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Old 11-30-2003, 01:59 PM   #1 (permalink)
 
KnifeMissile's Avatar
 
Location: Waterloo, Ontario
Resorting to physics problems as puzzles...

This is an interesting physics problem that I got as a high school student. I'm making the numbers up but I understand the problem so I can probably figure out the answer...

There is a 100kg aeroplane travelling at 200m/s landing on a 1000kg aircraft carrier floating in frictionless water. If the coefficient of friction (between the plane and carrier) is 0.1, then how long does the aircraft carrier need to be for the plane to land on it assuming that the normal force of friction (the weight of the plane) is constant throughout the landing procedure.

If I have forgotten anything, just make the typical highschool assumptions necessary to solve the problem or come back to the forum to ask.

I hope you have fun!
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Old 11-30-2003, 08:12 PM   #2 (permalink)
!?!No hay pantalones!?!
 
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Location: Indian-no-place
Are we going to assume that we're not being helped by a tailhook?


-SF
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Old 12-01-2003, 09:28 AM   #3 (permalink)
 
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Location: Waterloo, Ontario
Of course! The only thing stopping the plane is friction...
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Old 12-01-2003, 01:18 PM   #4 (permalink)
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Location: Atlanta, GA
If I did this right, and I haven't done a question like this lately, I got 20,000m (using g = 10 m/s^2) as for how long the aircraft carrier would have to be if the only thing to slow the aircraft down was the friction between its wheels and the carrier deck. This is why the use of a tailhook is vital to stopping the plane in time.

Hope I did it right.
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Old 12-01-2003, 09:08 PM   #5 (permalink)
 
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Location: Waterloo, Ontario
Wow, I picked some really bad numbers. I should have picked a coefficient of friction greater than 1, so the result wouldn't seem totally stupid (although I did deliberately pick unrealistic weights for the plane and carrier).

Yes, real aircraft carriers have tailhooks to shorten the necessary length of the boat. Still, real runways (the ones on land) aren't this long, so there's more stopping a real plane besides friction!

Oh, and your answer is wrong, shred_head...
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Old 12-01-2003, 11:39 PM   #6 (permalink)
Psycho
 
Location: Atlanta, GA
Ok I'm not very sure as to where I went wrong in my calculation. I realized I didn't take into account the momentum change between the plane itself and when it was on the carrier but even then it didn't come out to make any significant difference at all.

So what is the answer and how did you go about finding it?
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Old 12-02-2003, 12:51 AM   #7 (permalink)
 
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Location: Waterloo, Ontario
Hmm, it sounds like you're on the right track. How did you "take into account the momentum change?" It should have made a signifacant difference...

The answer is approximately 14876m. Can you see how this was derived? I should have picked a friction coefficient of 10! (that's an emphatic 10, not 10 factorial!)

Last edited by KnifeMissile; 12-02-2003 at 12:54 AM..
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Old 12-02-2003, 01:23 AM   #8 (permalink)
Banned
 
Location: UCSD, 510.49 miles from my love
remember too, that the carrier will move in the frictionless water with the plane creating a force due to its friction which will cause the carrier to move in the direction of motion of the landing plane.

either way, I have electricity and magnetism physics to work on right now, I have no idea what Im doing here...
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Old 12-19-2003, 02:50 AM   #9 (permalink)
Tilted
 
Location: Austin, TX
i believe the answer is a bit surprising. the length of the runway is actually less on the aircraft carrier than if it were to have been in an airport.

i don't feel like using the calculator, but i think the basic equation is as follows

momentum of airplane before is Pa =MaVa
momentum of airplane/carrier after is Pt=MtVt where t = total and Mt = Mc+Ma

so using energy equations
1/2Ma(Va)^2 = uMagx + 1/2Mt(Vt)^2

where u = coefficent of friction (which by the way is always between 0 and 1)

and g = gravity and x is the distance.

all that is easliy solved.

the interesting part is that if it were an airport instead, the final velocity would be 0, and so its just:
1/2mv^2 = umgx, where x in this airport solution would be larger than the x in the aircraft carrier solution.

it seems weird, but we have to remember that the airplane travels a distance longer than the carrier's length because the airplane pushes the carrier through the water as well.

in both cases, solve for x yourself. i don't have a calculator handy nor do i wish to use the accessories calculator.
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