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Homework Help: Dropping ball into spring (inelastic collision)

  1. Aug 9, 2015 #1
    1. The problem statement, all variables and given/known data: I asked a question on this topic before, but I want to make sure I know the material well. So, I looked up another question similar to it (and a little more complex) to check my understanding.

    Here is the practice problem: A board with mass M, when placed into a coil spring (with spring constant k), compresses the spring in a distance of x. A ball of mass m is dropped from rest onto the board from a height h. The ball sticks into the board.

    What is the restoring force made by the spring?

    Bonus: What is the amplitude made by the spring?

    2. Relevant equations: Conservation of energy, conservation of momentum, Fspring = -kx. vb, final2 = 2ad

    3. The attempt at a solution: I am given spring constant k, board mass M, ball mass m, and height h.
    I noticed that the ball sticks into the board, so it is an inelastic collision. Conservation of energy cannot be applied, but conservation of momentum can.

    At the instant the ball touches the board, m(vb, final) = (m + M)(v2b)
    I can find vb, final by using the kinematic equation, vb, final2 = 2ad, with d = h and a = g since it is moving downwards. vb, final is √(2gh).

    Then, (v2b) = (m*√(2gh))/(m + M))

    After collision, I can now apply the conservation of energy.

    But before that, I take note of how the board already tips the spring from equilibrium.
    Mg = kx, x0 is the first compression due to the board alone. (Mg)/k = x0.

    Now, for the energy. (0.5)(m+M)(v2b)2 = 0.5kx12 + (m+M)gx1
    x1 is the compression of the spring due to the ball and the board.
    Solving the quadratic gives me x1.

    To find restoring force, Fspring = -k (x0 + x1)

    For the bonus, amplitude is (0.5)((x0 + x1))

    Is this right? Did I make errors along the way?
    Thanks for all the help in advance!
    Last edited: Aug 9, 2015
  2. jcsd
  3. Aug 9, 2015 #2


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    Need to be careful here. What is the change in PE of the spring as the board descends by x1?
  4. Aug 9, 2015 #3
    The change in PE of the spring is 0.5kx12. But do you mean that it is supposed to be negative? That would make sense.

    I have been reading some more examples, and I think I am going to do a minor revision.
    I'm going to be more explicit, so any flaws in the logic can be corrected.

    After impact of m and M, conservation of energy can be applied.
    E1 = E2
    KE1 + PE1 + SpringPE1 = KE2 + PE2 + SpringPE2

    Substituting known values, I can now write this statement as
    0.5(m+M)(v2b)2 + (m+M)gx1 + 0.5(k)(x0)2 = 0.5(m+M)(0)2 + (m+M)g(0) + 0.5(k)(x1)2

    When the ball and the board compress the spring, the final velocity will be zero, and essentially as it goes downward x1, the "h" in PE2 will also be zero.

    This will be simplified as 0.5(m+M)(v2b)2 + (m+M)gx1 + 0.5(k)(x0)2 = 0.5(k)(x1)2
    0.5(m+M)(v2b)2 + 0.5(k)(x0)2 = 0.5(k)(x1)2 - (m+M)gx1
    0 = 0.5(k)(x1)2 - (m+M)gx1 - [0.5(m+M)(v2b)2 + 0.5(k)(x0)2]

    In here, x1 includes x0 right? Or is 0.5(k)(x1)2 really 0.5(k)(x1 + x0)2?
    That quadratic looks even more complicated. Solving with only variables is going to be complex, unless it cancels somewhere. Are there other principles I may have missed?

    Also, for the bonus question: Amplitude is actually only equal to (x0 + x1).
    Last edited: Aug 9, 2015
  5. Aug 9, 2015 #4


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    No it isn't. What is the PE of the spring before the board hits it? What is the PE when at extension x0+x1?
  6. Aug 9, 2015 #5
    Before the ball hits it, the board already has compressed it x0. So the SpringPE at that instant would be 0.5(k)(x0)2.
    When it is extended, then it will be 0.5(k)(x0 + x1)2.

    So, change in SpringPE is 0.5(k)(x0 + x1)2 - 0.5(k)(x0)2. This simplifies to:
    0.5k((x0 + x1)2 - (x0)2)
    0.5k((x12 + 2x0x1))

    In here, all variables are known except for x1.
    Last edited: Aug 9, 2015
  7. Aug 9, 2015 #6


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    Yes. Continuing with your preceding post...
    Here you correctly included the initial spring PE on the left, but you' ve only got the change in extension on the right.
    What equation do you have now?

    Your answer for the amplitude is wrong. Please post your corrected energy equation and subsequent working.
  8. Aug 9, 2015 #7
    I thought I was so close. Wouldn't the amplitude be equal to the distance the spring is compressed? From -kd = FSpring?

    Anyway, the new E1 = E2 will be as follows.

    0.5(m+M)(v2b)2 + (m+M)gx1 + 0.5k(x0)2 = 0.5k((x12 + 2x0x1))
    I am unsure how to solve for x1 here. My approach is to expand the equation first.
    0.5(m+M)(v2b)2 + (m+M)gx1 + 0.5k(x0)2 = 0.5kx12 + kx0x1
    0 = 0.5kx12 + kx0x1 - 0.5(m+M)(v2b)2 - (m+M)gx1 - 0.5k(x0)2
    0 = 0.5kx12 + [kx0 - (m+M)g]x1 - 0.5[(m+M)(v2b)2 - k(x0)2]
    So, this will be another quadratic. It looks very bad. :(

    Substituting values from the initial post (#1),
    0 = 0.5kx12 + [Mg - mg - Mg]x1 - 0.5m2[(2gh)/(m + M) - g2/k]
    Last edited: Aug 9, 2015
  9. Aug 9, 2015 #8


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    You're not being consistent over the spring PE. On the left you have the initial spring PE, on the right the increase in its PE.
    Either delete the initial PE on the left or change the right hand side to be the final spring PE.
    The amplitude is the maximum displacement above and below the equilibrium position. Where is the equilibrium position?
  10. Aug 9, 2015 #9
    Oh, ok. I will change it right now.
    0.5(m+M)(v2b)2 + (m+M)gx1 = 0.5k((x12 + 2x0x1))
    0.5(m+M)(v2b)2 + (m+M)gx1 = 0.5kx12 + kx0x1
    0 = 0.5kx12 + [kx0 - (m+M)g]x1 - 0.5(m+M)(v2b)2

    Again, I will be using substitutions which I solved for from initial post (#1),
    0 = 0.5kx12 + [Mg - mg - Mg]x1 - 0.5(2m2gh)/(m + M)
    0 = 0.5kx12 + [- mg]x1 - (m2gh)/(m + M)

    Slightly simpler, but still a complex quadratic for x1.

    With my calculations, this quadratic simplifies to [mg ± m √(g(1-2h)/(m+M)) ]/k
    So, x1 = [mg ± m √(g(1-2h)/(m+M)) ]/k

    If this is correct, then restoring force would be k*[mg ± m √(g(1-2h)/(m+M)) ]/k, which cancels out the k.
    This means that restoring force is [mg ± m √(g(1-2h)/(m+M)) ].

    The equilibrium position is without m or M. It is the "zero" position and (x1 + x0) is the displacement going down or up. My thoughts were that at half an oscillation, the compression would represent the peak on a sinusoidal graph.

    Here is my new attempt for the bonus: 2(x1 + x0)
    Amplitude = 2[mg ± m √(g(1-2h)/(m+M)) ]/k + Mg/k
    So amplitude is 2(Mg + [mg ± m √(g(1-2h)/(m+M)) ]) /k
    Last edited: Aug 9, 2015
  11. Aug 9, 2015 #10


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    1-2h makes no sense. It's dimensionally wrong. What would happen if h > .5 (in whatever units!)?
    Your quadratic was dimensionally ok, so the error must be in solving it.
    No, that's the spring's relaxed position. The equilibrium position for the oscillation is where M+m could sit at rest and not oscillate.
  12. Aug 9, 2015 #11
    I'll look at it again. I may have miscalculated or made a mistake on understanding my own handwriting (happens often).

    The equilibrium position is then when kxe = (m+M)g
    and xe is the distance from the zero position.
    So, amplitude is then x1 + x0 - xe
  13. Aug 9, 2015 #12


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  14. Aug 9, 2015 #13
    I'll look over everything again just to make sure.
    Alright! Now, I think I have improved on this topic. Thanks again, haruspex! :smile:
  15. Aug 10, 2015 #14
    Actually, I have one more question about the amplitude of this problem if I can. I looked up other ways to find amplitude and I saw how most strategies were to use angular frequency and maximum velocity. Is this solution different?

    It seems simple enough. But I am not sure if it is a shortcut, at least that I can apply here, maybe?
  16. Aug 10, 2015 #15


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    Both methods work. Which is simpler depends on what facts you start with.
  17. Aug 10, 2015 #16
    Oh, ok. In this case, relating restoring force with amplitude is easier done the original way.
    My class has not discussed that chapter yet though, so I may have to stick with the original.

    Again, thanks a lot! I appreciate the help!
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