Normal Force Question

[Shark 774]

Hey guys,

When a ball bounces on the floor why does the magnitude of the force that the ball exerts on the floor and vise versa change throughout the bounce??

Thanks!

[Whatlol]

Not entirely sure but this is all i can think of :

Well picture a ball hitting the ground at a speed of x. After a short amount of time the change in speed is small. As this time increases the change in speed is larger, therefore the forces will be larger.

[appianway]

This way of thinking of things might help as well:

As the ball falls, its centre of mass must approach the ground. However, this means that the dimensions of the ball must change - the centre of mass will come within a distance r of the ground, where r is the normal radius of the ball. This means that some form of deformation must occur. What does this mean physically? Well, we know that when we're dealing with springs (which are different, but will suffice for this analogy), when we deform (ie compress or extend) the spring, the spring exerts a force. This force is reliant on the extent of the deformation, and it therefore follows that the force that the ball exerts on the ground (which is the reaction pair to the normal force) is dependent on how "squished up" the ball is. As the centre of mass moves towards the ground, the extent of the deformation increases, and thus the force increases.