# Momentum/Kinematics (Contest Question).

1. Jul 9, 2015

### Ethan_Tab

1. The problem statement, all variables and given/known data
A 1000 kg 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; 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 50 m/s toward the front of the barge?

2. Relevant equations
F=ma
p=mv
Kinematic equations.

Although I'm not sure where this one will be used...
mPvP +mBvB = mPvP'+mBvB' Where ' represent final condition

Also possibly KE=KE'------ Not sure if elastic????

3. The attempt at a solution

The frictional force the plane and boat will experience is 2450N (by newtons third law, equal and opposite)
Lets set rear of boat to be - and the direction of travel of the plane to be +

By F=ma,
the plane will accelerate at -2.45m/s2
and the boat will accelerate at 1.225m/s2

Im not sure where to go from here. This is meant to be a momentum question yet I don't see where that would be applicable. Also the question does not specify weather the point at which the plane collides with the boat is elastic or inelastic.

2. Jul 9, 2015

### Nathanael

Certainly not. Friction produces thermal energy.

Are you questioning if the plane will bounce back up?

Anyway, you know the accelerations of both objects and you know the initial and final speeds of both objects. Can you then find the distance each object travels? Then can you use this information to find how long the barge must be?

3. Jul 9, 2015

### Ethan_Tab

Once the plane touches down, how do you know that there will be not transfer in momentum?

4. Jul 9, 2015

### Nathanael

Vertically? This doesn't affect the problem, as you are given the frictional force (not a coefficient of friction).

5. Jul 9, 2015

### Ethan_Tab

Here is the textbook solution. I cant make sense out of the collision part: They're taking the collision to be completely inelastic which is also confusing me.

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6. Jul 9, 2015

### Nathanael

Looking at the solution is a waste of a problem

They are not taking the collision to be inelastic, they are simply finding the final speed of the boat and plane.
The final speed is when they both have the same speed (because then the plane is at rest relative to the barge).

Edit:
It's very misleading that they wrote, "for the collision" :\ So I see how you got confused.
But this confusion would not have arisen if you didn't peek at the solution

7. Jul 9, 2015

### Ethan_Tab

I see what you saying, I think I'm just confusing myself with the frame of reference. However I don't see how they were able to find the final relative speed (16m/s) using momentum. Could you perhaps explain?

8. Jul 9, 2015

### Nathanael

Okay, you already have this equation:
We know that vB = 0 and vP = 50 m/s and we know both masses. We just don't know the final speeds of both objects.

The key thing to realize is that, in order for the plane to be at rest relative to the barge, they must be moving at the same speed.
When the plane "stops on the barge," it is moving at the same speed as the barge.

Therefore v'B=v'P which I will call vf (the final speed of the system)

So your conservation of momentum equation becomes mPvP=(mB+mP)vf which can now be solved for vf
(You need vf to find how far the plane and the barge each travel during this process of stopping.)

9. Jul 9, 2015

### Ethan_Tab

I understand know. Thanks for the help!