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Homework Help: Bullet shot into a block

  1. Mar 23, 2010 #1
    1. The problem statement, all variables and given/known data

    A bullet of mass m is fired into a block of mass M initially at rest on a frictionless table of height h. The bullet remains m in the block, and after impact the block lands a distance d from the bottom of the table. Determine the initial speed of the bullet.

    2. Relevant equations

    I used (m +M)gh for the potential energy and set that equal to the kinetic energy ½(m + M)v₂² and found that v₂ = √(2gh).

    And since it was a completely inelastic collision (right?) I used mv₁ = (m + M)v₂ → v₁ = (m + M)v₂ * 1/m

    Is this right?


    3. The attempt at a solution

    initial velocity = √(2gh)*((m + M)/m)
     
    Last edited: Mar 23, 2010
  2. jcsd
  3. Mar 23, 2010 #2
    Do you have an attempt at the problem? If not, tell me your thought process for the problem.
     
  4. Mar 23, 2010 #3
    I used (m +M)gh for the potential energy and set that equal to the kinetic energy ½(m + M)v₂² and found that v₂ = √(2gh).

    And since it was a completely inelastic collision (right?) I used mv₁ = (m + M)v₂ → v₁ = (m + M)v₂ * 1/m

    Is this right?
     
  5. Mar 24, 2010 #4
    Something about this chunk doesn't seem right to me for some reason. Everything else is correct, from my understanding. Is this a formula you received from your text, or is this something that you derived yourself?

    Check out the Wiki article, I think I see where you're going with this, but I see a slight error in your formula: http://en.wikipedia.org/wiki/Inelastic_collision
     
  6. Mar 24, 2010 #5
    It's from my textbook
     
  7. Mar 24, 2010 #6

    Borek

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    Staff: Mentor

    I think the formula here is OK, wiki describes different situation (initial velocity of M not being zero) and uses different symbols, hence the confusion.

    Problems are with the other part - v2 that you calculated from the energy conservation is a vertical component of the speed when block with a bullet hits the ground. That's not initial HORIZONTAL speed of the body.
     
  8. Mar 24, 2010 #7
    Elaborate please
     
  9. Mar 24, 2010 #8

    Borek

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    Staff: Mentor

    Once the bodies (m+M) left the table they behave just like something thrown norizontally - they maintain horizontal speed, but they start to fall down, till they hit the ground at distance d from the table border.
     
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