Let’s say a bullet has a mass of 147 grains (0.009526 kg = 0.021 lbs) and a speed of 2700 f/s (822.96 m/s). This means it has a kinetic energy of 2379.1 foot pounds (3225.7 J).
Does this mean the bullet’s energy is equal to the amount of energy it would take to raise a 2,379.1 lb boulder 1 foot in the air? Or equivalently, moving a 1 lb rock 2,379.1 feet in the air?
Now let’s say a 2379.1 lb (1079.13 kg) boulder is hung by a crane using a rope. I walk directly under the boulder, point the rifle straight up, and shoot the boulder with the above-mentioned bullet. The bullet lodges in the boulder. How much will the boulder rise?
Now, I understand momentum is conserved, and most of the bullet’s energy will probably be converted to heat. But what if the none of the bullet’s energy was converted to heat, and all of it was used to move the boulder? In other words, pretend the bullet lodged in the boulder and no heat was created due to friction. Would the bullet move the boulder up 1 foot?
The answer to Question 1 (both parts) is yes. However, that doesn’t mean that you’ll kick a tonne boulder up in the air.
Assuming that the collision is inelastic–that is, the bullet doesn’t bounce or deflect, but instead sticks to the boulder–total momentum of the system is conserved, such that:m[sub]bullet[/sub] * v[sub]bullet[/sub] = m[sub]boulder+bullet[/sub] * v[sub]boulder+bullet[/sub]
So your final velocity of the boulder+bullet system would be:v[sub]b+b[/sub] = 0.021 lbm * 2700 feet/sec / 2380 lbm = 0.024 feet/sec
(I’m going to dispense with the precise weight of the after system in favor of conserving only the necessary significant digits.)
Energy of the system is conserved in the global sense; however, in the inelastic case, kinetic energy is not conserved. Most of it goes into smushing the bullet into a pancake (since the collision is inelastic, the bullet must deform). If the bullet doesn’t deform, it’ll bounce off of the boulder having imparted a modicum of momentum in proportion to its mass (assuming the centers of mass are in line with the collision so that we can neglect angular momentum effects) and causing an almost impercievable velocity change upon the boulder. Total system momentum is always conserved, regardless of what happens to the bullet.
BTW, 2700 fps is way fast for a 147 grain 9mm Parabellum. Typical speeds for the 147 grain would be in the neighborhood of 900 fps.
No, this means the bullet has the same amount of momentum as a 2,379.1 pound boulder that is moving at 1 f/s. And if you shot the bullet at the boulder with no heat loss/deformity of either (and it was an inelastic collission, meaning the bullet did not bounce off, and the bullet did not add to the mass of the boulder despite this, and was not slowed by gravity on its way up) you would temporarily cause the boulder to start rising at 1 foot per sec. Unfortunately, gravity is also pulling the boulder down at 32 ft per sec. So–roughly (to do this for real you need to whip out some calculus)–your boulder starts to head upwards but by the time .03125 seconds have passed it is at a standstill. It will have travelled (again, roughly) .015625 feet before it starts to fall back earthward. That’s about .2 inches, and don’t blink or you’ll miss it :-).
The key thing here is that energy (1/2 MV[sup]2[/sup]) and momentum (MV) are both conserved. The bullet has a lot of energy, but comparitively little momentum, the slow moving boulder will have low energy, but comparitively more momentum. Momentum is always a function of your velocity, but energy isn’t. In order for the momentums to be conserved, a lot of the energy has to be lost as heat. There’s really no “what if” to discuss, it’s impossible for all of the energy to be transferred as kinetic energy via an impact of items with dissimilar masses. I don’t know if it’s possible to go through some rube goldbergian device to make it work, though
Suppose, however, that you converted the stored energy of the gunpowder into electrical energy and used that to run a motor that went through some pulleys to lift the boulder. Then, you could theoretically lift the boulder one foot before running out of fuel.
Shame on you all, forgetting about g. Since g is not 1 ft/s[sup]2[/sup], but more like 32, the energy in that bullet (if you could somehow use all of it, which as others have pointed out is problematical) would only be enough to lift that boulder 1/32 of a foot into the air.
Buggering English and their slug-ridden units. You are correct. Whenever I do impulse calcuations (using the in-lb-s system, because that’s usually what our data is in) I always have to remind myself to multiply (or divide, depending on which way I’m going) by 386.4. Very annoying, and I don’t have to worry about it in nms (at least, not until some injut decides to make a kilogram-force unit.)
At least something good came out of the French revolution.
Wait, … what? You were right the first time, Stranger.
We’re given a boulder weighing 2379.1 lbs. That’s a weight w=mg, not a mass; g is already included. (Or, if it’s pounds-mass, then it’s the mass that weighs that much on Earth. That means it masses about 2379.1/32=74.34 slugs.) The energy required to lift the boulder one foot is E=mgh = wh = 2379.1 ft-lbs. The mystery factor of 32 was already taken out by Crafter_Man when he calculated the bullet’s kinetic energy.
Or just do it in SI and convert; that’s less painful anyway.
:smack: So he was. I’m so used to the various forms of the metric system that I expect to see a boulder described in terms of its mass, not its weight. But of course, 2379.1 pounds is the boulder’s weight, not its mass.
Let’s pretend all of the bullet’s kinetic energy is somehow converted to kinetic energy after hitting the boulder. The bullet has 2379.1 foot pounds of kinetic energy. If the boulder weighs 2379.1 pounds, wouldn’t the bullet raise the boulder 1 foot??
Well, entropic law says that his mistake must be balanced by a net increase in the accuracy of other posters on the board. Everyone’s seemed pretty smart lately, this is probably why.
