#$&*
course phy 201
Coefficient of RestitutionA bead dropped on the tabletop was observed to rebound to abou 90% of its original height. A marble dropped on the floor rebounded to an estimated 35% of its original height.
`qx001. Symbolic Solutions. Most students will need to work through the details of subsequent specific problems before attempting the symbolic solution. However if you can get the symbolic solutions, you will be able to use them to answer the subsequent questions. If you prefer to work through the subsequent questions first, please do. if you get bogged down on this, move on to the subsequent questions.
* If the height from which the bead is dropped is h, and if it rebounds to height c * h, then what is the percent change in the bead's speed between the instant when it first contacts the table during its fall, and the instant when it loses contact on the way back up?
****
#$&*
* What is the ratio of the magnitude of the ratio of its corresponding momentum change to its momentum just before contact with the tabletop?
****
#$&*
* What is the ratio of the kinetic energy of the bead immediately after, to its kinetic energy immediately before striking the tabletop?
****
#$&*
`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?
Vf2=2(980cm/s2)(80cm)
Vf= -396cm/s
****
#$&*
* What was its velocity just after striking the tabletop?
02=Vo2+ 2(980cm/s2)(72cm)
Vo= 376cm/s
****
#$&*
* What is the ratio between the speeds just before, and just after striking the tabletop?
376cm/s / 396cm/s= 95%
****
#$&*
* What is the magnitude of the ratio of the ball's momentum just after, to its momentum just before striking the tabletop? The answer doesn't depend on the mass of the bead, but if you feel you need it you may assume a mass of .2 grams.
Just before momentum= 396cm/s/g
Just after= 234cm/s/g
234/396= 59%
cm/s/g would imply that you divided cm/s by grams.
However your numbers aren't consistent with that, and you wouldn't divide by the mass in any case.
****
#$&*
* The bead rose to 90% of its original height. What percent of the magnitude of its momentum did it retain? 376/ 396= 95%
****
#$&*
* What is the ratio of the kinetic energy of the bead immediately after, to its kinetic energy immediately before striking the tabletop?
KEafter= .5(.2g)(3.76m/s)2= 14137 g*m2/s2
KE before= .5(.2g)(3.96m/s)2= 15681 g*m2/s2
14137/15681= 90%
****
#$&*
`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?
Vf2=2*980cm/s2*120cm
Vf= -485cm/s
****
#$&*
* What was its velocity just after striking the floor?
Vo2= 2*980cm/s2*42cm
Vo=287cm/s
****
#$&*
* What is the ratio between the speeds just before, and just after striking the floor?
287/485= 59%
****
#$&*
* What is the magnitude of the ratio of the marble's momentum just after, to its momentum just before striking the floor? The answer doesn't depend on the mass of the marble, but if you feel you need it assume a mass of 5 grams.
Momentumafter=2425g*cm/s
Momentumbefore= 1435g*cm/s
1435/2425= 59%
****
#$&*
* The marble rose to 35% of its original height. What percent of the magnitude of its momentum did it retain? 59%
****
#$&*
* What is the ratio of the kinetic energy of the marble immediately after, to its kinetic energy immediately before striking the tabletop?
KEafter= 717.5g*m2/s2
KEbefore= 1212.5g*m2/s2
717.5/1212.5= 59%
You don't show how you got your results for the KE, and your results don't appear to be correct, nor is your ratio.
****
#$&*
`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?
****
#$&*
If you feel you have worked out the answers to all or most of the numerical questions correctly, but haven't yet worked out the symbolic solution, you should consider returning to the first question. However don't get bogged down for a long time on that solution.
Acceleration of toy cars due to friction
The magnitude of the frictional force on a rolling toy car is the product of the coefficient of rolling friction and the weight of the car.
The coefficient of friction can be measured by placing the car on a constant incline and giving it a nudge in the direction down the incline. It will either speed up, slow down or coast with constant velocity. If the incline is varied until the car coasts with constant velocity, then the coefficient of friction is equal to the slope of the incline.
As before the symbolic question at the beginning can be attempted before or after the subsequent numerical questions.
`qx005. Answer the following symbolically:
* If the length of the incline is L and its rise is h, then what is the symbolic expression for the magnitude of the frictional force on a car of mass m? What therefore is the expression for its acceleration when rolled across a smooth level surface? The Greek letter traditionally used for coefficient of friction is mu.
Mu= (h/L)*m
Mu=m*L
If the acceleration of the object when moving down the incline is zero, then mu = h / L. Its mass is not relevant to this result.
What would be the frictional force on the car as it rolls along the level surface?
What therefore would be its acceleration?
****
#$&*
* 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?
A= ds/dt
that would be its average velocity, not its acceleration
****
#$&*
`qx006. Give your data for this part of the experiment.
Ramp distance= 30.5cm
Height= .07cm
****
#$&*
`qx007. Show how you used your data to find the slope of the 'constant-velocity' incline.
A2+(.07)2=(30.5)2
A= 30.5cm
.12cm/30.5cm= 0.004=slope
I expect that the height was 7 cm. .07 cm would be less than a millimeter.
****
#$&*
`qx008. The weight of your car is given by the symbolic expression m g, where m is its mass. What therefore is the expression for the magnitude of the frictional force on the car?
****
#$&*
`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?
The car would not be accelerating. It would be rolling with a constant velocity or not moving at all.
****
#$&*
`qx010. You also timed the car as it coasted to rest along the tabletop, after having been given a nudge. What were the counts and the distances observed for your trials? Give one trial per line, each line consisting of a count and distance in cm, with the two numbers separated by a comma.
1.6sec, 20cm
2.6sec, 30.5cm
1.4sec, 15cm
****
#$&*
`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.
7.7cm/s2
The accelerations of the first and third trials were 7.8cm/s2 and 7.6 cm/s2. The second trial was an outlier so I left it out. The average acceleration of the two trials was 7.7cm/s2.
****
#$&*
`qx012. You have obtained two results for the acceleration of your car along a level surface, one based on the slope of an incline, the other on observed counts and distances. How well do they compare? Is there a significant discrepancy? If so, can you explain possible sources of the discrepancy?
****
#$&*
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)
ds1/T1=V1
ds2/T1=V2
V1/V2
****
#$&*
* What therefore should be the ratio of the masses of the two cars?
1:2
you aren't given any numerical quantities, so your results would be in terms of symbols; the numerical results would depend on the numerical values of those symbols
****
#$&*
* 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?
KE-friction
****
#$&*
`qx014. When the two cars were released, what were their approximate average distances in cm? Give your answers in a single line, which should consist of two numbers separated by commas.
35cm, 7cm
****
#$&*
`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?
Vo2=2(980cm/s2)(35cm)
Vo= -262cm/s for car 1
Vo2=2(980cm/s2)(7)
Vo=117cm/s for car 2
You got an acceleration of about 7 cm/s^2, not 980 cm/s^2, for the one car. You don't appear to have found an acceleration for the other, but you would use the two accelerations, not the acceleration of gravity.
****
#$&*
`qx016. What was the ratio of the speed attained by the first car to that of the second? Which car do you therefore think had the greater mass? What do you think was the ratio of their masses?
2.24:1, 2.24:1
you don't say how you got this, though given your assumption of equal accelerations it would follow from your data
****
#$&*
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?
K/2*x_1
****
#$&*
* What is the ratio of work done when the distance is x_2, to the work done when the distance is x_1.
If x_2 is double that of x_1 according to the formula k/2*x work done would be double, but it should be more than twice.
****
#$&*
* 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?
****
#$&*
* What is the ratio of the two coasting distances?
****
#$&*
`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?
The chair would also be expected to gain twice as much energy.
****
#$&*
"
Pretty good work on many of the questions. You were off the mark on others, and in a number of cases you didn't include enough explanation for me to tell what you were doing.
&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you. &#
#$&*