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Linear Momentum

  1. Nov 6, 2006 #1
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    "A bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length l and negligable mass.
    A) What is the minimum value of v such that the pendulum bob will barely swing through a complete verticle circle?
    Energy is conserved for the bob-Earth system between bottom and top of swing. At the top, the stiff rod is in compression and the bob nearly at rest.

    I'm not really sure how to do this problem, but i think that at the top, it would barely swing through because it is nearly at rest, but im not sure what to do with that.
     

    Attached Files:

  2. jcsd
  3. Nov 6, 2006 #2
    Since it is the minimum speed at which it will make it over, you calculate v such that the kinetic energy at the top is zero (no velocity)

    then:

    mv = m(v/2) + MV (conservation of momentum)

    1/2MV^2 = Mg(2l)

    that should be enough to solve it.
     
  4. Nov 6, 2006 #3
    I undestand the formula's but i'm really bad at physics so i'm not sure if i know what to do next. I never understood how to combine these two formulas.
    Can I solve for V in the conservation of momentum equation and then sub that into the conservation of energy equation? Are V and v supposed to be different or the same?
     
  5. Nov 6, 2006 #4
    V and v are different, one is velocity of the bullet, the big V is that of the bob. yes, you have 2 equations and 2 unknowns (v and V) so solve one and substitute as you suggested
     
  6. Nov 6, 2006 #5
    Thank you for all your help
     
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