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

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SUMMARY

The discussion focuses on calculating the minimum length of a barge required for a 1000 kg plane to safely land and stop on its deck. The plane approaches at a velocity of 50 m/s, and the only frictional force considered is one-quarter of the plane's weight, resulting in a braking force of 2450 Newtons. The final answer determined is that the barge must be at least 340 meters long to accommodate the plane's stopping distance.

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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.?
 
Last edited:
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
 
Last edited:
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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