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

AI Thread Summary
To determine the minimum length of a barge for a 1000kg plane landing at 50m/s, the momentum conservation principle is applied, resulting in a calculated momentum transfer that pushes the barge back at 25m/s. The frictional force acting on the plane is calculated to be 2450N, leading to an impulse time of approximately 20.4 seconds. Using this time and the barge's velocity, the required distance for a safe landing is found to be at least 510 meters. The discussion also raises a question about reviewing previous posts and clarifies the application of impulse in relation to mass. The calculations and logic presented appear sound, but confirmation of the final answer is sought.
sean-820
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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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