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Block of wood friction problem

  1. Oct 24, 2004 #1
    a 7 g bullet fires into a 1kg block of wood held in a vise, will penetrate the block a depth of 8.00 cm . this block of wood is placed on frictionless horizontal surface, and a 7.00 g bullet is fired from the gun into the block. to what depth will the bullet penetrate the block ?
    I bang my head in the wall, but I coudn't find a hint to solve this . Could you guy please give me a hint . Thank you
     
  2. jcsd
  3. Oct 25, 2004 #2

    arildno

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    a) Penetration depth is proportional to kinetic energy spent on penetration.
    Let us consider the case where a bullet penetrates a wall (i.e, the bullet doesn't cause the other object to move.)
    All the bullet's kinetic energy is used to deform the wall, and is subsequently removed from the system in the form of heat&sound (or remains as heightened temperature).
    In terms of the average force F acting on the bullet, we have:
    [tex]\frac{1}{2}mv_{0}^{2}=Fd[/tex]
    where the penetration depth d is seen to be proportional to the initial kinetic energy.

    b)In the case where the wooden block starts to move, only part of the system's (bullet+block) initial kinetic energy has been expended for penetration.
    Some remains as kinetic energy.
    Therefore, the new penetration depth is proportional to the difference between system's initial and final kinetic energies. To solve the problem, assume the proportionality constant to be the same.
     
  4. Oct 25, 2004 #3
    thank for your reply . So the above problem does not give me enough info to solve. Is it right? (because it does not give me V and F)
     
  5. Oct 25, 2004 #4

    arildno

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    Oh, yes it does!
    1) F is the proportionality constant that I said should be regarded as the same in both cases.

    2)Initial velocity is NOT necessary, you may find the new penetration depth as a fraction of the old penetration depth:
    a) In the first case, we have:
    [tex]\frac{1}{2}m_{bull}v_{0}^{2}=Fd_{0}[/tex]
    ([tex]d_{0}=8cm[/tex])
    In the second case, you have the equation:
    [tex]\frac{1}{2}m_{bull}v_{0}^{2}-\frac{1}{2}(m_{bull}+M_{block})V^{2}=Fd_{new}[/tex]
    Dividing the first equation on the last, yields:
    [tex](1-\frac{m_{bull}+M_{block}}{m_{bull}}(\frac{V}{v_{0}})^{2})=\frac{d_{new}}{d_{0}}[/tex]
    Since you may express V in terms of [tex]v_{0}[/tex] by conservation of linear momentum, you get determinate solution.
     
    Last edited: Oct 25, 2004
  6. Nov 2, 2004 #5
    yes, but how do you solve for the V(bullet + Block) in terms of V(bullet)? :uhh:
     
  7. Nov 2, 2004 #6

    arildno

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    This is given by conservation of linear momentum:
    [tex]m_{bull}V(bull)=(m_{bull}+m_{block})V(bull+block)[/tex]
     
  8. Nov 2, 2004 #7

    Andrew Mason

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    This is correct, of course. But the problem does not require a detailed mathematical approach. We know that the energy of the block will be about .7 percent of the energy of the bullet (since the recoil speed of the block is 7/1007 of the bullet speed and the mass is 1007/7 of the original mass, the recoil energy is 7/1007 of the bullet energy). The rest is dissipated by the bullet in penetrating the block. So, the bullet will penetrate 7.94 cm. (99.3 percent of the original cm depth).

    AM
     
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