My Blog for Physics in Film

I chose the scene in Eraser when the bumbling goon picks up John’s shotgun, hence causing the wielder of the rail gun to tag him as John. The bullet sends him flying backwards into the necessary glass window. I used the following site for conversions: http://www.translatorscafe.com/cafe/units-converter/velocity/calculator.

I will start this discussion with the momentum of the bullet, the star of the show. It was mentioned that the bullet was an aluminum slug traveling near light speed. A bullet usually weighs around 8 g or so. However, those bullets are usually of a heavier metal, so I will say the bullet masses around 5 g, or 0.005 kg. Lightspeed is approximately 3.8E8 m/s. Ignoring relativity, and denoting the path of the bullet to be the positive direction, the momentum of the bullet, or Pb= (0.005 kg)(3.8E8 m.s) = 1.9E6 kg*m/s.

Next, the victim’s momentum. Being the average goon, he likely has a mass around 90 kg. So what should his velocity be when he is hit by the bullet with a momentum of 1.9E6 kg*m/s. Since momentum is conserved his momentum, Pg, g for goon must be equal to Pb because he will be going in the same direction, assuming all the momentum of the bullet is transferred. So given that, the Mg*Vg=1.9E6 kg*m/s, so Vg= (-1.9E6 kg*m/s)/90 kg, kilograms cancel and his new velocity is 21,111 m/s or 47,223 mi/h. Actually, according to Dr. Fragile’s site, escape velocity for the earth is only 11,000 m/s. So the bullet should actually put him into orbit (not necessarily literally). However, the bullet did not transfer all its momentum to him as the bullet passed through him with relatively little visual evidence slowing down (but it’s traveling at the speed of light, how do you see it?). So let’s say the bullet only transferred 0.01 % of its momentum. He would still be going about 2.1 m/s. This would be more reasonable…if the bullet could even be fired at lightspeed. Hence the next subject is below.

The gun and the gunner MUST have the same momentum (albeit negative) of the bullet. Let us say the total mass of the gun combo is 100 kg (10 kg for the gun). Assuming the gunner is attached to the gun, with something like a heavy duty harness or by the sheer grip of his hands, the gun and him would be traveling at the same velocity, and in the opposite (negative direction) of the bullet. So –Pb = Mc*Vc, c for “gun combo”, so Mc = 100 kg and –Pb = -1.9E6*m/s, therefore Vc = 19,000 m/s. Enough to also send the gunner into orbit. The other scenario would be if the gun flew off on its own, transferring no momentum to the gunner. The 10 kg gun would then probably burn up in atmosphere as it slowly drifts towards the horizon at 190,000 m/s.