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course PHY 201
10/15 11:30 I haven't completed the sections with the symbolic questions/answers. Will we be reviewing this in class?
At this point the symbolic solutions are the quickest way to solve the problems, but few students will be able to get them. Having worked the questions through, though, you'll be in a position to understand as we go over some of them.
Coefficient of Restitution
A bead dropped on the tabletop was observed to rebound to abou 90% of its original height. A marble dropped on the floor rebounded to an estimated 35% of its original height.
`qx001. Symbolic Solutions. Most students will need to work through the details of subsequent specific problems before attempting the symbolic solution. However if you can get the symbolic solutions, you will be able to use them to answer the subsequent questions. If you prefer to work through the subsequent questions first, please do. if you get bogged down on this, move on to the subsequent questions.
If the height from which the bead is dropped is h, and if it rebounds to height c * h, then what is the percent change in the bead's speed between the instant when it first contacts the table during its fall, and the instant when it loses contact on the way back up?
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What is the ratio of the magnitude of the ratio of its corresponding momentum change to its momentum just before contact with the tabletop?
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1.05:1
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What is the ratio of the kinetic energy of the bead immediately after, to its kinetic energy immediately before striking the tabletop?
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2.7:1
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`qx002. Assume that the bead was dropped from a height of 80 cm and rebounded to 90% of this height. Analyze the motion on the uniform-acceleration interval of its fall, and then on the uniform-acceleration interval of its subsequent rise. Assume the net force during each interval to be equal to the bead's weight.
What was its velocity just before striking the tabletop?
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`dx=80 cm a=980 cm/s^2 v0=0
vf=+/- Sqrt (v0^2+2a*`dx)
vf=+/- sqrt (2(980 cm/s^2)*80 cm
vf= +/-sqrt(156800 cm^2/s^2)
vf=395.9 cm/s
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What was its velocity just after striking the tabletop?
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Rebounded to 72 cm.
vf=+/- Sqrt (v0^2+2a*`dx)
vf=+/- sqrt (2(980 cm/s^2)*72 cm)
vf= +/-sqrt(141120 cm^2/s^2)
vf=-375.7 cm/s
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What is the ratio between the speeds just before, and just after striking the tabletop?
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395.9 to 375.7 or 1.05:1
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What is the magnitude of the ratio of the ball's momentum just after, to its momentum just before striking the tabletop? The answer doesn't depend on the mass of the bead, but if you feel you need it you may assume a mass of .2 grams.
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1.05:1
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The bead rose to 90% of its original height. What percent of the magnitude of its momentum did it retain?
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90 %
momentum is m v. According to your (correct) calculations v changed by about 5%. So momentum changed by about 5%, not 10%.
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What is the ratio of the kinetic energy of the bead immediately after, to its kinetic energy immediately before striking the tabletop?
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KE=1/2mV2
Interval 1=1/2(0.2 g)(395.9 cm/s)^2
KE=37615.8 g*cm^2/s^2
Interval 2=1/2(0.2g)(375.7cm/s)^2
KE=14115 g*cm^2/s^2
Ratio 37615.8 g*cm ^2/s^2/14115 g*cm^2/s^2= 2.7:1
you have an arithmetic error; nothing wrong with the process
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`qx003. Assume that the marble was dropped from a height of 120 cm and rebounded to 35% of this height. Analyze the motion on the uniform-acceleration interval of its fall, and then on the uniform-acceleration interval of its subsequent rise. Assume the net force during each interval to be equal to the bead's weight.
What was its velocity just before striking the floor?
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`dx=120 cm a= 980 cm/s^2 V0=0
vf=+/- Sqrt (v0^2+2a*`dx)
vf=+/- sqrt (2(980 cm/s^2)*120 cm)
vf= +/-sqrt(235200 cm^2/s^2)
vf=484.9 cm/s
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What was its velocity just after striking the floor?
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`dx=42 cm
vf=+/- Sqrt (v0^2+2a*`dx)
vf=+/- sqrt (2(980 cm/s^2)*42 cm)
vf= +/-sqrt(82320 cm^2/s^2)
vf=286.9 cm/s
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What is the ratio between the speeds just before, and just after striking the floor?
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484.9/286.9= 1.7:1
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What is the magnitude of the ratio of the marble's momentum just after, to its momentum just before striking the floor? The answer doesn't depend on the mass of the marble, but if you feel you need it assume a mass of 5 grams.
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1.7:1
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The marble rose to 35% of its original height. What percent of the magnitude of its momentum did it retain?
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35%
this needs to be based on m v ratio; see my previous notes
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What is the ratio of the kinetic energy of the marble immediately after, to its kinetic energy immediately before striking the tabletop?
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KE-1/2mv^2
Interval 1=1/2(5g) (484.9cm/s)^2
KE=612315 g*cm^2/s^2 or 6.123 J
Interval 2=1/2(5) (286.9 cm/s)^2
KE=205779 g*cm^2/s^2 or 2.057 J
Ratio 6:2 or 3:1
good
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`qx004. How can you predict the percent of momentum retained from the percent of the original height to which an object rises after being dropped?
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I assumed it was relational to the change in distance after being dropped.
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If you feel you have worked out the answers to all or most of the numerical questions correctly, but haven't yet worked out the symbolic solution, you should consider returning to the first question. However don't get bogged down for a long time on that solution.
Acceleration of toy cars due to friction
The magnitude of the frictional force on a rolling toy car is the product of the coefficient of rolling friction and the weight of the car.
The coefficient of friction can be measured by placing the car on a constant incline and giving it a nudge in the direction down the incline. It will either speed up, slow down or coast with constant velocity. If the incline is varied until the car coasts with constant velocity, then the coefficient of friction is equal to the slope of the incline.
As before the symbolic question at the beginning can be attempted before or after the subsequent numerical questions.
`qx005. Answer the following symbolically:
If the length of the incline is L and its rise is h, then what is the symbolic expression for the magnitude of the frictional force on a car of mass m? What therefore is the expression for its acceleration when rolled across a smooth level surface? The Greek letter traditionally used for coefficient of friction is mu.
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If the car requires time interval `dt to come to rest while coasting distance `ds along a level surface, what is the expression for its acceleration?
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A=`dv/`dt. Find a by using `dx and `dt to calculate Vave and Vf.
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`qx006. Give your data for this part of the experiment.
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My car rolled down the incline at a constant rate when it was 1 cm high
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`qx007. Show how you used your data to find the slope of the 'constant-velocity' incline.
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Slope = rise/run The rise was 1 cm and the run was 30 cm. The slope =1 cm/30cm or 0.03
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`qx008. The weight of your car is given by the symbolic expression m g, where m is its mass. What therefore is the expression for the magnitude of the frictional force on the car?
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Ffriction=mg
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`qx009. While the car is coasted along a smooth level surface, the net force on it is equal to the frictional force. What therefore would be its acceleration?
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.03*980 cm/s^2=29.4 cm/s ^2
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`qx010. You also timed the car as it coasted to rest along the tabletop, after having been given a nudge. What were the counts and the distances observed for your trials? Give one trial per line, each line consisting of a count and distance in cm, with the two numbers separated by a comma.
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18, 66
17, 61
8, 33 (this result may have been skewed because the car ran into the meter stick)
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`qx011. Based on your data what is the acceleration of your car on a level surface? In the first line give your result in cm/s^2. Starting in the second line give a brief but detailed account of how you got your result from your data.
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12.8 cm/s^2
I used the first two sets of data to determine acceleration by first converting my counts to seconds. After re-calibrating my personal timer, each count is equal to about .18 s. Then I used the distance and time to have Vave and then doubled it to find my final velocity. Using my final velocity, I found acceleration. I did this for both the sets of data and then averaged the answers to arrive at 12.8 cm/s^2.
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`qx012. You have obtained two results for the acceleration of your car along a level surface, one based on the slope of an incline, the other on observed counts and distances. How well do they compare? Is there a significant discrepancy? If so, can you explain possible sources of the discrepancy?
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There is a large discrepancy between the two accelerations. My first guess would be that the timing mechanism used to time the distance coasted (myself) may be off.
there is of course a lot of uncertainty in that sort of timing
could also be that the car rolls less smoothly on the wooden surface
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Interaction between magnets mounted on toy cars
It should be very plausible from your experience that the two magnets exert equal and opposite forces on one another, so that at any instant the two cars are experiencing equal and opposite magnetic forces. This is not generally the case for frictional forces. However if frictional forces are considered to have negligible effect while the magnetic forces are doing their work, we can assume that the cars experience equal and opposite forces.
As before you may if you wish save the question of symbolic representations until you have worked the situation through numerically.
`qx013. When released from rest we observe that car 1, whose coefficient of rolling friction is mu_1, travels distance `ds_1 while car 2, whose coefficient of rolling friction is mu_2, travels distance `ds_2 in the opposite direction.
What is the expression for the ratio of the speeds attained by the two cars as a result of the magnetic interaction, assuming that frictional forces have little effect over the relatively short distance over which the magnetic interaction occurs? (obvious hint: first find the expressions for the two velocities)
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What therefore should be the ratio of the masses of the two cars?
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Assuming that the magnetic forces are significant for a distance that doesn't exceed `ds_mag, what proportion of the PE lost by the magnet system is still present in the KE of the cars when the magnetic forces have become insignificant?
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`qx014. When the two cars were released, what were their approximate average distances in cm? Give your answers in a single line, which should consist of two numbers separated by commas.
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36, 45.75
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`qx015. Assuming the accelerations you determined previously for the cars, and assuming that they achieved their initial speeds instantly upon release, what were their initial velocities?
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zero.
they didn't coast 40 cm or so starting with speed 0; the interval in question is the coasting interval
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`qx016. What was the ratio of the speed attained by the first car to that of the second? Which car do you therefore think had the greater mass? What do you think was the ratio of their masses?
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12.8 cm/s^2 and 11.8 cm/s^2
12.8 cm/s^2/11.8 cm/s^2=1.08:1 I think the car with the faster speed had the lesser mass. I would assume the ratio of their masses would be equivalent to 1.08:1.
it's not clear how you got the speeds; you've calculated the ratio of accelerations
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Bungee Cord and Chair
`qx017. Symbolic solution: Suppose the average force exerted by the bungee cord on the chair, as it moves between the equilibrium position and position x, has magnitude k/2 * x.
If the chair is pulled back distance x_1 from its equilibrium position and released, what is the expression for the work done by the cord on the chair as it is pulled back to its equilibrium position?
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What is the ratio of work done when the distance is x_2, to the work done when the distance is x_1.
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Assuming the net force on the coasting chair to be mu * m g, in the direction opposite motion, how far would the chair be expected to coast with each pullback?
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What is the ratio of the two coasting distances?
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`q018. When the bungee cord was pulled back twice as far and released, the chair clearly coasted more than twice as far. Assuming that the average force exerted by the bungee cord was twice as great when it was pulled back to twice the distance, how many times as much energy would the chair be expected to gain when released?
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Twice as much.
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Very good overall, but see my notes. You'll fill in any gaps when we go over these.
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