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A bullet embedded in a sliding block

  1. Dec 4, 2009 #1
    1. The problem statement, all variables and given/known data

    A bullet of mass m is fired with horizontal velocity u at a block of mass M. The bullet embeds itself inside the block. The force of resistance of the block is constant.
    Find the depths of penetration in the cases:
    a) The block is held fixed.
    b) The block is free to move on a smooth horizontal surface


    2. Relevant equations

    Conservation of momentum? Kinematic equations? Conservation of energy? I don't know.

    3. The attempt at a solution

    Using simple kinematics for a), with the resistive force [tex]F = -k[/tex] of the block on the bullet, I get the depth, x, as

    [tex]x = \frac{mu^2}{2k}[/tex]

    However, I'm having more trouble with part b)

    I've tried a conservation of energy/conservation of momentum approach:

    [tex]mu = (M+m)v \Rightarrow v = \frac{m}{M+m}u[/tex]

    Now, the work done by the bullet on the block is [tex]Fx[/tex], which translates into [tex]\frac{1}{2}Mv^2[/tex].

    [tex]Fx = \frac{1}{2}Mv^2[/tex]

    [tex]kx = \frac{1}{2}Mv^2[/tex]

    [tex]x = \frac{Mv^2}{2k}=\frac{M}{2k}\left(\frac{M}{M+m}u\right)^2[/tex]

    However, this isn't correct, what's the proper way to do this?

    Thanks
     
  2. jcsd
  3. Dec 4, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    This is good; v is the final speed of the 'bullet + block'.

    Hint: Compare the initial and final kinetic energy of the system. (How much is 'lost'?)
     
  4. Dec 4, 2009 #3
    Ah right!
    [tex]Fx = \frac{1}{2}mu^2 - \frac{1}{2}(M+m)v^2[/tex]

    Thanks Doc Al :)
     
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