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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?

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up vote 2 down vote accepted

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
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There's also some drag due to air resistance ... – McGarnagle Nov 17 '12 at 21:39
Thanks, I included that (and a few other things) now :) – Claudius Nov 17 '12 at 23:11
@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? – Mack Nov 18 '12 at 15:07
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. – Claudius Nov 18 '12 at 17:54

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