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Newton's third law says "for every action, there's an equal and opposite reaction." So why is it, say, that when a pool/billiards ball hits the wall of the table, the ball doesn't just turn around?

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    $\begingroup$ The actions and reactions Newton refers to are forces, not velocities. $\endgroup$ – dfan Dec 15 '14 at 23:55
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Note that in your suggested motion after the collision, momentum would not be conserved. The Law of Conservation of Momentum is kind of a big deal in Physics, and especially useful when analyzing collisions. It states that for a closed system, momentum is conserved. It's also sometimes restated that for a closed system experiencing a collision, the momentum before the collision is equal to the momentum after. In your example, object $A$ has some mass, $m_A$ and is traveling with some velocity $v$ (If we choose a coordinate system such that $A$ is moving in the positive x direction)and a second object, $B$ is at rest. So before the collision the total momentum $p$ is given by $p=m_Av$ However, if $A$ just "turned around" after colliding with $B$ it would now be moving in the negative x direction and so would have a velocity of $-v$. As a result the momentum after the collision would be given by $-m_Av$ which is decidedly not equal to $m_Av$ anymore than -5 equals 5.

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  • $\begingroup$ I think you're over simplifying or I'm missing something (will admit it's been quite a while since I studied physics). If B is the wall of the pool table and A hit it directly (as in the wall is perpendicular to v), A very much will be moving -v afterwards $\endgroup$ – Foon Apr 5 at 21:19
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The force from the ball on the wall is exactly equal and opposite to the force of the wall on the ball. Both forces however are perpendicular to the wall (and must be assuming the wall is frictionless) and not necessarily perpendicular to the ball's initial direction of motion. Being perpendicular to the wall the force on the ball has absolutely no effect on the motion of the ball in the direction parallel to the wall and can only change the motion of the ball in the direction perpendicular to the wall.

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Because in your example the action is not in the same direction than the velocity. The action and reaction are normal to the wall of the table, so the billiard ball only feels a reaction force (and thus a change in its velocity) in that direction. The component of the velocity of the ball parallel to the wall is not affected.

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