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Mechanics problem (Bullet penetration)

  1. Sep 1, 2008 #1
    1. The problem statement, all variables and given/known data
    A 7.00g bullet, when ired from a gun into a 1.00kg block of wood held in a vise, penetrates the block to a depth of 8.00cm. This block of wood is placed on a frictionless horizontal surface, and a second 7.00g bullet is fired from the gun into the block. To what depth will the bullet penetrate the block in this case?


    2. Relevant equations



    3. The attempt at a solution
    In the first case, by taking the final velocity as 0 and displacement as 8.00cm.
    I can get the initial velocity of bullet in terms of acceleration or time by using kinematics equations.
    or I can find the force of the bullet exerts on the block by equating EK = F[tex]\Delta[/tex]r
    But they are not seems to be useful in case 2.

    In case 2, I am not sure how the mechanics will occur. But, I think the bullet will travel with the same initial velocity, penetrate a shorter distance into the block and bring the block to travel with it with some velocity. I do not know how to put this by using physical principals. I guess I need to use conservation of energy again. By equating
    EK of bullet = F[tex]\Delta[/tex]r + EK of (bullet + block). But I do not think I have enough information to do that. Please help me.
     
  2. jcsd
  3. Sep 1, 2008 #2

    Doc Al

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    Staff: Mentor

    You're on the right track. Hint: What else is conserved during the collision of bullet and block in case 2?
     
  4. Sep 1, 2008 #3
    I think momentum is conserved which gives
    [tex]\frac{1}{2}[/tex]mv1 = (M+m)v2 ...........(1)
    where m=mass of bullet
    M=mass of block
    v1=initial velocity of bullet
    v2=velocity of bullet and block

    From the 1st case,
    I have [tex]\frac{1}{2}[/tex]mv12 = F[tex]\Delta[/tex]r1 ....... (2)

    From the 2nd case,
    [tex]\frac{1}{2}[/tex]mv12 = F[tex]\Delta[/tex]r2 + [tex]\frac{1}{2}[/tex]mv22.... (3)

    By equating (3) and (2)
    I have F([tex]\Delta[/tex]r1-[tex]\Delta[/tex]r2) = [tex]\frac{1}{2}[/tex](M+m)v22

    But I still have one extra unknown (r2 , F , v1) for 2 equations.
    I have one more question here, if I had assumed that the collision is inelastic, will (3) still correct since the kinetic energy is not conserved anymore.
     
    Last edited: Sep 1, 2008
  5. Sep 1, 2008 #4

    Doc Al

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    Staff: Mentor

    Good. But get rid of that 1/2.

    Good.

    Good, but that second "m" should be "M + m".

    You have 3 equations. In any case, noodle around with the equations, expressing everything in terms of F, r_1, & r_2. See what happens.
    The collision is most definitely inelastic. KE is not conserved--total energy is conserved. (Your equation 3 is a total energy equation, not simply a KE equation.)
     
  6. Sep 1, 2008 #5
    Oops... I apologise for all the typing errors.

    Wow, the unknowns canceled like magic. Thank you so much.
    The final answer I got is 7.994cm. Not much different from the first penetration.
     
  7. Sep 1, 2008 #6

    Doc Al

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    I get a slightly different answer, so you might want to recheck your arithmetic. But there's not much difference.
     
  8. Sep 1, 2008 #7
    I recheck my arithmetic. There is nothing wrong.

    The final expression I get is
    mv12 = mv12*[tex]\frac{\Delta r_2}{\Delta r_1}[/tex] + [tex]\frac{m^2}{M+m}[/tex]*v12

    Canceled off all the v and m gives
    [tex]\Delta[/tex]r2 = [tex]\frac{M}{M+m}[/tex]*[tex]\Delta[/tex]r1
    = 1/1.007 * 0.08
    = 0.07994m
     
  9. Sep 1, 2008 #8

    Doc Al

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    Staff: Mentor

    Right.
    According to my calculator, 1/1.007 * 0.08 = 0.07944 m.
     
  10. Sep 1, 2008 #9
    Yea.. Thank you so much for your help.
     
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