat them as two different bodies both have a speed of 2200m/s or should one be 2200m/s and the other be v-2200? Try to be easy on me considering I'm teaching momentum to myself out of the book

Nov 19 2007 5:05PM

0 0

kepler.

Posts: 8761 - Karma: 139

Karma: 139

Posts: 8761

Crazy Fool

Ok so this is to anyone who has looked at the problem. I got the velocities of the two objects. They are 6900 m/s and 4700m/s. But now how do I find the change in energy? I keep getting no change in KE

Nov 19 2007 5:20PM

0 0

kepler.

Posts: 8761 - Karma: 139

Karma: 139

Posts: 8761

Crazy Fool

bump. i'm not asking you to do it for me, just some help.

Nov 19 2007 5:42PM

0 0

noj1080

Posts: 355 - Karma: 69

Karma: 69

Posts: 355

Loner

Ok so heres what I just did:

First you find KE for the initial Rocket KE = 1/2mv^2

Then find the KE after the rocket splie KE = 1/2mv^2 + 1/2mv^2
you need to use the mass of half of the original rocket once with each velocity that you found in the first step. Then the diffenece in the to KE is your total energy change.

Good Luck

Nov 19 2007 5:47PM

0 0

thegreyarea

Posts: 3265 - Karma: 34

Karma: 34

Posts: 3265

Addict

haha right after I made a picture for you.

The change in kinetic energy is KEnew - KEorig.
KEorig = (976kg) * (5800m/s)^2 / 2
KEnew = (976kg/2)*(6900m/s)^2 / 2 + (976kg/2)*(4700m/s)^2 / 2
or simplified
KEnew = (976kg) * ((6900m/s)^2 + (4700m/s)^2) / 2

You could make this even easier and do it relative to the explosion. This would make...
KEorig = (976kg) * (0 m/s)^2 / 2
KEorig = 0
KEnew = (976kg/2)*(1100 m/s)^2 / 2 + (976kg/2)*(-1100/s)^2 / 2
KEnew = (976kg) * ((1100 m/s)^2 + (-1100 m/s)^2) / 2
KEnew = (976kg) * 2 * (1100 m/s)^2 / 2
KEnew = (976kg) * (1100 m/s)^2

So, your answer is (976kg) * (1100 m/s)^2

Nov 19 2007 5:49PM

0 0

thegreyarea

Posts: 3265 - Karma: 34

Karma: 34

Posts: 3265

Addict

Side note to the above answer, it doesn't matter if it is relative to the earth or to the explosion. The answer will be the same.

Nov 19 2007 5:50PM

0 0

noj1080

Posts: 355 - Karma: 69

Karma: 69

Posts: 355

Loner

^^ you explained that soo much better than me.... listen to him

Nov 19 2007 5:53PM

0 0

kepler.

Posts: 8761 - Karma: 139

Karma: 139

Posts: 8761

Crazy Fool

Oh my fucking god! I forgot to square the velocities! I feel like an idiot. Thank you so much for taking the time to write all that down. I probably would have never figured out how to solve it if you guys didn't.

Thanks again!

Nov 19 2007 5:56PM

0 0

shin-bang

Posts: 4100 - Karma: 6,997

Karma: 6,997

Posts: 4100

No Life

hahahahha just google searched this problem for some help with my physics homework and this came up.

thanks for the help guys (even tho its been 12 years from the original thread) glad to be a part of NS.

Feb 27 2020 10:39AM

5 0

Lonely

Posts: 6739 - Karma: 61,637

Karma: 61,637

Posts: 6739

No Life

kepler.So I have to present a solution to a physics problem in class on wednesday and I am a little confused on some of the terminology on the problem. I'll put the problem then ask my questions. any help would be greatly appreciated. FYI, I can't ask my teacher because I haven't been to school in over a week.

A 975 kg two-stage rocket is traveling at a speed of 5.8x10^3 m/s (or 5800 m/s) with respect to Earth when a pre-designed explosion separates the rocket into two sections of equal mass that then move at a speed of 2.20x10^3 m/s (2200 m/s) relative to each other along the original line of motion.

a)What are the speed and direction of each section (relative to Earth) after the explosion?

b)How much energy was supplied by the explosion? [Hint: What is the change is KE (kinetic energy) as a result of the explosion]

***The first road block I am at is the phrase "relative to each other along the original line of motion." Does this mean that the combined speed of the two rockets after the explosion is 2200 m/s? Or does this mean that it is a vector and the velocity up added to the velocity down is equal to 2200m/s I can't start the problem till I figure this out.

"Alright students we are just going to ignore air resistance in this problem"

Feb 27 2020 11:42AM

3 0

Evantheshredster

Posts: 429 - Karma: 935

Karma: 935

Posts: 429

Ridiculous

Lonely"Alright students we are just going to ignore air resistance in this problem"

the question didnt clarify if the rocket was in space yet at the time of detatch. Because if it's in space then there wouldn't be any air resistance. good point though- if it was still in the atmosphere, that would definitely come into play.

Feb 27 2020 1:49PM

0 0

Lonely

Posts: 6739 - Karma: 61,637

Karma: 61,637

Posts: 6739

No Life

Evantheshredsterthe question didnt clarify if the rocket was in space yet at the time of detatch. Because if it's in space then there wouldn't be any air resistance. good point though- if it was still in the atmosphere, that would definitely come into play.

I was more just making a joke about physics and engineering problems in school ignore air resistance.

Feb 27 2020 2:53PM

0 0