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Clay falling onto a pan attached to a vertical spring

  1. Feb 4, 2013 #1
    1. The problem statement, all variables and given/known data

    "A spring is hung from the ceiling. A pan of mass 100.0 g is attached to the end, which causes it to stretch 5.00 cm. Find the maximum distance the pan moves downward when a lump of clay of mass 120.0 g is dropped from a height of 40.0 cm onto the pan."

    The correct answer, according to my teacher is 0.233 meters.

    2. Relevant equations

    Conservation of momentum, perfectly inelastic collision.

    m1v1i + m2v2i = (m1 + m2)vf

    Conservation of mechanical energy.

    MEi = MEf

    PEg = mgh
    PEe = 1/2 kx^2
    KE = 1/2 mv^2

    Work kinetic energy theorem.

    Fnet d cosθ = 1/2 m (vf^2 - vi^2)

    3. The attempt at a solution

    I tried a perfectly inelastic conservation of momentum + work-KE theorem approach.

    m(ball)v0 + 0 = M(combined)v(final)

    Fnet d cos(θ) = 1/2 M(vf^2 - vi^2)

    where Fnet was the average spring force minus gravity, F = [k(0.05) + k(x)] / 2 - M(combined)g

    d was (x+0.05)

    and vf = 0, vi = root(2gh)

    UPDATE:

    OK so I tried conservation of mechanical energy after the collision to no avail. I did:

    1/2 (.22kg) (1.53 m/s)^2 + (.22 kg)(9.81) x = 1/2 (19.62 N/m) (x+0.05)^2

    Solving for x I got x = 0.225 m.

    I still can't get x = 0.233m. I'm sure it's not a rounding error of some sort. What am I missing?
     
    Last edited: Feb 4, 2013
  2. jcsd
  3. Feb 4, 2013 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Mechanical energy is not conserved during the collision, so that won't work.

    So far so good.

    You forgot about gravity.

    Hint: After the collision, mechanical energy is conserved.
     
  4. Feb 4, 2013 #3
    Actually I forgot to include gravity in my attempted solution post, but I did include it in my actual calculations already and I still got the wrong answer. : / But Fnet was the average spring force and gravity.

    Thanks for pointing out that mechanical energy is not conserved during the collision. Obviously I should have seen that with it being perfectly inelastic.

    Trying conservation of mechanical energy after the collision.

    EDIT: OK so I tried conservation of mechanical energy after the collision to no avail. I did:

    1/2 (.22kg) (1.53 m/s)^2 + (.22 kg)(9.81) x = 1/2 (19.62 N/m) (x+0.05)^2

    Solving for x I got x = 0.225 m.

    I still can't get x = 0.233m. I'm sure it's not a rounding error of some sort. What am I missing?
     
    Last edited: Feb 4, 2013
  5. Feb 4, 2013 #4

    Doc Al

    User Avatar

    Staff: Mentor

    The left side of your equation is incomplete. You forgot the spring potential energy.
     
  6. Feb 4, 2013 #5
    Derp...Ty!
     
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