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Concervation of Momentum: Bullet and Block of Mass

  1. Mar 17, 2012 #1
    A bullet of mass m = 19.7 g traveling horizontally at speed vo = 445 m/s strikes a block of mass M = 854 g sitting on a frictionless, horizontal table. The bullet comes out the other side of the block at speed v = 415 m/s. What is the speed of the block after the bullet exits the block?

    Using conservation of momentum, I have the following equation but there are two unknowns (v'1 and v'2) so I need 2 equations but am kind of lost as to what should be my second equation? Can someone please help?

    m1v1 + m2v2 = m1v'1 + m2v'2

    (.0197kg)(445m/s) + (.854kg)(0) = (.0197kg)v'1 + (.854kg)v'2

    8.7665 = .0197v'1 + .854v'2
     
  2. jcsd
  3. Mar 17, 2012 #2

    cepheid

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    You know v1', it is given in the problem.
     
  4. Mar 17, 2012 #3
    the velocity on the right side of the equation is 415, it is given in the problem. you can solve for V2 using the same equation
     
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