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If an object is acted on by equal and opposite forces then it will be in equilibrium, and it's acceleration or velocity (and so direction as well) will not be changed.

So when a ball bounces, it exerts a force on the floor, which matches the magnitude of the force in the opposite direction (the ball is bouncing perfectly vertical), up. So how is it's velocity/direction changed? If the forces are equal and opposite to each other. In order for it bounce, surely the force acting from the floor to the ball must be greater than the force acting from the ball to the floor?

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You've misunderstood the statement.

When A exerts $\vec{F}_{A \to B}$ on B, B exerts an equal and opposite force $\vec{F}_{B \to A} = - \vec{F}_{A \to B}$ on A.

The only forces acting on the ball a gravity and the normal force, and the floor experiences a force from the ball which is equal in magnitude and opposite in direction from the normal force on the ball.

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  • $\begingroup$ I don't understand you're last paragraph. If the forces are equal and opposite why is there a change in motion? $\endgroup$
    – Jonathan.
    May 24, 2012 at 20:06
  • $\begingroup$ One of the equal and opposite forces acts on the ball, and the other one acts on the floor. When you add up the forces acting on ball it includes only one of the pair. $\endgroup$
    – dmckee --- ex-moderator kitten
    May 24, 2012 at 20:08
  • $\begingroup$ That seems so obvious now you've said it :) $\endgroup$
    – Jonathan.
    May 24, 2012 at 20:13
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The easiest way to understand this is by flows of momentum, the ball has momentum down, and when it hits the floor, it transfers twice this amount of down momentum to the floor, and so is left with negative this much down-momentum, which is up-momentum.

The statement of Newton's laws are nothing more than the flows of momentum from object to object, and how the object moves when momentum is building up on it. Understanding it this way is helpful for clearing up the elementary confusions. This is a supplement of dmckee's answer, which is correct, but it doesn't give the intuition of momentum-flow, which makes the misunderstanding impossible to formulate.

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  • $\begingroup$ Assuming we didn't already know the answer, why would one intuitively guess that it transfers two times the momentum to the floor, and not, say, only it's momentum? That would make it much more intuitive for me. Thanks. $\endgroup$
    – kηives
    May 25, 2012 at 3:43
  • $\begingroup$ @knives: If the momentum transfer is reversible, the process looks the same back in time, so the magnitude of the post-bounce velocity must be the same (so the motion looks the same backwards in time). The details are exactly the same as usual, the ball transfers all the down momentum at the point of maximum compression, and then absorbs up momentum as long as it is compressed. The analogy with charge flows is for LC circuits which are reversible, not with lossy R circuits. If you have a capacitor with an inductor connecting the plates, the charge can oscillate forever back and forth. $\endgroup$
    – Ron Maimon
    May 25, 2012 at 3:49
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The problem is a common conflation of 1st and 3rd law forces. Yes, the normal force (from the floor up) and the contact force (from the ball down) are equal and opposite since they are 3rd law forces. However, to understand why the ball's motion changes we must focus on external first law forces. If the the normal force up exceeds the downward forces of Earth's gravity and drag from the air, the ball will accelerate upward.

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