# Collision between two non-deformable objects

Let say we have a non-deformable object and we release it into free fall and it hits the non-deformable floor. What would happen?

Here is the way I think.

Since the floor and the object are both non-deformable, there is no chance that the object will bounce. If we assume the Earth to be at rest before and after the collision (*), the object stops suddenly (like in 0,0000000000.. second) and that means the acceleration and the force are infinitely big.

This reasoning makes me think that there are no such things as perfect non-deformable bodies but let say we use the closest object and the floor to conduct the experiment. What would happen both in terms of momentum and energy conservation?

*In reality the Earth would be moving towards the object during free fall so that the center of mass of the system stays unchanged.

• You are correct that such nondeformable objects are not physical. I'm not sure why you don't think "the closest object" wouldn't bounce... – Trevor Kafka Sep 30 '18 at 21:34

Collisions between non-deformable objects violate Newton's laws. By Newton's second law and third laws, we have $$a_1/a_2=m_2/m_1$$. Then if $$m_1\ne m_2$$, the accelerations have to be unequal. But they can't be unequal during the time when the two rigid objects are in contact, so this is a contradiction.

• Why cannot the accelerations be different during a rigid-body collision? They can be different during a non-rigid collision. What is so different about rigid-body collisions? ... If a rigid body is defined as a deformable body with a spring constant which tends towards infinity, there is no point at which Newton's Laws break down. Or is there another way of defining a rigid body? – sammy gerbil Oct 2 '18 at 17:38
• @sammygerbil: If the two objects are touching and are both rigid, then their accelerations are equal. (This is neglecting possibilities such as rotation of the objects while they're touching.) Non-rigid bodies deform, so that their centers of mass can have unequal accelerations while they're in contact. – Ben Crowell Oct 3 '18 at 0:15
• If the bodies are non-deformable even in the limit, then they cannot accelerate at all while in contact, whether with the same acceleration or a different acceleration, because that would mean the distance between the centres changes. Unless you examine the collision as the limit of infinitely stiff objects, I don't see how contact time and acceleration can have any meaning. Then the issue of violating Newton's laws does not arise, because acceleration cannot be defined. – sammy gerbil Oct 3 '18 at 11:44
• If the bodies are non-deformable even in the limit, then they cannot accelerate at all while in contact, whether with the same acceleration or a different acceleration, because that would mean the distance between the centres changes. I'm not sure what you have in mind here. Suppose e.g. that both objects are spheres of radius $r$, and let their positions be $x_1=(1/2)at^2$ and $x_2=(1/2)at^2+2r$. Then they are accelerating while in contact, and the distance between the centers is constant at $2r$. – Ben Crowell Oct 3 '18 at 22:52
• Unless you examine the collision as the limit of infinitely stiff objects, I don't see how contact time and acceleration can have any meaning. In my example above, no limit is required, and the contact time and acceleration are perfectly well defined. – Ben Crowell Oct 3 '18 at 22:52

What would happen both in terms of momentum and energy conservation?

When the ball hits the floor, the molecules of the ball will get real close to the molecules of the floor. At some point, they will start repelling each other. Whether the deformation at the point of contact significant or not, a sufficient reaction force will be acting on the ball for some period of time to stop it from going through the floor.

At the moment the ball will stops, its kinetic energy will be converted to the potential energy of the deformed surfaces and, in case of a non-elastic collision, to heat. The momentum of the ball will be reduced to zero by the impulse of the reaction force.

If the collision is elastic, the deformed surfaces will fully recover and their potential energy will be converted back to the ball's kinetic energy. As a result, the magnitude of the ball's velocity and momentum will be the same as they were just before the collision, while their direction will change to the opposite. The impulse of the reaction force, acting on the ball after the full stop, will be equal to the new momentum.

If the collision is not elastic, the recovered kinetic energy will be smaller than the kinetic energy before the collision. As a result, the magnitude of the velocity and the momentum after the collision will be smaller as well. The lost energy will be turned into heat.