course phy 201 vGwqzΆ{Hassignment #017
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20:57:10 `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 --> mass= 10kg v0=5m/s, vf = 3m/s mass2 = 2kg v02 = 0 dt = .03sec s F = ma vf-v0/dt = a a = -66.7m/s^2 F = 10 * 66.7 = 667 confidence assessment: 1
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21:00:55 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 --> it would be -force because it the initial velocity decreased because of the resistance by the second obj. self critique assessment: 2
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21:04:20 `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 --> same equation fnet * dt = mvf-mv0 except mv0 is the same as the mvf of the first object and mv0 = 0. Time stays the same. fnet = 2-0 / .03 = 667 confidence assessment: 1
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21:07:37 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 --> Fnet *dt = dmv 667 * .03 = 20kg m/s of the 2kg obj p=mv 20 = 2kg * dv dv = 10m/s I know there would be an equal and opps. force during the collision and should have gone on to use the same equation used before. self critique assessment: 2
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21:10:44 `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 --> Ke = .5mv^2 it would be less because there are two systems which absorb the KE confidence assessment: 2
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21:14:24 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 --> KE = 125 first mass alone The kinetic energy will be greater because there is two masses involved and each will have a seperate KE. I did think that KE would be lost due to friction and since KE would be transfer. between the two objects but once looking at the problem it is plain to see how we end up with more KE self critique assessment: 2
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21:17:05 `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 --> p1 = mvf - mv0 p = 50 - 30 = 20 m = 10kg v0 5m/s vf=3m/s p2=mvf - mv0 p = -20 m= 2kg v0 = 10m/s vf=0 confidence assessment: 2
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21:18:49 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 --> total p was the same before and after the collision it was conserved self critique assessment: 2
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21:23:07 `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 --> by Newtons third law because forces exerted by two interacting objects on one another must be equal and opposite. Because momentum is based on Force application then it follows this law for any closed system. confidence assessment: 2
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21:23:19 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 --> self critique assessment:
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21:23:21 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 --> self critique assessment:
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