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Conservation of Momentum or Energy in a collision?

  1. Mar 8, 2015 #1
    1. The problem statement, all variables and given/known data
    A ball with mass 2kg is travelling 2m/s. It impacts and sticks to a hanging, stationary 4kg pendulum. How high will the pendulum + ball go?

    2. Relevant equations
    KE = 1/2mv2
    PEg = mgh
    mava = m(a+b)v(a+b)

    3. The attempt at a solution
    I tried to solve this question assuming conservation of energy in the collision, while a friend tried to solve it using conservation of momentum, and we got two separate answers. The question itself dos not say whether the collision is elastic or inelastic (is there some way to figure it out from the given information?).

    My attempt:

    Let the ball be represented by a
    Let the ball + pendulum be represented by (a+b)

    KEa = PEga+b
    1/2mava2 = m(a+b)gh
    1/2(2)(2)2 = 6gh
    4 / 6g = h
    0.066 = h

    My friend's attempt:

    mava = m(a+b)v(a+b)
    2(2) = 6V(a+b)
    V(a+b) = 2/3

    KE(a+b) = PEg(a+b)
    1/2m(a+b)v(a+b)2 = 6gh(a+b)
    1/2(6)(2/3)2 = 6gh
    1.33 = 6gh
    1.33/6g = h
    0.022 = h

    Neither of us can see where the other went wrong; can someone please shed some light on which way is correct?
  2. jcsd
  3. Mar 8, 2015 #2


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    The question states that the ball "impacts and sticks", which automatically makes the collision inelastic. Conservation of energy does not work for inelastic collisions.
  4. Mar 8, 2015 #3
    Thanks, that's pretty clear-cut. A few followup questions:

    1) I had been under the impression that the sole criterion for a collision to be considered inelastic is that there is some permanent deformation of the objects involved. Is that incorrect? Or is it correct and the sticking together of the ball and pendulum is considered a 'deformation'?

    2) The difference in height is a factor of 3 between our answers. Interestingly, this is the same as the difference between the ball's mass and the ball + pendulum's mass. Is there a generalizable relationship between the mass of the pre-collision object, the mass of the post-collision object, and the energy lost in the collision? Like energy pre-collision / energy post-collision = mass pre-collision/mass post-collision?

    3) The principle of conservation of energy must still hold true, I'd imagine. So that means the 2/3rds of the ball's kinetic energy that are not in the ball + pendulum's potential gravitational energy have gone... somewhere. Do they just dissipate as sound and heat in the collision? Or is there some other mechanism by which energy is not conserved during inelastic collisions?

    Thanks a bunch!
  5. Mar 8, 2015 #4


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    Needn't be permanent. There's mechanical energy lost every time you hit a tennis ball.
    You could obtain some general rule for simple linear collisions, but here there's rotation involved.

    Edit: Rereading the question, it is effectively a simple linear collision here.
    Last edited: Mar 9, 2015
  6. Mar 8, 2015 #5


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    Staff: Mentor

    There is a range of collisions between "perfectly elastic" and "perfectly inelastic" where energy can be lost to different processes, including deformations. When you have a chance, look up the "coefficient of restitution".
    Sounds like something you could investigate using the equations for collisions. I haven't looked into it myself.
    Energy will find its way to any process it can reach. Sound and heat are typical sinks, but other mechanical vibrations or displacements/compressions are possible too. If there's magnetic or electric fields involved they can act as a way to move energy around too. In the end though, pretty much all the lost energy will degrade to waste heat unless it is somehow directed into a potential energy form (for example a collision might power a small generator and store a charge on a capacitor, or move a piston to compress a gas).
  7. Mar 9, 2015 #6


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    The ball sticks to the pendulum, so the collision is completely inelastic. The initial PE of the ball is not the same as the final PE of the ball+pendulum. Your approach is wrong. .

    That is correct. The momentum or angular momentum conserves during the collision.
    After the collision, the energy conserves during the motion of the pendulum.
    The unit (m) is missing, otherwise it is correct.
  8. Mar 9, 2015 #7


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    Yes. You should be able to prove this yourself:

    If ##m_1## is the incident mass and ##m_2## is the (stationary) target mass, show that the proportion of kinetic energy retained in a perfectly inelastic collision is ##\frac{m_1}{m_1 + m_2}##
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