Answer Happy

III. On Center Bullet/Block Calculations About the Bullets: The bullets used in the videos you are analyzing were manufactured by CCI. As you can see in Figure 2, the hollow point copper plated lead bullet compresses into a ball shape as it enters the wood. Bullet Mass (w/o casing): m=238=23x10 kg Fig. 5: A hollow point bullet and its Muzzle Speed of the Bullet: casing containing air and gunpowder 15=384 m/s. Note that this is probably an average or a maximum. There will obviously be differences from bullet to ballet because the amount of gunpowder in the casing that ignites to push the bullet upward may vary. After Bullet Entry: The mass of a rising block, M, is the sum of the initial mass of the block, the mass of the suspension nails (shown in Fig. 3), and the mass of embedded bullet a: shown below: Mmblock + Muile + mulle = 222g + 3.8g + 2.3g = 228 g=0.228 kg In the introduction to this assignment, we suggested that you use momentum and energ calculations to determine whether the energy the bullet loses while changing shape an breaking chemical bonds is large enough to explain why height differences between spinning and non-spinning block are not significant. Let's start by analyzing an on-cente shot like that shown in Fig. 6. Plexiglas Block Suspension Point Muzzle Fig. 6: Photo of the bulletblock setup showing a suspended block that has just been hit on-center by a bullet and rison te haine fashout 6.83 m. NOTE: The height was calculated using the known length of 1.25 m for the extended support table as a scale object. Scale Object 1.25 m a) Linear Momentum of the Bullet-Block System Just After Penetration: Use the fact that linear momentum is always conserved in a collision to calculate the magnitude of the bullet block momentum, Panjtest after the bullet stops plowing into the block. Assume that the bullet of mass mo = 2.3g enters the block at the manufacturer's reported speed of v) = 384 m/s.
b) The Kinetic Energy of a Typical Bullet Before it Enters the Block: Use the manufacturer's data given in the last section for a bullet's mass and muzzle speed in m/s to show that a typical bullet's kinetic energy just before it enters the block is about KE-170 J. c) Potential Energy of the Bullet/Block System: Recall that the mass of our block with its embedded bullet and support nails has a mass of M=0.228kg and rises to a height of h 0.83m (as depicted in Fig. 6). Show that at its highest point the block & bullet has a potential energy, PE, of 1.86). d) Initial Kinetic Energy of the Bullet/Block System: If air resistance is negligible, the potential energy of the block and bullet at the top of their 0.83m rise should equal the kinetic energy of the block and bullet just as the bullet stops plowing into the wood. What is the kinetic energy of the block and bullet system, KEs, just as the bullet stops plowing into the wood? e) Consider a hypothetical situation in which none of the bullet's energy was wasted breaking chemical bonds in the wood. Approximately what would the maximum height, Hmax that the block and bullet would rise in this case? NOTE: Don't be surprised if the value you calculate for Hmar is much higher than the height that a typical block rises! 1) What percentage of bullet energy is transformed to gravitational Potential Energy: You should be able to show that the actual rise of the block is only a small percent of what it would be if the bullet/block system retained all the kinetic energy that the bullet had before penetration into the block. Show your calculations.