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    two balls of equal mass are thrown horizontally with the same initial velocity. they hit identical stationary boxes resting on a frictionless horizontal surface. the ball hitting box 1 bounces back, while the ball hitting box 2 gets stuck.

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    get two balls of equal mass are thrown horizontally with the same initial velocity. they hit identical stationary boxes resting on a frictionless horizontal surface. the ball hitting box 1 bounces back, while the ball hitting box 2 gets stuck. from EN Bilgi.

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    SOLVED:The next two questions pertain to the situation described below: Box 1: Ball Bounces Box 2: Ball Sticks Two balls of equal mass are thrown horizontally with the same initial velocity: They hit identical stationary boxes resting on a frictionless horizontal surface_ The ball hitting box bounces back, while the ball hitting box 2 gets stuck to the box. You can assume that there is no motion in the vertical direction before or after the collisions_ 11) Which box has the greater speed after the collision? pts: ) Box 2 Both boxes have the same speed Box 12) For Box 1, the mass of the box is 1.8 kg and the mass of the ball is 0.45 kg; Suppose the ball has an initial horizontal velocity of 14 m/s and bounces off the box with a horizontal velocity of

    let's take a look at a double collision example and ask some things about conservation. So what we have is a bullet that goes through one block fired horizontally, goes through the first block. M one which starts to move after the impact and then the bullet moves further and impacts a second block, embedding itself in there. So there are two collisions and let's ask the question, can we conserve momentum. Total momentum conserved, is it? And I would say yes, simply because we have to think through but makes total momentum conserved. Are there external forces acting in any of those collisions collisions? Um And what we mean by external is external to the system. So we here we are defining the system To be the bullet plus the two blocks. Um And the only possibilities of course are small external forces such as air drag or friction perhaps in the surface, even though we are going with a fairly frictionless surface. So those external forces are either not present. You know? Remember we are perpendicular to gravity, that would the one that could come into play. Um So gravity is perpendicular, it does not play a role. And other forces are either not present or so small. You can neglect especially with a bullet doing things. You can't really worry too much about small air drag. So we're going to go with Yes, simply because the external forces are negligible. What about energy. So let's think about energy and certainly not mechanical energy in this case, kinetic energy will not be conserved. And we can see that because as the bullet went through the first block and we'll worry about the second block too. Um but there was a huge force of um friction acting on that bullet. It's an internal force, so it doesn't impact the momentum, but a huge amount of friction um And therefore energy gets deposited in the 2nd block. If we think about uh the first block, I mean, and if we think about that bullet embedding in the second block similarly, there was a lot of friction making that bullet stick and therefore energy was deposited most likely converted from kinetic into internal thermal energy. So, uh a thermal energy due to friction, a bullet with both blocks and there could be other forms of energy that the kinetic was converted to for example, sound. Uh even if the surface was frictionless, probably the bullet did some things uh inside of the blocks making some sound. So let's go ahead and take a look. But in other words, we would expect the kind of kinetic energy final to be less than the kinetic energy initially of the free bullets. But let's take a look at a case, We will let the bullet be going an initial speed of 355 meters per second And it will have a mass of 4g. Well convert kilograms. Not necessary, but I just like to do that Uh massive block. one, 1.1, kg And be one final Is one half actually .550 meters per second. Second block has a mass of one 53 a kilogram. And the final velocity of that last situation is what we're after. Okay, so momentum conservation p initial the extraction equals p final results in the relationship the mass of the bullet. Thanks for you not has to equal as one the one final Plus the combination of mass two plus the bullet With the two final And it is V two final that we are looking for. Okay, so putting in some numbers and I will leave off units for the time being. But we could have just kept the masses and grams because the units grams will cancel but we'll go ahead and get things in kilograms. I like that better. And we'll put in the units at the very end. So we get that the final velocity of the block and the bullet is .513 meters per second. Now let's consider energy. We want to compare the final kinetic energy to the initial. Okay. Mhm Yeah, I would call this a highly inelastic collision. But the initial kinetic energy, His 1/2 the mass of the bullet times its initial speed squared. And here we do need our mass in kilograms if we are going to wind up with a a good unit of jewels. But again, we're probably more interested in the ratio of initial to final. So again, the kilograms are not necessary. The final kinetic energy is a some of the two blocks. Yeah. And each block has its own mass and its own speed. Again, I will calculate these in terms of jewels. And let's see we get a very tiny amount of jewels. Yeah if we take the ratio we see that the final kinetic energy is way way less than the initial. It's less than one jewel. Finally and several 100 jewels to begin with. So yeah, a lot of energy is lost in the interaction. So uh only a small fraction of a percent of the original kinetic energy is left.

    ZE

    Zachary E.

    Physics 101 Mechanics

    5 months, 3 weeks ago

    The next two questions pertain to the situation described below: Box 1: Ball Bounces Box 2: Ball Sticks Two balls of equal mass are thrown horizontally with the same initial velocity: They hit identical stationary boxes resting on a frictionless horizontal surface_ The ball hitting box bounces back, while the ball hitting box 2 gets stuck to the box. You can assume that there is no motion in the vertical direction before or after the collisions_ 11) Which box has the greater speed after the collision? pts: ) Box 2 Both boxes have the same speed Box 12) For Box 1, the mass of the box is 1.8 kg and the mass of the ball is 0.45 kg; Suppose the ball has an initial horizontal velocity of 14 m/s and bounces off the box with a horizontal velocity of -9.1 mls. If the ball is in contact with the box for 0.23 $, what is the average horizontal force acting on Box 1 during the bounce? pts: ) 17.8 N 45.2 N 180.8 N

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    Linda W.
    University of North Carolina at Chapel Hill

    Answer

    Each of these problems consists of Concept Questions followed by a related quantitative Problem. The Concept Questions involve little or no mathematics. They focus on the concepts with which the problems deal. Recognizing the concepts is the essential initial step in any problem-solving technique. Concept Questions Part aa of the drawing shows a bullet approaching two blocks resting on a horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the first block (an inelastic collision) and embeds itself in the second one, as indicated in part bb. Note that both blocks are moving after the collision with the bullet. (a) Can the conservation of linear momentum be applied to this three-object system, even though the second collision occurs a bit later than the first one? Justify your answer. Neglect any mass removed from the first block by the bullet. (b) Is the total kinetic energy of this three-body system conserved? If not, would the total kinetic energy after the collisions be greater than or smaller than that before the collisions? Justify your answer. Problem A 4.00 -g bullet is moving horizontally with a velocity of +355m/s+355m/s, where the + sign indicates that it is moving to the right. The mass of the first block is 1150g1150g, and its velocity is +0.550m/s+0.550m/s after the bullet passes through it. The mass of the second block is 1530g1530g. (a) What is the velocity of the second block after the bullet imbeds itself? (b) Find the ratio of the total kinetic energy after the collision to that before the collision. Be sure your answer is consistent with that in part (b) of the Concept Questions.

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    let's take a look at a double collision example and ask some things about conservation. So what we have is a bullet that goes through one block fired horizontally, goes through the first block. M one which starts to move after the impact and then the bullet moves further and impacts a second block, embedding itself in there. So there are two collisions and let's ask the question, can we conserve momentum. Total momentum conserved, is it? And I would say yes, simply because we have to think through but makes total momentum conserved. Are there external forces acting in any of those collisions collisions? Um And what we mean by external is external to the system. So we here we are defining the system To be the bullet plus the two blocks. Um And the only possibilities of course are small external forces such as air drag or friction perhaps in the surface, even though we are going with a fairly frictionless surface. So those external forces are either not present. You know? Remember we are perpendicular to gravity, that would the one that could come into play. Um So gravity is perpendicular, it does not play a role. And other forces are either not present or so small. You can neglect especially with a bullet doing things. You can't really worry too much about small air drag. So we're going to go with Yes, simply because the external forces are negligible. What about energy. So let's think about energy and certainly not mechanical energy in this case, kinetic energy will not be conserved. And we can see that because as the bullet went through the first block and we'll worry about the second block too. Um but there was a huge force of um friction acting on that bullet. It's an internal force, so it doesn't impact the momentum, but a huge amount of friction um And therefore energy gets deposited in the 2nd block. If we think about uh the first block, I mean, and if we think about that bullet embedding in the second block similarly, there was a lot of friction making that bullet stick and therefore energy was deposited most likely converted from kinetic into internal thermal energy. So, uh a thermal energy due to friction, a bullet with both blocks and there could be other forms of energy that the kinetic was converted to for example, sound. Uh even if the surface was frictionless, probably the bullet did some things uh inside of the blocks making some sound. So let's go ahead and take a look. But in other words, we would expect the kind of kinetic energy final to be less than the kinetic energy initially of the free bullets. But let's take a look at a case, We will let the bullet be going an initial speed of 355 meters per second And it will have a mass of 4g. Well convert kilograms. Not necessary, but I just like to do that Uh massive block. one, 1.1, kg And be one final Is one half actually .550 meters per second. Second block has a mass of one 53 a kilogram. And the final velocity of that last situation is what we're after. Okay, so momentum conservation p initial the extraction equals p final results in the relationship the mass of the bullet. Thanks for you not has to equal as one the one final Plus the combination of mass two plus the bullet With the two final And it is V two final that we are looking for. Okay, so putting in some numbers and I will leave off units for the time being. But we could have just kept the masses and grams because the units grams will cancel but we'll go ahead and get things in kilograms. I like that better. And we'll put in the units at the very end. So we get that the final velocity of the block and the bullet is .513 meters per second. Now let's consider energy. We want to compare the final kinetic energy to the initial. Okay. Mhm Yeah, I would call this a highly inelastic collision. But the initial kinetic energy, His 1/2 the mass of the bullet times its initial speed squared. And here we do need our mass in kilograms if we are going to wind up with a a good unit of jewels. But again, we're probably more interested in the ratio of initial to final. So again, the kilograms are not necessary. The final kinetic energy is a some of the two blocks. Yeah. And each block has its own mass and its own speed. Again, I will calculate these in terms of jewels. And let's see we get a very tiny amount of jewels. Yeah if we take the ratio we see that the final kinetic energy is way way less than the initial. It's less than one jewel. Finally and several 100 jewels to begin with. So yeah, a lot of energy is lost in the interaction. So uh only a small fraction of a percent of the original kinetic energy is left.

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