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

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Homework Help Overview

The discussion revolves around a physics problem involving a plane making a forced landing on a barge. The scenario includes considerations of mass, velocity, and frictional forces, specifically focusing on the minimum length of the barge required for the plane to stop safely after landing.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore concepts of momentum conservation and the effects of friction on the plane's landing. There are attempts to clarify the final velocities of both the plane and the barge, as well as the implications of inelastic collisions.

Discussion Status

Some participants have provided guidance on the conservation of momentum and the calculation of frictional forces. However, there remains uncertainty regarding the final speed of the plane and the necessary calculations to determine the barge's length. Multiple interpretations of the problem are being explored.

Contextual Notes

Participants are grappling with the correct application of physics principles, particularly regarding the definitions of forces and the setup of the problem. There is an emphasis on understanding the relationship between the plane and barge during the landing process.

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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