# Ball hits curve of same curvature [closed]

I was doing some physics problems for homework and, while procrastinating, I came up with a theoretical scenario that I couldn't figure out the result of.

The following is from a side view and in a frictionless, ideal environment:

A ball moves toward a wall at a constant velocity. At the bottom of the wall, connecting the wall and the floor, there is a curve with the same radius as the ball (i.e. the wall and floor form two sides of a rectangle with rounded corners). What will happen when the ball comes into contact with the wall? Will it bounce off the wall like the curve isn't there, or will it roll up the wall?

• Assume no drag, no friction, g=10m/s/s, it will roll up until all the kinetic energy is converted into potential energy. Dec 6, 2019 at 6:13
• Is the ball rolling or sliding? Dec 6, 2019 at 16:08
• @ja72 I don't think rolling would have any effect as the surface is frictionless.
– user238497
Dec 6, 2019 at 16:34
• In the "second" case (top right diagram) it is not even clear that it will go upwards. Depending on the speed, radii, it would very likely be initially deflected upwards at say 10° - 15°, and then because of that, it would fly (clear) anbd then hit the vertical or near vertical wall, and be sent backwards at a shallow angle. Note that in the real world with cases like "B", it is impossible to make it go "straight up". (Unless the ball is extremely smaller than the curve, even then.) Dec 6, 2019 at 18:38
• This question should be closed because it does not show any prior work, ask about any specific physics concepts, etc. Please edit to make the question more on-topic for this site. Dec 7, 2019 at 18:19

You don't get an answer in an ideal world. This is the boundary between 2 different outcomes. The answer depends on details that have been idealized away.

When a ball hits an obstacle, it deforms. Likewise, the wall is deformed at least a little by the ball. These changes in shape don't matter much for the first two cases, but they can have a bigger impact in the following scenario;

Suppose the ball squashes horizontally and becomes taller. That will tend to lift the ball. Also the ball will spring back as it regains its shape.

--- Edit

In response to Lamar Latrell and Graham I would like to make a point that various comments and answers have alluded to. The first two cases have different behaviors.

When a ball bounces off the wall, the interaction takes place over a very short time. The force is very large. The ball deforms and springs back hard enough to fly off the wall. In the case of an elastic collision, it springs back with the same velocity. This is often ignored when people are only interested in the final velocity. It is abstracted away as an instantaneous collision.

When a ball slides up a gentle curve, the interaction takes much longer. Forces are relatively gentle, and deformation is smaller. The spring back is approximately $$0$$. The normal forces from the wall slow the ball and lift it.

When the radius of corner is slightly larger than the ball, deformations become important. The point of contact might quickly shift from bottom to side. But it might also expand from a point to a patch. You have to think in 3 dimensions. A sphere is rolling up a cylinder. The patch has area, not just length.

When the radii match, the point of contact definitely becomes a patch of at least 90 degrees. It also flattens onto the wall.

Forces depend on the degree of flattening. Some forces will be upward. It depends on details like the shape of the patch. This is controlled by properties of the ball.

In an ideal world, the ball is approximated as infinitely rigid. In that case, you might apply normal forces along a line of contact all at once as AccidentalTaylorExpansion and others have done (+1). But you should not be surprised if a better model of the interaction gives a different answer. In particular, the normal forces do not directly cause the ball to fly off the wall. They cause deformation, and internal forces in the ball cause it to spring back into shape and fly off.

• Not to mention there isn't a frictionless surface, and hence the ball is going to roll which means it will tend to ride up the corner bend. Dec 6, 2019 at 19:52
• Could you elaborate more on 'You don't get an answer in an ideal world' ? That is the question that the OP actually asked. Dec 6, 2019 at 20:24
• My guess is real life is "softer", where the conditions change gradually near the critical angle, and perhaps it's less repeatable run to run. Dec 7, 2019 at 7:12
• You've got that the wrong way round. You only get an exact answer in an ideal world, which is that the vertical component from the curve is exactly zero. Any tighter radius and it only hits the wall; any wider radius and you get progressively more vertical component. With deformation of the ball and surface in the real world, of course in reality it won't exactly do that - but this is the theoretical boundary condition. Dec 7, 2019 at 10:39
• @LamarLatrell - I updated my answer. Dec 7, 2019 at 15:44

If you want to make your life more complicated sit down and buckle up because I'm going full in.

First we must take a look at how the ball bounces back in the first place, since this is rather tricky for instant bounces. Consider the following situtation:

A ball with some initial momentum $$\vec p$$ moves to the right and hits a slope. Let's ignore gravity for now because this problem is going to be complicated enough. During the collision a force will act on the ball and we don't how that force looks exactly, but we know two things

1. The force is a normal force, so during the collision the force will be along a line joining the contact point and the center of the ball (see the red arrow in the picture)
2. Energy will be conserved (ideal environment) so $$|\vec{p}\,'|=|\vec p|$$ where $$\vec p\,'$$ is the momentum after the collision.

From this we can conclude the momentum will change in the following way:

The momentum is rotated by $$2\theta$$ since $$\Delta p$$, the red arrow, has to be parallel to the normal force. Now apply this to the two edge cases you mentioned to see this makes sense. For the head on collision we have $$\theta=\pi/2$$ so the momentum will be rotated by $$\pi$$. This means the particle will bounce back directly like expected. For the case with the large arc we have a continuously varying curve. Let's break up the curve in many line segments and take the limit towards a continuous curve. In this limit the angle between two sections goes to zero so the angle at which the ball bounces away also goes to zero. The ball keeps hugging the curve like expected.

Now let's look at the case where radius of curvature equals the radius of the ball. This case is tricky and we have to make some choices. Let's consider a short time interval during which the collision happens. The entire lower-right quarter of the ball experiences a force at the same time, but we don't actually know how this force is distributed. I'll make the following assumption so we can actually calculate something: the force at each contact point is proportional to $$\hat r\cdot\vec v$$, where $$\vec r$$ is the vector joining the center of the ball and the contact point and $$\hat r=\tfrac 1 r\vec r$$. The force is also in the direction of $$\hat r$$ since it is a normal force. Convince yourself that this agrees with the case I mentioned first. I define $$\phi$$ such that $$\phi=0$$ at the right-most contact point and $$\phi=\pi/2$$ at the lowest contact point.

The total change in momentum can now be written as an integral over all the contact angles ($$\phi$$). Since we don't know the exact magnitude yet I introduce a factor $$c$$ to be determined later. $$\Delta \vec p\propto\int (-\hat r\cdot \vec v)\hat r\mathrm{d}\phi\\ =c\int_0^{\pi/2} (-\cos\phi)\begin{pmatrix}\cos\phi\\-\sin\phi\end{pmatrix}\mathrm{d}{\phi}\\ =c\begin{pmatrix}-\pi/4\\1/2\end{pmatrix}$$ The last line used $$\int_0^{\pi/2}-\cos^2(\phi)\,\mathrm{d}\phi=-\pi/4,\int_0^{\pi/2}\cos(\phi)\sin(\phi)\,\mathrm{d}\phi=1/2$$

To determine $$c$$ I use conversation of enery again. So $$|\vec p+\Delta \vec p|=|\vec p|$$. Plugging this equation for $$c$$ in Mathematica gives $$c=\frac{8\pi mv}{4+\pi^2}.$$ An ugly expression, but an answer at least. The fact that it is proportional to $$\vec p$$ makes the angle always the same as you will see shortly. To determine the angle between two vectors you can use $$\cos\theta=\frac{\vec a\cdot \vec b}{|\vec a||\vec b|}.$$ Plugging this in for $$\vec p$$ and $$\vec p\,'=\vec p+\Delta\vec p$$ finally gives (using Mathematica again I'm not crazy) $$\theta=\arccos\left(\frac{4-\pi^2}{4+\pi^2}\right)\approx 115.037^{\circ}$$ Or about this angle:

I would like to stress again that I made some assumptions about how the force is distributed during the collision so your answer might be different if you made different assumptions. You have to make these assumptions since this problem is impossible to define exactly.

TLDR - under some assumptions you can calculate the angle at which the particle bounces back. My calculations give $$115^{\circ}$$ ( $$65^{\circ}$$ with respect to the ground)

EDIT - In the comments it was suggested that the restoring force for a particular angle is proportional to $$-\cos^2\phi$$ instead of $$-\cos\phi$$. I'm not convinced entirely but in that case the calculation comes out to be $$\theta=\arccos(-3/5)\approx 2.21$$ radians or $$53^{\circ}$$ from the ground.

• Yay! Finally a plausible treatment. I personally would have used the limit of a compressible (Hook's law) surface as the spring constant goes to infinity, so you'd get a restoring force of $-\cos ^2 (\theta) \cdot \left( \cos(\theta) \sin(\theta) \right)$, but for a physically unrealistic scenario the difference isn't important enough for me to care to rederive it. Dec 6, 2019 at 20:39
• Sounds interesting, how would you get $-\cos^2\theta$ from Hooke's law? I kind of don't want to rederive the whole thing either haha. Dec 6, 2019 at 20:57
• @AccidentalTaylorExpansion - A displacement of $dx$ forwards results in an edge that has moved inwards $\cos(\theta) dx$ for a restoring force of $-k \cos^2(\theta) dx^2$ Dec 6, 2019 at 21:30
• @AccidentalTaylorExpansion hey nice username Dec 6, 2019 at 23:17
• @EricDuminil Under these assumptions my answer is exact. It's like giving $\pi/2$ or 1/3 as an answer. Dec 7, 2019 at 10:49

The ball will rise up the curve if it's horizontal velocity is not stopped or reversed. It's because if it's horizontal velocity becomes $$0$$ it won't be able to proceed further to climb up because of gravity. It might just for a while if the initial velocity is sufficient.

## First case

Here, as you mention, it won't climb up because as soon as it hits the wall the ball will receive a force in the opposite direction (assuming elastic collision). Moreover it won't receive any net force in the vertical direction because the radius of curvature of the wall is smaller than that of the ball.

## Second case

In this case, the ball will rise up because it's kind of an incline which is varying it's slope. Moreover since in this case the ball doesn't lose all its velocity in the $$x$$ direction at once it can move above the curve until all its potential energy is converted into kinetic energy.

## Third case

Here, the ball will receive a net force which will be above the floor in the $$-x$$ direction as shown in the diagram. Now this force could raise the ball up but it may not be able to do so because of gravity. If there were no gravity the ball would go up and to the left.

• This is just a claim without substantiation, isn't it? The nice sketch not withstanding ;-). Dec 6, 2019 at 13:05
• @Peter-ReinstateMonica Thank you very much for your response. :-) I respect your as well as the user's views who considered deformation while answering the question. If we consider deformation, I think the analysis may become a bit messy. I'm not sure I'll be able to handle that. Moreover, the question asks to consider ideal situations so I considered a very rigid ball and walls etc. Dec 6, 2019 at 16:18
• I like this answer. Imagine a cube instead of a ball, or a prism with many sides (in fact so many that the number of sides tends to infinity) - how would we expect that to react? When the number of sides is low, like a cube we expect it to just bounce and go left, so why would increasing the number of sides affect its behaviour? Dec 6, 2019 at 17:36
• This is not a good answer, unfortunately. Dec 6, 2019 at 18:33
• This sort of force diagram doesn't work to determine whether the ball will wind up going backward or not (and if so, how fast). It isn't clear whether the backward force is more than is needed to stop the ball. Dec 6, 2019 at 21:03

The ball will bounce away at an initial angle of 64.96° to the horizontal.

At the point of contact, the incident force is distributed over the lower right quadrant of the ball; a reasonable assumption is that the incident force (and thus the impulse) at a point is proportional to the cosine of the angle between the incident velocity and the surface normal (so zero at the base of the ball, rising to its greatest amount at the right side of the ball). Integrating and discarding constant factors (since we only care about the direction of the net impulse), we get:

$$J = \int_0^{\pi/2}\begin{pmatrix}\cos^2\theta \\ \cos\theta \sin\theta \end{pmatrix} d\theta \propto \int_0^{\pi/2}\begin{pmatrix}\cos 2\theta + 1 \\ \sin 2\theta \end{pmatrix} d\theta \propto \left[\begin{pmatrix}\sin 2\theta + 2\theta \\ -\cos 2\theta \end{pmatrix} \right]_0^{\pi/2} \propto \begin{pmatrix}\pi \\ 2 \end{pmatrix}$$

This net impulse then has an angle to the horizontal of $$\arctan 2/\pi = 32.48°$$.

By conservation of energy (since the collision is ideal) the ball must exit with a speed equal to its incoming speed, so by parallelogram addition of vectors the angle to the horizontal of the change in velocity is half that of the exit velocity, so the exit angle must be $$2 \arctan 2/\pi = 64.96°$$ to the horizontal.

• You can improve your answer by supplying mathematical expressions that support your claims. Instead of invoking Occam's razor try justifying your assumption by some physical input Dec 6, 2019 at 15:06
• I dont see why there would be ANY incident force from the "extreme bottom" of the semi-circle of contact. Dec 6, 2019 at 18:34
• @Fattie yes you're right, I think cosine of the angle between the normal and the impact velocity would make more sense? Dec 6, 2019 at 20:31
• And this is the same result as @AccidentalTaylorExpansion above. Dec 6, 2019 at 20:58

This can be answered by considering the extreme cases

• The ball is always in contact with the floor at a single point with normal reaction from the wall acting in direction opposite to that of the ball. This case would arise (at the time of hitting the wall) I when the radius of curvature there is smaller than that of the ball.

• The ball is always in contact with the surface and has a single point of contact with the normal reaction always being perpendicular to that of the direction of motion. This case would arise (at the time of hitting the wall) I when the radius of curvature there is greater than that of the ball.

Now the case you specified is an intermediate one where normal reaction is acting in perpendicular as well as opposite direction to the motion hence the ball would fly off from the wall at an angle of $$45°$$ from the horizontal.

• The impulse applied by the ramp will be 45deg, but you forgot to add the momentum of the ball moving right... Dec 9, 2019 at 16:25

Ok, here's my guess after a good forty seconds' thought. I think it will bounce away from the corner at an angle that will depend upon its speed. Reasoning is as follows. Assume perfect conditions, eg zero friction, perfect curvature, perfect elasticity, etc. In that case the impact will be instantaneous at all points along the lower right quadrant of the ball. The magnitude of the reaction at each point will have two components, one due to the effect of gravity the other due to the impulse of the collision. The gravitational component will be independent of the speed of the ball, the other component will increase with speed. If we assume the speed is such that the gravitational component can be ignored, then in the absence of friction the reaction must everywhere be normal to the surface of contact, ie toward the centre of the ball. The sum of the forces must therefore have an upward component and a component to the left, so the ball will recoil at some angle.

• The effect of gravity depends on the elasticity of the material that the ball is rolling on, though. For an "ideal" material, I would think the effect of gravity gets zeroed out. Dec 6, 2019 at 21:17

To get an answer one must look at the normal contactforce distribution at the time of impact. It should be obvious that the contactforce will be the highest at the height of r on the wall and 0 at the bottom contactpoint. Inbetween those two points the force depends on the angle between the velocity and the normalvector. Since I have no information about the amplitude of the force I will neglect it. The resulting angle if $$\alpha=0$$ is where the force is $$0$$ and $$\alpha=\pi/2$$ where force has its maximum can be calculated by $$\beta=\frac{\int_0^{\pi/2} sin(\alpha)\alpha d\alpha}{\int_0^{\pi/2} sin(\alpha)d\alpha} =1~rad = 57.3^{\circ}$$. Basicly the Ball will bounce of and fly back in the direction where it came from with an angle of $$\gamma=90^{\circ}-57.3^{\circ}=32.7^{\circ}$$ from the ground.

Edit: In this picture the normalforce distribution is shown and the resulting force vector at impact is also shown. The angle to the ground is $$32.7^{\circ}$$. Black are the normal contact forces and red is the resulting force.

Gravitation is not included in this calculation since there is no information about the speed and the weight of the ball. But one could therefore do the same calculation and change the assumptions to be force maximum from gravity is at the maximum at $$\alpha=0$$ and has to be $$0$$ at $$\alpha=\pi/2$$. Therefore the resulting force vector of the gravitational contactforce has to be at $$\gamma=57.3^{\circ}$$.

Edit2: I forgot to take into account that gravity acts in a differnt direction then the other force. Superposition has to take this into account.

Edit3: Corrected some spelling errors and adjusted the formula above so the anwser has correct units. The formula itselfs comes from the calculation of the point of attack for a line load or distributed load. Just looking at the angle as a "normal coordinate". I also did some numerical calculations and the resultant is given as $$32.52^{\circ}$$

• You might want to expand your derivation, as it is, it is not exactly clear how your argument holds, observe the problem is ideal, there is no friction and has a high degree of symmetry for which an angle of 32.7º doesn't fit. Dec 6, 2019 at 15:11
• Actually it is not that symmetric. The velocity vector is parallel to the floor and therefore you cant say that at the contact point the conditions are symmetric. Dec 9, 2019 at 13:09
• It is along a 45º line, without gravitation involved. You can apply such a transformation to all coordinates in the problem and your problem doesn't change. So all I am saying is, if you write down the details perhaps is clearer for the OP to understand how the angle of 32.7º can be obtained. I was too quick to say it doesn't fit. Dec 9, 2019 at 13:45

I think the answer is fairly simple. The ball will rise up and rises until all of its kinetic energy has been converted to potential energy. If we try to visualise, if the radius of curvature of the corner in collision is less than the radius of the ball, it will obviously collide before being able to rise up (although it might bounce due to slight deformation on collision ). Otherwise, it can rise. That should be sufficient :)

• Funny that you call the answer simple. The OP's case is between the two simpler ones which is exactly what makes it hard. Dec 6, 2019 at 13:04