#$&* course Phy 121 12/15 9 am Mr. Smith, I have been working on physics very much the last few weeks, and I found it more effective to skip over the labs and come back to them. I did not want to turn in my work out of order, as to seeing that my labs would be turned in later than everything else, but I believe that would be best to do so now, so that you won't be bombarded with all of my work at once. I realize that I am turning in everything late in the semester, but due to the fact that I really didn't know the nature of this course till I got into the first assignment, I believe I made it harder than what it should be. I worried that I was not understanding something or that I got something wrong, so I would take a lot of time concerned with each question and was looking up how to do each question, which in turn got me very far behind. I apologize for this, I was very disappointed in myself in this course this semester. If I am able to take the second part to this class next semester I am almost 100% certain that I will have all of my assignments turned in on time, if not earlier than the due dates. I realize I was probably a pain to have in this course, but I thank you for your kindness and patience with me. I have really worked hard in this course, though I don't have much to show for it. All that being said, I am going to start turning in the assignments that I have done, I am done with all the qa, seeds, and queries for the course, and then I will go back and do my labs, I think I only have 8 or 9 of them left. So sorry about everything, but thank you! 018. `query 18
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Given Solution: `a** We treat the vertical and horizontal quantities independently. We are given vertical displacement and initial velocity and we know that the vertical acceleration is the acceleration of gravity. So we solve the vertical motion first, which will give us a `dt with which to solve the horizontal motion. We first determine the final vertical velocity using the equation vf^2 = v0^2 + 2a'ds, then average the result with the initial vertical velocity. We divide this into the vertical displacement to find the elapsed time. We are given the initial horizontal velocity, and the fact that for an ideal projectile the only force acting on it is vertical tells us that the acceleration in the horizontal direction is zero. Knowing `dt from the analysis of the vertical motion we can now solve the horizontal motion for `ds. This comes down to multiplying the constant horizontal velocity by the time interval `dt. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Couldn't you also find the `dt by taking the `ds and dividing it by the vAve in the first step? ------------------------------------------------ Self-critique rating: ********************************************* Question: `qQuery class notes #17 Why do we expect that in an isolated collision of two objects the momentum change of each object must be equal and opposite to that of the other? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because every force has an equal and opposite force acting against it. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a Briefly, the force exerted on each object on the other is equal and opposite to the force exerted on it by the other, by Newton's Third Law. By assumption the collision is isolated (i.e., this is a closed system); the two objects interact only with one another. So the net force on each object is the force exerted on it by the other. So the impulse F_net `dt on one object is equal and opposite the impulse experienced by the other. By the impulse-momentum theorem, F_net `dt = `d ( m v). The impulse on each object is equal to its change in momentum. Since the impulses are equal and opposite, the momentum changes are equal and opposite. **COMMON ERROR AND INSTRUCTION CORRECTION: This is because the KE change is going to be equal to the PE change. Momentum has nothing directly to do with energy. Two colliding objects exert equal and opposite forces on one another, resulting in equal and opposite impulses. i.e., F1 `dt = - F2 `dt. The result is that the change in momentum are equal and opposite: `dp1 = -`dp2. So the net momentum change is `dp1 + `dp2 = `dp1 +(-`dp1) = 0. ** STUDENT QUESTION Are impulses the same as momentum changes? INSTRUCTOR RESPONSE impulse is F * `dt momentum is m v, and as long as mass is constant momentum change will be m `dv by the impulse-momentum theorem impulse is equal to change in momentum (subject, of course, to the conditions of the theorem) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `qWhat are the six quantities in terms of which we analyze the momentum involved in a collision of two objects which, before and after collision, are both moving along a common straight line? How are these quantities related? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I am not sure what this answer is. I know momentum = m * v and that there must be an initial velocity and a final velocity. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** We analyze the momentum for such a collision in terms of the masses m1 and m2, the before-collision velocities v1 and v2 and the after-collision velocities v1' and v2'. Total momentum before collision is m1 v1 + m2 v2. Total momentum after collision is m1 v1' + m2 v2'. Conservation of momentum, which follows from the impulse-momentum theorem, gives us m1 v1 + m2 v2 = m1 v1' + m2 v2'. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understand what this is saying now. ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `1prin phy and gen phy 6.47. RR cars mass 7650 kg at 95 km/hr in opposite directions collide and come to rest. How much thermal energy is produced in the collision? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I believe that thermal energy is the Fnet_noncon * `ds I don't see anything about friction or anything, so I am just going to try and solve it using conservative forces. The mass of the cars is 7650 and the rate at which they are moving is 95 km/hr which needs to be converted to m/s. 95 km/hr * 1000 m/km * 1 hr / 3600 s = 26.4 m/s. From this we can find the KE of them since they both have a mass of 7650 kg it will be the same for both. KE = 1/2 m v^2 = 1/2 * 7650 kg * (26.4 m/s)^2 = 2,665,872 Joules. Since this is true for both of them, I guess the total would be them added together. 2665872 Joules * 2 = 5,331,744 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThere is no change in PE. All the initial KE of the cars will be lost to nonconservative forces, with nearly all of this energy converted to thermal energy. The initial speed are 95 km/hr * 1000 m/km * 1 hr / 3600 s = 26.4 m/s, so each car has initial KE of .5 m v^2 = .5 * 7650 kg * (26.4 m/s)^2 = 2,650,000 Joules, so that their total KE is 2 * 2,650,000 J = 5,300,000 J. This KE is practically all converted to thermal energy. STUDENT QUESTIONS Why is the kinetic energy multiplied by two? And why is all of the kinetic energy practically converted to thermal energy? Is thermal energy simply two times the kinetic energy? Is this what happens to all kinetic energy in real life? INSTRUCTOR RESPONSE You've calculated the KE of one of the cars. There are two cars, which is why we multiply that result by 2. Some of the KE does go into producing sound, but loud as the crash might be only a small fraction of the energy goes into the sound. Practically all the rest goes into thermal energy. A lot of the metal in the cars is going to twist, buckle and otherwise deform, and warm up some in the process. They probably won't become hot to the touch, but it takes a lot more thermal energy that that involved in this collision to achieve an overall temperature change we would be likely to notice. If two cars of unequal mass and equal speeds collide they don't come to rest, so they have some KE after the collision. It the cars were perfectly elastic they would rebound with their original relative speed. A perfectly elastic collision is one in which kinetic energy is conserved. No energy would go into thermal energy and there would be no sound. This is an idean and cannot actually be achieved with railroad cars (nor with steel balls, or marbles, or pool balls, etc.). However the collisions of molecules in a gas are perfectly elastic, and analyzing the statistics of those collisions allows us to explain a lot of what we observe about gases. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!