Introduction (+ Disclaimer)

This isn’t meant to be a serious answer to the question “Which would win, an immovable object, or an unstoppable force?”. It’s more meant as a thought experiment, to try and answer it in a way that I didn’t really see others try to before. I always saw this problem as an interesting paradox, but I never gave it much thought before now. I welcome any thoughts on what you think about the following mess of writing.

Part 1: A Big Helping of Newton

There are two important equations in this scenario:

F = m a and F 1 = − F 2 F=ma F_1=-F_2 F=ma and F1​=−F2​

The first states that a Force is equal to the mass, multiplied by the acceleration. This comes from Newton’s Second Law of motion, which states that an object’s force is dependent on its acceleration and mass.

The second is the infamous Newton’s Third Law of motion, which states that for each force, there’s an equal and opposite force (or was it action?)

Then there’s momentum. Momentum is described as an object’s mass, multiplied by its velocity, and its a property that any moving object has. There’s a law called the conservation of momentum, which says that in a closed system, the total momentum must remain constant. In equation form, this looks like m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2 m_1u_1+m_2u_2=m_1v_1+m_2v_2 m1​u1​+m2​u2​=m1​v1​+m2​v2​, where u u u is the starting velocity, and v v v is the resulting velocity.

Then, there’s the collision. There are two kinds of collision, inelastic and elastic. A perfectly inelastic collision means that the two things end up with the same motion - effectively sticking together. A perfectly elastic collision means that no kinetic energy. This results in the things possibly bouncing off of each other.

Think of it like this. If two sticky boxes collide, they would stick together, forming one mass - meaning that their resulting velocity after the collision would be the same. If two rubber boxes collide, they would bounce off each other, each going at their own velocity. The former is the inelastic, and the latter is the elastic.

Each one has its own equation:

Inelastic Collision ⇒ v = m 1 m 1 + m 2 u 1 v = u_1 Inelastic Collision⇒v=m1​+m2​m1​​u1​ Elastic Collision ⇒ v 1 = ( m 1 − m 2 m 1 + m 2 ) u 1 + ( 2 m 2 m 1 + m 2 ) u 2 ⇒ v 2 = ( m 2 − m 1 m 2 + m 1 ) u 2 + ( 2 m 1 m 2 + m 1 ) u 1 v_1 = ()u_1 + ()u_2 \[5pt] v_2 = ()u_2 + ()u_1 Elastic Collision⇒v1​=(m1​+m2​m1​−m2​​)u1​+(m1​+m2​2m2​​)u2​⇒v2​=(m2​+m1​m2​−m1​​)u2​+(m2​+m1​2m1​​)u1​

Note that for an inelastic collision, we’re using the equation for when one mass is stationary. We can use this because the immovable object is, well, immovable.

Now that we have all of this out of the way, it’s time to start the reasonings.

Part 2: The Fight

Now the unstoppable force clearly has a force size of ∞ ∞, since it’s unstoppable and all. Therefore, its equation would be:

∞ = m a = ma ∞=ma

The immovable object is, well immovable. Mass is a measure for inertia, and if something is immovable, then its inertia must be ∞ ∞. To add, its acceleration has to be 0, otherwise it’s not immovable anymore. So its equation ends up being:

F = ∞ ∗ 0 F = 0 F=∞∗0

Now we have some variables to think about. If the unstoppable force has an infinite mass, then it would just be another immovable object. Since that would just solve the problem right there (neither would move), we’ll assume it’s a trivial solution. As a result, we conclude that its mass must be non-infinite.

However, you can’t multiply two non-infinite numbers to get an infinite one. By process of elimination, this means that a a a must equal ∞ ∞. So the force’s new equation becomes:

∞ = m ∗ ∞ = m * ∞=m∗∞

…where m m m is a real number. However, any number is trivially small when compared to infinity, so in essence, the m m m becomes effectively 0 0 0. This leads us to the final equation: ∞ = 0 ∗ ∞ = 0 * ∞=0∗∞

Okay, so this might not be making much sense, so lets recap quickly.

An unstoppable force needs to be infinite right? Otherwise, there would always be a chance for a stronger force to stop it. For it to be infinite, there has to be at least one infinite variable. Why? Because you can’t grab two numbers whose product is infinitely big! If mass was infinite, then it would just be another immovable object. That is not only boring, but it’s also the opposite of unstoppable, since it’s always stopped. As a result, mass has to be finite. This leaves us with an infinite acceleration. Look at Rule 2 for why. Since acceleration is now infinite, mass is now trivially small. As a result, we can effectively treat it as 0 0 0 .

So now we end up with the following.

An immovable object with force: F = ∞ ∗ 0 F = 0 F = ∞ ∗ 0 .

with force: . An unstoppable force with force: ∞ = 0 ∗ ∞ = 0 * ∞ = 0 ∗ ∞ .

Now you might be tempted to say that the force of the immovable object has to be infinity as well, but remember that right now, we have a blaring middle finger of common sense staring at us. Particularly, that infinity multiplied by zero is also infinity. Since we are in some newfound territory, we’ll assume that 0 ∗ ∞ 0 * 0∗∞ can be different from ∞ ∗ 0 0 ∞∗0.

So. Now for the confrontation. Our first weapon of choice is the Third Law. This means that when the unstoppable force acts on the immovable object, there’s a reaction of equal size. This basically fills in our F F F for us, leaving us with:

∞ = ∞ ∗ 0 = 0 ∞=∞∗0

See? Didn’t need common sense to get through that hurdle — it was solved nigh instantaneously. Now comes the momentum. Remember the function for the conservation thing?

m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2 m_1u_1+m_2u_2=m_1v_1+m_2v_2 m1​u1​+m2​u2​=m1​v1​+m2​v2​

So, let’s fill in what we already now so far:

∞ u 1 + 0 u 2 = ∞ v 1 + 0 v 2 u_1+0u_2= v_1+ 0v_2 ∞u1​+0u2​=∞v1​+0v2​

…yes, this hurts my brain too. I know you’re used to 0 0 0… zero-ifying everything, but I’m afraid that’s not possible in infinityland.

But now we have some more reasoning to do. See, we can use the extreme accelerations to our advantage. An immovable object has to have a velocity of 0 0 0, since its acceleration is 0 0 0, and it never had any velocity to begin with.

As for the unstoppable force, it has an infinite acceleration. This means that after the smallest unit of time, the velocity immediately jumps up to infinity. So we can practically replace those values in the equation now.

∞ ∗ 0 + 0 ∗ ∞ = ∞ v 1 + 0 v 2 0 + 0 * = v_1+0v_2 ∞∗0+0∗∞=∞v1​+0v2​

We can then group them up, leaving us with:

∞ + ∞ = ∞ v 1 + 0 v 2 + = v_1 + 0 v_2 ∞+∞=∞v1​+0v2​

…oh god what have we done.

Yep, we’ve hit a dreaded deadend. I’m afraid we can’t see which thing moves where using this equation. However, we have reasoned out the velocities of the things now, so it’s collision time!

For now, let’s use the formula for an inelastic collision, because it’s less threatening and also it makes sense - because we have a thing that ain’t moving:

v = m 1 m 1 + m 2 u 1 v = u_1 v=m1​+m2​m1​​u1​

So, let’s just plug our newly reasoned variables in:

v = 0 ∗ ∞ 0 + ∞ = ∞ 0 + ∞ v = = v=0+∞0∗∞​=0+∞∞​

Now, we can use some theorems on 0 and infinity. The useful ones that come to mind are the following:

∞ + x = ∞ + x = ∞+x=∞

0 + x = x 0 + x = x 0+x=x

Note that those two theorems are very compatible. If you do ∞ + 0 + 0 ∞+0, both the infinity theorem, and the zero sum theorem suggest the same answer. As a result, we can take the answer to be a valid infinity, resulting in:

∞ ∞ ∞∞​

…well. This is a deadend. See, the solution to this is undefined. In essence, it means that there really isn’t any meaning to it, and we can’t exactly attach one. Far smarter people than I have tried to find a solution, but they keep running into “solutions” that break all of mathematics. So, all in all, we can’t continue from here.

So let’s try elastic collisions…

v 1 = ( m 1 − m 2 m 1 + m 2 ) u 1 + ( 2 m 2 m 1 + m 2 ) u 2 v_1=()u_1+()u_2 v1​=(m1​+m2​m1​−m2​​)u1​+(m1​+m2​2m2​​)u2​

This turns into:

v 1 = ( 0 − ∞ 0 + ∞ ) ∞ + ( 2 ∞ 0 + ∞ ) 0 v_1 = () + () 0 v1​=(0+∞0−∞​)∞+(0+∞2∞​)0

Subtraction is a bit weirder, but we can tackle it by saying that:

0 − x = − x 0 - x = -x 0−x=−x

x − ∞ = − ∞ x - = - x−∞=−∞

…so using these two we can find that 0 − ∞ 0 - 0−∞ would be − ∞ - −∞.

However, I’m going to stop there, because we just ended up with the same “infinity divided by infinity” problem. Not only that, but the result is now more complicated, because we suddenly have ∞ ∗ ∞ ∞∗∞, and ∞ ∗ 0 0 ∞∗0, both of which are undefined.

So, what’s the solution?

The solution is, that the final velocity is undefined. More concretely, it’s:

v = ∞ ∞ = undefined v = = v=∞∞​=undefined

Which, unfortunately, means that there just simply isn’t a solution to this problem. We can play over what the undefined result can be all we want, but since there’s no way to prove it, it’ll just be a thought experiment.

So let’s just guess and think about options.

The answer is positive. This makes sense with two intuitions - anything divided by itself results in 1 1 1 usually, and anything divided by infinity ends up being infinitesimal, which although trivially small, is still a positive number. This would mean that the unstoppable force wins, since it manages to move the object. The answer is negative. Considering the answer is undefined, and there’s no exact way to invalidate any answer, it can just so happen to be negative. It makes no sense, and I’m not sure how exactly a negative can appear from thin air, but hey, that’s just undefined-edness for you. The answer is zero. This can possibly make sense, since if you divide anything by infinity, each bit ends up so small it might as well be zero. However it feels like a wrong version of outcome 1 when you think about it.

Well. The conclusion sure is anticlimactic. But on this journey we managed to learn some physics, barely learn a scrap of math, reason out something quite stupid (subjectively), and try to treat a paradox with some real life tools (foolishly might I add).

I hope you had fun reading, and may your journeys into the infinite be less… weird than this one!