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I Kinematics 1D

  1. Feb 21, 2017 #1
    A bullet is fired into a block of wood at a speed of 200 m/s. Upon hitting the block of wood the bullet decelerates with magnitude 640,000 m/s2 and the block accelerates with magnitude 360,000 m/s2. When the bullet and block obtain the same speed the accelerations stop. At this point, the bullet is inside the block, moving with the block.
    (a) What is the speed of the bullet and block after the accelerations stop?
     
  2. jcsd
  3. Dec 19, 2017 #2
    I was browsing through older posts when I saw your question...
    Let v denotes velocity of the bullet
    v = v_o + a_o*t, v_o = 200 m/s, a_o = - 640,000m/s2
    Let V denotes velocity of the wood block
    V = V_o + A_o*t, V_o = 0 m/s, A_o = 360,000 m/s2
    at some time t, v = V, we get
    200 - 640,000t = 360,000t
    1,000,000t = 200
    t = 2x10^-4s
    Substituting t into V equation, you should get V = 72m/s, which has same value as the speed of the bullet and block after the accelerations stop.
     
  4. Dec 19, 2017 #3

    jim mcnamara

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    Staff: Mentor

    The OP is no longer active. Please do not give complete answers when others ask homework questions. This thread is closed.
     
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