just want to clear something up.

Take a ball that's been dropped to the ground. Gravity acts and this ball as it has mass and then the ball now moves to the ground with a constant force of say ($X$).

Now when the ball makes contact with the ground, Newtons 3'rd Law takes effect (no air resistance):

If object A (the ball) exerts a force on object B (the floor), then object B will exert an equal force on object A in the opposite direction. (Action has equal opposite reaction).

Now here's where I get confused.

If the ball (which has a constant force when it hits the ground ($X$)) experiences the same constant force in the opposite direction ($-X$, minus indicating opposite direction), then the total force acting on the ball should be net ZERO ( $X + (-X) = 0$).

So there are no net forces acting on the ball, so why does it BOUNCE BACK? What am I missing?

Shouldn't the ball just stay on the ground? Bouncing back means a force greater than (-X) was applied to the ball giving it upward motion. Where did it come from?

## Best Answer

First, $X=mg$, force of Earth's mass pulling on the ball. Second, Newton's 3rd Law (N3L) is always in effect; the ball is pulling on the Earth

while the ball falls.Objects do not

haveorpossessor carryforce. A force results from an interaction of two things (ball/Earth or ball/floor) and acts on an object. While the interaction of the ball with Earth manifests in a constant force on the ball, that force hasvery littleto do with the force magnitude between the floor and ball, beyond the velocity which results from the acceleration of $g$. I believe that is the big mistake you are making.The force which the floor exerts upward on the ball cannot be X. If it was X, the ball would keep moving at constant velocity. You don't observe that, so you can conclude it isn't true. The force of the floor on the ball is electromagnetic in nature and results in elastic or plastic deformation, and must be larger than X for some time interval unless the ball crashes through the floor.