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Inelastic collision and conservation of energy

  1. Nov 4, 2005 #1
    I've been working on this problem for a long time now, and my answer's off by about 0.1 m. I've looked at the problem several times, and I can't figure out why my answer's off. I would appreciate any help. Here's the problem, and my work.

    "A 0.2 kg frame, when suspended from a coil spring, is found to stretch the spring 0.05 m. A 0.2 kg lump of putty is dropped from rest onto the frame from a height of 0.3 m. Find the maximum distance the frame moves downward."

    My coordinate system is up - positive, down - negative

    First I solved for k with the given information.

    -Weight = Force of spring
    -mg = kx
    -(0.2)(9.8) = k (-0.05)
    k = 39.2 N/m

    Next, I solved for the final velocity of the putty blob at the instant that it strikes the bottom of the frame. I used conservation of energy; gravitational potential energy is turned into kinetic energy.

    mgh = .5 m v^2 <- m cancels
    2gh = v^2
    2(9.8)(0.3)= V^2
    v= -2.43 m/s

    Since the problem says that 'putty' is dropped, I assumed that they were implying an inelastic collision will occur. If there's an inelastic collision, then not all of the kinetic energy gained from the loss of height is converted into elastic potential energy; some K must be lost to stick the putty to the frame.

    m1v1 + m2v2 = (m1 + m2)v <- the frame is initially at rest, so v2=0
    m1v1 = (m1 + m2)v
    (0.2)(2.43) = (0.2 + 0.2)v
    v = 1.215 m/s

    The combined kinetic energy of the blob and frame is what I presume will move the spring down. Since the blob and frame are initially moving, then they posess kinetic energy that must eventually become elastic potential energy. I used the above velocity to find the kinetic energy of the two frame and putty combined.

    .5mv^2 = .5 k x^2 <- .5 cancels
    mv^2 = k x ^2
    (0.4)(1.215)^2 = (39.2)x^2

    Here's where some confusion comes in. How do I factor in the weight of the blob? Is it already included in the above calculations somehow? Otherwise, I would have to add the displacement due to the added weight separately. The problem states that a 0.2 kg mass displaced the spring 0.05m, so I can add 0.05m to my answer. Even then, the answer becomes .173 m, and that still isn't what the book has as the answer.

    The answer in the back of the book says that the whole shebang is displaced by 0.182m.

    What am I doing wrong? I feel that I must've overlooked some subtlety somewhere, but I can't see what.
  2. jcsd
  3. Nov 4, 2005 #2


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    Homework Helper

    Not really subtle ... you ignored the spring stretch during the putty-fall process and ignored gravity during the spring stretch process.

    In condition C, just after the inelastic collision has completed,
    there is gravitational PE and spring PE as well as KE, nonzero.

    You have a choice as to where to measure height from, for mgh,
    but the spring has a prefered reference point ("relaxed" length)
    and is non-linear (so adding a constant to it DOES make a difference)

    Sorry to break it to you, but you've got a linear term, too.

    There will be 2 roots to the quadratic ... condition D is
    the bottom root, where it stops on the way down
    (the thing will later stop at the other root, at the top of oscillation).
    Last edited: Nov 4, 2005
  4. Nov 5, 2005 #3

    You're right, I completely neglected the work done by gravity during all this; I added an (Fd cos 0) to the right side of the conservation of energy equation to account for the work done by gravity and solved the quadratic on my calculator. Phew, glad that's over.
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