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Impulse momentum and morequestion

  1. Oct 28, 2007 #1
    The problem statement, all variables and given/known data

    A block of mass 200g is suspended through a vertical spring.The spring is stretched by 1cm when the block is in equilibrium.A particle of mass 120g is dropped on the block from a height of 45cm.The particle sticks to the block after the impact.Find the max extension of the spring.

    The attempt at a solution

    Well in the beginning my aim was to find the force constant k.But i am stuck ..should i use k|x| = mg or work done by spring = loss in gravitational PE i.e (kx^2)/2 = mgx.Getting diff result in both. After that I must find out the velocity of the system due to the impulse but how?Then i can apply conservation of mechanical energy I guess i.e KE just after the collision + spring PE_1 = mgh + spring PE_2. (where spring PE_1 is the spring PE before the impact and spring PE_2 is spring PE when it is max stretched.)

    Plz help
    Last edited: Oct 28, 2007
  2. jcsd
  3. Nov 5, 2007 #2
    You should not use (kx^2) / 2 = mgx because, if the weight is allowed to fall freely under gravity, the GPE will be translated into spring PE (SPE) plus KE (as you said when considering conservation of mechanical energy):
    GPE + SPE + KE = constant

    It may help to recall the definition of impulse.
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