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Conservation of Work and Energy

  1. Nov 13, 2015 #1
    1. The problem statement, all variables and given/known data
    Hello, I am having trouble understanding the logic behind the solution to the posted problem. How did they deduce the equation for "Just after impact" ? I don't see how
    $$v_y = \sqrt{2g(0.6)}$$

    What assumptions did they make or how did they get this simplified equation?

    I understand how they got the formula for "Just before impact"

    In the y direction:
    $$mg(1.6)+0.5m(0)^2 = mg(0)+0.5mv_f^2$$
    $$v = \sqrt{2g(1.6)}$$

    But I don't see how they applied this formula for "just after impact"

    Any guidance is really appreciated.

    Thank you for your time.

    kMJDUvb.png
     
  2. jcsd
  3. Nov 13, 2015 #2
    They are applying the formula for motion under constant acceleration because there is a resultant force acting on the plate. It is actually v2=u2+2as
     
  4. Nov 13, 2015 #3
    So what happened to u2 ? Why is it assumed to be 0 here?
     
  5. Nov 13, 2015 #4

    Student100

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    Education Advisor
    Gold Member

    Apply the same logic for the energy of the system after collision.
     
  6. Nov 13, 2015 #5
    u is not assumed to be 0. Look back at the previous diagram and u will know why =)
     
  7. Nov 13, 2015 #6

    ehild

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    Homework Helper
    Gold Member

    Use the same equation for conservation of energy mgh+0.5mvy2=constant, but with initial height 0 and velocity unknown and final velocity zero at height 0.6 m.
     
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