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anonymous

  • one year ago

A 150 kg loaded toboggan is gliding along a flat surface at 2.0 m/s [forward] when one of the 50 kg passengers jumps off towards the back with a velocity of 3.0 m/s [backwards] with respect to the ground. What is the resulting speed of the toboggan and remaining passengers.

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  1. ganeshie8
    • one year ago
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    do we use conservation of momentum ?

  2. anonymous
    • one year ago
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    The initial velocity of the toboggan is calculated as 150kg*2.0m/s=300N/s Since one of the passengers weighing 50kg jumps off back of the boat at 3.0m/s with the frame of reference being the ground, 3.0m/s+2.0m/s=5m/s Since the passenger acquires a momentum of 250N/s to the opposite direction 300N/s-250N/s=50N/s [forward] Now the only momentum acting on the boat gliding along is 50N/s Given the mass of the boat 150kg, and momentum acting on the boat being 50N/s 150kg*v=50N/s v=0.333m/s Therefore final velocity of the boat is 0.33m/s

  3. anonymous
    • one year ago
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    Yeah we do

  4. anonymous
    • one year ago
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    Anyway that's my quick shot but I could be wrong.

  5. anonymous
    • one year ago
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    @ganeshie8 I could be wrong.

  6. ganeshie8
    • one year ago
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    that doesn't look correct because the initial speed of boat is 2 m/s how can it reduce if the boat gets a push by unloading a passenger ?

  7. anonymous
    • one year ago
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    Oh that's true

  8. anonymous
    • one year ago
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    Perhaps add the 300N/s+250N/s=550N/s Then increased momentum results in supposedly greater magnitude of velocity so 550N/s=v(150kg) v=3.66m/s The final velocity 3.7/s

  9. anonymous
    • one year ago
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    That seems more or less right

  10. anonymous
    • one year ago
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    The key is when the passenger jumps towards the back with velocity of 2.0m/s to the ground I get it as 2.0m/s+3.0m/s because of the frame of reference being the ground and also the fact that the boat is gliding along the water to the opposite direction means increased velocity with respect to the boat.

  11. ganeshie8
    • one year ago
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    looks good to me!

  12. anonymous
    • one year ago
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    I got it wrong I realized

  13. anonymous
    • one year ago
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    The momentum that originates from the boat was initially 300kgm/s and more so according to the conservation of momentum originating from one body, which in this case is the boat.

  14. anonymous
    • one year ago
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    So I bet I should do 300N/s=100kg(v)+50kg(-2m/s)

  15. anonymous
    • one year ago
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    300N/s=100kg(v)+50kg(-3m/s)

  16. anonymous
    • one year ago
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    100kg(v)=450kgm/s v=4.5m/s?

  17. anonymous
    • one year ago
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    I guess the wording of this question failed.

  18. anonymous
    • one year ago
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    Boat=momentum of this boat is conserved at all times.... regardless of whatever comes out of it the initial momentum with respect to what comes out of it in one dimensional setting will be equal. Hence addition of -150kgm/s+300km/s=300kgm/s= that's initial momentum of this boat .

  19. anonymous
    • one year ago
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    I interpreted newton's third law as canceling each other out... which was my fallacy.

  20. ganeshie8
    • one year ago
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    initial momentum = 150*2 = 300 a mass of 50kg disappears at a speed of 3--2 = 5 m/s with respect to boat, pushing the boat forward so final momentum is given by 100v and it must satisfy : 100v = 300 + 50*5 solving gives me final speed = 5.5

  21. anonymous
    • one year ago
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    Right.. I forgot with respect to the ground part.

  22. ganeshie8
    • one year ago
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    don't trust me on these though im going by my gut.. its been a very long time me touching physics..

  23. anonymous
    • one year ago
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    Yeah physics is a full of trickeries.

  24. anonymous
    • one year ago
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    Frame of reference is such an arbitrary concept.

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