course phy 121 ý¼ùïÕšx’ÿ½|û•™èŽ•ƒóÓìâƒassignment #017
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12:58:24 `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 --> m1 = 17kg v01 = 5m/s vf1 = 3m/s m2 = 2kg v02 = 0 dt = .03s Fnet = ? Fnet * dt = mvf - mv0 Fnet * .03s = (17kg)(3m/s) - (17kg)(5m/s) Fnet * .03s = 51-85 Fnet = -34 / .03 = -1133 So for exerted on the first object is 1133 confidence assessment: 1
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13:05:20 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 --> The main difference is that I read that the object was 17kg, not 10 kg. I think I understand why it is negative instead of positive. The force exerted by the smaller stationary object is in the opposite direction of the object that is larger and moving. So it is automatically going to be negative, just becasue it is in the opposite direction of the motion. self critique assessment: 2
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13:15:03 `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 believe that the Fnet of the 1st object has to be equal and opposite to the Fnet of the 2nd object, but to be sure... a = dv/dt a = 2m/s / .03s a = 66.66m/s^2 Fnet = m*a Fnet = 17kg * 66.66 m/s^2 Fnet = 1133N Yes, it is equal and opposite. The force exerted on the 2nd object by the first is 1133N. Fnet * dt = m* dv 1133N * .03s = 2kg * dv 34/2 = dv = 17m/s Of course you came up with a different answer because you were using a mass of 10kg for the first object... If I did my calculations correctly using 10kg for the 1st object the dv comes out to 10m/s Hhmmmm, the change in velocity for the 2nd object both times came out to the mass of the first object. Very curious confidence assessment: 2
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13:15:22 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 --> self critique assessment: 3
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13:25:15 `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 they should be equal because the energy is just passed from one object to another, but should be otherwise conserved. There is nothing adding additional energy or subtracting energy along the way (unless you count friction or air resistence) KE = 1/2m*v^2 For the first object I think I use the change in velocity which is 2 m/s KE = 1/2 (10kg) (2m/s)^2 KE = 5 * 4 = 20N For the second object I use the change in velocity which is 10 m/s KE = 1/2 (2kg) (10m/s)^2 KE = 1 * 100 = 100N That didn't meet my expectation. If I use the intial velocity for the 1st object (5m/s) I would come up with KE of 125N. Maybe that is right because the first object didn't stop, it just slowed down, so it kept some of the KE for itself. Total KE should stay the same, but the KE of the two objects will each have some KE, so the KE of the 2nd object will be lower than the KE for the 1st object prior to the collision confidence assessment: 1
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13:27:33 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 --> OK, yes the total kinetic energy is higher, but I was right that it is different than I expected. It violates the conservation of energy law. I didn't calculate the KE of the 1st object after collision, but I was on the right track. self critique assessment: 2
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13:30:18 `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 --> The total momentum is conserved. Prior to collision: Object 1 momentum = 10kg * 5m/s = 50 Object 2 momentum = 2kg * 0 = 0 So total mometum = 50 After collision Object 1 momentum = 10 kg * 3m/s = 30 Object 2 momentum = 2kg * 10m/s = 20 Total momentum = 50 confidence assessment: 2
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13:30:43 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 --> self critique assessment: 2
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13:34:58 `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 think it has to do something with the force staying the same and the masses staying the same, but I'm not sure what. confidence assessment: 0
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13:36:15 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 --> equal and opposite force = equal and opposite momentums. Since force is equal and opposite then the momentum will be unchanged also. self critique assessment: 1
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