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Homework Help: Initial velocity of bullets - collisions

  1. Jan 6, 2006 #1
    a 11.9g bullet is fired vertically into a 5.49kg block of wood. the bullet gets stuck in the block, and the impact lifts the block 0.11m up. given g=9.8m/s^2. what was the initial velocity of the bullet?

    Pi=Pf is this right?
  2. jcsd
  3. Jan 6, 2006 #2


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    Yes, that's right. Where Vf is the initial velocity of the block/bullet combination just before it rises up.
    You now have to figure out Vf from the motion of the block/bullet.
  4. Jan 6, 2006 #3

    in order find Vf of the block/bullet do i use the same formula and that will equal Vf=M1Vio/(m1+m2) if so, i now have two unknowns
  5. Jan 6, 2006 #4
    use energy to find the velocity of the combination
  6. Jan 6, 2006 #5
    ok..........is it m2gh=1/2(m1+m2)Vf^2 ?
  7. Jan 6, 2006 #6
    the bullet gets stuck in the wood, and the equation would probably be:

    [tex] KE_i = GPE_f [/tex]
  8. Jan 6, 2006 #7
    is it 1/2m1Vo^2=m2gh ?
  9. Jan 6, 2006 #8
    alright, since the bullet gets stuck in the wood, the block and the bullet would rise together, so you need to use both masses

    the equation would be [itex] \frac{1}{2}(m_1 + m_2)(v_i)^2 = (m_1+m_2)(g)(h) [/itex]
  10. Jan 6, 2006 #9
    ok...........now is this correct?



    but isnt that final velocity?
  11. Jan 6, 2006 #10


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    The block rises 0.11m.

    Using v² - u² = 2as

    with u = Vf, v = 0, a = -g, then

    Vf² = 2*9.8*0.11 = 2.156
    Vf = 1.468 m/s

    Plug that into your original momentum eqn,


    to get v1.
  12. Jan 6, 2006 #11
    its the final velocity after the collision occurs, but also the initial velocity as the block+bullet move upwards. so yeah, you're right in a way.
  13. Jan 6, 2006 #12

    thank you so much for your help. i actually worked it out myself but i thought it was such a big number that it couldnt be correct. thanks again for everything
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