# Why do balls bounce in terms of force, momentum and energy?

From what I understand, as a ball is released from some height above a surface, it accelerates downward at 9.81m/s^2 due to the Earth's gravitational field. Once it reaches contact with the solid surface, the surface exerts a normal force on the ball due to electromagnetic interactions with the ball, and because of Newton's 3rd law, the ball also exerts an equal and opposite force onto the surface. As the forces are exerted, the force on the ball results in the downward velocity of the ball to decrease very quickly, and it must exceed the ball's weight too in order to make it accelerate upwards. My problem with this is the following: why must the normal force exert itself onto the ball at a high enough magnitude and high enough duration to switch the direction of the ball's velocity? Why doesn't the normal force simply decelerate the ball until it is at rest on the surface? What properties influence how much the object is accelerated upwards and why?

Please avoid the standard and overly simple "conservation of energy" explanation because it doesn't satisfy me, as it somewhat ignores the process of the ball's contact with the ground, and only considers the ball before and after collision with the ground.

In terms of energy conservation, what makes a ball conserve all its energy before and after bouncing if energy conservation only means that the energy in the closed system (the surface and the ball), must remain constant, and not that the energy of the ball before collision must be equal to the energy of the ball after collision?

Also, I have heard of the term "elastic" and "inelastic" collision, but I don't really understand what properties make an elastic collision perfectly elastic? What, physically, makes the energy of the ball be equal to the energy after?

Please comment on the fact that the surface exerts a force on the ball as it bounces over some distance. Would that mean that the ground does some work on the ball, ther3efore changing its energy?

I've heard some people saying that the ball's kinetic energy is transformed into elastic potential energy as it deforms, and is transformed back into kinetic as the ball bounces, but this doesn't seem to make much sense to me because it assumes the the ball's energy must be conserved, even though it shouldn't be, given that the surface exerts a force on it over some distance to accelerate it upwards.

I might be wrong, so please correct me. Thank you!

• Do you have a physics and math background? Mar 2 at 1:48
• I'm an IB physics student, so my physics knowledge is quite limited. I only know the IB SL physics syllabus. I also haven't learned calculus yet since I'm in grade 11. Mar 2 at 1:53

The difference between boucing and less bouncing objects is how the kinetic energy is dissipated.

The impact of a solid object in the ground can be divided in 2 steps. At first, there is a quickly increase of the compressive stress in the region of contact as the ball decelerates.

A short fraction of second after the impact, the ball is at rest, but under compressive stress. The ground continues to make an upward force in the ball in reaction to that stress. But now, as it is at rest, it starts to acelerate upward until the relieve of the stress, what happens when it loses contact.

But there are other ways to dissipate the stress. The most important is vibration, and that is why there is sound associated to the impact.

Bouncing and sound are competitive. Hard ground like concrete and a ball full of air are less noise and better to bounce than wood ground and a ball not full. The last couple is more prone to air vibrations inside the ball and wave mechanical vibrations along the wooden board.

Assuming a perfectly elastic collision, the force applied by the ball on the ground will not really be equal to it's weight. It will be Δp/t, where p is momentum. Let the velocity of the ball just above the ground be v. The ball doesn't sink into the ground for even a millimeter (in a perfectly elastic collision). Applying equations of motion,

s = vt +1/2a*t^2.

vt = -1/2a*t^2

a = -2v/t

Now, final velocity = u + at = v - 2v = -v (velocity equal in magnitude but opposite in direction). That's how elastic bouncing works. Now you see that acceleration depends on the toughness of the surface (you might've noticed that a ball bounces higher when dropped on a marble surface than when dropped on a soft surface). That's because the resisting force applied by the surface on the ball depends on the toughness of the surface. Thus, even though momentum of the ball changes from mv to 0 in both cases, acceleration of the ball is different as time taken for the momentum to change is different.

why must the normal force exert itself onto the ball at a high enough magnitude and high enough duration to switch the direction of the ball's velocity? Why doesn't the normal force simply decelerate the ball until it is at rest on the surface? What properties influence how much the object is accelerated upwards and why?

You fail to notice that while the ball is decelerating, it is not at rest, so it must move somewhere. If the ground allows it to move downwards, it will, and it will eventually come to rest without bouncing (drop a fridge, it won't bounce, just sink into the ground). Notice that force exerted by the ball is change in momentum/ time. Since the ball's momentum is not enough, it doesn't sink, the time of collision is low, and force applied by the ground on the ball is enough to make the ball bounce. Also, the duration is quite low, not high. Also, throw the ball at the speed of a bullet, it will just get embedded into the ground. I hope I have answered your question.

• Ps: I've ignored energy dissipation and all other factors. Mar 2 at 7:20

I think what you need is to look at the interaction between ball and ground in terms of gravity and Electromagnetic force. When the ball is released, earth pulls it. It keeps pulling untill ball reaches ground. What happens there is that when ball reaches close enough to ground then electromagnetic force enters into the picture. When two bodies come close to each other, the electrons start repelling each other. And at a certain point the two forces balance and ball has to stop and then accelerates in opposite direction.And since energy is also lost within bodies during this process it's not elastic. Here's same thing explained by Feynman.

https://youtu.be/P1ww1IXRfTA