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Say we have a ball. We throw it at the wall at an angle. The ball hits the wall with a force and a direction away from you. The wall then applies a force equal in magnitude and opposite in direction. So the direction was away from you at an angle and now the opposite of that should be back at you right? Then why does the ball bounce off with the same angle on the other side of the perpendicular like a ray of light. Am I missing something here?

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    $\begingroup$ Note: answers below neglect the angular momentum of the ball and, how it changes as a result of contact with the wall. They answer as if the contact was frictionless. If you bounce a superball on a hard surface, giving it a good spin in various directions, you can get an intuitive feel for how angular momentum and friction change the direction of the bounce. Then, let it bounce more than once, and you can start to see how the direction of the bounce changes the spin. $\endgroup$ Jan 21, 2023 at 14:09
  • $\begingroup$ To see what @Solomon-Slow is alluding to, check out this "Back & Forth Bouncing Ball" demo from St. Mary's University. $\endgroup$ Jan 21, 2023 at 21:12
  • $\begingroup$ If you give the ball an appropriate spin, it will come back to you. $\endgroup$ Jan 22, 2023 at 2:04

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The ball hits the wall with a force and a direction away from you. ... the opposite of that should be back at you right?

Indeed, if the force were in the direction you suggest then the motion would be as you describe. This implies that the initial assumption must be incorrect. The force is roughly perpendicular to the wall. It is not directly along the line of motion of the ball.

The wall provides a constraint force. Than means a force that constrains the motion. In the case of a smooth wall, it constrains the motion to not go into the wall, but does not constrain motion away from the wall or along the wall. Such a constraint force is directed normal to the wall and is of sufficient magnitude to prevent any inward motion.

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Neglecting friction, the wall applies a force that is orthogonal to its surface (and is equal and opposite to the normal component of the force from the ball). Therefore, the velocity of the ball in the plane parallel to the wall is unchanged.

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The ball does not apply the force in the direction in which it hits the wall. The force exterted is in the direction perpendicular to the surface of the wall.. So it exerts a force that is opposite to only the normal component of the force.. Which makes the ball to recieve equal and opposite force (only in the direction perpendicular to it).. This can be better understood if you resolve the momentum or force when the ball decelerates at contact into its parallel and normal components.. And in addition, The ball thrown has only a fixed velocity, it has no force in principle, The force comes into play only when there is a momentum change, a ball in motion has a fixed momentum, The momentum description is more correct than a description based on forces.

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enter image description here

Suppose the ball has a velocity vector v1 and hits the floor.

If you think in terms of hard body objects, neither the ball nor the floor can deform, so the ball has to instantly flip its velocity vector to bounce off the floor: collision with the floor will take zero time, and a velocity vector flipping direction in zero time implies infinite acceleration and therefore an infinite force. That doesn't work. So it's better to not use the hard bodies paradigm: the ball will squish a little bit. Even if it's a steel ball bearing, it'll squish.

Bouncing happens because the kinetic energy of the ball is stored as potential energy in the compressed "spring" formed by the elastic material (or the air pressure inside, if it's inflatable). Then this energy is released and converted back into kinetic energy as it jumps off the floor.

Anyway, if the floor has very low friction, for example a steel ball bearing bouncing on a greased polished steel plate, then the floor's reaction will be normal to the surface and the horizontal component of the ball's velocity vector will not change. So it will bounce off with the vertical component of its velocity vector flipped, and the horizontal component will remain identical (minus losses).

If the floor has friction, as I drew in the second illustration, then it will indeed exert a reaction in the opposite direction as the force applied by the ball, and as you said this reaction will push the ball back to the person who threw it.

However this reaction force does not apply to the center of gravity of the ball, but to the contact area with the floor, so it will apply torque to the ball and tend to make it turn (ie, gain angular momentum).

So the ball's velocity vector does lose a bit of its horizontal component, and also a bit of magnitude, as some of the ball's kinetic energy and momentum are converted into rotational energy and angular momentum.

If the ball was turning before it hit the floor, then that will also influence the direction of the bounce. If you turn it fast enough, and if the angle with the floor is not too low, it is possible to make it bounce back towards you.

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  • $\begingroup$ wow, what an insightful answer! the torque part made it even more clear and also the conservation of energy part. thank you $\endgroup$
    – Domeoryx
    Jun 29, 2023 at 10:43
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Lets look at this figure enter image description here

( http://wandinger.userweb.mwn.de/TMD/v2_4.pdf)

you throw a ball ,with the velocity $~v_A~$ and the angle $~\alpha_A~$ towards a wall

the ball hit the wall and come back, in case of inelastic collision (left figure) with the velocity $~v_A~$ and the angle $~\alpha_A~$

if the throwing angle $~\alpha_A~$ is zero the ball will come exactly back to you , but a soon as the throwing angle not exact zero the ball will not come exactly back to you.

in "reality" you have the semi elastic case so even if the throwing angle $~\alpha_A~$ is zero the ball will hit the wall an come back with the angle $~\alpha_B\ne 0~$ not exactly to you

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    $\begingroup$ Can you add proper attribution for the source of the images? $\endgroup$
    – Kyle Kanos
    Jan 22, 2023 at 2:58

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