# How to characterize a rubber ball moving horizontally and bouncing off of a vertical wall?

• B
So imagine we had a rubber ball and we threw it perpendicular to the surface of wall. The ball will obviously experience some “throwing” force — horizontal with respect to floor (let’s assume the ball is in perfect vacuum and there is no net torque). As it hits the wall, the normal force will balance the force exerted by the ball and the ball will be at rest for some tiny time interval — deformed. However, since the ball is made of rubber, the molecular forces ought to return the ball to its initial shape — thus some force will be exerted on the wall — also on the ball, that will cause it to “bounce off” (technically fall parabolically because there is also gravity).

Correct me if and where I am wrong. As a high school student I sometimes find it a bit complicated to apply things to more advanced examples that I see in every day life… maybe it’s just me, I don’t know.

Also, if possible let’s try to keep this problem in domain of Newton’s laws, please don’t talk about it from the conservation of energy perspective, it won’t help me much

OK, so you're throwing a ball without spin, upward to the wall where it hits at the peak of its parabolic trajectory. It applies a varying amount of force to the wall during the time in which it is in contact. So far so good. It seems to be all you have to say. Was there a question involved?

The initial throw is at an angle, not horizontal, else it would not hit the wall perpendicularly. I'm not sure if you meant it to, but you said no torque.

If the throw is horizontal at first, the gravity will have it hit the wall at an angle, in which case in addition to the horizontal force on the wall, the vertical friction force with the wall will impart torque to the ball, and thus spin. It will rebound from the wall spinning.

You’re right about the angle part, I almost forgot about gravity as if I was imagining a bullet! (in which case horizontal velocity is high enough that we can technically neglect vertical component).

However I was mostly concerned about the elastic force part. I didn’t actually read it from a textbook but concluded it myself so it’s important for me to know if I am thinking the right way.

As it hits the wall, the normal force will balance the force exerted by the ball and the ball will be at rest for some tiny time interval — deformed.
Yes. A simple model of a ball hitting a wall is a point mass and a spring. The spring contacts the wall, compresses, and pushes the mass away. This is an elastic impact. In a perfect elastic impact, the ball bounces back at the same velocity that it hit the wall. In the real world, it bounces back slower than it hit. In the theoretical physics world, it can bounce back at the same velocity.

The spring-mass model does a good job of describing elastic impact. That model also allows calculating the duration of the impact from the peak deformation, or the peak deformation from the duration of impact if the impact velocity is known. Search terms high speed video golf ball compression will find some interesting videos. Or possibly other types of balls.

A steel ball peen hammer hitting a heavy steel object has duration of impact about 0.001 second. This was a problem on a project where I needed to hit something with an impact duration less than 50 microseconds.

To simplify things, replace the ball with a short elastic rod and neglect gravity.

The key to understanding what happens in impact of the rod against the wall is to recognize that the rod is not a rigid body and so it does not have to stop all at once. Different parts of the rod deform at different times during the collision, and different parts of the rod stop and rebound at different times.

At the initial contact of the leading edge of the rod with the wall, the very front surface of the rod comes to a stop, while the remainder of the rod continues moving, unaware that anything has even happened at the leading edge. As time progresses, a compression zone forms in the front part of the rod. The material within this compression zone has a compression force acting within it, but it is no longer moving. Rearward of the compression zone, there is no compression, and the rod material is still moving at the initial velocity. The size of the compression zone increases very rapidly (at the speed of sound within the rod material), and, eventually the entire rod within the compression zone is no longer moving. The entire rod is stationary.

In the next instant, the compression releases from the trailing edge, and material at the trailing edge begins moving away from the wall, but at the original speed of the rod. As time progresses, the size of the compression release zone grows. The material within the compression release zone is all traveling away from the wall at the same original speed, while the material within the compression zone is still not moving. The size of the compression release zone grows with time, at the speed of sound within the rod material. Eventually the compression release zone encompasses the entire rod, and, at this point the rod loses contact with the wall.

A bit of a different approach, why don't you look into something like LIGGGHTS or Granoo to model the required behavior.

As it hits the wall, the normal force will balance the force exerted by the ball
This sounds like a common error.

Yes, the normal force of wall on ball is equal and opposite to the force of ball on wall. That is Newton's third law. But that is not a balance. The two forces act on different objects. Nothing is being held stable. The ball is accelerating away from the wall. The wall is accelerating (slightly) away from the ball.

Newton's third law applies at all times. Not just when things come to rest or come to an equilibrium. It is not (despite the popular wording) a statement about action and reaction. It is the statement that interaction forces come in pairs. The force of A on B and the force of B on A are both part of the same interaction. Neither is cause. Neither is effect. Two sides of the same coin if you will.

In Newton's original words (albeit translated from Latin), "the mutual actions of two bodies upon each other are always equal, and directed to contrary parts"

and the ball will be at rest for some tiny time interval — deformed. However, since the ball is made of rubber, the molecular forces ought to return the ball to its initial shape — thus some force will be exerted on the wall — also on the ball, that will cause it to “bounce off” (technically fall parabolically because there is also gravity).
Yes. The repulsive force between ball and wall that acted as the ball smushes against the wall and the stress of compression builds continues to act as the ball un-smushes and the stress is released. The ball has an away-from-the-wall acceleration throughout both phases of that interaction. That is Newton's second law. The wall also accelerates in the opposite direction, but at a much less noticible rate due to its much larger mass. Again, that is Newton's second law in action.

As you know, a reduction in toward-the-wall velocity over time is an away-from-the-wall acceleration. Just as an increase in away-from-the-wall velocity over time is an away-from-the-wall acceleration.

Last edited:
Ibix
jbriggs444 said:
This sounds like a common error.

Yes, the normal force of wall on ball is equal and opposite to the force of ball on wall. That is Newton's third law. But that is not a balance. The two forces act on different objects. Nothing is being held stable. The ball is accelerating away from the wall. The wall is accelerating (slightly) away from the ball.

Newton's third law applies at all times. Not just when things come to rest or come to an equilibrium. It is not (despite the popular wording) a statement about action and reaction. It is the statement that interaction forces come in pairs. The force of A on B and the force of B on A are both part of the same interaction. Neither is cause. Neither is effect. Two sides of the same coin if you will.

In Newton's original words (albeit translated from Latin), "the mutual actions of two bodies upon each other are always equal, and directed to contrary parts"

Yes. The repulsive force between ball and wall that acted as the ball smushes against the wall and the stress of compression builds continues to act as the ball un-smushes and the stress is released. The ball has an away-from-the-wall acceleration throughout both phases of that interaction. That is Newton's second law. The wall also accelerates in the opposite direction, but at a much less noticible rate due to its much larger mass. Again, that is Newton's second law in action.

As you know, a reduction in toward-the-wall velocity over time is an away-from-the-wall acceleration. Just as an increase in away-from-the-wall velocity over time is an away-from-the-wall acceleration.
Yes I understand that about Newton’s third law. I might have formulated it poorly. Force exerted by the ball acts perpendicular to the wall and on the wall. The normal reaction of the wall will act on the ball. So the ball cannot get through the wall (due to normal force), but the wall will still experience force exerted by ball. These two are action reaction pairs so F=N. They indeed act on different bodies. It is correct to say that normal reaction will not allow the ball to get through the wall.

jbriggs444
It is correct to say that normal reaction will not allow the ball to get through the wall.
Yes. It is (in normal circumstances) as large as it needs to be to prevent the bodies from interpenetrating.

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