course Phy 121 þú¯¨µ½ÕØù¼¬²æöœÀï«o±žRÍDå¦ág¡”{assignment #017
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22:10:21 `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 --> The farthest I got was solving for momentum of the first object which was 10 kg*-2m/s = -20m/s. I got this using the collision time which I'm not sure you're supposed to do. I don't really know what to do as far as the 2nd object on the first. confidence assessment: 1
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22:13:52 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 take it that net force and average force are the same. I get how to solve the equation here but just wasn't sure about which mass to use, from the first object or the second. I see that the only force would be from the other object in this particular situaiton. self critique assessment: 2
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22:16:45 `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 --> I set up the equation as Fave = m`dv/`dt. Since it was for the the second object, I used Fave = 2kg(-2)/.03 sec to get -133.33. confidence assessment: 2
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22:21:20 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 now that an equal force must be applied to the second object from teh first as the second exerted on the first for the first problem. Also, if I knew that it was a momentum we were solving for I think I could have gotten this one right. self critique assessment: 2
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22:22:26 `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 --> I think that the total KE after the collision would be less that what it was before the collision because I think that the collision itself would make the system lose some KE. confidence assessment: 3
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22:26:24 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 didn't think there was an actual math explanation for this, but now I see how easy it would have been to solve. IF the total KE is now higher, do we expect that the 2 objects traveled farther backward than when they started? self critique assessment: 2
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22:28:50 `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 --> Before the 2kg object had no momentum and the 10kg object had 50 kg/sec momentum. After the collision the 2kg had 20 kg and the 10 kg had 30. These two are equal momentums before and after. confidence assessment: 3
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22:30:18 `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 was actually just wondering why the two were equal before and after. I am thinking that since impulse and momentum equal eachother this has something to do with one be proporiotnal to the other. confidence assessment: 2
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22:31:12 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 --> It makes sense to think about it from the point that the forces have to be equal and opposite so the impulses do and then the momentum does. self critique assessment: 2
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