Plane landing on barge (momentum)

[plutonium]

A plane with mass 1000kg lands on a stationary 2000kg-barge at an initial velocity of 50m/s. The only force to consider is the braking force which is 1/4 of the plane's weight (2450N). How long does the barge have to be if the plane lands at one end of the barge and stops at the other end?

Here's what my friend did:
1. calculate the acceleration (or rather deceleration) to be -2.45 m/s^2.
2. used v2 = v1 + at to determine time, which works out to be about 20.4s.
3. used conservation of momentum to determine the velocity of the plane-barge system, which is 16.67m/s.
4. used the velocity of the plane-barge system and time from step 2 to determine the distance, which is 340m (the correct answer BTW).

My question is, how does it work? I understand step 1, and step 2 sort of makes sense. Step 3 makes sense too, but the velocity of the plane-barge system is moving together, so from the way I understand it, the plane is at rest relative to the barge. Step 4 works out if I assume that the barge does not move. I'm just really confused by all of this.

 

[NateTG]

I think that there are some canceling errors.
Consider:
If the plane spends 20 seconds slowing down, then it's final velocity will be 0 m/s, but you've got the barge moving at 16.67 meters per second, so the two aren't at rest w.r.t. each other.

 

[Andrew Mason]

A plane with mass 1000kg lands on a stationary 2000kg-barge at an initial velocity of 50m/s. The only force to consider is the braking force which is 1/4 of the plane's weight (2450N). How long does the barge have to be if the plane lands at one end of the barge and stops at the other end?

...

My question is, how does it work? I understand step 1, and step 2 sort of makes sense. Step 3 makes sense too, but the velocity of the plane-barge system is moving together, so from the way I understand it, the plane is at rest relative to the barge. Step 4 works out if I assume that the barge does not move. I'm just really confused by all of this.

I think you have to assume that the barge is free to move and is on a frictionless water-body. The impulse of: the braking force x time gives the barge momentum and reduces the plane's momentum. The changes in momentum of each are equal and opposite.

Use overall conservation of momentum to determine the speed of the plane and barge together. That will give you the change in momentum of the plane and of the barge. The impulse (Ft) must equal that change. The time will then enable you to work out the distance covered by the plane in landing (force/acceleration is constant so how is time related to distance and acceleration?). In that time period, how far has the barge moved? Subtract the two to find the barge's minimum length.

BTW, this question bears no resemblance to reality. For the plane to land on the barge it must be a long barge. It would have a mass much greater than double the mass of the airplane. A 2 T. barge would be about the size of a large motor boat.

AM

 

[plutonium]

So you are saying that:
$$F \Delta t = \Delta p_{plane} = -\Delta p_{barge}$$ ?

I worked out the change in momentum of the plane to be -3333.33, and t to be 13.61, then used d = v2t - 1/2 at^2, and determined distance to be 227m.

 

[Andrew Mason]

So you are saying that:
$$F \Delta t = \Delta p_{plane} = -\Delta p_{barge}$$ ?

I worked out the change in momentum of the plane to be -3333.33, and t to be 13.61, then used d = v2t - 1/2 at^2, and determined distance to be 227m.

$$p_{p} = m_pv = p'_{pb} = (m_p+m_b)v'$$

$$v' = \frac{m_p}{m_p+m_b}v$$

so:
$$\Delta p_{p} = m_pv(1 - \frac{m_p}{m_p+m_b}) = \frac{2}{3}m_pv$$

$$\Delta p_p = \frac{1}{3}1000*50 = 33333$$ Kg m/sec

The rest of your answer is correct. Distance = 227 m.

AM