course Phy 121 015. Impulse-MomentumPhysics II
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18:47:23 `q001. Note that this assignment contains 5 questions. . Suppose that a net force of 10 Newtons acts on a 2 kg mass for 3 seconds. By how much will the velocity of the mass change during these three seconds?
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RESPONSE --> Remembering the equation F = m*a, we need to determine the acceleration first in order to find out the change of velocity, so: a = F/m a = 10 N/2kg a = 5 m/s^2 Since the force persisted for 3 seconds, we would multiply the acceleration by time (5 m/s^2 * 3 sec.) = 15 m/s confidence assessment: 2
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18:47:30 The acceleration of the object will be accel = net force / mass = 10 Newtons / (2 kg) = 5 m/s^2. In 3 seconds this implies a change of velocity `dv = 5 m/s^2 * 3 s = 15 meters/second.
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RESPONSE --> OK self critique assessment: 2
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18:48:15 `q002. By how much did the quantity m * v change during these three seconds? What is the product Fnet * `dt of the net force and the time interval during which it acted? How do these two quantities compare?
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RESPONSE --> ? confidence assessment: 1
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18:49:27 Since m remained constant at 2 kg and v changed by `dv = 15 meters/second, it follows that m * v changed by 2 kg * 15 meters/second = 30 kg meters/second. Fnet *`dt is 10 Newtons * 3 seconds = 30 Newton * seconds = 30 kg meters/second^2 * seconds = 30 kg meters/second. The two quantities m * `dv and Fnet * `dt are identical.
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RESPONSE --> I think I could have answered correctly on the second question--I just didn't have enough confidence in myself. self critique assessment: 2
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19:01:08 `q003. The quantity m * v is called the momentum of the object. The quantity Fnet * `dt is called the impulse of the net force. The Impulse-Momentum Theorem states that the change in the momentum of an object during a time interval `dt must be equal to the impulse of the average net force during that time interval. Note that it is possible for an impulse to be delivered to a changing mass, so that the change in momentum is not always simply m * `dv; however in non-calculus-based physics courses the effective changing mass will not be considered. If an average net force of 2000 N is applied to a 1200 kg vehicle for 1.5 seconds, what will be the impulse of the force?
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RESPONSE --> Assuming that the mass of the vehicle doesn't change (unless a whole bunch of stuff is thrown inside), then we would multiply the net force (2000 N) by the amount of time that elapsed (1.5 seconds) to find an impulse of 3,000 N. Change in momentum of an object is equal to the impulse of the average net force during the same time frame. It appears to me that the impulse that an object feels is like an equivalent reaction to the change in momentum on that same object. confidence assessment: 2
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19:02:10 The impulse of the force will be Fnet * `dt = 2000 Newtons * 1.5 seconds = 3000 Newton*seconds = 3000 kg meters/second. Note that the 1200 kg mass has nothing to do with the magnitude of the impulse.STUDENT COMMENT: That's a little confusing. Would it work to take the answer I got of 3234 N and add back in the weight of the person at 647 N to get 3881? INSTRUCTOR RESPONSE: Not a good idea, though it works in this case. Net force = mass * acceleration. That's where you need to start with problems of this nature.Then write an expression for the net force, which will typically include but not be limited to the force you are looking for. *&*&
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RESPONSE --> Oops, forgot to notate it as 3000 kg m/s (forgot my unit of measure in the answer!) self critique assessment: 2
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19:06:05 `q004. If an average net force of 2000 N is applied to a 1200 kg vehicle for 1.5 seconds, what will be change in the velocity of the vehicle?
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RESPONSE --> Again, to find the change in velocity of the vehicle, we would need to solve for the acceleration, so I would dig out the equation a = F/m again and solve: a = 2000 N/1200 kg a = 1.67 kg m/s^2 Since the force pushed on our vehicle for 1.5 seconds, then: 'dv =1.67 kg m/s^2 * 1.5 seconds 'dv = 2.50 m/s confidence assessment: 2
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19:07:15 The impulse of the 2000 Newton force is equal to the change in the momentum of the vehicle. The impulse is impulse = Fnet * `dt = 2000 Newtons * 1.5 seconds = 3000 Newton*seconds = 3000 kg meters/second. The change in momentum is m * `dv = 1200 kg * `dv. Thus 1200 kg * `dv = 3000 kg m/s, so `dv = 3000 kg m/s / (1200 kg) = 2.5 m/s. In symbols we have Fnet * `dt = m `dv so that `dv = Fnet * `dt / m.
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RESPONSE --> Well, I got the right answer, but went by the wrong route (should have used impulse equation--must memorize that) self critique assessment: 2
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19:11:46 `q005. Use the Impulse-Momentum Theorem to determine the average force required to change the velocity of a 1600 kg vehicle from 20 m/s to 25 m/s in 2 seconds.
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RESPONSE --> Hope I'm right on this: Fnet = m * 'dv/'dt Fnet = 1600 kg * (25 m/s-20 m/s / 2 seconds) Fnet = 1600 kg * (5 m/s/ /2 seconds) Fnet = 1600 kg * 2.50 m/s Fnet = 4000 N confidence assessment: 1
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19:12:05 The vehicle changes velocity by 5 meters/second so the change in its momentum is m * `dv = 1600 kg * 5 meters/second = 8000 kg meters/second. This change in momentum is equal to the impulse Fnet * `dt, so Fnet * 2 sec = 8000 kg meters/second and so {} Fnet = 8000 kg meters/second / (2 seconds) = 4000 kg meters/second^2 = 4000 Newtons. In symbols we have Fnet * `dt = m * `dv so that Fnet = m * `dv / `dt = 1600 kg * 5 m/s / ( 2 s) = 4000 kg m/s^2 = 4000 N.
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RESPONSE --> :) self critique assessment: 2
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V?????????p???? assignment #015 015. Impulse-Momentum Physics II 11-08-2007