#$&* course Phy 231 2/28 9:30pm 015. Impulse-Momentum
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Given Solution: The acceleration of the object will be accel = net force / mass = 10 Newtons / (2 kg) = 5 m/s^2. In 3 seconds this implies a change of velocity `dv = 5 m/s^2 * 3 s = 15 meters/second. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. By how much did the quantity m * v change during these three seconds? What is the product Fnet * `dt of the net force and the time interval during which it acted? How do these two quantities compare? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: vf=v0+15m/s m*vf-m*v0=2kg*(v0+15m/s)-2kg(v0) =(2kg*v0)+30kg*m/s-(2kg*v0) =the quantity m*v changes by 30kg*m/s. Fnet*`dt=10N*3s=10kg*m/s^2*3s = 30kg*m/s These two quantities are the same. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: impulse=Fnet*`dt =2000N*1.5s =2000kg*m/s^2 * 1.5s =3000kg*m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The impulse of the force will be Fnet * `dt = 2000 Newtons * 1.5 seconds = 3000 Newton*seconds = 3000 kg meters/second. Note that the 1200 kg mass has nothing to do with the magnitude of the impulse.STUDENT COMMENT: That's a little confusing. Would it work to take the answer I got of 3234 N and add back in the weight of the person at 647 N to get 3881? INSTRUCTOR RESPONSE: Not a good idea, though it works in this case. Net force = mass * acceleration. That's where you need to start with problems of this nature.Then write an expression for the net force, which will typically include but not be limited to the force you are looking for. *&*& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q004. If an average net force of 2000 N is applied to a 1200 kg vehicle for 1.5 seconds, what will be change in the velocity of the vehicle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Impulse=Fnet*`dt =2000N*1.5s =3000kg*m/s So momentum changes by 3000kg*m/s. Momentum=m*v 1200kg*vf - 1200kg*v0 = 3000kg*m/s. Divide everything by 1200kg vf-v0=2.5m/s The vehicle will decrease in velocity by 2.5m/s. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): For some reason I have a very hard time simply mutliplying by the quantity `dv. My instinct is to break it down into vf and v0, and it leads me to the right place but adds one more step where it would be easy for me to make an error. I have the same problem when looking at changes in KE. I want to calculate initial and final of EVERY quantity, but sometimes we're not given every quantity, but it's possible to do the problem anyway just knowing what's constant and what's changing. This is difficult for me to remember. ------------------------------------------------ Self-critique rating:
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK, that was good practice for just dealing with ""change"" rather than final and initial. ------------------------------------------------ Self-critique rating: If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties. ********************************************* Question: `q006. ‘Each time they thought they had ‘im, his engine would explode. He’d go by like they was standin’ still on Thunder Road.’ Good song. If you don’t know it you might want to look it up and listen to it (the name is 'Thunder Road'; the video is from a movie nearly 60 years old and is very dated so if it comes up, listen but watch it. Your imagination will create a much better picture.). His car, including him and his load, had a mass of 2500 kg. To escape, he had to speed up from 35 m/s to 45 m/s. How much impulse did he need from his engine? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `dv=10m/s Change in momentum=10m/s * 2500kg= 25,000kg*m/s Impulse = 25,000kg*m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): So it doesn't matter how much time he has to speed up?