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Bullet and a block

  1. Oct 22, 2006 #1
  2. jcsd
  3. Oct 22, 2006 #2

    OlderDan

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  4. Oct 22, 2006 #3
    Okay... i equated .5kx^2 and .5mv^2 for the block, and got the velocity, v. Then i used mv=MV (conservation of momentum) to find the final velocity of the bullet as it emerges out of the block. But that is not the correct answer.
    Help me.
     
  5. Oct 22, 2006 #4

    OlderDan

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    Let's be clear on M vs m and V vs v. Your energy calculation involves the spring and the block, so i assume m is mass of block and v is its velocity when the spring starts to compress. The momentum conservation problem can be assumed complete before the spring compreses.

    Look at the momentum problem again. You have a bullet moving with known velocity toward a stationary block. After the collision you know the velocity (from the energy calculation) of the block and you need the final velocity of the bullet. Your momentum equation does not correspond to this situation. See if you can fix it.
     
    Last edited: Oct 23, 2006
  6. Oct 23, 2006 #5

    Andrew Mason

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    Interesting problem.

    Since the bullet is travelling at 500 m/sec, it passes through the block before the block really begins to move. So we can treat this as if the block acquires an instantaneous kinetic energy which is then transferred into the spring. We know what the energy transferred to the block is [itex]E = kx^2/2[/itex] where v0 is the speed of the block after collision. So you can work out the speed of the block after the bullet passes through. I think that is what you have done.

    Since momentum has to be conserved,

    [tex]mv_{f-bullet} + Mv_{block} = mv_{i-bullet}[/tex]

    That may be where you are having some difficulty.

    AM
     
    Last edited: Oct 23, 2006
  7. Oct 23, 2006 #6
    Thank You. I finally solved it. siggghhhh
     
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