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Homework Help: Stuck on conservation of energy problem

  1. Jun 6, 2004 #1
    "On a frictionless, horizontal surface there is a spring with one end fixed and the other attached to a board of mass 1 kg. The spring coefficient is 150 N/m. 500g of clay moving at 3.0 m/s on this horizontal surface collides squarely with the board. As a result, the board and clay stick together, and the spring is compressed. Determine by how much the spring has shortened."

    Since the clay sticks to the board, the coefficient of restitution is probably 0. The kinetic energy of the clay prior to collision is 1/2(500g)9=2.25 J while its potential energy is (I think) 0. The potential and kinetic energy of the spring prior to collision are both 0.
    I'd appreciate a hint on how to start.
    Many thanks
  2. jcsd
  3. Jun 6, 2004 #2


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    "The kinetic energy of the clay prior to collision is 1/2(500g)9=2.25 J while its potential energy is (I think) 0."

    Be careful with your thinking; potential energy is a quantity that can ONLY be shared by two bodies. Clay by itself cannot have a potential energy.

    Since this is an inelastic collision, you cannot use energy methods throughout the duration of the collision, and must resort to conservation of momentum instead. Make your system the board and the clay for the purpose of conserving momentum before and after contact. This will give you the velocity of the board, clay system after contact. AFTER the collision you can start thinking energy and simply use the work energy theorem to calculate the compression of the spring. I don't know if you've gotten to the ballistic pendulum problem yet, but this problem is pretty much analogous. This should help get you started; try it out and if you get stuck again, i'd be happy to help you work through it in more detail.
  4. Jun 6, 2004 #3
    So, if the momentum of the clay is mv= 0.5kg(3 m/s)=1.5 kg/ms, do you mean that all this energy will be conserved by the spring?

    What exactly is the work energy theorem?

  5. Jun 6, 2004 #4
    Did some more reading.
    Since this is a conservation of energy problem, E(initial) = E(final)
    Now, E(initial) is the kinetic energy of the clay 1/2mv^2 = (1/2)(.5kg)(3m/s)^2 = 2.25 J
    E(final) is the potential energy stored in the compressed spring 1/2kx^2 = 1/2(150)x^2
    Therefore... 2.25 = 75x^2
    solving for x gives 0.173 cm

    Does this look right?
  6. Jun 6, 2004 #5
    Just realized I forgot about the board. So since, as Gza said, momentum is concerved here, p(initial) = p(final)
    so, (0.5)3 = (1 + 0.5)V'
    Solving for v' gives 1 m/s (the velocity of the clay/board system)
    Plugging this into E(initial) = E(final) gives
    1/2mv^2 = 1/2kx^2
    x = 0.1 meter
    That looks better!
  7. Jun 6, 2004 #6


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    That's correct. I would recommend studying up on the work energy theorem:
    [tex]W_{ext} = \Delta E [/tex]; since the conservation of energy equation you used in your solution is simply a consequence of this theorem when the external work is zero. Understanding it is powerful in regards to solving problems where the energy may not be conserved (ie: problems involving friction.)
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