Why do elastic balls bounce?

Can someone please explain the process of a ball bouncing in terms of Forces?

First, it doesn't make sense to say that the normal force is greater than that of the upwards force. When the ball hits the ground and gradually starts exerting the force of its weight, the ground exerts the same equal force just in the opposite direction. The ball starts compressing and ultimately, when it reaches $$0 \ \ m/s$$, it stops.

Now, the common explanation here is that, due to this compression, the ball pushes back creating a bigger upwards force and hence an upwards acceleration but I think this explanation is simply false or not adequate enough. The ball, due to its compression, pushes the floor. Meaning the force downwards now is $$F_g+F_e$$ (Weight and the elastic force), from this then follows that the upwards force is exactly the same $$F_g+F_e$$ just in different directions. This implies that since there is no net force in the upwards direction, there should be no acceleration in the upward direction and thereby the velocity of the ball will stay at $$0 \ \ m/s$$.

Is it even possible to explain this without mentioning energy and momentum?

• I don't want to shamelessly self-promote but in this answer physics.stackexchange.com/a/541440/93729 you can see an animation of a bounce in slow motion. The important takeaway of that animation is that when the velocity is zero the normal force is at its highest. Commented Feb 27, 2022 at 21:59
• Commented Feb 28, 2022 at 0:40
• Each force exists from one object another. Your question would be easier to follow if you were to state what those objects are. For instance, when you say "Meaning the force downwards now is Fg+Fe ", what object is applying that force, and on what object? Commented Feb 28, 2022 at 7:52

First, it doesn't make sense to say that the normal force is greater than that of the upwards force.

I agree. The normal force is the upward force. If the normal force is greater than the weight of the ball, it will be accelerating upward.

The ball due to it's compression pushes the floor. Meaning the force downwards now is $$F_g+F_e$$ (Weight and the elastic force), from this then follows that the upwards force is exactly the same $$F_g+F_e$$ just in different directions.

Sort of. Just be sure you understand what the forces are acting on. The downward force from the ball is acting on the ground. The upward force from the ground is acting on the ball. Also it is probably simpler to describe the contact/elastic force as a single force not added to gravity, $$F_e$$. It just so happens that at maximum compression, $$|F_e| > |F_g|$$. When the ball is just sitting on floor the two magnitudes are equal.

This implies that since there is no net force in the upwards direction

No. We consider only the forces acting on a single object to determine its acceleration. The only forces on the ball are the upward/normal force from the floor (which I describe as $$F_e$$) and gravity ($$-F_g$$). This means there is a net force on the ball because the two forces are not equal.

• Yep I get it. The force the ball exerts due to elasticity is not the downwards force experienced by the ball. I'm so mentally ill sometimes lol Commented Feb 26, 2022 at 23:46
• @Peter Your statement that upwards and downward forces balance is true for the bottom part of the ball, but not for the rest of the ball. Commented Feb 27, 2022 at 17:34

Your mistake is thinking that action and reaction pairs cancel out. When the ball first hits the ground it starts to compress. That sets up a growing elastic force inside the ball which slows the speed of the centre of mass of the ball until the centre of mass stops- thereafter the effect of the elastic force is to accelerate the centre of mass upwards, which causes the bounce.

Acceleration equal to zero does not mean that the velocity must be zero.

Consider the system as a mass, an unextended spring, with spring constant $$k$$ and the Earth.
What happens when a stationary mass, $$m$$, is placed at the end of the spring?
The spring extends and at an extension $$x_{\rm o}$$, the extension when the system is in static equilibrium.
When the extension is $$x_{\rm o}$$ the net up force on the mass is $$kx_{\rm o}-mg=0$$.
However if these is no damping the mass does not stop there as it during its fall the system gains kinetic and spring potential energy at the expense of a net decrease of gravitational potential energy.
So the mass continue past the static equilibrium position and only stops when the extension is $$2x_{\rm o}$$.
In this state there is a net upward force of $$k(x_{\rm o})-mg\ne 0$$ on the mass so it will start to accelerate upwards and with no damping execute simple harmonic motion about the static equilibrium position.

Moving on to doing the same thing with a spring which is compressed when the mass is placed on it the analysis is still the same with the mass oscillating about the static equilibrium position.

Now consider the mass being dropped onto the compression spring.
In this case the compression of the spring will be greater than $$2x_{\rm o}$$ mbut eventually there will come a time when the mass stops moving but has a net upward force on it.
So the mass will start moving upwards, there will be a rebound.

Nearly there.
Finally consider the ground as the compression spring with damping present (mechanical energy converted to sound and heat and work done permanently deforming the ground (and mass), an inelastic collision.
The mass will rebound but not to the height it was released from.

An elastic collision between a ball and the ground is rather like two perfectly efficient springs colliding end to end. An easily compressed spring representing the ball and a very hard to compress spring representing the Earth.

As the ball approaches the ground, to it the coils of the Earth are slightly shorter than its own due to their relative speed towards one another, and to the Earth, the coils of the ball are equally shorter than its own.

During the coming together this must even up (because when their relative speed is zero their relative wavelengths must be the same for any given type of wave), and the coils of the ball, being more easily compressed bear the lion's share of the evening up (the ball does most of the decelerating). After the evening up stops (point of zero relative speed) the coils start to return to their original states as is natural for a perfectly efficient spring. Of course this causes a relative speed that's equal in magnitude and opposite in direction to the initial one.

The coils of each are now longer from the viewpoint of the other, but are locally the same as they were to start with.

In Newton's equation $$F=ma$$, it is only the force on the object being accelerated at"$$a$$" that matters. In this case the force that the floor exerts on the ball is "$$F$$". The force that the ball pushes the on floor with is irrelevent to the ball. The latter force is accelerating the earth.

• Yeah, I see. I switched frames of reference interchangeably for some reason. The force the ball exerts on the wall due to elasticity is not the force the ball feels. The ball feels the reaction of the elastic force and the normal force which combined overpower gravity. Commented Feb 26, 2022 at 23:42
• Great! You got it! Commented Feb 26, 2022 at 23:43

First of all, there is no way to explain this without energy or momentum. In simple terms, when a ball hits the ground, potential energy builds up. That potential energy has to be released and there is no other way than up because of Newton's laws.

Just imagine a stiff spring falling in its length onto the ground. The spring gains potential energy when it's compressed. If all kinetic energy is conversed in compressing the spring, the spring, will push itself back and takes of with the same velocity it fell onto the ground. It bounces back like the ball, excerting a short force onto the ground greater than the force caused by its mass.

So, it touches, the normal force rises to a value exceeding the force of gravity acting on the ball. This causes the ball to stop. Because the force upward is greater than the pull of gravity, the ball accelerates upwards. After the motion reverse the force on the ground decreases to zero. When the downward force equals the upward force, the ball has already an upward velocity. Almost the same as the velocity with which it fell on the ground.