#$&* course phy 201 10/20 3:20 am If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: The acceleration of the object will be accel = net force / mass = 10 Newtons / (2 kg) = 5 m/s^2. In 3 seconds this implies a change of velocity `dv = 5 m/s^2 * 3 s = 15 meters/second. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q002. By how much did the quantity m * v change during these three seconds? What is the product Fnet * `dt of the net force and the time interval during which it acted? How do these two quantities compare? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Quantity change: estimated at 30 m/s (mass= 2 kg, dv = 15 m/s, mass * dv = 30 m/s). Fnet: 10 Newtons (assuming no friction); dt: 3 seconds; quantity of net force by time interval: 30 Newton seconds; confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I messed up with the units, and forgot to put compare the quantities, but I did get the math correct to come up with the right answers. ------------------------------------------------ Self-critique rating: 3 ********************************************* 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: Fnet: 2000 Newtons; dt: 1.5 seconds; impulse: 3000 Kg m/s confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The impulse of the force will be Fnet * `dt = 2000 Newtons * 1.5 seconds = 3000 Newton*seconds = 3000 kg meters/second. Note that the 1200 kg mass has nothing to do with the magnitude of the impulse.STUDENT COMMENT: That's a little confusing. Would it work to take the answer I got of 3234 N and add back in the weight of the person at 647 N to get 3881? INSTRUCTOR RESPONSE: Not a good idea, though it works in this case. Net force = mass * acceleration. That's where you need to start with problems of this nature.Then write an expression for the net force, which will typically include but not be limited to the force you are looking for. *&*& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q004. If an average net force of 2000 N is applied to a 1200 kg vehicle for 1.5 seconds, what will be change in the velocity of the vehicle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Force: 2000 newtons; mass: 1200 kg; accelration: 1.67 m/s^2; change in velocity = roughly 2.5 m/s confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I utilized the old method of finding the acceleration, so though the answer was correct, I did not find it using the new equations given. ------------------------------------------------ Self-critique rating: ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Impulse = change in momentum; impulse = net force * change in time; change in momentum = dv * mass; change in momentum = 8000 kg m/s; 8000 kg m/s = net force * 2 seconds; net force = 4000 Newtons; confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The vehicle changes velocity by 5 meters/second so the change in its momentum is m * `dv = 1600 kg * 5 meters/second = 8000 kg meters/second. This change in momentum is equal to the impulse Fnet * `dt, so Fnet * 2 sec = 8000 kg meters/second and so Fnet = 8000 kg meters/second / (2 seconds) = 4000 kg meters/second^2 = 4000 Newtons. In symbols we have Fnet * `dt = m * `dv so that Fnet = m * `dv / `dt = 1600 kg * 5 m/s / ( 2 s) = 4000 kg m/s^2 = 4000 N. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q006. ‘Each time they thought they had ‘im, his engine would explode. He’d go by like they was standin’ still on Thunder Road.’ Good song. If you don’t know it you might want to look it up and listen to it (the name is 'Thunder Road'). His car, including him and his load, had a mass of 2500 kg. To escape, he had to speed up from 35 m/s to 45 m/s. How much impulse did he need from his engine? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Dv: 10 m/s; mass: 2500 kg; change in momentum: 25,000 kg m/s; Change in momentum = impulse; impulse = 25,000 kg m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* 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'). 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: 10 m/s; mass: 2500 kg; change in momentum: 25,000 kg m/s; Change in momentum = impulse; impulse = 25,000 kg m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!