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course PHY241
10/24 4
Give Chapter 8 a quick readCoefficient 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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1-(v(ch)/vh)
What are the two velocities, in terms of c and h, and of course the acceleration g of gravity?
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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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(p2-p1):p1, p1 is the momentum down before it hits the table and p2 is the momentum back up
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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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.5(mv2)^2:.5(mv1)^2
m isn't part of the expression that gets squared
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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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395.98cm/s. vf^2=0^2+2(980)(80)
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* What was its velocity just after striking the tabletop?
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375.66cm/s. 0^2=v0^2+2(-980)(72)
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* What is the ratio between the speeds just before, and just after striking the tabletop?
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aprox 1:.95
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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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v2:v1
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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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approx 95%
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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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9:10
that would be 90%, for easier comparison with your result 95%
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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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484.97cm/s
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* What was its velocity just after striking the floor?
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286.91cm/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:.59
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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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.59:1
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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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59%
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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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1:3.5
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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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Its the sqrt of the percent of original height
Good work on the numerical calculations, and your conclusion here is correct. Can you now follow the same sequence of reasoning with the symbols?
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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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Ffric=(mu)mg
a=(mu)g
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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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2((`ds-v0`dt)/`dt^2)=a
The solution needs to be in terms of `ds and `dt; v0 is not known. However you do know that vf = 0.
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`qx006. Give your data for this part of the experiment.
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a=18cm/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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my board was 30cm long and was raised by 5 washers that are .08cm tall. so slope = .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?
****
(mu)*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?
****
a=(mu)g
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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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8count,94.6
5count,22
11count,110
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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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54.5cm/s^2
took the ave of all of my accelerations
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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?
****
I think the dicrepancy lies with my ability to just ""nudge"" the car.
I should bring in short pool cues. Much easier to control than just your hand by itself.
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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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`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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`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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W=((k/2)*x1)*(x_1)
Good. Simplified this is k / 2 * x1^2.
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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)*x2)*(x_2):((k/2)*x1)*(x_1)
right, but this simplifies to (x2 / x1)^2
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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?
****
W/(mu*mg)=`ds
good; if you express W as 1/2 k x^2 you'll have the expression in terms of the assumed quantities
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* What is the ratio of the two coasting distances?
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Work1/(mu*mg):Work2/(mu*mg)
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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
how does this compare to your ratio (x2 / x1)^2?
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you've got some good work here
you're making progress with the symbolic solutions, but be sure to see my notes
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