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course PHY 231
10/19 3:37 pmLot's of unknowns for me on this one, not sure why
Coefficient of Restitution
A 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? ****
The ratio is v1(√c)=v2 so percentage wise = √c * 100
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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?****
Ratio of the mag of the ratio??? ∆p is v1/v2 or percentage= √c*100 as above since m cancels.
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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?****
Substituting √c*v1=v2 then (√c*v1)^2/v1^2 and this is actually just c.
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You seem to have assumed that the ratio of velocities is the same as the ratio of the heights. This is not the case, as the fourth equation of motion shows.
Your numerical results show the correct ratios. Had you followed the same reasoning with the symbols, your symbolic result would have been consistent with your numerical results.
`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?****
3.9 m/s
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What was its velocity just after striking the tabletop?****
3.7 m/s
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What is the ratio between the speeds just before, and just after striking the tabletop?****
3.7/3.9 = .95 ----> 95%. Here I will not that depending on your value of g this may actually be closer to 90% in your calculation but
don't actually think the h:ch ratio is the same as the v1:v2 ratio. (I did this backward so 1/.95 = 1.05)
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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.****
same as velocity since p1=mv1 and p2=mv2 so p1/p2 = mv1/mv2 = v1/v2 or, since p is a vector either p1 or p2 would be negative,
depending on pos dir, but that doesn't necessarily affect the %change. So I'm not completely certain of this.
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The bead rose to 90% of its original height. What percent of the magnitude of its momentum did it retain? ****
Seems this would be the same as just above but since you included the 90% statement I'm not sure. I know p is a vector but like my
answer above I don't think the direction matters and especially not since we're talking magnitude. The thing is that at both h and ch p=0 but since
ch p2 where p1 is measured just before and p2 just after. So still 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?****
3.7^2/3.9^2 = 0.9 which is 90%. The m and 1/2 drop out here so that KE ratio is the ratio of the square of v1 to the square of v2 this
is closer to 0.9h than p1/p2 ratio is an seems to suggest that more momentum was conserved than KE, but maybe not.
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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? ****
v1=4.9 m/s
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What was its velocity just after striking the floor? ****
v2=2.8 m/s
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What is the ratio between the speeds just before, and just after striking the floor?****
4.9/2.8= 1.5
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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.****
p2/p1= m2.8/m4.9 = 0.57 This is the reciprocal of above since m cancels out you just have v2/v1.
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The marble rose to 35% of its original height. What percent of the magnitude of its momentum did it retain? ****
As I replied above, I don't know how that 35% figures in b/c the momentum conserved is 57% and that value doesn't depend on how high
the marble bounced and in fact how high the marble bounces is dependant on how much momentum is conserved, I think.
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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?****
2.8^2/4.9^2 = 0.33
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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?
****
This is found by using the formula above that shows that p1(√c)=p2.
This would imply that, for example, the 90% height corresponds to 90% of the momentum.
However as your numerical results show, this is not the case. Nor does the 35% height imply 35% of the momentum.
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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. ****
With h being small enough that the cos is extremely close to 1 then (I think this should show up right as it's mu on my keyboard) µ=
L/h Here L is assumed to be the hyp though for very small h values it doesn't make much difference. I'm not sure about this second part but it seems
that if it's a level surface then h=0 and then µ=L or the distance the car travelled. Well if L = ∆x and it rolls from rest to rest then a = 1/2L
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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? ****
I think this is something like this: a=2(∆x/∆t^2)
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`qx006. Give your data for this part of the experiment.
****
I don't seem to have any data on this aside from the measurements of the ramp. It was 24 cm long and h= 0.6 cm
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`qx007. Show how you used your data to find the slope of the 'constant-velocity' incline.
****
The slope here is .6/24 = 0.025 the cos of this angle is like 0.999687 so very very close to 1.
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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?
****
Ffric= µmg on a flat surface but for the incline it is -mg(cos theta) however since cos theta is very close to 1 it is still very close to µmg.
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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= -µg
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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.
****
I don't have any data on this in conjunction with what we did above.
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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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`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?
****
There should not be a significant discrepancy but there would be a detectable difference and part of that has to do with that very small h value since
not quite all of mg goes into the frictional force on the incline.
If you have the car you can set it on a smooth level surface, give it a poke and time it to rest. Doesn't take long to get a result. Then see if the acceleration is consistent with what you observed from the slope.
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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) ****
v1^2=2(1/2L1)(L1)= L^2---> v1=L1 and v2=L2 This is from above with a=1/2L, I think this works out okay using this expression here. Also
a1=-µ_1 so v1/v2=L1/L2=µ_1/µ_2
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the cars have different accelerations and accelerate through different distances
for each you know vf, `ds and a, so you can calculate v0 for each car
What therefore should be the ratio of the masses of the two cars?****
m1= µ_1/µ_2 m2
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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?****
W=F*d so W_mag=F_mag*∆s_mag and the ∆PE is -Work and ∆KE = W. I'm not sure about this any further than this for some
reason I just can't get this pictured right in my head. I'm not sure about the ratio of ∆PE to ∆KE.
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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.
****
15, 29
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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?
****
15, 29
If this is referring to the data obtained acceleration I don't know since I don't seem to have that. So I'm using the 1/2L as acceleration, I'm not
certain that this is acceptable here but anyway.
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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?
****
15/29 = 0.5 The left hand car, data on the left car that is, has the greater mass and I think the ratio of the masses is the same as the velocities
like this= (m1v1)/(m2v2)----> m1= v1/v2*m2
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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? ****
W=k/2 * x_1^2
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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. ****
x2^2/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? ****
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I don't know about this since W_Fnet is negative and the work done by the bungee cord on the chair is positive I'm not sure how this
works out exactly, b/c as I picture it Fnet = Fbungee - µmg so I don't see how Fnet can just be µmg. I know the chair is slowing down so the Fnet must
be in that direction but I don't know how it's the whole of µmg and not just part of it.
friction dissipates the energy released by the bungee cord
so from release to rest the second car should travel (x2 / x1)^2 as far
What is the ratio of the two coasting distances? ****
This doesn't work out for me since I don't understand the previous question completely.
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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?
****
(2*F)(2*d) = 4W so 4 times as much energy.
Right. In relation to your previous question, the car would therefore travel 4 times as far, from its point of release.
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&#Good responses. Let me know if you have questions. &#
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