If every action has an equal and opposite reaction, how come an object dropped from a certain height doesn’t bounce back to the original height?



If every action has an equal and opposite reaction, how come an object dropped from a certain height doesn’t bounce back to the original height?

In: Physics

So first of all, when you drop an object it accelerates equal to the gravity of earth minus the slowdown caused by air friction. When it hits the ground the object will, under perfect circumstances, impart all of its momentum into the ground. If the ground can give way then some of this momentum is lost. If it can’t then all of the momentum will be pushed back the other direction; it’ll bounce.

Now, when the object is going back up though, gravity AND air friction are both slowing it down.

So the end result is that the action does have an equal and opposite reaction, you just didn’t consider some of the opposite reactions.

Some of the kinetic energy is converted into heat as the object’s material and that of the impacted surface flexes, during impact.

Some is converted into vibrations in the air.

There’s also the loss from air resistance, if the scenario isn’t performed in a vacuum. 🙂

I might be forgetting to list a few minor losses, but the pithy answer is that it’s more accurate to say, “for every action there is a set of reactions that total up to being equal, with the average vector thereof being the opposite direction from the action’s force.”

^(American high schools still teach the obsolete Newtonian model due to it being closer to the flawed intuitive model possessed by the human brain, and thus easier to learn. At least it isn’t Aristotle’s model!)

^( Edit: and, as pointed out, it is close enough for many things.)

^(Edit 2: Yeah, the USA isn’t the only nation that still teaches the Newtonian model in public school. My sidepoint was that the USA does, and uses an oversimplified version of it to get the idea across. This massive oversimplification gets stuck in the mind when the rest falls away…. leading to questions like this. 😉)

Gravity is constant. When you drop the object, gravity is pulling it down. When the object strikes the ground and rebounds, gravity is STILL pulling it down.

The equal and opposite force of earth’s gravity pulling the ball is the ball’s gravity pulling the earth. The bounce and the falling are different forces with different third law pairs.

It is not the same expression of energy, even though it is the same amount of energy. A large amount is absorbed by the impact on the ground, and again by gravity as it attempts to rise.

Quite simply because it isn’t about the action and reaction. That is a sadly widespread misworded of Newtons 3rd law. That wording focuses your mind on the results, which isn’t what is equal. It’s the forces that are equal. For every force exerted by an object there is an equal and opposite force acting back on the object.

There ball strikes the ground with 100 Newtons of force. The ground strikes back at the ball with 100 Newtons (equal and opposite FORCES)

The result is dictated by the f=ma. The ball bounces back a lot, as the mass is small. The Earth bounces away as well… But it’s huge so you don’t notice the extremely tiny acceleration.