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Conservation of momentum problem with a bullet and stationary block

  1. Dec 22, 2008 #1
    1. The problem statement, all variables and given/known data
    A bullet with a mass of 6.00 g is fired through a 1.25 kg block of wood on a frictionless surface. The initial speed of the bullet is 896m/s, and the speed of the bullet after it exists the block is 435 m/s. At what speed does the block move after the bullet passes through it?


    2. Relevant equations
    I think I have the corect momentum, but the answer for the acceleration of the block seems a bit too easy. Is it right, or am I way off?


    3. The attempt at a solution
    First I drew a free body digram. The diagram has the bullet on the left, with its velocity and mass (Mass=.006 Kg Velocity=896 m/s), and the stationary block of wood with its mass and velocity (Mass=1.25 Kg Velocity=0 m/s) .

    Since this is a collision problem, I figured I should find the momentum using the formula Mass*Velocity=Momentum. So, for the bullet... .006 Kg * 896m/s= 5.376 Kg*m/s (initial momentum) and... .006 Kg * 435 m/s= 2.610 (final momentum). For the block of wood... 1.25 Kg * 0 m/s= 0 Kg*m/s (Initial momentum) and... 1.25 Kg * 461 m/s (the difference of the bullet's Initial and Final velocities)= 576 Kg * m/s

    I realize this is momentum not the block's velocity. I'm not sure what else to do because the only other logical answer is 896-435= 461 m/s and that just seems way to easy for a physics problem. Or am I over thinking this?
     
  2. jcsd
  3. Dec 22, 2008 #2

    Doc Al

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    Staff: Mentor

    This is good.
    That last step isn't correct. Why would you multiply the change in the bullet's velocity by the mass of the block?

    Instead, set up a conservation of momentum equation:
    Intial momentum (of bullet and block) = Final momentum (of bullet and block)

    Initially, you just have the bullet moving; After the collision, the bullet and the block move. Set up the equation and solve for the speed of the block.
     
  4. Dec 23, 2008 #3
    So your saying that what I shoud do is this:

    5.367+0=2.610+x
    5.367=2.610+x
    2.757 Kg*m/s=x

    Where X equals the blocks momentum.

    2.206(Kg*m/s)/1.25(Kg)=
    2.2056m/s.

    Using significant figures it would be 2.206.

    Or am I still wrong?
     
  5. Dec 23, 2008 #4

    Doc Al

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    Staff: Mentor

    Your method is exactly correct. Two small problems:
    (1) Check the digits that I highlighted; you have them transposed. (Fix that and redo the calculation.)
    (2) Your final answer should be rounded to 3 significant figures, since your data is only given to 3 digits.
     
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