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?(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Homework Help: Plane landing on barge (momentum)

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