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Homework Help: Hard tutorial question can anyone help

  1. Oct 21, 2006 #1
    A torpedo boat with a mass of 50 tons is moving at 10 knots when it fires a 250-kg torpedo horizontally with the launch tube at the 30 degree angle. The torpedo has a relative velocity to the boat of 10m/s as it leaves the tube
    1) Determine the momentary reduction v in forward speed of the boat?

    it is meant as a linear impuse question, can anyone help me please!
  2. jcsd
  3. Oct 22, 2006 #2


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    I can't help with the math at all, unfortunately. My first question would be regarding the launching method. Not all of them would impart the same reaction to the boat.
  4. Oct 22, 2006 #3


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    Have you done linear impulse questions before?

    Start by converting all the units to SI and it might become a bit clearer.
  5. Oct 22, 2006 #4


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    The sub and torpedo as a whole has a combined linear momentum. Whe the two separate........
  6. Oct 29, 2006 #5
    conservation of momentum
    the mass of the sub + torpedo times its velocity before firing=
    the mass of the sub*vsub +mass torpedo*vtorp.

    or you could say that the sub is at rest altogether and fires the torpedo which then has momentum of 250*10=2500
    so the sub must loose this amount of momentm
    so vsub= -.05m/s
    the sub looses .05m/s when it fires the the torp.

    it aint rocket science, but wait, maybe it is
  7. Nov 1, 2006 #6


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    Unfortunately, phlegmy, you miseed an important portion of the problem: the torpedo is launched at a 30* angle. This will slightly decrease the momentum loss from the boat.
  8. Nov 3, 2006 #7
    in that case loss of mom in x direction = 250*10*cos(30)= 2165
    sub loosese .043 m/s in the forward direction! and starts to decend at
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