Checking a question involving impact/impulse forces

  1. 1. The problem statement, all variables and given/known data
    A ceramic pot of mass 0.5 kg will break if it strikes the floor with an impact force equal
    to, or greater than, 20 N. The collision lasts for 0.05 s. Find the minimum vertical height
    from which, if dropped from rest, the pot will break upon impact with the floor (you
    may ignore air resistance).

    2. Relevant equations
    Impulse=FΔt=Pfinal-Pinital
    Fd=ΔKinetic energy
    PE=mgh
    KE=1/2mv2

    3. The attempt at a solution
    At a height h, the object has KE=0 and PE=mgh=0.5gh
    At impact, the object has KE=1/2mv2=0.25v2 and PE=0
    FΔt=Pf-Pi
    However Pi=0 as the velocity is 0.
    equation 1: 20*0.05=Pf=mv=0.5v
    to find v:
    1/2mv2=mgh
    rearrange to give v=√2gh
    back into equation 1:
    1=0.52*2gh
    h=1/0.52*2*9.81
    h=0.2m

    Is this correct? That doesn't seem big enough in order for the pot to break (thinking in real world terms)
     
  2. jcsd
  3. You need to use the fact that force = change in momentum per second.
    They give you the force so you need to get momentum information
    Do you know how?
     
  4. That's what I attempted to do with the Pf-Pi with P being momentum (mv) ... is that not the right equation?
     
  5. You are correct and if you look at your equation 1 you see that v = 2 m/s ?
    So 2 = √2gh.
     
  6. Perfect - thanks :)
     
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