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B How will the two balls move after the collision?

  1. Feb 4, 2017 #1
    There are two identical ball,ball A with velocity v and ball B is at rest. Ball A hit ball B directly. How will the two balls move after the collision?
    In my mind there are three cases:
    1. Ball A stops and ball B moves with the velocity v.
    2. Two balls move together with same speed v/2.
    3. Two ball moves with different velocity that B is faster.
    Case 1 is elastic collision
    Case 2 is perfectly inelastic collision
    Case 3 is inelastic collision
    What is the factor affecting the resulting movement of the two balls? How do I know if it is case 1 2 or 3 ? Since the difference between them is energy loss, I'm guessing if it is based on the source of energy loss. For example, it is likely to be case 1 if the collision is in space since no sound is produced and the energy loss is reduced. You could say there is still energy loss in heat or radiation but there are ways to reduce them. My main point is that the movement of the balls is determined by the energy loss while the momentum is still conserved. Am I correct?

    If I am correct,then something strange happened. If two balls have glue on them and they must move together after the collision, which means it has to be case 2, does that mean more energy is lost because of the glue? Before sticking them together, may be it is case 3 and some energy is lost by heat or radiation. After adding the glue, more radiation and heat is produced? Why? Or the energy is lost in the glue? Where is the energy? That confused me. Please help.
  2. jcsd
  3. Feb 4, 2017 #2
    Let's assume that the glue does not absorb heat. There can be some improvement in the question such that the "glue is heated up "theory cannot explain the energy loss.
  4. Feb 4, 2017 #3


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    Sound is a miniscule mode of energy loss. You should ignore it entirely.

    In an elastic collision, the balls deform slightly, rebound and then separate. Some kinetic energy is lost in vibrations associated with the deformation, but with very elastic balls, over ninety percent of the kinetic energy is present in the rebound and less than ten percent is lost into vibrations.

    But you've thrown a wrench into the works. The glue prevents the balls from separating and carrying that kinetic energy away. Instead, the balls stretch apart, attached by the glue and rebound back together again. To a reasonable approximation, at every impact/rebound event and at every stretch/rebound event another ten percent of the remaining energy is lost into vibration. Quite quickly (almost instantly), the result is that 100% of the original energy is lost as vibrations which will, in turn, quickly dissipate into heat energy.

    You've turned an elastic collision into an inelastic collision.

    Edit: If you've ever bounced a ping pong ball between two paddles while bringing those paddles together on the ball, you may have a faint idea of how rapid a series of bounces can become.
  5. Feb 6, 2017 #4
    Why do you classify Case 2 or even Case3 as necessarily inelastic? The kinetic energy before the collision is exactly the same as the kinetic energy after the collision in Case 2, therefore the collision must be elastic. This does however require a third process which causes the two balls to attach to each other. (One can imagine one ball coated with velcro hooks and the other coated with the velcro fibresmat for example). Are you perhaps confusing the elasticity (or inelasticity) of the bodies taking part in the collision with the elasticity or inelasticity of the collision itself. They are separate concepts.

    An inelastic collision requires a mechanism for transferring some of the kinetic energy to another form of energy not associated with the motion of the particles. This could, for example, be a deformation process in which some of energy is converted to heat or the deformation does not return to its former state. (In molecular or atomic bodies, energy can be transferred into the creation of an excited molecular or atomic state or even to the creation of an excited nuclear state and in these cases the collision is inelastic).

    A better description of the three cases you describe is perhaps

    Case 1 All the momentum and energy is transferred from body A to B
    Case 2 Body A and B form one body after the collision
    Case 3 Part of the momentum and energy of body A is transferred to body B

    Whether the collisions are elastic or inelastic is a separate issue.

    Case 1 would occur when the material of the bodies is perfectly inelastic (as distinct from the collision being inelastic) e.g. billiard balls.

    I am not sure, but I don't think case 3 can occur in an elastic collision between inelastic bodies but it may be possible in an elastic collision between elastic bodies or an elastic collision of an elastic body with an inelastic one.
    Last edited: Feb 6, 2017
  6. Feb 6, 2017 #5


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    In case 2, the balls have different speeds before the collision and the same speed after. Given that momentum is conserved, kinetic energy is necessarily lost.

    This can most easily be seen by shifting to a frame of reference where the center of mass is motionless. Prior to the collision the system has non-zero kinetic energy in this frame. After the collision the system has zero kinetic energy in this frame.
  7. Feb 6, 2017 #6


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    Yes. That is the starting rule for any of these problems. The details of how the energy may be transferred from Kinetic to other forms are secondary.
    I remember being introduced to The Coefficient of Restitution in my A Level Dynamics classes. It was a useful quantity for solving basic A level problems but I don't remember the Energy situation being discussed as much as it should have been.
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