Technology The physics in firearms.

This is the research I had on the actual physics of the firearm. From a thermodynamic point of view, a firearm is a special type of piston engine, or in general heat engine where the bullet has a function of a piston. The energy conversion efficiency of a firearm strongly depends on its construction, especially on its caliber and barrel length. The equation is simple.

MV+mv=0. The forces of m (mass) The force of the bullet must be equivalent or the same of the shooter.

In Hallywood, the depictions of the firearm are simply just inaccurate. A .44 Remington Magnum with a 240-grain (0.016 kg) jacketed bullet is fired at 1,180 feet per second (360 m/s) at a 170-pound (77 kg) target. Here is my basic question: What velocity is imparted to the target (assume the bullet remains embedded in the target and thus practically loses all its velocity)? To find this out, we are going to be using another of our equations. Let mb and vb stand for the mass and velocity of the bullet, the latter just before hitting the target, and let mt and vt stand for the mass and velocity of the target after being hit. Our equation is going to be looking like this: mbvb = mtvt. Solving for the target's velocity gives us the beutiful equation we see as vt = mbvb / mt = 0.016 kg × 360 m/s / 77 kg = 0.07 m/s = 0.16 mph. Easy as pie. Hope this had helped!



P.S In you want me to tell you about the physics of something else, PM me or suggest in the comments below!

 

Yes. My teacher just questioned me hard.

 

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They go hand and hand.

Thanks!

Should I make another one of these?