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course PHY 201
10/16 5:40pm
Coefficient of RestitutionA bead dropped on the tabletop was observed to rebound to about 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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It would be whatever ratio c*h is to h.
This will turn out not to be the case.
Your own results for the next question demonstrate this. In that example c was .90. You calculated speeds of 400 cm/s and 379 cm/s, which appear correct. However .90 * 400 cm/s = 360 cm/s, not 379 cm/s. So 90% of the height doesn't correspond to 90% of the speed.
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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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* 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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`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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400cm/s
`ds=80cm
A=9.8cm/s^2
V0=0
Using the 3rd or 4th equation, we can solve for `dt.
`ds=v0`dt+1/2a`dt^2
80cm=0*`dt+1/2*(9.8m/s^2)*`dt^2
80cm=1/2*(980cm/s^2)*`dt^2
`dt^2=80cm/490cm/s^2
`dt^2=.163/s^2
`dt=sqrt of .163/s^2
`dt=.4s
Now we have:
`dt=.4s
`ds=80cm
A=980cm/s^2
V0=0
So, to get the velocity just before striking the tabletop, we get
VAve=80cm/.4s=200cm/s
Since vAve is 200cm/s, then vF=400cm/s
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* What was its velocity just after striking the tabletop?
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379cm/s
v0=0
`ds=72cm (90% of 80cm)
a=980cm/s^2
Using the 3rd or 4th equation, we can solve for `dt.
`ds=v0`dt+1/2a`dt^2
72cm=0*`dt+1/2*(9.8m/s^2)*`dt^2
72cm=1/2*(980cm/s^2)*`dt^2
`dt^2=72cm/490cm/s^2
`dt^2=.147/s^2
`dt=sqrt of .147cm/s^2
`dt=.38s
Now we have
`dt=.38s
v0=0
`ds=72cm (90% of 80cm)
a=980cm/s^2
Vave=72cm/.38s=189.5cm/s
Since Vave=189.5 and v0=0, then vF=379cm/s
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* What is the ratio between the speeds just before, and just after striking the tabletop?
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1.05 to 1
Just before speed-400cm/s
Just after speed-379cm/s
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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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Plus or minus 1.11 to 1 depending on the positive direction
On this, I think that because the ball rose to 90% of its original height, then regardless the distance, the ratio would be the ratio of 100% compared to 90% which is 1.11 to 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%
It rose to that percentage of its original height
72cm was 90% it’s original height which was 80cm
The bead retained about 95% of its original speed, as your calculations verify.
Momentum is mass * velocity, so it retained 95% of the magnitude of its original momentum, not 90%.
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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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1.11 to 1
Just before=16000g/cm^2/s^2
Just after=14364g/cm^2/s^2
Just before
KE=1/2mv^2
=1/2(.2g)*400^2=.1g*160,000cm^2/s^2
=16000g/cm^2/s^2
=16J
Just after
KE=1/2mv^2
=1/2(.2g)*379^2=.1g*143,641cm^2/s^2
=14364g/cm^2/s^2
=14.4J
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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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686cm/s
`ds=120cm
a=980cm/s^2
v0=0
Again using the 3rd or 4th equation, we can solve for `dt.
`ds=v0`dt+1/2a`dt^2
120cm=0*`dt+1/2*(9.8m/s^2)*`dt^2
120cm=1/2*(980cm/s^2)*`dt^2
`dt^2=120cm/980cm/s^2
`dt^2=.122/s^2
`dt=sqrt of .122/s^2
`dt=.35s
Now we have
`dt=.35s
`ds=120cm
a=980cm/s^2
v0=0
vAve=120cm/.35s=343cm/s
vF=686cm/s
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* What was its velocity just after striking the floor?
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420cm/s
`ds=42cm (35% of 120cm)
a=980cm/s^2
v0=0
Again using the 3rd or 4th equation, we can solve for `dt.
`ds=v0`dt+1/2a`dt^2
42cm=0*`dt+1/2*(9.8m/s^2)*`dt^2
42cm=1/2*(980cm/s^2)*`dt^2
`dt^2=42cm/980cm/s^2
`dt^2=.04/s^2
`dt=sqrt of .04/s^2
`dt=.2s
Now we have
`ds=42cm (35% of 120cm)
a=980cm/s^2
v0=0
`dt=.2s
vAVE=42cm/.2s=210cm/s
vF=420cm/s
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* What is the ratio between the speeds just before, and just after striking the floor?
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1.63 to 1
Just before=686cm/s
Just after=420cm/s
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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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Plus or minus 2.86 to 1 depending on the positive direction
On this, I think that because the ball rose to 35% of its original height, then regardless the distance, the ratio would be the ratio of 100% compared to 35% which is 2.86 to 1
The ratio of speeds isn't 2.86 to 1; more like 1.7 to 1, according to your results (which I believe are correct).
So ratio of momentum is about 1.7 to 1, not 2.86 to 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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Plus or minus 35% depending on the positive direction
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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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2.67 to 1
Just before=686cm/s
Just after=420cm/s
KE=1/2*5g*(686cm/s)^2
=1176490g/cm^2/s^2
=1176J
KE=1/2*5g*420cm/s
=441000g/cm/s^2
=441J
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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 would think that the mass of the object would tell a lot of how much momentum will be retained when dropped. So, in order to predict the retention of momentum, you could throw it up in the air and catch it to see how fast and how hard it hits your hand.
to get the percent of the original height you drop the ball and see how far it rebounds
tossing it up in the air and catching it won't give you any additional information about this question
what did you do in your solutions?
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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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Incline ????
smooth surface (mu=a/m) where a=acceleration of car and m=mass of car
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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=change in velocity/change in clock time
To solve, since v0=0, we 1st find average velocity and then vF. After that we use a=`dv/`dt.
you would want to obtain this expression in terms of the given quantities `ds and `dt; `dv was not a given quantity
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`qx006. Give your data for this part of the experiment.
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`ds=43cm
`dt=1.8s (12 counts)
v0=0
vAVE=43cm/1.8s=23.9cm/s
vF=47.8cm/s
a=(47.8cm/s)/1.8s=26.5cm/s^2
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`qx007. Show how you used your data to find the slope of the 'constant-velocity' incline.
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We measured the height of the ramp as well as the length of the ramp. Height of ramp was 1cm and length of ramp was 30cm. This gave us a slope of .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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Frictional force=plus or minus m*a
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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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a=Fnet/m
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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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12 counts, 43cm
10 counts, 32cm
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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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26.5cm/s^2
`ds=43cm
`dt=1.8s (12 counts)
v0=0
vAVE=43cm/1.8s=23.9cm/s
vF=47.8cm/s
a=(47.8cm/s)/1.8s=26.5cm/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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I need to redo the acceleration for the ramp and resubmit this part of the assignment.
you got slope .03
so frictional force would be .03 * m g, and acceleration would be .03 m g / m = .03 g.
what is that in cm/s^2, and how does that correspond to your observed accelerations?
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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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I am completely stumped on the introduction of the coefficient of friction. The hint said to first find the 2 velocities but how can I find the acceleration when given mu and `ds?
the acceleration is mu * g (see also my previous note)
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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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left side 34,right side 30
left side 56, right side 26
left side 60, right side 33
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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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For left side car
`ds=34cm
vF=0
`dt=
a=26.5cm/s^2
This is the information that I have but am not clear how to find the initial velocity. Can the initial velocity be negative? Also, is vF=0 accurate? Or did I not measure something correctly?
You have values for `ds, a and vf, which you just specified. You can use these to find vAve, v0 and `dt. Reason them out, then use the equations. You should get the same result both ways.
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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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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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k/2*x_1
that would be the average force; still one more step to get work
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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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(k/2*x_2)/(k/2*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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Whatever the distance pulled back is*(mu*m g)
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* What is the ratio of the two coasting distances?
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The ratio of the 2 would be the 1st distance divided by the 2nd distance compared to the 1st distance.
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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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4 times as much
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Very good work. You're a step shy on some of the problems, but just a step, and that's well up to expectations at this stage of the course.
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