Minimum barge length for plane landing with friction

[excel000]

Homework Statement

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

Homework Equations

F=ma
m1v1 +m2v2 = m1v1' + m2v2'

The Attempt at a Solution

first i found the friction force
Ff=1/2mg
=1/2(1000kg)(9.8)
Ff=2450N

then i found the acceleration of the plane
F=ma
-2450/1000=a
a=-2.45m/s^2

then i used kinematics to find the displacement
V2^2 = V1^2 + 2ad
0 = 50^2 + 2(-2.45)d
d=510m

i don't know if this is the right answer, but since this is a question from a momentum assignment, is there a step involving momentum that i missed?

 

[Doc Al]

i don't know if this is the right answer, but since this is a question from a momentum assignment, is there a step involving momentum that i missed?

Yep. What's the final speed of the plane + barge?

Also realize that while the plane moves forward, the barge moves backward.

 

[excel000]

shouldnt the planes final velocity be 0?

mpvp + mbvb = mpvp' + mbvb'
1000(50) + 0 = 0 + 2000vb'
vb'=25m/s
the equation says that the barge is moving forward

 

[Doc Al]

shouldnt the planes final velocity be 0?

The final velocity of both barge and plane had better be the same. Redo your momentum equation.

 

[Senjai]

exactly, v1' = v2' because th exlane and the barge will move at the same velocity after the landing, as one object persay.. (for simplicity) so the equation is m1v1 = m1v' + m2v'

m2v2 = 0 as the barge is at rest

 

[excel000]

ok so its like this?
mpvp + mbvb = mv(barge and plane)
1000(50) + 0 = (1000+2000)v
v= 16.67m/s

 

[Doc Al]

oh, its all relative to the water

Exactly.

 

[excel000]

ok so using that velocity from above i subbed all the info into this equation
v2^2 = v1^2 +2ad
16.67^2 = 50^2 +2(-2.45)d
d=453m

 

[Doc Al]

ok so using that velocity from above i subbed all the info into this equation
v2^2 = v1^2 +2ad
16.67^2 = 50^2 +2(-2.45)d
d=453m

Careful. You found the distance the plane moves with respect to the water. But as I said earlier, don't neglect the movement of the barge.