Why do objects rebound after hitting the ground? [duplicate]

Question

When an object, say a shoe, falls from a height (under the influence of gravity), it rebounds after hitting the ground. For an object to move upwards, it requires a force to overcome its weight. When the shoe hits the ground some of its energy is lost and the ground pushes back with a force less than its weight, so why does it rebound, since the upward force is not large enough to overcome its weight?

Answer 1

Whatever the object lands on and the object itself acts as a spring and in compression the objects store elastic potential energy which comes from the downward motion (kinetic energy) of the objects.
That elastic potential energy is then converted into kinetic energy due to the upward motion of the object which was originally falling.

In general such collisions are inelastic and so not all the kinetic energy due to the downward motion becomes the kinetic energy of upward motion.

So it is the springiness of the objects which result in the force to slow the falling object down and then to exert a force greater than the weight of the object to propel the object upwards.

Update as a result of @CortAmmon ‘s comment to show the storage of elastic potential energy.

The granddaddy of them all?

Answer 2

It appears from the tags of your question that you want an answer using Newton's Laws, rather than using energy or elasticity.

Let us assume there are two forces on the object:

1. The weight of the object, pulling down.

2. The ground pushing on the object.

Let's look at the possibilities for force #2:

• There is no force between the ground and the object (or the amount of this force is zero). In that case, the object remains in free-fall even after it hits the ground, and passes through the ground, still accelerating. We know from experience that this doesn't happen.

• The ground pushes downward on the object. In that case, the object accelerates downward even more than it did in free-fall, and passes through the ground. We know from experience that this doesn't happen.

• The ground pushes upward on the object, but with less force than the weight of the object. In that case, there is a downward net force on the object. The object will still accelerate downward -- although less than it had in free-fall -- and will pass through the ground. We know from experience that this doesn't happen.

• The ground pushes upward on the object, with equal force to the weight. In that case, the object is in dynamic equilibrium, for which the 1st Law applies. The object in motion remains in motion with the same velocity, and passes through the ground. We know from experience that this doesn't happen.

• The only remaining possibility is that the ground pushes upward on the object with a greater force than the weight. This makes the net force upward, the acceleration upward, and slows the object down. This is what we observe from experience. Objects that hit the ground slow down, whether or not they bounce.

So, we've just demonstrated that the ground can and does produce a greater force on the object than its weight. If this effect continues after the object's velocity reaches zero, then the object will rebound (i.e. it is an elastic collision). On the other hand, if the force reduces to the object's weight once it reaches zero velocity, the object will stop at the ground (i.e. an inelastic collision).

There are other, more elaborate, and more robust ways of looking at the problem. As we say in physics, "there's more than one way to do it." However, this is the answer using only Newton's Laws, and there is no need to make the answer more elaborate than it needs to be.

Answer 3

There are also many objects that do not rebound when they got the ground but rather they get deformed . So the total potential energy stored in the body at a height is used to deform the body and some energy is lost in the form of heat or sound energy. The objects which are elastic in nature have a tendency to rebound and these objects don't get deformed or get a little bit deformed.