course Phy 201 ??????}?[?x?assignment #017
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18:45:39 `q001. Note that this assignment contains 5 questions. . A mass of 10 kg moving at 5 meters/second collides with a mass of 2 kg which is initially stationary. The collision lasts .03 seconds, during which time the velocity of the 10 kg object decreases to 3 meters/second. Using the Impulse-Momentum Theorem determine the average force exerted by the second object on the first.
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RESPONSE --> Fave*`dt = m`dv Then we find that Fave = m `dv/`dt = 10kg * (-2m/s)/(.03 seconds) = -667 N confidence assessment: 2
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18:46:16 By the Impulse-Momentum Theorem for a constant mass, Fave * `dt = m `dv so that Fave = m `dv / `dt = 10 kg * (-2 meters/second)/(.03 seconds) = -667 N. Note that this is the force exerted on the 10 kg object, and that the force is negative indicating that it is in the direction opposite that of the (positive) initial velocity of this object. Note also that the only thing exerting a force on this object in the direction of motion is the other object.
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RESPONSE --> I was afraid my answer was wrong because it was negative, but I understand now that that only means the force is moving in the other direction (as I learned about other situations, too). self critique assessment: 2
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18:56:12 `q002. For the situation of the preceding problem, determine the average force exerted on the second object by the first and using the Impulse-Momentum Theorem determine the after-collision velocity of the 2 kg mass.
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RESPONSE --> Since before we knew that -667N force was pushing against the object, 667N must be pushing in the direction of the object too, since the forces must be equal and opposite. So, we can use the equation F*'dt=667N*.03 seconds=20 k m/s to an object of mass 2kg confidence assessment: 2
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18:57:29 Since the -667 N force exerted on the first object by the second implies and equal and opposite force of 667 Newtons exerted by the first object on the second. This force will result in a momentum change equal to the impulse F `dt = 667 N * .03 sec = 20 kg m/s delivered to the 2 kg object. A momentum change of 20 kg m/s on a 2 kg object implies a change in velocity of 20 kg m / s / ( 2 kg) = 10 m/s. Since the second object had initial velocity 0, its after-collision velocity must be 10 meters/second.
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RESPONSE --> I see that in order to find the actual momentum change I should have divided the Impulse force by the mass, yielding a change of momentum of 10m/s (the kg's cancel). I also see that one can't just assume that this number will be the actual momentum change--one must compare it to the initial velocity and note the change between the two. self critique assessment: 2
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21:54:03 `q003. For the situation of the preceding problem, is the total kinetic energy after collision less than or equal to the total kinetic energy before collision?
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RESPONSE --> To find this, we can use the equation KE=.5mv^2, the first time filling in the intiial velocity and the second filling in the final velocity and then see which had the highest KE. .5(10kg)(5m/s)^2=125 --KE0 10kg object .5(2kg)(0^2)=0J --this is initial KE of 2kg object 125J is the total energy before collision. .5(10kg)(3m/s)^2=45 J --KE0 10kg object .5(2kg)(10^2)=100J --this is final KE of 2kg object So, the total for ""before"" is 125J and the total after is 145J, so more KE is present afterwards. confidence assessment: 2
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21:56:12 The kinetic energy of the 10 kg object moving at 5 meters/second is .5 m v^2 = .5 * 10 kg * (5 m/s)^2 = 125 kg m^2 s^2 = 125 Joules. Since the 2 kg object was initially stationary, the total kinetic energy before collision is 125 Joules. The kinetic energy of the 2 kg object after collision is .5 m v^2 = .5 * 2 kg * (10 m/s)^2 = 100 Joules, and the kinetic energy of the second object after collision is .5 m v^2 = .5 * 10 kg * (3 m/s)^2 = 45 Joules. Thus the total kinetic energy after collision is 145 Joules. Note that the total kinetic energy after the collision is greater than the total kinetic energy before the collision, which violates the conservation of energy unless some source of energy other than the kinetic energy (such as a small explosion between the objects, which would convert some chemical potential energy to kinetic, or perhaps a coiled spring that is released upon collision, which would convert elastic PE to KE) is involved.
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RESPONSE --> I understand the problem but also that some other force must have somehow impacted the KE. self critique assessment: 3
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22:00:20 `q004. For the situation of the preceding problem, how does the total momentum after collision compare to the total momentum before collision?
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RESPONSE --> 2kg object: 2kg*0m/s=0 kg m/s 2kg*10 m/s=20 kg m/s 10Kg object: BEFORE: 10kg*5m/s=50 kg m/s AFTER: 10kg*3m/s=30 kg m/s So, for the first object, the momentum increases; for the second, it decreases. confidence assessment: 2
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22:01:19 The momentum of the 10 kg object before collision is 10 kg * 5 meters/second = 50 kg meters/second. This is the total momentum before collision. The momentum of the first object after collision is 10 kg * 3 meters/second = 30 kg meters/second, and the momentum of the second object after collision is 2 kg * 10 meters/second = 20 kg meters/second. The total momentum after collision is therefore 30 kg meters/second + 20 kg meters/second = 50 kg meters/second. The total momentum after collision is therefore equal to the total momentum before collision.
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RESPONSE --> Oh, I see that overall the momentum does equal out from beginning to end. self critique assessment: 2
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22:16:42 `q005. How does the Impulse-Momentum Theorem ensure that the total momentum after collision must be equal to the total momentum before collision?
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RESPONSE --> I guess because the forces acting on the objects should be equal and opposite too. confidence assessment: 1
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22:17:01 Since the force is exerted by the 2 objects on one another are equal and opposite, and since they act simultaneously, we have equal and opposite forces acting for equal time intervals. These forces therefore exert equal and opposite impulses on the two objects, resulting in equal and opposite changes in momentum. Since the changes in momentum are equal and opposite, total momentum change is zero. So the momentum after collision is equal to the momentum before collision.
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RESPONSE --> I understand. self critique assessment: 3
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