What makes a ball bounce back up?

Question

Suppose, I'm dropping a rubber ball on the floor and it bounces back up. I'm trying to understand this by considering the forces, and not using momentum, collision, or energy.

Eventually, the ball would come to rest, due to dissipation of energy, and when that happens, the normal reaction of the floor on the ball would be equal to the weight of the ball, due to electrostatic force.

However, I'm wondering what happens before that. When the ball drops for the first time, it would bounce back up. There are three forces involved here, that are relevant. First is the normal force exerted on the floor by the ball (indirectly due to gravity). The second force is the normal reaction of the floor on the ball (due to electrostatic repulsion). The third is the force of gravity exerted on the ball by the Earth.

Normally it is expected that the normal force should be equal to the weight, as is the case when the ball comes to rest. However, in this case, the initial normal force exerted on the ball, by the floor, is obviously more than the weight — hence the ball bounces back up. My question is, how is that so?

How can the normal reaction force be initially greater than the weight? Please tell me this on the basis of the acting forces, and not using momentum conservation and energy conservation.

My intuition is that there is a fourth force working here, one that deforms the ball when it hits the ground. This force increases the normal reaction of the ground toward the ball, and makes it greater than the weight, and is the reason why the ball bounces back up. Each time the ball loses energy, so this deformation would be lesser and lesser, and the reason why the normal reaction would decrease after every bounce, until it becomes equal to the weight.

Please tell me if this explanation is correct for this situation. Moreover, please tell me, if this explanation is correct for collisions in general.

I'm trying to look at collisions from the force perspective, rather than the momentum perspective.

Answer 1

First is the normal force exerted on the floor by the ball (indirectly due to gravity).

Gravity is not the only effect here. The ball is moving downward when it reaches the floor. In order to stop moving downward, the ball has to undergo an acceleration upward. This acceleration requires a force: $a = F/m$. The strength of this force and acceleration depends on the speed of the ball (how much acceleration is required) and the materials (how far the ball can deform).

It may be helpful to think of the ball as a spring. The force the ball exerts depends on the deformation, just like the force exerted by a spring depends on the deformation.

When the ball first touches the floor, the normal force is small, so the ball is still accelerating downward. After a certain amount of deformation, the ball is exerting a force equal to its weight. But this doesn't stop the ball, it just stops the acceleration. The ball continues to move downward.

As it does it compresses further. The normal force exceeds the weight and the ball begins to slow down. This continues until the acceleration removes all of the ball's downward velocity. The spring forces are still greater than gravity though, so the upward acceleration continues. This is sufficient to give it an upward velocity that allows it to bounce off the floor.

So, the normal reaction should always be more than the weight of a falling body, to account for that extra velocity that remains, even after the net force acting on the body is 0.

Well, at the instant the net force is zero, the normal force would be equal to the weight. But the downward velocity that remains will cause the ball to move closer to the floor and for the forces to increase with time.

For a stationary object kept on a floor, the normal force is exactly equal to weight, as there is no 'extra downward velocity'. Are these assertions correct ?

Right. The system is in equilibrium. The weight is exactly opposed by the normal force, so the object does not accelerate.