Sunday, September 6, 2015

Mission Impossible III

The year is 3015. I believe the month is September.

Those of us examining the physics of old movies gathered together last week in order to watch the first one we have found, tucked away in the back corner of a crumbled down building that I believe used to be a museum.

Mission Impossible III.

Staring Tom Cruise as action hero/spy/general awesome guy Ethan Hunt, the plot revolves around megalomaniac black market bad guy trying to get his hands on ambiguous McGuffen in order to bring about the end of the world.

Before going into the physics, I feel the need to point out that as far as this movie is concerned, no one can have a good death. Early in the movie, important side character that Hunt is sent to save dies mid-sentence, seconds before they actually had the means to save her, leaving us with the most gruesome death face I believe I have ever seen. Main bad guy gets run over by a jeep. Other main bad guy dies without even a single word, shot by Hunt's wife who has never held a gun before. Even Ethan Hunt dies saving his own life, but is brought back from the brink only a minute later.

Now, on to the physics. Early in the movie, Ethan Hunt is charged with rescuing one of his old pupils (the one with the death face I mentioned earlier) from a the previously mentioned megalomaniac. During the resulting (mandatory) gun fight on the way out, Hunt is using an MP5 sub machine gun. As one of my fellows pointed out, at one point in the gunfight you can see one of the enemy mooks running towards Hunt, however is killed and falls forward, face first onto the ground. The question is, would the forward movement of the mook be enough to over come the force of the bullet impacting into his chest? Clearly, to answer this question, we need to know a few different quantities. We want to know his final velocity after the impact of the bullet, if it was positive or negative in regards to his direction of movement. To figure this, obviously we will need to know the mook's initial forward velocity, which I will call v(i,m). We will also want to know the initial velocity and final velocity of the bullet, which i will call v(i,b) and v(f,b) respectively. The mass of both the mook and the bullet will also be needed, which I will call m(m) and m(b), respectively.
Now, the values of these I found:
v(i,m) We only get to see him moving for a short time. After some lengthy research on an old, clanky, binary computer we found, I found the National Council of Strength and Fitness' listing of the average human run speed, at 15 mph. After converting to m/s, we get v(i,m) = 6.7 m/s
v(i,b): Using the average muzzle velocity of an MP5K sub-machine gun shooting 9mm full metal jacket rounds, and assuming negligible loss of velocity during flight time, we arrive at an initial velocity of v(i,b) =  375 m/s
v(f,b): this is just 0, because for this we are assuming that the entire energy of the bullet entered the mook, in other words the bullet stopped inside him.
m(m): just taking the average weight of a white male, we get m(m) = 82 kg
m(b): Just using a generic, 115 grain 9mm FMJ round, we get m(b) = 7.45 g
NOTE: Since the bullet and the mook are moving in opposite directions, one of the velocities needs to be negative. I choose to set the mook's direction of travel as the positive direction, and the bullet's velocity as negative.

After doing the quick calculations using the inelastic collisions equations, we find:

([82 kg * 6.7 m/s] + [.00745 kg * -375 m/s])/82.00745 kg = 6.665 m/s

In fact the movie was perfectly accurate, as the mook's final velocity decreased by a whopping 0 m/s, after sig figs are applied.

Later on in the movie, Ethan Hunt has the job of infiltrating the Vatican. Presumably a very well gaurded location. And of course he makes it look easy. Though in the process, he is laying on top of a wall, hooked to some sort of winch, and rolls off and the winch kicks in to slow him down and stop him just before we get to see what a Tom Cruise pancake looks like. Obviously, this looks awesome on film. What I want to know is, how fast was he going when the winch yanked him to a stop at the bottom of his fall? To answer this, we need to know the distance fallen, d. We also need the time it takes him to cover said distance, t, and how fast he was going at the top (hint, that one is easy)
d=16.55 m as shown in the film. However, Hunt stops just short of hitting the ground, so we shall assume an even 16 m for simplicity's sake.
t: by analyzing the footage we have, I was able to calculate 4 seconds from start to finish on his fall. t=4 s
v(i)= 0 m/s, as he starts from rest at the top of the wall.

Just by using one simple equation, we arrive at

v(f) = 2d/t

Plug and chug reveals: v(f) = 8 m/s at the bottom of his fall. Which that jerk to a stop would hurt. A lot. (quick and dirty calculations give me almost 4 g's, but that is outside the scope of this problem)

And finally, we reach the rope swing. Ethan Hunt has to infiltrate a super secure highrise in Shanghai. Of course this wouldn't be an action/spy movie without plenty of guards, lots of traps, and only one way in. In this case, the only way in is through the roof. As in, swinging across from a nearby roof on a rope. Yes, 21st century Spiderman style. Except Ethan Hunt isn't Spiderman, and this is totally just as bad of an idea as it sounds. At the end of the swing, Hunt disconnects the rope, and falls down onto the glass below. Hard enough to crack the (assumedly) plexiglass. Ouch. How far did he fall, and how fast was he going when he hit the glass? Well, luckly we only need to know a few things. The time it took him to fall, the acceleration of his fall, and how fast he was going when he started to fall.
Those values are:
t: this shot has several cuts, but seems to go straight through, so t = 3 s
a: Free fall baby, ignoring air resistance is the best. So a=-9.81 m/s^2
v(i): He releases the rope right at the apex of his swing, so v(i) = 0 m/s

Based on these calculations, Hunt fell 44 meters, and was going 29 m/s when he hit. Can someone say ouch? For those of you keeping up at home, that is 144 feet, and 65 mph. Maybe he is Spiderman. Or better yet Wolverine. The man is unkillable.

That is it for this week's physics. I think someone managed to find a copy of "Eraser" and maybe I will finally get to see this legendary Schwarzenegger character.

1 comment:

  1. This is a great post! It's entertaining to read and I learned something (I had never heard of a MacGuffin before)! I thought I'd point out, and you may have realized that for the Vatican fall problem, what you actually calculated was Hunt's average velocity, not his instantaneous velocity at the bottom of the fall. For that, you would use the equation d = v_i*t + 1/2 a*t^2 with a = -9.81 m/s^2.