14
Sep

Eraser – Conservation of Momentum


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The conservation of momentum is probably one of the most commonly laws of physics that is ignored for the sake of theatrics. The truth is that if someone is shot and completely absorbs the momentum of the bullet, they will be pushed back with the same force as the gun pushed back on the person that fired the weapon (ignoring momentum loss due to wind resistance).

The movie Eraser wildly ignores this law of physics especially surrounding the railgun featured in the movie. In order to calculate the relevant quatities we need to make some estimates about the gun and its victims. We don’t get to see the projectile, but we do get to see the cartrage that loads them into the gun. Looking at this and using the actor’s hand and other know cartrages as references; we can estimate (conservatively) that the projectile is 2 inches long and ½ inch in diameter. The movie also tells us that the projectile is aluminum. Using the density of aluminum (2.7g/cm3) we can calculate the mass of the projectile to be 17g. The movie aslo says that the projectile is fired at near the speed of light. This will give astronomical values so we’ll do the calculations both with c/2 and the velocity of the navy’s version.

With a mass of 0.017 Kg and a velocity of 1.5×108 M/s, the projectile would have a momentum of 2.6 million KgM/s! If this momentum was completely absorbed by a stationary freely moving 80 Kg man, he would then be moving backwards at 32,000 M/s. This is with the characterization of the weapon provided by the movie. This is absurd not only because of the uncontrollable recoil of such a weapon, but the energy envolved would probably just vaporize a human.

A more probable railgun muzzel velocity would be 3500 M/s. With this velocty, the same projectile would have a momuntum of 60 KgM/s. If this was absorbed by a stationary freely moving 80 Kg man, he would then be moving back at 0.75 M/s. This calculation makes the recoil of a handheld gun manageable, but not the impact on a victim. In the first scene where someone is shot, the ex-boyfriend is sent back about 20 ft in less than 1 second. We’ll aproximate this to 10 M/s. This is much faster than the momentum of the round can provide.

This problem is based on Newton’s third law of motion which basically says that for every action there is an equal and opposite reaction. As a gun is fired, the bullet pushes on the gun and on the person firing the gun. This is the recoil and is due to starting the bullet in motion. The equal force of stopping the bullet is shown in pushing on the victim. In short, if you shoot someone, the force of the bullet hitting them will equal the force of the gun’s recoil on you as you fire it.

07
Sep

Speed 2

The boat crashing into the dock scene:

1) With the rate at which the boat slowed down, how far on shore would it go before it stopped?

2) I would need to know how fast it was going when it first hit the dock and how much time it took for the boat to stop.

3) The last speed give before the boat hit the dock was 7.1 Knots. This converts to 3.65 M/s. The final velocity of 0 M/s was achieved after 200 seconds. Using the equation d=1/2(vi+vf)t, the distance was calculated to be 365 yards. The final after the boat stopped there is a shot that showed it on shore by about 1 boat length which is probably on the order of 365 yards.

The failed evacuation scene:

1) When the 15 min are up and the boat starts moving again, the lifeboat becomes unstable and our hero has to rescue the people on the life boat. He does this because the cable that supports the lifeboat is about to fail. I want to know what is the weight of the lifeboat with passengers and is it likely that it will fail.

2) I need to know the weight of the boat, how many people can get into it, and what kind of support it has.

3) I was difficult to get this info from the actual scene so I did some research on cruise websites to get what I needed. A fully equipt 150 person capacity lifeboat weighs in around 27,000 lbs. At 165 lbs each, passengers push that up to 52,000 lbs. These boats look larger than the ones in the film, but the testing is the same. For the boats I found the factory tested the lowering mechanism at over twice the maximum load it will ever have to work with.

Water Displacement of Ship:

1) Early in the movie, we see the cruise ship that they are getting on. I want to know what volume of water it has to displace to float.

2) I need to know the weight of the boat and the density of water.

3) They mentioned that the boat has 5 passenger decks. Comparing to several other ships, I guessed that it weights 50,000 tons. This converts to 4.5×1010 g. Since the weight of the boat equals the weight of the water displaced, I know that 4.5×1010 g of water is displaced. I looked up the density of sea water to be 1.025 g/cm3. This allows me to calculate that the displacement is 4.6×1010 cm3 or 46,000 m3.




January 2012
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