# What are the Normal forces at instantaneous point of collision?

Related to my previous question.

When two colliding objects are at rest, say, a book on a table, the normal force the table exerts on the book will exactly cancel out the force of gravity (or other forces) the book exerts on the table. Due to that, I've been subconsciously assuming that these forces are present in that exact manner as soon as the two objects collide, assuming a perfectly elastic collision.

However, thinking about it more, that clearly seems wrong: If the normal force at the exact time of impact was only equal to the downwards force, the net force acting on the book would be 0, and by Newton's 1st Law, the book would continue travelling at a constant speed through the table - which is wrong.

Therefore, I assume the normal force exerted on the book at the time of impact is much higher than the downwards force, and reduces over time to be equal to it.

The question I have is as follows: What is the normal force at the time of impact? How do I calculate it? Where does it come from?

Maybe something to do with conservation of linear momentum? The problem there is that I don't really have a "time of impact" if I consider my impact to be an instant (Or, alternatively, extremely long given that the book stays on the table) I'm honestly confused.

• What normal force are you talking about? There is a normal force at the interface between the two colliding bodies and another one at the interface between each body and the surface supporting them (gound, table, whtever).
– nasu
Nov 13 '20 at 2:02

## 2 Answers

You are right to think its has to do with momentum. Know that$$\overrightarrow{F_{\rm net}} = \dfrac{{\rm d} \overrightarrow{p}}{{\rm d} t}$$ where $$\overrightarrow{p}$$ is the momentum.

Since collisions happen almost instantly, the force exerted by each object on the other will be extremely large for a very small time, and then go back to zero when the objects move away from one another.

Because of that, it's hard to measure the magnitude of this force as function of time. There exist many models to approximate it, depending on the type of collision, material, etc. A basic way is by using the following equation $$\overrightarrow{F}_{\text{net, average}} = \dfrac{ \Delta \overrightarrow{p}}{\Delta t}$$

Another way to think about collisions is imagining springs on each object's exterior, that compress as the objects collide, and then due to the large potential energy stored in these 'mini springs,' the springs will decompress and both objects will move away from each other. Do note the following, as stated on Wikipedia:

An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies remains the same. In an ideal, perfectly elastic collision, there is no net conversion of kinetic energy into other forms such as heat, noise, or potential energy. During the collision of small objects, kinetic energy is first converted to potential energy associated with a repulsive force between the particles (when the particles move against this force, i.e. the angle between the force and the relative velocity is obtuse), then this potential energy is converted back to kinetic energy (when the particles move with this force, i.e. the angle between the force and the relative velocity is acute).

• Thank you for your answer. I did try to model this using the impulse-momentum equation you mentioned, but I must have made a conceptual mistake as my calculations didn't work out. Take the example of my post, book & table: How would I begin to choose the time of collision, since the book stays on the table and the table doesn't budge? Is it near-0, making F immense, or is it near-infinite, making F 0? Both cases don't seem realistic nor make my calculations work out, unfortunately... Nov 13 '20 at 0:09
• If the book and table are already at rest, why consider collisions? Are you trying to simulate dropping a book on the table? Nov 13 '20 at 0:20
• Yup, pretty much. Not a book but rather a system of particle, but same stuff. What I'm having trouble with is the moment of collision - not the falling, nor the "resting"; which is why I'm having trouble figuring out, for example, what value I should use for t in the impulse... Nov 13 '20 at 0:26
• @AnotherStudent Well you could use the speed of sound of the material and the size of the objects to get a back-of-the-envelope $\Delta$t. But to do better than that, you’ll need to seriously up your game with detailed time-dependent structural mechanics simulations. Nov 13 '20 at 4:02

The force between the objects is almost entirely electrostatic repulsion between the electrons of the atoms on the outside of the objects. If you just had two electrons coming towards each other, then the force would be rather simple, increasing as they get closer, inversely proportional to the square of the distance to each other. However, it's greatly complicated by the fact that the electrons and protons of an atom produce a dipole force, and you have two surfaces of atoms coming towards each other, and all of them are exerting force on each other.

Once the force becomes significant, the atoms of one object will be pushed away from the other object, which will create stress. The atoms already have bonds with the atoms next to them in the object, and those bonds will be strained by the new forces, and the force of the collision will be distributed through the object by the inter-atoms forces.

How this plays out depends on the nature of the object. For instance, with a rubber ball, the portion of the ball close to the collision will experience significant deformation, allowing the center of masses of the two objects to come much closer than otherwise. This extends the collision time and reduces the force. For more rigid objects, the collision time is lower, giving a greater force. Thus jumping onto concrete hurts more than jumping into a pit of rubber balls.

With a high speed camera, you can see that there is a finite duration of the collision. Example. The force will be varying over time, and unless you have sensitive instruments, you'll more have average force than instantaneous force. Here is an example of the force varying over time. Webpage this is from.