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course phy 201
One possible suggested method: Use the short ramp to get the ball started, and let it roll from the short ramp onto the longer ramp, with the longer ramp inclined so the ball rolls up, rather than down. Get the ball started on the short ramp, either by inclining it toward the long ramp or giving it a push (before it reaches the long ramp). Click the TIMER at the instant the ball hits the 'bump' between the two ramps, again when the ball comes to rest for an instant before accelerating back down the long ramp, and once more when the ball again hits the 'bump'.Everyone will tend to anticipate their 'clicks', and to try to compensate for their anticipation. If the effect of anticipation is the same for all three timed events, then uncertainties in your results will all be due to the TIMER. If the degree of anticipation differs, as it inevitably will, then the uncertainties are compounded. If the degree of anticipation (and/or compensation) tends to be either greater or less for the second event (the ball stopping for an instant) than for the first and third (the 'clicks' made when the ball hits the 'bump'), then a systematic error is introduced (you will have a tendency to 'short' one interval while extending the other).
If there exists a difference between the times up and down the ramp, and if the difference is great enough to show through the uncertainties and systematic errors, then you might get a useful result (e.g., either the actual times are the same, or they differ).
Give a brief report of your data and your results:
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I let the ball roll down the short ramp with a two domino onto the longer ramp the longer ramp is inclined with a one domino and with the timer program I got the following times.
3.432,3.454,3.567
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Displacement and force vector for a rubber band
The points (2, 1) and (4, 6) have the following characteristics:
* From the first point to the second the 'run' is 4 - 2 = 2 and the 'rise' is 6 - 1 = 5.
* The slope of the line segment connecting the points is therefore 5 / 2 = 2.5.
* A right triangle can be constructed whose hypotenuse is the line segment from (2, 1) to (4, 6), and whose legs are parallel to the coordinate axes. The leg parallel to the x axis has length 2, and the leg parallel to the y axis has length 5. The hypotenuse therefore has length sqrt(2^2 + 5^2) = sqrt(29) = 5.4, approx..
* The area of the trapezoid formed by projecting the two points onto the x axis, whose sides are the projection lines, the segment [2, 4] of the x axis and the line segment between the two points, has 'graph altitudes' 1 and 5, giving it average 'graph altitude' (1 + 6) / 2 = 3.5, and width 5 - 1 = 4, so its area is 14.
* The vector from the first point to the second has x component 2 and y component 5. Its length is sqrt(20) and its direction is specified by the ratio of the y to the x component.
Any time you see two points on a graph you should be aware that you can easily construct the right triangle and use it to calculate the slope of the segment joining them, the distance between the points, the area of the associated trapezoid and the components of the vector. Depending on the nature of the graph, some of these quantities will sense and be have useful interpretations, while others probably will not.
In the particular example of a rubber band stretched between the two points, with the coordinates in centimeters, the most important quantities are the components and length of the vector. The vector from the first point to the second is in this case a displacement vector. The displacement vector has x component 2 cm, y component 5 cm and length sqrt( 2 cm)^2 + (5 cm)^2 ) = sqrt( 29 cm^2) = sqrt(29) cm, about 5.4 cm.
* We denote the displacement vector from (2 cm, 1 cm) to (4 cm, 6 cm) as <2 cm, 5 cm>
If we divide the this vector 2 we get a vector of length 2.15 cm. The x component will be 1 cm (half of the original 2 cm) and the y component 5 cm /2 = 2.5 cm (half of the original 5 cm). The ratio of the components is the same, so this vector will be in the same direction as the original vector. If we multiply this vector by 2 we get a vector of length 8.6 cm. Its components will be double those of the original vector, and it will also be in the same direction as the original vector.
If we divide our vector by 5.4 cm, then the resulting vector has length 5.3 cm / (5.3 cm) = 1. The x and y components will be 2 cm / (5.4 cm) = .37 and 5 cm / (5.4 cm) = .92. As before, the ratio of the components is the same as for the original vector (accurate to 2 significant figures),
The vector <.37, .92> has magnitude 1 (calculated to 2 significant figure), as you can easily verify using the Pythagorean Theorem. Its direction is the same as that of the displacement vector <2 cm, 5 cm>.
Now we invoke a rule to find the tension in the rubber band. Assume that for this rubber band the tension is .5 Newtons for each centimeter in excess of its 'barely-exerting-force' length of 4.8 cm.
* The 5.4 cm length of the stretched rubber band is therefore .6 cm in excess of the 'barely-exerting-force' length of 4.8 cm.
* According to the rule, we conclude that the tension is therefore .6 cm * .5 N / cm = .3 N.
The vector <.37, .92> has magnitude 1, and its direction is the same as that of the displacement vector. A rubber band can exert a force only in the direction of its displacement vector.
* If we multiply our vector <.37, .92> , which we again mention has length 1, by .3 N, we will obtain a vector whose magnitude is .3 N. (whatever quantity we multiply by a vector of length 1, we end up with a vector whose length is equal to that quantity, or more correctly whose length represents that quantity)
* The direction of this vector will be the same as that of our original vector.
* <.37, .92> * .3 N = <0.11 N, 0.28 N>
* This is the tension vector. It tells us that the tension of this rubber band has x component 0.11 N and y component 0.28 N.
Using your data from the rubber band experiment, check your previous calculations of the displacement vector, the unit vector and the force vector for each rubber band. If your original work on this experiment was not done correctly, please correct it and resubmit. Please acknowledge that you have read this instruction and understand it.
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OKAY
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Linear force function and work\energy
If the graph below indicates the tension in Newtons vs. length in cm for a rubber band:
At what length does the rubber band begin exerting a force?
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-1
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What is the average force exerted between the lengths x = 5 cm and x = 7 cm?
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.5
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As the rubber band is stretched from length 5 cm to length 7 cm, through what distance is the force exerted?
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2 cm
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How much work is therefore done in stretching the rubber band from 5 cm to 7 cm length?
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W = F x D
W=.5*2
w=1
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How much work is done in stretching the rubber band from the 5 cm length to the 8 cm length?
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w=f’ds
W=.5*3
w=1.5
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How much work is done in stretching the rubber band from the 6 cm length to the 8 cm length?
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w=f’ds
W=.5*2
w=1
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What are the areas beneath the graph between each of the following pairs of lengths:
* x = 5 cm and x = 7 cm
* x = 5 cm and x = 8 cm
* x = 6 cm and x = 8 cm
* x = 7 cm and x = 8 cm
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Verify that the equation of the straight line in the given graph is F(x) = (x - 5) * .5, where F(x) is force in Newtons when x is position in centimeters.
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Nonlinear force function and work\energy
The graph given previously was linear. That graph would be realistic for a well-made spring, but not for a rubber band.
The graph given below is more realistic.
Estimate the average force exerted by this rubber band between the x = 5 cm and x = 6 cm lengths. Give your estimate in the first line, and explain how you made the estimate in the second:
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-.6
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Repeat, for the x = 6 cm to x = 7 cm length interval.
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-.7
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Repeat, for the x = 5 cm to x = 5.5 cm length interval.
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-.4
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Repeat, for the x = 5.5 cm to x = 6 cm length interval.
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-.7
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Estimate the work done in stretching the rubber band for each of these intervals. Give your four estimates in the first line below, separated by commas. Starting in the second line explain how you got your estimates:
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How much work do you estimate is done between the x = 5 cm length and the x = 7 cm length? Give your estimate in the first line, your explanation starting in the second.
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W = F x D
W=.5*2
w=1
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If a trapezoid was constructed by projecting the x = 5 and x = 7 points of the graph, what would be its area, and would this result be an overestimate or an underestimate of the actual area beneath the graph between these two points?
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Area 6*2=12
Overestimate
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If a trapezoid was constructed by projecting the x = 5 and x = 8 points of the graph, what would be its area, and would this result be an overestimate or an underestimate of the actual area beneath the graph between these two points?
The area would be 6.5*3=19.5
overestimate
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Report your counts for the five trials with the toy car going in the first direction in the first line below, separated by commas. Report you counts for the five trials with car going in the opposite direction in the second line below, separated by commas.
Up
9counts, 23cm
10counts, 25cm
11counts, 24cm
11counts, 25cm
12counts, 26cm
Down
11counts, 18cm
10counts, 19cm
13counts, 20cm
10counts, 19.5
11counts, 19.5
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Report the five resulting accelerations for the first direction in the first line below, separated by commas, and use the same format to report in the second line the five for the second direcion. Starting in the third line show the details of how you found one of your accelerations.
.56 counts/cm, .23 counts/cm, .19 counts/cm, .20 counts/cm, .18 counts/cm
.14 counts/cm, .19 counts/cm, .11 counts/cm, .195 counts/cm, .16counts/cm
Vave=‘dx/’I’d
A=‘dv/’ dt
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How far from the lower end of the ramps did you have to position the two balls in order to synchronize their consecutive time intervals with the time interval for the third ball released from rest at the top of the third ramp?
6cm
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You didn't time the intervals for either trial. Suppose that the ball on the third ramp required 3 seconds to travel the 30-cm length of that ramp. What was the acceleration of that ball? Report the acceleration in the first line. Explain how you found it starting in the second line.
The car went 30 cm in 3 second
‘s=30cm
‘dt=3seconds
30/3=10
v=10cm/s
Vf=0
‘dv=vf-vo
‘dv=0-30
‘dv=-30cm/s
A=‘dv/’ dt
A=-30cm/s/3s
a=10cm/s^s
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Assuming the 3-second interval for the ball on the third ramp, what was the time interval for ball released on the first ramp, and what was the resulting acceleration? Report the acceleration in the first line, and explain how you calculated it starting in the second line.
6cm/3sec
2cm/sec
a=4/3
1.33cm/s
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Are your two results for the accelerations reasonably consistent? Why would you or would you not expect them to be so?
They are somewhat close and yes I would expect them to be some what consistent because it was a short ramp and you would expect it to be closer and more accuet.
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Are your results in any way relevant to the problem of ordering v0, vf, `dv, v_mid_x, v_mid_t and vAve? If so, what conclusions can you draw?
Yes because you can solve for them in the same way
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