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The swinging block

  1. May 18, 2012 #1
    1. The problem statement, all variables and given/known data
    : A bullet of mass 25g is fired at 425m/s into a block of wood , mass 1.200kg,hanging at the end of a 1.0m long string which is attached to the ceiling .The bullet becomes embedded in the block causing it to swing back along an arc.Through what angle does the block swing back?


    2. Relevant equations



    3. The attempt at a solution
    0.5*0.025*(425)^2=(0.025+1.2) * 9.81h +0.5(0.025+1.2)v^2
    h=L-Lcosθ
    you cannot use the conservation of momentum because its not a linear motion
    the height needs the angle
    i feel the many unknowns please help
     
  2. jcsd
  3. May 18, 2012 #2
    It looks like you took the initial kinetic energy of the bullet and set it equal to the potential energy of the bullet/block system plus that systems kinetic energy.

    When I first saw this problem, I though that that was what you did, and then set the final velocity to 0; essentially saying the KE of the bullet goes into gravitational potential of the system. But you can't say this, as energy is lost in the collision.


    So you would have to take the kinetic energy of the bullet/block system immediately after the collision and then set that equal to the gravitational potential to find height.
     
  4. May 18, 2012 #3
    So could i use conservation of momentum

    (mv)1 =(m1 + m2)v
    v= (0.025*425)/(0.025+1.2)= 8.67m/s

    0.5mv^2 = (m1 +m2) gh
    0.5*(0.025 +1.2) * (8.67) = (0.025+1.2)*(9.8)h
    h=3.8m which cannot be true it cannot be greater than 1m
     
  5. May 18, 2012 #4
    I got the same answer.

    Maybe my approach is also flawed, but maybe the answer is just saying the block will have enough energy to complete a loop around the wire.
     
  6. May 18, 2012 #5
    so the angle will be L-LCos(theta)=h
    L=1 1-cos(theta)=3.8
    theta is undefined
     
  7. May 18, 2012 #6
    Because the system will have so much energy, the energy will never transfer completely to gravitational and some will remain kinetic.
     
  8. May 18, 2012 #7
    so we have to include the kinetic energy at the top?
     
  9. May 18, 2012 #8
    I'm not sure. When it gets to the top it will continue to rotate around in circles forever (ignoring any dampening effects).
    Unless the block is moving too slowly at the top to keep the string taught, in which case you might have to find the angle at which the string slacks and that is what the question is asking for? But that seems a little too involved for what should be a simple problem.

    Is it possible the question was written wrong?


    The total energy here is 46.08J, and at the top the kinetic energy is 22.07J, which means the speed of the system is 6m/s.
    Ac = v2/r, so g = v2, and .5v2 is ~36, meaning the block will complete the loop and swing indefinitely.
     
    Last edited: May 18, 2012
  10. May 18, 2012 #9
    Chances are yes,but thanks for helping I will consult my lecturer.
     
  11. May 18, 2012 #10
    I agree with all your calculations..... I think there is something wrong in the original question.
     
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