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What is elastic collision

  1. Jul 24, 2014 #1

    A collision is said to be elastic if the total kinetic energy of all the bodies involved in the collision remains constant.

    Most collisions are NOT elastic.

    Conservation of momentum applies to ALL unrestrained collisions.


    For a two body collision,

    [tex]m_1v_1^2 + m_2v_2^2 = m_1(v^\prime)_1^2 + m_1(v^\prime)_2^2[/tex]

    Extended explanation

    When two bodies with known velocities collide, two (or more) equations are generally needed to calculate their velocities after the collision.

    Conservation of momentum and of angular momentum are always valid equations for this purpose (but see below for restrictions on their use in restrained collisions).

    Conservation of energy is not, unless the problem specifies that the collision is elastic.

    For example, although conservation of energy applies to a body sliding or rolling down a curved path, it does not apply if the path has a sharp angle, such as where a ramp meets the ground, since that is a collision, and there is no reason to assume that it is elastic.

    Restrained collision:

    A collision is restrained if an external impulsive force acts on one or more of the bodies involved.

    For example, consider an unattached ball colliding with a ball hanging on the end of a string.

    If the collision is from below, then the string will go slack, and there is no force from the string to consider.

    If the collision is from above, then the tension in the string will force the first ball to move along an imaginary spherical surface … this tension force is external and impulsive, and cannot be ignored. However, it is along the direction of the string, which is vertical, and so conservation of momentum still applies in any horizontal direction.

    Conservation of angular momentum also applies, about any axis which passes through the string … but in this case it gives the same result as conservation of momentum, which is easier to use.

    If, instead of a ball on a string, there is a rigid rod suspended from a hinge, then in all cases the hinge forces the rod to move so that (obviously :rolleyes:) the hinged end is stationary. So there is an external impulsive force at the hinge.

    Unlike the string, the direction of this impulse at the hinge is unknown, and so there is no known direction in which momentum is conserved. However, the impulse has no torque about any axis through the hinge, and so angular momentum is conserved about any such axis.

    If one ball is constrained to move along a physical track, or on a physical surface, then there will be an external impulsive reaction force (there may also be a friction force, but it should not be impulsive). Momentum will be conserved in the direction of the track, or in any direction along the surface.


    Impulse is force times time.

    From Newton's second law, total impulse = change in momentum

    By comparison, work done is force times distance, and total work done = change in energy.

    A force which imposes a sudden change in momentum (a "jerk") is impulsive.

    A force which imposes a smooth change in momentum is not impulsive.

    This is all in the context of a collision … any collision, if examined with a fast enough "camera", will be smooth, but we prefer to examine it on a longer time-scale, in which it is jerky by comparison with other forces.​

    For example, friction and gravity are forces which are determined smoothly by the distance moved, and so are not impulsive.

    Why doesn't conservation of energy apply to an inelastic collision?

    The problem is that we are focusing on kinetic energy. In an inelastic collision, kinetic energy is converted into other forms. Common examples include heat, sound, and shock waves.

    * This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
  2. jcsd
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