# Why does a metal ball not trace back its original path if it hits a wall?

If we have a metal ball, why does it follow the law of reflection after it hits a wall? My point is that if it applies a force on the wall at an angle say 30deg below the horizontal (the wall is the vertical itself), shouldn't the wall apply a force back which is completely opposite to the original i.e. 150deg above the horizontal? As far as I know the direction of the motion of the ball flips, only around the vertical i.e. it is 150deg below the horizontal. Why is this so?

And a sub question which came up while I was writing this, how do you even calculate the force on the wall if you are only given the mass and velocity of the ball. I know we can calculate the momentum but what more do you need to know the force on the wall?

• An easy (despite superficial) answer would be: who says that the force exerted by the ball on the wall should be parallel to the ball's velocity? In fact it's not. However this raises a very interesting question that I add to yours: how do we microscopically describe a scattering event between a rigid body and a very stiff solid wall? Commented Jun 6, 2023 at 13:28
• Why metal? Why not just a ball. Commented Jun 8, 2023 at 4:08
• @Mazura a ball made of peanut butter would behave differently :) Commented Jun 8, 2023 at 13:42
• As would one made out of a lattice of carbon. I'm presuming they picked metal because they believed it has a low enough elasticity to be ignored, which wouldn't even be true of diamonds. But the Q is of the law of reflection so has nothing to do with an imaginary inelastic material. Commented Jun 9, 2023 at 0:53
• @Mazura Exactly. Thats what I thought lol Commented Jun 9, 2023 at 9:55

For this kind of problem, objects are idealized to make the problem focus on how forces affect objects. You can concentrate on things like decomposing forces into components, and not worry about how complex friction really is.

A wall is idealized as a perfectly rigid object. This is not far from the truth. Bonds are imagined to hold the walls atoms a fixed distance from each other. If something pushes on some atoms, they push back with a reaction force just hard enough to hold their position. This ignores the momentary compression that goes on during a collision between two rigid objects.

A wall may or may not have friction, and friction is simplified. Friction is either static or proportional to the force holding the objects together. Again not far from the truth. A wall of ice is pretty close to frictionless. A wall of concrete is not. Friction prevents objects in contact from sliding along each other.

Without friction, a wall just keeps a sphere from penetrating the surface. It pushes back perpendicular to the surface. With no component along the surface, the component of velocity along the surface would not change. In an elastic collision, the final speed is the same as initial. So the normal component of velocity must have the same magnitude before and after. From this, you can see the angle does not change.

By contrast, friction pushes back when the sphere tries to slide along the surface. This would change the component of velocity along the wall, and would change the angle.

You can see this if you hit a tennis ball with spin against a concrete wall. You can make it bounce off at various angles. This does not work with a wall of ice.

Okay, this is a very fun question for which we can test our knowledge of conservation laws. Here in your situation we can use the law of conservation of linear momentum.

To test it let's assume that the ball is moving with a velocity $$\vec{v}$$ making a angle of $$\alpha$$ with the horizontal then we can associate its momentum components as \begin{align} p_x & = mv \cos\alpha \\ p_y & = -mv \sin\alpha \end{align}

What happens then is that the normal force will only act in the y-direction. Why?

I try to look at it from this perspective. Don't consider the ball to be moving at an angle. Imagine that the ball is going downways and going rightwards at the same time. Why do we say that? It is because both of the momentum or I should particularly say the velocities are independent of each other. So we can break down the motion in two different kinds of motion and what we do is to combine the two motions. So when we talk about the vertical case it is very obvious that the normal force is perpendicular to the surface. The second case shows us that the ball will not apply any kind of force on the surface so there is no normal force along the horizontal direction. And now we can apply conservation of momentum to say that it must reflect at the same angle with which it was incident.

I hope you are satisfied by the answer (although I must say that you might have learned this approach to find projectile motion of a object so I don't see it as a new thing to you).

• This is of course correct, but I would argue that in case the wall-ball system has a non-zero friction coefficient, then the horizontal motion of the ball would result in a horizontal force applied to the wall. So does this happen only for a frictionless wall-ball system? Commented Jun 6, 2023 at 13:38
• So I won't argue that the ball will lose some of it's speed. It may also start rotating so yeah. Commented Jun 6, 2023 at 13:57
• The answer isn't saying anything wrong, but OP's question isn't really answered. It can now be rephrased as asking why you choose to deconstruct it into two forces, one of which is exactly perpendicular to the wall (and the other parallel)? Why not an arbitrarily different angle? Commented Jun 7, 2023 at 3:36
• Momentum of the ball is not conserved through the bounce. The direction of motion is changed, which means the direction of the momentum has changed. Momentum is only conserved when you also count the tiny motion of the wall and the Earth. Commented Jun 8, 2023 at 3:01

I think it's simple geometry: A line normal to any point on the surface of a sphere passes through the center of the sphere. If the contact between the (presumably spherical) ball and a (presumably planar) wall is frictionless, then the contact force between them must be directed along that line. It gets a little trickier if you want to account for friction and, for the ball's moment of inertia.

what more do you need to know the force on the wall?

You can only understand the force as a function of time. Accurately modeling that function will take you pretty deep into understanding the material properties of the ball and, of the wall. Probably either have to solve some differential equations or else gen up a numerical simulation of the collision. Way above my pay grade.

• I am not sure this is pure geometry, because it truly depends on the physical nature of the wall. If, for instance, the "wall" is a liquid, then it's a totally different story. So my extra question is: how can we give a microscopic explanation for the fact that a very stiff solid wall exerts on the ball a purely perpendicular force? Commented Jun 6, 2023 at 13:45
• @Matteo, The actual direction of the force does, in fact, depend on the physical nature of the wall and the ball, but if that physical nature can be described as "frictionless," then the contact force must be normal to both surfaces. Commented Jun 6, 2023 at 13:52

In reality, the case where the angle of reflection equals the angle of incidence is only one of the possibilities. It happens if the contact is frictionless.

It is more easily understandable if instead of a wall we think of a table tennis racket. Suppose we are in the reference frame of the racket on the contact instant, and the player is applying a top spin. The situation is similar to a ball hitting a wall at an angle. But it is perfectly possible that the ball returns back in the same incidence line.

Of course it will spin, and the magnus effect will curve the trajectory, but that is another story.

The average contact force depends not only on the momentum change but also on the time of contact, which is small but not zero. $$F_{av} = \frac{\boldsymbol{\Delta p}}{\Delta t}$$.

User Charu _Bamble has already introduced the notion of splitting the velocities of the ball into $$X$$ and $$Y$$ component vectors, with $$X$$ being parallel to the wall, and $$Y$$ being perpendicular.

Let's view the edge cases:

• For Y=0, the ball is just moving parallel to the wall. Assume that it is going as close to the wall as you can imagine, i.e. barely not touching at all. Hence, the ball and the wall do not interact at all. Very important: the concept of "touching", on atomar levels, is very complicated. Atoms are not balls; and no atoms ever touch in real life situations anyways - they get repelled by the fundamental forces, mostly the electromagnetic force, long before the actual particles get anywhere near each other, on atomic scales. While of course atoms passing each other by would influence each other, the aspect of them being in the macro iron ball lattice is much much more relevant; so some very small forces ocuring here while the ideally smooth ball flys along the ideally smooth wall are neglectable.
• For X=0, the ball goes straight into the wall and bounces back, but this is also uninteresting because this special case indeed is the what if case of your question (i.e., the ball goes back exactly on its original, reversed path). What happens atomically in this case is that the atoms of the ball get ever closer to the atoms of the wall; and at some point (long before the actual atoms would "overlap") are repelled by electromagnetic forces. This in turn can deform both participants depending on their elasticity, but in our particular case, both the ball and the wall are rather rigid, and presumably the ball is slow enough not to punch a hole into the wall. This means that all those forces occuring at the very border between the wall and the ball just reflect the ball right back where it came from.

Now with that mental model, let's handle the case where there is both an X and Y component; let's just say it hits the wall at 45° (X=Y).

Individual iron atoms will indeed maybe get close to individual wall atoms, and those individual interactions, if they were heads-on (from the point of view of the atoms) will of course in a way lead to a force pushing straight back. So a "straight-back" force will work on that one iron atom (and the opposite force on the "wall atom"). Both of these atoms are firmly held by their neighbours; the wall atom cannot go anywhere, so stays there more or less; the iron atom will impart a tiny little force on its neighbours, and will want to travel back right where it came from.

But the issue is that we cannot really only watch individual atoms here; neither the ball nor the wall are gasses, but are already packed relatively densely. That means the previous case can - statistically - almost never happen. Even if there is an "outlier" atom in the wall which gets very close to the wall before all of its neighbours, all the electromagnetic influences of the wall atoms closest to it will themselves form a field which is almost like a plane or a wall itself. It will not, on average, receive just the spherical nature of a single wall atom, but will interact with all the little spheres of all the little wall atoms, statistically smeared into a big boiling mass of electromagnetism.

While the X motion along those "wall of electromagnetism" will have some effect, this will be extremely less than the effect of the Y motion which would make the iron atom move into the field of the atoms. It is that Y vector which will quickly lead to much increased repelling force, and this force will be perpendicular to the (idealized) relatively smooth planar field made up by all the electromagnetic clouds of all the wall atoms. Hence the overall, statistical, macro force on all the iron atoms is very much predominantly straight along the -Y vector.

The rest follows from conservation of momentum. As there is only very little affecting the X momentum, it will be unchanged. Watched from straight on, the ball will move along the wall just as before the collision. But the Y vector must reverse since those materials do not intersect easily, and we didn't use enough force to make the iron ball actually disturb the wall. So it bounces straight up in the Y direction, while flying unchanged in the X direction, which leads to the effect you are asking about.

Let's simplify things a bit and assume that the wall is a perfect mirror and the ball is a photon of light. That way all complications of friction and the like can be left out.

Relative to an observer at rest, the motion of the photon can be deconstructed into having an x-component parallel and a y-component perpendicular to the mirror (v(x) and v(y)) (refer to the fine drawing by @Charu_Bamble in another answer). If you then imagine, for a moment, that the mirror is also moving in the same x-direction as the photon and with the same speed v(x) then only the y-component of the photon's motion (perpendicular to the mirror) remains. And only that component changes direction upon the bounce.

After the bounce, the photon's speed v(x), parallel to the mirror, has not changed and if you now imagine it's path through space from the perspective of the observer, then you'll have to conclude that the angle of the photon's path is the same but inverted around the perpendicular of the mirror.