How Long Must the Barge Be for a Safe Plane Landing?

AI Thread Summary
A 1000 kg plane needs to land safely on a 2000 kg barge, requiring a minimum barge length of 340 meters to stop from a speed of 50 m/s. The only friction considered is a constant force equal to one-quarter of the plane's weight. The momentum conservation equation is applied, leading to the conclusion that both the plane and barge must move together after landing to prevent the plane from falling off. The force of friction calculated is 2450 Newtons, which is crucial for determining the stopping distance. The discussion emphasizes the importance of understanding momentum and friction in this forced landing scenario.
musicfan31
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1.

A 1000 kg plane is trying to make a forced landing on the deck of a large 2000 kg barge at rest on the surface of a calm sea. The only frictional force to consider is between plane's wheels and the deck, and this braking force is constant and equal to one-quarter of the plane's weight. What must the minimum legnth of the barge be, in order that the plane can stop safely on the deck, if the plane touches down just at the rear end of the deck with a velocity of 50 m/s towards the front of the barge? (SIN.'76)

**PLEASE write out the full solution!**

Additional Details
The answer is 3.4*10^2 meters.




2.
work = force applied * displacement

momentum = mass*velovity

m1v1 + m2v2 = m1v1' + m2v2'



3.
Plane
m = 1000 kg
Vi = 50m/s
Vf = 0 m/s

Barge
m = 2000kg
Vi = 0 m/s

Force of friction = force of normal = mass*g


m1v1 + m2v2 = m1v1' +m2v2'

1000(50) + 2000(0) = 1000(0) + 2000v2'

1000(50)
________ = v2'
2000


25 m/s = v2' = speed of barge in oppsite direction

** I DUNT EVEN NOE IF THE STUFF I DID ABOVE IS RIGHT*** BUT NOW I"M LOST**** HELP MEEE***


 
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musicfan31 said:
Force of friction = force of normal = mass*g
This isn't correct. Reread what the problem tells you about friction.


m1v1 + m2v2 = m1v1' +m2v2'

1000(50) + 2000(0) = 1000(0) + 2000v2'
The collision is inelastic--the plane and barge end up moving with the same speed. (The final speed of the plane isn't zero.)

Find the final speed of the barge and plane.

What's the acceleration of the plane? What distance does it take to bring it to its final speed?

What's the acceleration of the barge? What distance does it move during its acceleration?
 
I have no clue as to what you are speaking of... the final speed of the plane is not Zero?

Then what is the number that needs to be subbed in as final velocity of the plane.?
 
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musicfan31 said:
I have no clue as to what you are speaking of... the final speed of the plane is not Zero?
That's right. The plane ends up moving at the same speed as the barge. (It better, or it will just fall off the barge!) Use momentum conservation to find the final speed of barge and plane.
Here is a solution but I can't make any sense of this you can help me figure this out.
That solution is precisely the one I'm trying to get you to figure out for yourself.
 
okay the Ff = 1/4M*g
= 1/4(1000)*9.8
= 250*9.8
= 2450 Newton is the force of friction
 
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The conservation of momentum formula is :

m1v1 + m2v2 = m1v1' + m2v2'
 
I go it ! I GOT IT!
Thank YOU DOC AL
 
I'm still stuck. Can you give more help?
 
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