Momentum (airplane landing, word problem)

[mickeymouseho]

Homework Statement

A 1.0 x 10^3 kg plane is trying to make a forced landing on the deck of a 2.0 x 10^3 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; this braking force is constant and is equal to one-quarter of the plane's weight. What must the minimum length of the barge be for the plane to stop safely on deck, if the plane touches down just at the rear end of the deck with a velocity of 5.0 x 10^1 m/s toward the front of the barge?

Homework Equations

F = ma
F x delta t = delta p
m1 x delta v1 = - m2 x delta v2

The Attempt at a Solution

I don't know how to approach this problem but this is just a guess.

friction force = 250N (1000x0.25)

force = ma
= 1000kg x 9.8 m/s^2
= 9800N

impulse = mv
= 1000kg x 50m/s
= 50000 kg m/s

delta t = impulse / force
= 50000 / 9800
= 5.1s

mass1 x delta v1 = mass2 x delta v2

1000kg x 50m/s = 2000kg x (delta d / delta t)

[(1000kg x 50m/s) / 2000kg] x delta t = delta d

[(1000kg x 50m/s) / 2000kg] x 5.1s = delta d

127.5m = delta d ??

I think this is completely wrong and I don't know what to do with the friction force . Any help would be appreciated. Thanks in advance!

PS: The answer is 3.4x10^2m according to the answer key.

 

[PhanthomJay]

Newton's laws applied to the boat/plane system don't work very well with relative motion problems when the 2 objects have different accelerations, so look at the plane and boat separately and apply Newton 2 to a FBD of each. What's the net force acting on the plane, and in which direction? What's is the magnitude and direction of the plane's acceleration with respect to the calm sea? What's the net force acting on the boat? What is the acceleration and direction of the boat with respect to the calm sea? What then is the plane's acceleration with respect to the boat? (NOTE: friction force is uN, not um, as you have indicated).

 

[thomate1]

Firstly, it is important to note that momentum is a vector quantity, meaning it has both magnitude and direction. In this problem, the direction of the momentum is towards the front of the barge.

To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum of a system remains constant unless acted upon by an external force. In this case, the external force is the frictional force between the plane's wheels and the deck.

We can start by calculating the initial momentum of the system, which is equal to the momentum of the plane just before it touches down on the deck. This can be calculated using the formula p = mv, where p is momentum, m is mass, and v is velocity.

Initial momentum = (1000 kg)(50 m/s) = 50000 kg m/s

Next, we can calculate the final momentum of the system, which is equal to the momentum of the plane and barge after the plane has come to a complete stop. This can be calculated using the same formula as above, but with the final velocity being 0 m/s since the plane has stopped.

Final momentum = (1000 kg + 2000 kg)(0 m/s) = 0 kg m/s

Since momentum is conserved, the initial momentum must be equal to the final momentum. Therefore, we can set up an equation: initial momentum = final momentum.

50000 kg m/s = 0 kg m/s

Now, we can use the formula for impulse, F x delta t = delta p, to solve for the time it takes for the plane to come to a stop. We know that the force acting on the plane is equal to the frictional force, which is one-quarter of the plane's weight. So, we can set up the equation: (1/4)(1000 kg)(9.8 m/s^2) x delta t = 50000 kg m/s.

Solving for delta t, we get delta t = 20.4 seconds.

Now, we can use the formula for displacement, delta d = v x delta t, to calculate the minimum length of the barge. We know that the initial velocity of the plane is 50 m/s and the final velocity is 0 m/s. So, we can set up the equation: (50 m/s)(20.4 s) = delta d.

Solving for delta d, we