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I A question about conservation of momentum

  1. Apr 30, 2017 #1
    Let's say you've got a boat filled with n frogs of mass m each. Let's also say that all the frogs jump simultaneously out of the boat in the same direction with velocity v, then the boat will get a velocity v1 in the opposite direction because of conservation of momentum.

    Now if we instead let each frog jump separately with velocity v, the boat will get a velocity v2 that is bigger than v1.

    I did see the formula and the derivation for why the boat gets a different velocity, but I'm still not getting it intuitively. If the same mass is leaving the boat at the same velocity, why does the boats velocity differ depending upon when the mass leaves the boat?
  2. jcsd
  3. Apr 30, 2017 #2

    Buzz Bloom

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  4. Apr 30, 2017 #3


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    What happens if all the frogs do an identical jump (it terms of the mechanics of their jump)?
  5. Apr 30, 2017 #4


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    Those two velocities are relative to what?
  6. Apr 30, 2017 #5
    Yes I know what impulse is. If the frogs jump simultaneously, then the impulse time will be short but the force impacted on the boat will be large. If they jump one after the other, the force of each jump will be small but they will do it over a longer time. In the end they should add up to be the same impulse, I think.

    Sorry I don't understand your question.

    Yeah I should have stated that. v is relative to the boat while v1 and v2 are relative to the water.
  7. Apr 30, 2017 #6


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    So what is the velocity of the frogs with respect to the water in the two cases?
  8. Apr 30, 2017 #7


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    That still leaves v ambiguous -- relative to the boat pre-leap, post-leap or mid-leap?
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