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AsOkg block of wood is suspended by 4 thin masslcss rope from thc cciling AASgbulkt is fired horizontally at $0 that it becomcs cmbcddkd in thc block. Ifthc block s...

Question

AsOkg block of wood is suspended by 4 thin masslcss rope from thc cciling AASgbulkt is fired horizontally at $0 that it becomcs cmbcddkd in thc block. Ifthc block swings lo 4 hcight of 4.76 cm ulter the bullet strikes i, find the Initial speed of the bulket when it was firedFind the coordinales of the center of ma-s olthc Syslct shown (In Skem " 24g.und n; Kg) ( Asuma jch blck / Icm Xhcm)

AsOkg block of wood is suspended by 4 thin masslcss rope from thc cciling AASgbulkt is fired horizontally at $0 that it becomcs cmbcddkd in thc block. Ifthc block swings lo 4 hcight of 4.76 cm ulter the bullet strikes i, find the Initial speed of the bulket when it was fired Find the coordinales of the center of ma-s olthc Syslct shown (In Skem " 24g.und n; Kg) ( Asuma jch blck / Icm Xhcm)



Answers

400 m 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 $190 \mathrm{~m} / \mathrm{s}$. The block slides a distance of $72.0 \mathrm{~cm}$ along the surface from its initial position. (a) What is the cocfficicnt of kinctic friction between block and surface? (b) What is the decrease in kinctic cnergy of the bullet? (c) What is the kinctic cnergy of the block at the instant after the bullet passes through it?

So here we have our block that is being struck by a horizontal the fired bullet. The final speed of the block bullet system is 2.7 m/s. That is the speed just after the bullet gets embedded and we want to know what the speed of the bullet was before it struck the block. So be non equals question mark. So for this we can use conservation of momentum. This is a inelastic collision. So we have the momentum before the momentum before the collision is just the mass of the bullet times its initial speed and then the momentum after the collision is the combined mass of the block and the bullet times the final speed. So to find this initial speed, we just divide this by mm and we get let's make that a little bit Queener. So we have the combined mass of the block and the bullet times the final velocity divided by the mass of the bullet. And so we'll get the value of 1.4 Four times 10 to the 3rd m/s, or 1440 m/s.

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.

First part here it is given must offer it fall into 10 to the power minus three kg. And he said Dispute, I'll put it 6 15 m per second. Mask of blocks. Yes. 0.95 kg. Initial velocity on the block. Find it by the City of Blog. 23 m per second. We have to find find and veracity of law. Mm hmm. Applying conservation of momentum. Yeah. Initial momentum is called vital Momentum. My separate initial velocity of public mass of food and law and the simplicity your poor develop Marcelo Barreto. Final velocity or correct? Marcel Bloch. Final velocity of block. My separate is four into 10 to the power minus four and each of publicity is given 6 50 m per second. Marcia Brady Sorry. Mood and block Initial velocity zero Marsa bullet 14 to 10 To the apartment is for find a very positive We are finding Masa Block is given 1095 kg and final velocity is 23. So final velocity of the bullet you will get 103 175 We top our second the sun The answer of a part not to be part since the energy is lost. Dooley Religion. Uh huh. Yeah, in the pharma, the formation and shit. So I love you. Tech energy. It's interesting. And he should run. That's not

We have massive bullet small MV equals to 0.060 kg. And massive block. MV equals 2. 3 Kg. And speed of a bullet is equals two. B equals to 600 m/s. And we do is equal to We do is equal to 400 m/s after striking from the clock. Okay. And we is the initial velocity of the block. So we can write from the conservation of linear momentum that MB driven. This will be equals two MB V two plus MB multiplied by week. Okay, so substituting values, we get 0.60 Multiplied by we beat, Which will be equal to 600. This is equal to 0.060. multiplied by 400 plus three. Multiplied baby. So from here we comes out to be four m per second. Okay, Since the block is sliding at a distance of as equals to 2.70 m before stopping. So kinetic friction will be equals two density equals to you. So we can write that one by two M we square minus zero. This will be equals two mu K multiplied by MG multiplayer B S. Okay, so from here, um UK comes out to be, we swear by two G. S. So substituting values, we get these equals to four m per second, so foursquare divided by two, multiplied by 9.81 Molecular BS is 2.70. So from here, Coefficient of Canada collection comes out to 0.02. Okay, so this is the answer for the question. Okay.


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