course Phy 231 first of a two part submission ðm¯±ð¸EƒÑÎÀNv‰Èµª…ÞµÜ}òassignment #017
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07:28:22 `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 --> change in velocity = 3 m/s - 5 m/s = -2 m/s = 'dv1 therefore, change in momentum = 'dp1 = m1 * 'dv1 = 10 kg * -2 m/s = -20 kg m/s average force exerted by 2nd object on this object is therefore: Fave = 'dp / 'dt = -20 kg m/s / .03 s = -666.67 N confidence assessment: 3
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07:28:36 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 --> OK! self critique assessment: 3
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07:44:40 `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 --> we start by noting the force of the 2nd object on the first, will be equal and opposite to the force of the 1st object on the 2nd - therefore, the force would equal 666.67 N. then we find change in momentum of 2nd object: 'dp = Fnet * 'dt = 666.67 N * .03 s = 20 kg m/s (which is the exact opposite of the change in momentum of the 1st object) now we can find the change in vel: 'dv2 = 'dp / m = 20 kg m/s / 2 kg = 10 m/s now we can find the after collision vel: vf2 = 'dv2 + v0 = 10 m/s + 0 m/s = 10 m/s confidence assessment: 3
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07:44:54 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 --> OK!! self critique assessment: 3
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07:52:32 `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 --> total KE before collision: KE = 1/2 m v^2 = 1/2 * 10 kg * (5 m/s)^2 + 1/2 * 2 kg * (0 m/s)^2 = 5 kg * 25 m^2/s^2 + 0 = 125 Joules total KE after collision: KE = 1/2 mv^2 = 1/2 * 10 kg * (3 m/s)^2 + 1/2 * 2 kg * (10 m/s)^2 = 5 kg * 9 m^2/s^2 + 1 kg * 100 m^2/s^2 = 45 Joules + 100 Joules = 145 Joules therefore, more KE after collision...although, this really doesn't make sense. I would think you'd always have more KE before a collision then after, but perhaps not? confidence assessment: 3
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07:53:05 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 --> Ah, OK!! Thanks for the additional explanation... self critique assessment: 3
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07:56:55 `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 --> total momentum before collision: ptotal = m1v1 + m2v2 ptotal = 10 kg * 5 m/s + 2 kg * 0 m/s ptotal = 50 kg m/s total momentum after collision: ptotal = m1v1 + m2v2 ptotal = 10 kg * 3 m/s + 2 kg * 10 m/s ptotal = 30 kg m/s + 20 kg m/s = 50 kg m/s they are the same! confidence assessment: 2
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07:57:07 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 --> OK!! self critique assessment: 3
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08:03:15 `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 --> Because when two objects collide, they exert equal and opposite forces on each other, we are therefore able to show that the total momentum after collision is equal to the total momentum before collision confidence assessment: 2
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08:03:36 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 --> OK!! self critique assessment: 3
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