Minimum Length for Safe 1000kg Plane Landing on 2000kg Barge

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SUMMARY

The minimum length required for a 1000kg plane to land safely on a 2000kg barge is calculated to be 510 meters. This calculation is based on the conservation of momentum and the impulse-momentum theorem, where the plane's momentum of 50000 kg·m/s is transferred to the barge. The frictional force acting on the barge, calculated as 2450N, plays a crucial role in determining the time of landing, which is approximately 20.4 seconds. The discussion confirms the logical steps taken in the calculations, emphasizing the importance of understanding impulse as the change in momentum.

PREREQUISITES
  • Understanding of conservation of momentum principles
  • Familiarity with impulse-momentum theorem
  • Basic knowledge of frictional forces and their calculations
  • Ability to perform kinematic equations for distance and time
NEXT STEPS
  • Study the conservation of momentum in inelastic collisions
  • Learn about impulse-momentum theorem applications in physics
  • Explore frictional force calculations in various contexts
  • Investigate kinematic equations for motion analysis
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Physics students, engineers, and anyone involved in mechanics or dynamics, particularly those interested in collision analysis and motion on surfaces with friction.

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Homework Statement


1000kg plane lands on a 2000kg barge on in a calm ocean. The only frictional force is equal to one quarter of the planes weight. What must the minimum barge length be to land safely if it hits the deck at 50m/s. Also there is no water friction/resistance.

Homework Equations



mv=mv1+ mv2 (conservation of momentum)

FT=m (v2-v1) (impulse=change in momentum)

distance=VT

The Attempt at a Solution



I think i may have the right answer now, but I am not totally sure.

The momentum of the plane = 1000x50=50000
With twice the planes mass, the 50m/s plane will transfer all of its momentum to pushing the barge back at 25m/s.

Using friction=0.25mg the frictional force =250x9.8=2450N

T=(m(v2-v1))/F
=(1000(-50))/-24500
=20.4seconds

Then using D=VT=25 (20.4) i found the distance traveled must be at least 510m.

What I'm asking does this seem logical and correct? i think it may be, but I'm not totally sure. I wasn't provided the correct answer




Also, is there some way you can go back and read all your own posts?
 
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The main thing that I'm not sure on is the impulse = change in momentum. For the mass would i have to do m1 or the total m1+m1 mass?
 

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