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Conservation of momentum and energy

  1. Sep 27, 2007 #1
    1. The problem statement, all variables and given/known data

    A pendulum consists of a 0.5 kg bob attached to a string of length 1.6 m. A block of mass m rests on a horizontal frictionless surface. The pendulum is released from rest at an angle of 53° with the vertical and the bob collides elastically with the block. Following the collision, the maximum angle of the pendulum with the vertical is 5.73°. Determine the two possible values of the mass m.


    2. Relevant equations

    Conservation of momentum and energy


    3. The attempt at a solution

    I got 2.16 kg for one...which isn't right.
     
  2. jcsd
  3. Sep 27, 2007 #2

    Mentz114

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    Hi, you must show how you got that result, so we can fix it if it's wrong.
     
  4. Sep 27, 2007 #3
    Ki(objec1t)+Ki(object2)+Ui(object1)+Ui(object2)=Kf(object1)+Kf(object2)+Uf(object1)+Uf(object2)

    K= kinetic energy------U= potential energy-------i=initial------f=final
    K=1/2mv^2-----------U=mgh

    0+0+0+cos(53)*1.6*.5*9.8=sin(5.73)(1.6)+(.5)(cos(5.73)*1.6)+m2vf^2

    Change in kinetic energy = 0 so,

    (m1v1i^2+m2vi^2)+(m1v1f^2+m2v2f^2)=0 algebra...

    m1(v1f^2-v1i^2)=m2(v2i^2-v2f^2) where m2=3.76/v^2 from the conversation of momentum part

    then substituted

    v^2=.8/3.76
     
  5. Sep 27, 2007 #4

    Mentz114

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    OK, the LHS is the potential energy at the start. That should be sin(53) ( easy to remember because as the angle increases so does the pot.)
    On the right we should have the final potential energy of the bob plus the kinetic E of the mass.

    sin(5.73)*1.6*.5*9.8 + m2v2f^2
    To get the momentum equation, work out the velocity of the bob at the lowest point, and you've got it ( after a bit of algebra)
     
  6. Sep 27, 2007 #5
    Sorry I'm a little confused. I'm not sure how to get the velocity at the lowest point other than it being sin(0)*1.6 which would be 0. And doesn't " sin(5.73)*1.6*.5*9.8 + m2v2f^2" go into the momentum equation?

    If it's easier to talk on aim, my sn is:

    Royguitarboy246

    Edit: Wouldn't that be cos(0)*1.6 which would then equal 1.6?
     
    Last edited: Sep 27, 2007
  7. Sep 27, 2007 #6

    Mentz114

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    You get the velocity at the lowest point by equating the initial pot energy of the bob with its KE at the lowest point.

    sin(57)*1.6*0.5*9.8 = 0.5*0.5*vf^2

    You can get the velocity of the bob after the collision using the same approach.

    No, that's energy ! The momentum equation is

    0.5*( vi-vf) = m*v2 ( v2 is the speed of the mass after the collision, vi, vf are bob speeds before and after.)

    I have to go now. You can do it. Use a pencil, paper and eraser, and list your quantities carefully.
     
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