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Bullet is fired into block at at offset, conversation of kinetic energy?

  1. Oct 25, 2014 #1
    1. The problem statement, all variables and given/known data
    Hello, first year engineering student here :)https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-xpf1/v/t35.0-12/10744040_871781156179440_107104578_o.jpg?oh=73c4261941e7ef6b7c682558c0925a63&oe=544F5CA4&__gda__=1414428985_0415c28b8b2e194bf0b57f065267fe49

    The question is

    in Case B, what percentage of the bullet's initial KE is conserved as KE in the impact?
    2. Relevant equations

    KE=(1/2)*m*v^2 + (1/2)*I*w^2 where I=mass moment of inertia, w=angular velocity
    (angular momentum) H=m*v*r*sin(theta) (maybe? Not sure if this is revelant...)
    (angular momentum) H=I*w (also not sure if relevant)
    3. The attempt at a solution
    I found the mass moment of the block using I=(1/12)*m*(a^2+b^2) and the bullet I=m*r^2,
    then I used H=m*v*r*sin(theta) to find H, then use w=H/I to find angular velocity, where I is I(bullet)+I(block)

    then I plugged in all the values into the first KE equation above, but got the wrong answer. (BTW, I found final v of the block+bullet by Conservation of Momentum)
    Last edited: Oct 25, 2014
  2. jcsd
  3. Oct 25, 2014 #2
    I've had enough conversations of kinetic energy for one night :)
  4. Oct 25, 2014 #3


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    About what point will bullet+block rotate? What is the moment of inertia about that point?
  5. Oct 25, 2014 #4
    I think the bullet+block will rotate about its centre of mass? If that's the case then the moment of inertia can be found using I=1/2*m*(a^2+b^2). But the question hints that we should include the bullet's effect on the moment of inertia (the bullet's moment of inertia would be I=m*r^2, yeah?).
  6. Oct 26, 2014 #5


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    The centre of mass of what, precisely?
    Where r is what?
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