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Kinetic Energy and Momentum

  1. May 23, 2004 #1
    I am having problems with the kinetic energy formula KE = 1/2 mv^2.
    If an object of 1kg travels at a speed of 1ms its kinetic energy is 1/2 * 1 * 1^2 = 0.5J.
    But if it collides with an object of 0.5kg which is stationary, to conserve momentum which is equal to mass * velocity = 1 * 1 = 1kgms the new speed of the 0.5kg object = momentum / mass = 1/0.5 = 2ms. The kinetic energy of this new object is equal to 1/2 * 0.5 * 2^2 = 1/4 * 4 = 1J. This is double the energy of the initial object which is impossible. It does not make sense to me that the faster an object travels a disproportionate amount of energy is required. I think that the origin of KE (1/2 mv^2), the integration of F=ma, is something that cannot be integrated in reality, only in theory, if momentum is to be conserved.
    Can anyone shed any light on this?

    Stephen Lewis
  2. jcsd
  3. May 23, 2004 #2


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    What you're doigng is describing what is called an inelastic collsion, in an inelsatic collsion both kinetic energy and momentum cannot be conserved here's the one-diemsnional proof

    Imagine two objects of masses [itex]m_1[/itex] and [itex]m_2[/itex], travelling with velocities of [itex]v_1[/itex] and [itex]v_2[/itex] which after collding inelastically form a new object of [itex]m_3[/itex], travelling at velocity [itex]m_3[/itex], which due to the conservations of mass must equal [itex]m_1 + m_2[/itex]

    We can say this due to the conservation of momentum:

    [tex]m_3v_3 = m_1v_1 + m_2v_2[/tex]


    [tex]v_3 = \frac{m_1v_1 + m_2v_2}{m_3} [/tex]

    We can also say that due to the conservation of energy:

    [tex]\frac{1}{2}m_3{v_3}^2 = \frac{1}{2}m_1{v_1}^2 + \frac{1}{2}m_2{v_2}^2[/tex]


    [tex]v_3 = \sqrt{\frac{m_1{v_1}^2 + m_2{v_2}^2}{m_3}} [/tex]

    combing the equations we get:

    [tex]\frac{m_1v_1 + m_2v_2}{m_3} = \sqrt{\frac{m_1{v_1}^2 + m_2{v_2}^2}{m_3}} [/tex]

    square and mutiply [itex]{m_3}^2[/itex] byboth sides,substitue in [itex]m_3 = m_1 + m_2[/itex] and mutiply out:

    [tex]{m_1}^2{v_1}^2 + {m_2}^2{v_2}^2 + 2m_1m_2v_1v_2 = {m_1}^2{v_1}^2 + {m_2}^2{v_2}^2 + m_1m_2{v_1}^2 + m_1m_2v_2^2[/tex]

    Simply eliminate and you get:

    [tex]2v_1v_2 = v_1^2 + v_2^2[/tex]

    Which can be re-arranged as:

    [tex]v_1^2 - 2v_1v_2 + v_2^2 = 0[/tex]

    using the quadratic formula we can solve for [itex]v_1[/itex]

    And we find that:

    [tex]v_1 = v_2[/tex]

    So for an inelastic collision the intial velcoties of the two colliding objects must be the same, hence no collision.

    Therfore the nergy must take some other form rtaher than kinetic enrgy.
  4. May 23, 2004 #3

    Doc Al

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    Says who? During the collision the total momentum is conserved. In this case, the total momentum equals 1 kg-m/s. The situation you describe, where the struck object gets all the momentum, cannot happen--it would violate conservation of energy (assuming there is no energy source--like an explosive--involved). It could happen, but only if the two objects were the same mass.
    Right. It won't happen.
    It works just fine, done correctly. :smile:
  5. May 23, 2004 #4

    Doc Al

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    super-elastic collision

    One more note. Collisions are often classified as elastic or inelastic. In a purely elastic collision, the KE is conserved. In a more realistic inelastic collision, some of that energy is lost to thermal energy and deformation of the objects: so the total KE after the collision is less than what it was before the collision.

    The example you gave--in which the total KE increased--would be a super-elastic collision. This would require an additional source of energy.
  6. May 26, 2004 #5
    You need to consider Newton's coefficient of Restitution

    e = Relative Speed of separation/Relative Speed of Approach

    Where e is the the coefficient of resitution. This is used when there is a energy loss. i.e. the collision is inelastic.

    Yes youth!
  7. Jun 14, 2004 #6
    Thanks to all those who replied.
    Sorry for the long delay in this reply.

    One final thing, could two objects possess the same momentum but hold different kinetic energies, for example an object of 1kg mass travelling at 1ms, KE = 1/2 * 1 * 1^2 = 0.5J and momentum = 1 * 1 = 1kgms and an object of 0.5kg mass travelling at 2ms KE = 1/2 * 0.5 * 2^2 = 1J and momentum = 0.5 * 2 = 1kgms.
    Is this possible? If so, can you explain why?

    Many thanks

    Stephen Lewis
  8. Jun 14, 2004 #7

    Doc Al

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    It's not only possible, but if the objects have different masses it's inevitable. Momentum and kinetic energy are two entirely different concepts.
  9. Jun 15, 2004 #8
    Thanks Doc Al, very interesting.
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