Sunday, September 13, 2015

Eraser

The year is 3015. Someone found a calendar in the rubble, so I'm pretty sure the date is the 13th of September.

We watched a movie called "Eraser" this week. Arnold Schwarzenegger plays a U.S. Marshal who "erases" people's past, making them disappear for their own safety, basically a super beefy, super muscular witness protection agent. He is charged with protecting  a woman who is the only witness in a case against a big tech company who is creating weapons for enemies of America. But *gasp* they are after her! Who could have ever guessed? They ambush her at her home, using one of the aforementioned weapons, a railgun. A railgun evidently capable of firing at almost the speed of light.

RIP all of physics ever
The blue trails are the tears of physicists
who watched this movie.


So, of course, we need a first hand example of the power of these railguns, right? The witness' ex Darryl is there at her house, a nice character who is unimportant to the overall plot, and the watchers have not connected to at all. The chance of him surviving this encounter? If you think there is any you have never watched an action movie. The third round hits him right in the stomach, flinging him backwards at least 7 feet through the air, slamming him into the wall behind him. Prognosis? Dead on arrival. So, that's cool and all. But, really? We saw the shooter fire this round, and he certainly didn't go flying backwards. And he even looks a little bit smaller than Darryl. So what's up with this?

Well, to figure out what should have happened, lets do some physics. In the movie, it is stated that these railguns fire at "almost the speed of light." Let me just say, we have a railgun stashed in the armory, and it sure isn't shooting at almost the speed of light. Upwards of mach 6 yes, speed of light? No way. Well, we will humor the movie (and ourselves) and figure out what should have happened.

Incredibly professional and detailed drawing of what happened

From the amazing drawing above, we can see that the shooter (hereon called mook) fires his bullet, and the bullet impacts Derryl and he is going to go flying backwards. Thanks to the Law of Conservation of Momentum, we know that the initial and final momentum of the system are both equal to 0. And thanks to a little clause in that law, we get to ignore friction and other non-conservative forces. So what we end up with is:

For the shot:
P(f,m)=P(f,b)

For the impact:
P(i,b)=P(f,b+d)

Where f and i indicate final and initial, b refers to the bullet, m refers to the mook, and d refers to Darryl
Also, because we ignore air resistance, P(f,b) in the shot is equal to P(i,b) in the impact.

What I want to know, is how fast was Darryl's now lifeless body going after that impact, and how fast was the mook going after the recoil of the gun blew his shoulder out and carried him with it? They should be similar, as they are about the same size.

As far as values:
m(m): he looks a little on the light side (bulky jacket) so i said m(m) = 80 kg
m(b): our railguns fire are .26 g aluminum slugs.I have nothing else to go with so, m(b)=.00026 kg
m(d): he seems just a little on the heavy side, so m(d) = 85 kg
Note: according to Einstin, calculating the momentum of objects going that close to the speed of light requires special calculations, which upon doing, we get:
p(b)=161,000 kg*m/s

so, v(f,m) = p(b)/m(m)

And we get: 
2012.5 m/s
or
4501.83 miles per hour
or
almost mach 6

I think he is dead. And I am not sure he knows it.

Now, doing those calculations for poor Darryl, we get:

v(f,d)=p(b)/m(d)

and we get:
1894.1 m/s,
4236.98 mph,
about mach 5.5

Yeah, he is dead too. Faster than you can say "Bad movie physics"

1 comment:

  1. Well written entry, but there are some issues with your analysis. First, you need to remember that conservation of momentum states that the initial momentum of the system equals the final. In your first equation, you have the final momentum of one object equaling the final momentum of the other object. That's not possible if the initial momentum of the system is zero. You also should have been more explicit about what speed you assumed for the momentum. One other point is that in your last calculation, you need to divide by the mass of Daryl plus the mass of the bullet. In this case that's a tiny correction, but I want to make sure you know how to do these problems correctly.

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