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I am asking a related question here : Newton's Third Law Angular Bouncing

Why, given newton's third law would a ball bounce at an opposite angle to it's original, if an equal and opposite force should push it back through it's original trajectory?

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    $\begingroup$ The equal and opposite force is exerted on the wall and only reverses the velocity component perpendicular to the wall. There is no force parallel to the wall, which means that the ball keeps moving without any change in that direction. Is that what you are asking? $\endgroup$ – CuriousOne Oct 16 '14 at 15:19
  • $\begingroup$ Yes, that's what I was asking. Unfortunately I can't accept a comment as an anwser. But that makes sense. $\endgroup$ – AlphaModder Oct 16 '14 at 15:23
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The force between the ball and the wall is normal to the wall, and therefore its direction is not opposed to the initial motion of the ball but rather to the motion of the ball relative to the wall.

Forgetting for a moment rotation, the velocity of the ball perpendicular to the wall is reversed; the result is that the angle of incidence = angle of reflection (under the assumption of no rotation, no losses, etc...)

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