A wagon is coasting at a speed v A along astraight and level road. When 17.9% of

Question

A wagon is coasting at a speed vA along astraight and level road. When 17.9% of the wagon's mass is thrownoff the wagon, parallel to the ground and in the forward direction,the wagon is brought to a halt. If the direction in which this massis thrown is exactly reversed, but the speed of this mass relativeto the wagon remains the same, the wagon accelerates to a new speedvB. Calculate the ratiovB/vA. I have no clue how to even start this. Any help is muchappreciated. A wagon is coasting at a speed vA along astraight and level road. When 17.9% of the wagon's mass is thrownoff the wagon, parallel to the ground and in the forward direction,the wagon is brought to a halt. If the direction in which this massis thrown is exactly reversed, but the speed of this mass relativeto the wagon remains the same, the wagon accelerates to a new speedvB. Calculate the ratiovB/vA. I have no clue how to even start this. Any help is muchappreciated.

Explanation / Answer

Use conservation of momentum First part - total momentum of the system before any mass isthrown: m Va 0.11m is thrown with velocity of v1. the conservation of momentum for this situation: m Va = 0.11m v1 +0.89m(0) (Where did 0.89 came from? 100-11=0.89 which is the mass left inthe wagon.) Or it can be m Va=0.11 mv1 If you change this equation it gives you v1 = Va/0.11 Second part-resversed direction We can write initial momentum as again m Va 0.11m is thrown backwards with v1 speed and the wagonaccelerates to a speed Vb And we have to used the equation we just got before. m Va = -0.11m v1 + 0.89 m Vb m cancels out and whatleft is: Va + 0.11 v1 = 0.89 Vb Remember that v1 = Va/0.11, plug this into equation above: Va+0.11(Va/0.11)=0.89Vb 2Va=0.89Vb Now rearrange. Vb/Va=2/0.89=2.25 Hope this helps! Source:http://answers.yahoo.com/question/index?qid=20091031154833AAMZ2MN

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