Don’t Underestimate the Forces, Mandalorian

I’m really excited about this new Star Wars spin-off show, The Mandalorian. It’s not part of the overcooked Skywalker saga, so pretty much anything can happen. And there’s the whole mystery of the armored Mandalorian: Who is he? What’s his motivation? Can he see where he’s going through those tiny eye slits?

But I just watched the trailer for the show, and right now I’m enthralled by another mystery: How can the Mandalorian deliver a massive body shot to an assailant (at about 0:20 in the clip), sending him flying through the air, without himself being thrown backwards?

Assuming this far, far away galaxy is still in the same universe we live in, subject to the same rules, it’s hard to explain. I’m thinking it might involve special shoes. But no spoilers! First let’s break down the physics of the fight.

The Nature of Forces

In the beginning of the sequence, the Mandalorian uses a long stick-like weapon as a staff to parry the attacker’s blows. So I must confess, I first thought he delivered the finishing blow by ramming the guy, and that shaped how I approached the analysis.

Later I realized it’s an Amban phase-pulse blaster (d’oh!), and he actually probably shot him in the belly at point-blank range. But that doesn’t matter—the analysis is the same. So let’s jump into it!

There are two big physics ideas we need. One is the nature of forces. A force (not the Force) is a way to describe interactions between two objects, and forces always come in pairs. So when the Mandalorian hits (or shoots) this guy, he exerts a force on him. But then there has to be a reaction force back on the Mandalorian.

That’s just how it works. If you push on a wall, the wall pushes back on you. If you drop a ball, a gravitational force from the Earth pulls down on the ball, but there is also a gravitational force from the ball pulling up on the Earth. I know, it sounds like some loopy zen thing Yoda would say, but it’s true. All of it.

In general, if A and B are two interacting objects, and A is pushing on B with a force FA-B, then there will also be a force from B onto A (FB-A), which is of equal magnitude but opposite in direction. As an equation it looks like this:

Illustration: Rhett Allain

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