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A bullet of mass 0.062 kg traveling horizontally at a speed of 150 m/s embeds itself in a block of mass 90 kg that is sitting at rest on a nearly frictionless surface

What is the speed of the block after the bullet embeds itself in the block?

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asked by sara
  1. conserve momentum:

    .062*150 + 90*0 = (.062+90)v
    v = 0.103 m/s

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    posted by Steve
  2. Teri 9211

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    posted by Lolu

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