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Dropping a Block on a Spring with Completely Inelastic Collision

  1. Aug 12, 2017 at 2:08 PM #1
    1. The problem statement, all variables and given/known data
    A block of mass, m, is dropped from height, h (above the plate), onto a plate of mass, M, which is attached to a spring with spring constant, k. The block sticks to the plate and the system starts to oscillate. What is the amplitude of the oscillations.

    2. Relevant equations
    Etotal=K + Ugravitational + Uspring
    Uspring=0.5kx2
    K=0.5mv2
    Ugravitational=mgh
    Conservation of Energy

    3. The attempt at a solution

    I decided to approach this problem from an energy perspective, setting the equilibrium position of the spring (w/ plate) as 0 gravitational potential energy.

    The initial gravitational potential energy is Ug=mgh. Just before it hits the plate, all of this gravitational PE is converted to KE.

    mgh = 0.5mv2

    v= (2gh)0.5

    The problem specifically states that the block "sticks" to the plate, suggesting a completely inelastic collision. I can use the conservation of momentum to find the velocity after the collision; after the collision, the velocity of the plate and block is:

    vf= m(2gh)0.5 / (m+M)

    The kinetic energy of the system is now:

    0.5(m+M)vf2 = Kf

    The spring will then be compressed from its equilibrium position until the velocity of the plate/block becomes zero, which is the max compression/amplitude. At this point, all of the kinetic energy has been converted to elastic potential energy.

    0.5(m+M)vf2 = 0.5kx2

    I can then plug in my equation for vf and solve for x, which is the maximum compression/amplitude.

    Is this approach okay? Did I make any incorrect assumptions?

    Thaaaaaanks!
     
  2. jcsd
  3. Aug 12, 2017 at 4:05 PM #2

    haruspex

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    Did gravity mysterously turn off at point of impact?
     
  4. Aug 12, 2017 at 4:38 PM #3
    I didn't think about that since I set my reference point to be the equilibrium position of the spring.

    So after collision, the gravitational potential energy from the height, h, was either converted to kinetic energy or lost in the inelastic collision. The spring will compress until there is no kinetic energy, but some of this energy will be stored up in gravitational potential energy?

    0.5(m+M)vf2=(m+M)gx + 0.5kx2 + 0.5(m+M)(0)2

    This would then be a quadratic formula and I could solve for x.

    I'm a little hesitant on this approach now since I appear to be gaining gravitational potential energy as the plate/block goes downward. Is this approach at all possible or should I just restart and set the ground as potential energy zero and work from there?

    Thanks for the reply!
     
  5. Aug 12, 2017 at 4:52 PM #4

    haruspex

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    Er.. no.
    As the spring compresses the masses descend. So what is happening to GPE?
     
  6. Aug 12, 2017 at 5:00 PM #5
    It would continue to decrease normally. Would it be converted to elastic potential energy as well? I'm getting tripped up since I defined the top of the spring in equilibrium position to be 0 GPE. I suppose it would also decrease in this case too since the height is decreasing and that's what gravitational potential energy does.
     
  7. Aug 12, 2017 at 5:09 PM #6

    CWatters

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    Correct. The total height for the purpose of calculating GPE is h plus however much the spring compresses.
     
  8. Aug 12, 2017 at 5:48 PM #7
    So to recap, the kinetic energy that remains is converted to elastic PE. As the masses descend on top of the spring, their potential energies are converted to elastic potential energies as well:

    m2gh/(m+M) + max = 0.5kx2
    (the m2gh/(m+M) is a simplified kinetic energy term)

    This is a quadratic equation I can solve.
     
  9. Aug 12, 2017 at 6:20 PM #8

    haruspex

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    Yes (where max=?).
     
  10. Aug 12, 2017 at 8:17 PM #9
    My bad; I made a typo. Max should be mgx (autocorrect)
     
  11. Aug 12, 2017 at 8:26 PM #10

    haruspex

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    So, the block is passing through the plate?
     
  12. Aug 12, 2017 at 10:38 PM #11
    Oh the block and the plate both move downwards. So the amount of potential energy converted is actually (m+M)gh. Thanks for catching that!
     
  13. Aug 13, 2017 at 12:20 AM #12

    haruspex

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    (m+M)gx.
     
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