# Minimum Barge Length for Forced Plane Landing

• richievuong
In summary: The final momentum of the barge is not 3000(16.667) = 50000. The final momentum of the system is 3000(16.667) = 50000.deltaP = (Fnet)t <== This is correct.50000 = 2452.5t <== This is still correct.t = 20.387s <== This is still correct.m1v1 + m2v2 = m1v1' + m2v2' <== This is still correct.(1000)(50) + 2000(0) = (1000)(16.667) + (2000)(16.667
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...

richievuong said:
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.

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

richievuong said:
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.

richievuong said:
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.

## 1. What is the minimum barge length required for a forced plane landing?

The minimum barge length required for a forced plane landing is dependent on several factors, including the size and weight of the plane, the type of water body it will be landing on, and the weather conditions. Generally, a barge length of at least 100 feet is recommended for most forced plane landings.

## 2. How is the minimum barge length determined?

The minimum barge length for a forced plane landing is determined by various calculations and simulations. These take into account the weight of the plane, the angle of approach, and the potential impact forces on the barge. These calculations are then used to determine the minimum length needed for a safe and successful landing.

## 3. Can a smaller barge be used for a forced plane landing in an emergency?

In certain emergency situations, a smaller barge may be used for a forced plane landing. However, this is not recommended as it can increase the risk of damage to the plane and potential injury to passengers. It is always best to use a barge that meets the recommended minimum length for a forced plane landing.

## 4. Are there any regulations or guidelines for minimum barge length for forced plane landings?

Yes, there are regulations and guidelines set by aviation authorities and marine safety agencies for minimum barge length for forced plane landings. These regulations and guidelines are constantly reviewed and updated to ensure the safety of passengers and crew during such emergency situations.

## 5. Are there any alternatives to using a barge for forced plane landings?

In some cases, alternative methods such as using an inflatable raft or air cushion landing system may be considered for forced plane landings. However, these alternatives may not be suitable for all types of water bodies and may not provide the same level of safety as using a barge. It is important to always follow the recommended guidelines for forced plane landings to ensure the safety of all involved.

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