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A $4.00-\mathrm{g}$ bullet, traveling horizontally with a velocity of magnitude 400 $\mathrm{m} / \mathrm{s}$ , is fired into a wooden block with mass 0.800 $\mathrm{kg}$ , initially at rest on a level surface. The bullet passes through the block and emerges with its speed reduced to 120 $\mathrm{m} / \mathrm{s} .$ The block shides a distance of 45.0 $\mathrm{cm}$ along the surface from its initial position. (a) What is the coefficient of kinetic friction between block and surface? (b) What is the decrease in kinetic energy of the bullet? (c) What is the kinetic energy of the block at the instant after

the bullet passes through it?

a. 0.222

b. $-291 \mathrm{J}$

c. $0.784 \mathrm{J}$

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University of Michigan - Ann Arbor

University of Washington

Simon Fraser University

Hope College

{'transcript': "problem. 8.85. We have a bullet hitting and then passing through a block. The bullet slows down after passing through a block on the block slide 72 centimeters before coming to arrest again. So the first question is, what is the kinetic? The coefficient of kinetic friction between the block and surface on which it's sliding. Second is what is the change in kinetic energy of the bullet. And the third is what is the initial kinetic energy of the block right after the bullet passes through it and it starts like so. First, we need to model the collision of the bullet and the block and use conservation of momentum. And we assume that the collision entails the bullet passing through. Yes, um, so the speed of walk is going to be equal to the mass of the bullet times a bullet initial minus B. Well, it's final, the the negative of the change of momentum of the bullet. Fight it by massive block. And so putting the numbers into that, we get 1.0 five meters per second. Now we have to consider the work done by friction opposing the motion of the block and that will drain down its kinetic energy. So this is, ah, conservation of energy question. Now, so have I, Master of the bar. Uh, sorry. Times the speed it initially has squared. Many will the work done by friction, which is going to be the coefficient of friction times the normal force on the block, which is just it's wait and then the distance through it to travels with 72 centimeters. Master of the block has nothing to do. It's coefficient of friction. Answer coefficient of friction with the equal speed squared provided by two g Yes, um, and putting all those things in. This works out 1/8 or 0.12 five. So now the change and kinetic energy of the bullet is pretty easy. We know what his masses and we know what its initial and final speeds are. So delta cable it. I have times the mass of the bullet doesn't lose any mass or gain any mess. And 1/2 doesn't change. It would be in real trouble if that happens. So it's final speed squared, but in its initial speed squared and plug numbers in and we find that the bullet. The energy of the bullet has decreased by 200 ends, decreased 248. Jules see the kinetic energy of the block before it starts losing energy from friction doing work on it. It's 1/2 times its mass times square of it's speed, and it's for except for for what? Children? Not a lot. So you can see the bullet lost 248 jewels of energy. But Marco, I gained 441. So this is not an elastic collision. It's not completely inelastic because the bullet didn't lodge in the block. So that's what these sort of partially in the elastic types of collisions."}