Minimum Barge Length for Forced Plane Landing

[richievuong]

A 1000kg plane is trying to make a forced landing on the deck of a 2000 kg 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 1/4 of the plane's weight What must the minimum length of the barge be, in order that the plane can stop safely on the deck, if the plane touches down at the rear end of the deck with a velocity of 50m/s towards the front of the barge?

My work:

m1 = 1000kg
m2 = 2000kg
Ff = 1/4Fg
V1' = V2'

Fg = mg
=(1000)(9.81)
=9810N

Ff = 1/4Fg
= 1/4(9810)
=2452.5N

Fnet=Ff

P1(plane) = mv
= (1000)(50)
= 50000

P2(plane) = mv
= (1000)(0)
= 0

deltaP = (Fnet)t
50000 = 2452.5t
t = 20.387s

m1v1 + m2v2 = m1v1' + m2v2'
(1000)(50) + 2000(0) = 0 + (3000)v2'
50000 = 3000v2'
v2' = 16.667 m/s

a = V2-V1 / T
= (16.667-50) / 20.387
= -1.635m/s²

d = v1t + 1/2at²
= (50)(20.387) + 1/2(-1.635)(20.387)²
= 1019.35 - 339.777
=679.573m

Thats what i did but the answer is 340m...wondering what I did wrong...

 

[OlderDan]

My work:

P2(plane) = mv
= (1000)(0) <== First mistake I see. It contradicts what follows.
= 0

deltaP = (Fnet)t
50000 = 2452.5t
t = 20.387s

m1v1 + m2v2 = m1v1' + m2v2'
(1000)(50) + 2000(0) = 0 + (3000)v2' <== What are you really saying here?
50000 = 3000v2'
v2' = 16.667 m/s <== What does this tell you?

Thats what i did but the answer is 340m...wondering what I did wrong...

See the annotations in the quote.

 

[richievuong]

i'm saying that the final velocity of the barge+plane which is 3000kg would be 16.667m/s by using the conservation of momentum equation...

m1v1' would be 0 because the plane loses all momentum and transfers it to the barge right? if not please clarify because I'm starting to get confused...

i don't really understand why the first part is wrong can you explain that also? thanks

 

[OlderDan]

i'm saying that the final velocity of the barge+plane which is 3000kg would be 16.667m/s by using the conservation of momentum equation...

m1v1' would be 0 because the plane loses all momentum and transfers it to the barge right? if not please clarify because I'm starting to get confused...

i don't really understand why the first part is wrong can you explain that also? thanks

See the annotations in the following quote from your earlier post.

m1v1 + m2v2 = m1v1' + m2v2' <== this is correct

(1000)(50) + 2000(0) = 0 + (3000)v2' <== this 0 suggests that v1' is zero and that m2 has changed to 3000kg. From what you did earlier, and from what you said in the most recent post, it appears that you really believe v1' is zero. It is not. And m2 does not change. What is true is that v1' = v2'; m1 and m2 have the same final velocity.

50000 = 3000v2' <== because m1v1 + 0 = (m1+m2)v2'

v1' = v2' = 16.667 m/s <== The final velocities are the same.
You need to use this v1' result in your earlier calculation. v1' is not zero. The final momentum of the plane is not zero.