Question from CSSA trial - acceleration of a rocket (1 Viewer)

[chickenntaters]

17.b>
A rocket of initial mass 2000 tonnes, produces a constant thrust of 80000000N during liftoff by expelling 1500kg of exhaust gases per second.
Calculate: the net force acting on an 85kg astronaut 30s after liftoff


I tried to use the equation a=(T-mg)/m so that a=(8x10^7 - 1955000*9.8)/1955000. Then used F=am so that F=31.12*85 to get a final answer of 2645.2N. However the actual answer is 3478.26N.

Can anyone tell me what i've done wrong?

 

[Matthaeus]

If you did the question using the formula a = T/m (ie without the -mg) and then did what u did, you would get the final answer.

To understand why, think of it this way, first u said a = (T - mg)/m , when you plug this back into the f=ma for the astronaut you effectively forget about the force due to gravity because you get rid of it in the a = (T - mg)/m with the -mg bit.

so for the second part, for the acceleration on the astronaught if u did f = 85 * (your a + 9.8) you get the correct answer.

 

[chickenntaters]

Oh, ok, i see - thanks

What about with a question like this: [from the NEAP trial]
A rocket engine provides a thrust of 380,000N to a rocket with total mass of 29830kg prior to lift off and would use 25315kg of fuel during the flight. Assuming the thrust is kept constant, calculate the final acceleration of this rocket immediately before the fuel has been used. [then in part c i have to find the g-force experienced by the astronaut]

In this case, would i use a = (T - mg)/m to calculate the acceleration, or should i just stick with a=T/m. The Jacaranda physics text book uses the first equation for an example just like this one [page 29], but i want to make sure i can actually use that or if the book is wrong. If i CAN use it, how do i tell which situation i should use a = (T - mg)/m for, and which i should use a=T/m for.

Sorry for being such a pain, but i want to make sure i can do this.

 

[zeropoint]

17.b>
A rocket of initial mass 2000 tonnes, produces a constant thrust of 80000000N during liftoff by expelling 1500kg of exhaust gases per second.
Calculate: the net force acting on an 85kg astronaut 30s after liftoff


I tried to use the equation a=(T-mg)/m so that a=(8x10^7 - 1955000*9.8)/1955000. Then used F=am so that F=31.12*85 to get a final answer of 2645.2N. However the actual answer is 3478.26N.

Can anyone tell me what i've done wrong?

I think you can safely call the writer of that test a moron. You arrived at the correct answer, albeit for incorrect reasons. The net force on the astronaut is the sum of the forces acting on him, that is, the normal force exerted by the chair minus his weight.

ΣF = n - mg

In this case the magnitude of the normal force is equal to magnitude of the weight force plus the inertial force due to the acceleration a of the spacecraft, therefore

ΣF = ma + mg - mg = ma

where m is the mass of the astronaut and a is the acceleration, calculated as you described. The correct answer turns out to be 2645.26 N.

 

[zeropoint]

Oh, ok, i see - thanks

What about with a question like this: [from the NEAP trial]
A rocket engine provides a thrust of 380,000N to a rocket with total mass of 29830kg prior to lift off and would use 25315kg of fuel during the flight. Assuming the thrust is kept constant, calculate the final acceleration of this rocket immediately before the fuel has been used. [then in part c i have to find the g-force experienced by the astronaut]

In this case, would i use a = (T - mg)/m to calculate the acceleration, or should i just stick with a=T/m. The Jacaranda physics text book uses the first equation for an example just like this one [page 29], but i want to make sure i can actually use that or if the book is wrong. If i CAN use it, how do i tell which situation i should use a = (T - mg)/m for, and which i should use a=T/m for.

Sorry for being such a pain, but i want to make sure i can do this.

If you just use Newton's Second Law properly you will never get it wrong. For the rocket

ΣF = T - Mg = Ma

so

a = (T - Mg)/M

the correct answer is 74.3639 m/s/s. The equation a = T/m makes no sense, unless the rocket is in free space without any gravitational force.
For the astronaut (who is accelerating at the same rate as the rocket)

ΣF = n - mg = ma

so

n = mg + ma = m(a + g)

 

[xiao1985]

zero point: i believe the problem is much more complicated than this... after 30 seconds of acceleration, the rocket can be sufficiently far form the earth's gravitational field and hence resultin in a different g...

 

[Rorix]

i cant be bothered to plug the numbers in but from looking at it the gravitational attraction would still be pretty strong

plus that rocket has a lot of nonfuel mass

imo the question is just a bit crap

 

[zeropoint]

zero point: i believe the problem is much more complicated than this... after 30 seconds of acceleration, the rocket can be sufficiently far form the earth's gravitational field and hence resultin in a different g...

Considering that the acceleration due to gravity decreases by something like 1 part per million for every 3 metres of altitude, the difference is almost negligible.

 

[xiao1985]

Considering that the acceleration due to gravity decreases by something like 1 part per million for every 3 metres of altitude, the difference is almost negligible.

uhm yea true... in fact i was thinkin to neglect the g change at the first place... when i got the wrong answer, i thought the marker must have taken it into account, so i was stuck with a bunch of mathematics eqn with a bunch of unknowns... =p

 

[Matthaeus]

I think you can safely call the writer of that test a moron. You arrived at the correct answer, albeit for incorrect reasons.

If youre talking to me bout the a = T/m, i knew that the equation implicitly took care of the gravitation forces and thrust on the astronaut, and explained that as such.

I should've stated for other questions it would be much more useful to draw force diagrams to answer the questions rather than quote a formula and DO NOT use a = T/m for acceleration of a rocket.

If there is anymore confusion then just ignore my first post, i tend to take great shortcuts.

And about the decrease in gravity idea, its a HSC question, i dont think any marker would take the decrease of gravity into account.

 

[zeropoint]

If youre talking to me bout the a = T/m, i knew that the equation implicitly took care of the gravitation forces and thrust on the astronaut, and explained that as such.

I should've stated for other questions it would be much more useful to draw force diagrams to answer the questions rather than quote a formula and DO NOT use a = T/m for acceleration of a rocket.

If there is anymore confusion then just ignore my first post, i tend to take great shortcuts.

And about the decrease in gravity idea, its a HSC question, i dont think any marker would take the decrease of gravity into account.

Sounds to me as though you've got it completely wrong. The correct answer is 2645.2N, not 3478.26N.

 

[Rorix]

the ANSWER is 2645.2N


whether its CORRECT OR NOT is arguable

 

[kheir]

in my test i got 2465 but the teacher wrote i should have taken the change of g into consideration

 

[zeropoint]

the ANSWER is 2645.2N


whether its CORRECT OR NOT is arguable

What are you talking about? Didn't chickenntaters just say the ANSWER was 3478.26N? Which is it?

 

[chickenntaters]

yeh, that was the answer that CSSA gave

 

[zeropoint]

yeh, that was the answer that CSSA gave

Well, CSSA have clearly stuffed up as their supposed model answer is clearly erroneous.

 

[shazzam]

hmm, cssa has made many errors in the past as well, maybe ppl can post pmore errors if they find any.