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Using conservation of energy.

  1. Dec 12, 2016 #1
    1. The problem statement, all variables and given/known data

    PHYSICS.png 2. Relevant equations


    Inital energy=Final energy
    K.Ei+P.Ei=K.Ef+P.Ef
    3. The attempt at a solution

    Physics_2.png
     
  2. jcsd
  3. Dec 12, 2016 #2

    gneill

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    Describe the state of m3 (position, velocity) at the instant m5 hits the floor. What do you expect to happen to m3 after that instant?
     
  4. Dec 12, 2016 #3
    The position of m3 when m5 hits the ground is 4 m high above the ground and m3 will come to rest after that.
     
  5. Dec 12, 2016 #4

    gneill

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    It's the "m3 will come to rest after that" part that you need to look into. How fast and what direction is m3 moving at that instant? Can you think of a way to use energy conservation to find out how high m3 will go?
     
  6. Dec 12, 2016 #5
    m3 is moving up and with velocity 4.43m/s at the instant m5 hits the ground
     
  7. Dec 12, 2016 #6

    gneill

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    So it starts this phase of its journey at a height of 4m and moving upward with velocity 4.43 m/s. Thus it has some initial PE and KE. What happens to the PE and KE from there?
     
  8. Dec 12, 2016 #7
    So you mean in the second case, the m3 will start from height 4m and velocity 4.43m and go some distance above 4m?

    This is how I understood it:
    Inititaly the m3 is at height 4m and inital velocity 4.43m.
    so Initial energy= m3gh+1/2 m V2 (1)

    Then it will move some distance above 4m until it reaches its maximum distance and comes to rest.
    so Final energy of the whole system(both bodies)=m3gh(new highet)+ 0 (2)

    So from (1) and (2)
    m3gh+1/2 m V2=m3gh(new height)
     
  9. Dec 12, 2016 #8

    gneill

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    That looks good. What do you find for the new height?
     
  10. Dec 12, 2016 #9
    height approximatley equal 5.001 m.
     
  11. Dec 12, 2016 #10

    gneill

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    Remember to round to the correct number of significant figures. But otherwise, that looks great!
     
  12. Dec 12, 2016 #11
    Thanks for taking the time to help me. Appreciated :).
     
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