The specifics of a question I am working on are, "After a 0.280-kg rubber ball is dropped from a height of 1.80 m, it bounces off a concrete floor and rebounds to a height of 1.45 m."

Why doesn't the ball return to the same height?


1 Answer 1


The ball is deformed while bouncing off. In theory, this can be modelled as an entirely elastic process as a relatively good approximation, however, it actually is not, as some energy is lost in the process and radiated away as heat (try deforming a ball a few hundred times, it will heat up).

The process is therefore not entirely elastic, which reduces the kinetic energy of the ball.

Additionally, a number of other forces affect the ball, listing those mentioned above again for completenes and ordered roughly by the magnitude of the effect:

  • Energy lost due to inelasticity of the ball-earth interaction (ball heats up)
  • Friction of the ball with the air, causing it to slow down
  • Friction of the ball with the ground ("stuck to the ground")
  • Roughness of ground causing the ball to start spinning or change direction
  • Forces stemming from the fact that the earth rotates, although this should mostly affect horizontal velocity
  • Momentum transferred to earth
  • 1
    $\begingroup$ There's also some drag due to air resistance ... $\endgroup$
    – McGarnagle
    Nov 17, 2012 at 21:39
  • $\begingroup$ @Claudius For your last point, that momentum is transferred to the earth: is this so, because the system is only the rubber ball; and if earth was included in the system, there would be no transfer of momentum out of the system? $\endgroup$
    – Mack
    Nov 18, 2012 at 15:07
  • $\begingroup$ Exactly. The system Earth + ball has the same momentum as before, and since the change in momentum of the Earth is basically negligible, you can model (to a good approximation) the Earth as a solid plane simply imposing a restriction on the movement of the rubber ball, but not gaining any momentum or energy whatsoever. $\endgroup$
    – Claudius
    Nov 18, 2012 at 17:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.