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Phy 121
Your 'conservation of momentum' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Conservation of Momentum_labelMessages **
11/6 3:30 pm
** **
Conservation of Momentum
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Experiment 19. When two objects moving along a common straight line collide and maintain motion along the same line as before collision, the total of their momenta immediately after collision is equal to the total immediately before collision.
See CD EPS01 for a general overview of Lab Kit Experiment 19. Note, however, that this experiment has been revised since those videos were recorded, and any specific instructions given on the videos are superceded by the instructions given here, and by more recent practices in observing horizontal range.
The basic setup for this experiment is pictured below. The small metal ball (referred to here as the 'target ball') from the kit is supported on a short piece of drinking straw, and the large ball rolls down a ramp (the end of this first ramp appears at far right), then along a horizontal second ramp until it reaches the end of the ramp and collides with the target ball.
The small ball is positioned near the edge of the table according to the following criteria:
The large ball collides with the small ball just a very short distance after it has lost contact with the ramp.
The collision occurs when the centers of mass of the two balls are at the same height above the table.
After collision both balls fall uninterrupted to the floor. Their velocities after collision are easily determined by the distance to the floor and their horizontal ranges. The velocity of the first ball prior to collision is determined by removing the second ball and allowing it to run down the inclined track and across the horizontal track before falling uninterrupted to the floor.
In the picture above, the 'tee' holding the smaller ball consists of a piece of plastic tubing inserted into a glob of hot glue.
A piece of the drinking straw that came with your lab kit slips over the tubing, allowing you to easily adjust the vertical position of the ball.
The base of the 'tee' is flat and keeps the system stable, while providing no significant interference with the motion of either ball.
This 'tee' was not included in the kit for Spring 2006; if you are a Spring 2006 student you may submit your email address and the instructor will be glad to send you a few.
The alternative to using the 'tee' is to balance the ball just on the drinking straw section, which is feasible (students have been doing it for years) but which can be unnecessarily time consuming.
If you use the tee, you should trim about 1 cm, maybe just a little more (if you trim it a little too long it's easy to shorten it), from the end of the straw. The flat end of the straw should be the one that supports the ball; the end you trim will tend to be less straight and the ball might tend to roll off.
The straw might or might not fit tightly enough that the weight of the ball does not cause it to slide up or down. If this is not the case, you can insert a thin strip of paper into the straw, before slipping it over the plastic tubing in order to 'shim' the inside and ensure a tighter fit.
Set up the system:
The motion of the balls before and after collision should be horizontal:
To analyze the projectile motion of the balls it is very helpful if their initial velocities are both in the horizontal direction. Due to slight irregularities in the shapes of the balls and to the fact that the first ball is spinning, this cannot be completely assured, but if the centers of the two balls are at the same vertical height when they collide, the two will come out of the collision with their initial velocities very close to horizontal.
To adjust the height so this will be the case, proceed as follows:
Place a piece of carbon paper in contact with a piece of white paper just past the end of the ramp, as shown below. The ball should strike the paper just after it loses contact with the ramp.
The smooth edge of the white paper should be contact with the tabletop. The mark made on the paper when the balls collide will lie at a distance from this edge which is equal to the height of the point of collision above the tabletop.
If you place a hard, flat solid object (which should also be either cheap or unbreakable and not subject to denting) behind the paper, oriented so that its flat side is vertical, then when the ball strikes it will leave a mark indicating the height of its center above the table.
You should do this three times, positioning the system so that the ball will collide just after leaving the ramp. The centers of your three marks should all line along or very close to a single straight horizontal line, all at very nearly the same distance from the edge of the paper. If necessary repeat your trials until you are sure the system is properly set up to give you consistent results.
If you place the smaller ball on the tee behind the paper, then the collision will produce a mark at the point of contact of the two balls.
If the center of the mark made by colliding the balls is at the same height as the centers of the marks made by the ball against the flat object, then the centers of the balls will be at equal heights. If not, adjust the length of the 'tee'.
Use the long 'track' as the incline, and the short piece of track as the horizontal section. The high end of the long incline should be about 5 cm higher than the short end. This vertical distance should be measured and should be kept the same throughout your trials, but it doesn't have to be exactly 5 cm. For example 4 cm, or 6 cm, or 5.37 cm would be fine, as long as it is measured accurately and checked repeatedly to be sure it doesn't change after being set up.
The height of the top of the 'tee' supporting the target ball should also be checked throughout your trials to ensure that it doesn't change.
You should therefore check the height of the end of the ramp, and the height of the top of the straw, periodically throughout the experiment. You should note these 'maintenance checks' in your lab notebook, noting when they were done and verifying that the slope of the ramp and the height of the 'tee' has not changed.
The large ball must be just out of contact with the ramp at the instant of collision, being no more that a couple of millimeters from the point where it leaves the ramp, and after collision both balls should 'clear' the edge of the table as they fall. If they don't, then the system can be moved closer to the table's edge, and/or the slope of the inclined ramp could be increased to give the ball greater velocity.
Proceed to adjust the height of the 'tee' until the balls collide with their centers at the same vertical altitude. In the space below, give in the first line the measurement from the edge of the paper to the mark made by the ball as it strikes the vertical object, and from the edge of the paper to the mark made by the collision of the two balls. In the second line give the height of the top of the 'tee' above the tabletop. In the third line give the distance from tabletop to floor. Make both all measurements as accurate as possible, and indicate in the fourth line the uncertainty in each of your measurements and how these were estimated:
-------->>>>>>>> collision pt 1 ball against vert and coll pt 2 balls, ht of top of tee above tabletop, tabletop to floor, uncertainties
Your answer (start in the next line):
2.0, 2.0
1.35
77.3
All are distances in cm. Uncertainty is +-.05cm based on the ruler having .1cm gradations.
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Run your first set of trials:
Now you will remove the 'tee' and release the ball from the rest at the high end of the sloped track. You will use the same procedures as in previous experiments for observing the horizontal range of the ball as it falls to the floor.
Be sure the ramps remain well aligned, and if necessary 'shim' the end of the inclined ramp to ensure that there is no 'bump' when the ball moves from one ramp to the next.
Conduct 5 trials, and in the first line give 5 horizontal ranges; in the second line give the mean and standard deviation of the range of the ball. Starting in the third line explain in detail how you got your results.
-------->>>>>>>> 5 ranges uninterrupted, mean & sdev, explanation
Your answer (start in the next line):
29.0, 28.7, 29.8, 28.9, 29.9
29.3, .551
All values are in cm. First line is horizontal distance from origin. Origin was established as the point on the ground directly below the tee. To do this, I used a plumb bob to find the point on the ground and adjsusted back .5cm to compensate for the fact that the tee was .5 cm from the edge of the table. With this origin established I measured the distance from origin to the carbon marks left from the ball hits. The second line is the mean and standard deviation determined by using the Data Program.
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Now place the target ball at the edge of the table, as described earlier. Measure the distance in cm from the edge of the ramp to the closest point on the straw.
Align the target ball so that after the collision, the 'forward' paths of both balls are in the same direction as that of the uninterrupted first ball. That is, make sure the collision is 'head-on' so that one ball doesn't go to one side and the other to the opposite side of the original path.
Divide the carbon paper into two pieces, and position the two in such a way that after collision the two balls will leave clear marks when they land. Do this until you get marks for five trials. Be sure to note which second-ball position corresponds to which first-ball position (e.g., number the marks).
Using your marks, determine the horizontal ranges of the two balls after collision.
In the first line of the space below, give the five horizontal ranges observed for the second ball, using comma-delimited format. In the second line give the corresponding first-ball ranges. In the third line give the mean and standard deviation of the second-ball ranges, and in the fourth line give the same information for the first ball. Starting in the fifth line specify how you made your measurements, and as before specify the positions with respect to which you found your ranges, as well as how you measured those positions.
-------->>>>>>>> five ranges target ball, five ranges first ball, mean and std second, mean and std dev first ball, details
Your answer (start in the next line):
41.3, 41.8, 41.6, 40.7, 42.1
15.2, 14.8, 15.9, 15.8, 15.8
41.5, .534
15.5, .480
All values are in cm. First line is horizontal range of second ball over five trials. Range is measured from the origin described above to the carbon marks made where the balls hit. Second line is horizontal range of first ball over five trials. Range is measured from the origin described above to the carbon marks made where the balls hit. Third and fourth lines are the mean and standard deviation for the second and first ball trials respectively.
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Do not disassemble the system until you are sure you are done with it. General College Physics and University Physics students will use the system again in subsequent activities, and should leave it as it is.
Analysis of Results from First Setup:
Give in the first line below the vertical distance through which the two balls fell after collision, and in the second line the time required to fall this distance from rest. Starting in the third line, explain precisely how you determined these distances, how you determined the time of fall and what assumptions you made in determining the time of fall:
-------->>>>>>>> vertical fall, time to fall, explanation
Your answer (start in the next line):
78.7
.402
First line is the distance of fall in cm. This was measured from the height of the tee to floor and I calculated this by adding the height of the tee from the table top to the height of the table top from the floor. Using this as my height gives a slight error since the large ball falls slightly shorter distance since its bottom is slightly lower than the bottom of the small ball. The time of the fall was calculated using the 4th equation of uniform acceleration to determine final velocity and then the average velocity from the final velocity and initial velocity. Finally I determined the time since I knew the vAve and `ds. Here are my calculations.
vf^2 = v0^2 + 2a*`ds
vf^2 = 0 + 2 * 980cm/s^2 * 78.7 cm
vf^2 = 154,449 cm^2/s^2
vf = +- 393 cm/s
take the positive velocity
vAve = (v0 + vf)/2 = 393cm/s /2 = 196cm/s
`dt = ‘ds/vAVe = 78.7cm/196cm/s = .402s approx.
I also timed 5 drops from the height of the tee and found their average to be .513 s. I decided however to use the calculated time since, with such a short interval, I think my own reaction speed was likely to cause errors. The assumption I made was that the acceleration was indeed 980cm/s^2 in the building where I ran the trials.
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In the space below give in the first line the velocity of the first ball immediately before collision, the velocity of the first ball immediately after collision and the velocity of the second ball immediately after collision, basing your calculations on the time of fall and the mean observed horizontal ranges. In the second line give the before-collision velocities of the first ball based on (mean + standard deviation) and (mean - standard deviation) of its uninterrupted ranges. In the third line do the same for the first ball after collision, and in the fourth line for the second ball after collision.
-------->>>>>>>> velocity first ball before, first ball after, second ball after collision; mean +- std dev first ball before, after, 2d ball after
Your answer (start in the next line):
72.9, 38.6, 103
74.3, 71.5
39.8, 37.4
105, 102
All values are velocities in cm/s
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The masses of both balls are unknown. Using momentum conservation, you will determine the ratio of their masses:
Let m1 stand for the mass of the large ball and m2 the mass of the small ball. In terms of m1 and m2 write expressions for each of the following:
The momentum of the first ball immediately before collision, using the velocity you reported above (the velocity based on the mean range and distance of fall). Be sure to use both the numerical value of the velocity and its units. This will be reported in the first line below.
The momentum of the first ball immediately after collision, using the velocity you reported above. This will be reported in the second line below.
The momentum of the second ball immediately after collision, using the velocity you reported above. This will be reported in the third line below.
The total momentum of the two balls immediately before collision. This will be reported in the fourth line below.
The total momentum of the two balls immediately after collision. This will be reported in the fifth line below.
The total momentum immediately before collision is equal to the total momentum immediately after collision. Set the two expressions equal to obtain an equation. Report this equation in the sixth line below.
-------->>>>>>>> equation for momentum conservation
Your answer (start in the next line):
72.9cm/s *m1
38.6cm/s *m1
103cm/s *m2
72.9 cm/s * m1 + 0 cm/s *m2
38.6cm/s *m1 + 103cm/s *m2
72.9 cm/s * m1 + 0cm/s *m2 = 38.6cm/s *m1 + 103cm/s *m2
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Rearrange your equation so that all terms containing m1 are on the left-hand side, and all terms containing m2 are on the right-hand side. Report this equation in line 1 below.
Divide both sides by the appropriate quantity so that m1 appears by itself on the left-hand side. Report the resulting equation in line 2.
Divide both sides of the equation by m2, and report the resulting equation in line 3.
Simplify the right-hand side, if you have not already done so, to obtain a single number. If you have done your calculation correctly, the units will cancel out. Report the resulting equation in line 4. The left-hand side will be m1 / m2 and the right-hand side will be a single decimal number or, if you prefer, a reduced fraction.
Starting in the fifth line discuss the meaning of the ratio m1 / m2.
-------->>>>>>>> equation solution in steps, meaning of ratio m1 / m2
Your answer (start in the next line):
72.9 cm/s * m1 - 38.6cm/s *m1 = 103cm/s *m2 + 0cm/s *m2
m1 =3.00*m2
m1/m2 =3.0
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Measure and report the diameter of ball 1 and the diameter of ball 2, in comma-delimited format in the first line below.
Calculate the volumes of the two balls and report them in the second line.
-------->>>>>>>> diameters, volumes
Your answer (start in the next line):
2.6, 1.9
9.2, 3.6
First line is the diameter of the first and second ball in cm respectively. Second line is the volume of the first and second ball in cm^3 respectively.
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Physics 121 students are not required to continue, but may do so
Error Analysis for First Setup:
If at collision the center of the first ball is higher than the center of the second, how will this affect the magnitude and direction of the velocity of the first ball immediately after collision? Will the speed be greater or less than if the centers are at the same height? Will the direction of the after-collision velocity differ, and if so how?
In the space below answer this question, and also answer the same questions for the second ball.
-------->>>>>>>> if first ball higher what is effect on its motion, same question for second ball
Your answer (start in the next line):
If the first ball is is higher it will deflect slightly upward after the collision. This should give it a longer time in the air and perhaps a farther horizontal range. The velocity might be faster than when it hit the smaller ball head on because it may be encountering less of the mass of the smaller ball. This set up may also drive the smaller ball downward, thus giving it a faster downward velocity and therefore shorter horizontal range.
If the second ball is higher we may see the opposite, the second ball will deflect slightly higher and therefore possibly acheive a farther horizantal range. The larger first ball may be driven downward, thus acheiving a shorter horizontal range.
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How do you think this will affect the horizontal range of the first ball? How will it affect the horizontal range of the second?
-------->>>>>>>> effect on horizontal ranges
Your answer (start in the next line):
If the first ball is higher its range increases and the second ball's range will decrease.
If the second ball is higher, its range will increase and the first ball's range will decrease.
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For the first ball before collision you reported an interval of velocities based on mean + std dev and mean - std dev of observed horizontal ranges. You did the same for the first ball after collision, and the second ball after collision. Each of these intervals includes a minimum and a maximum possible velocity.
What do you get for the ratio of masses if you use the minimum before-collision velocity in the interval reported for the first ball, the maximum after-collision velocity for the first ball, and the minimum after-collision velocity of the second? Report how you determined this ratio in the space below:
-------->>>>>>>> mass ratio using min before, max after 1st ball, min after 2d
Your answer (start in the next line):
71.5 cm/s * m1 + 0cm/s *m2 = 39.8cm/s *m1 + 102cm/s *m2
71.5 cm/s * m1 - 39.8cm/s *m1 = 102cm/s *m2 + 0cm/s *m2
31.7 cm/s *m1= 102cm/s * m2
m1 =3.22*m2
m1/m2 =3.2 approx
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What percent uncertainty in mass ratio is suggested by comparing this result to your original result?
-------->>>>>>>> % uncertainty suggested by previous
Your answer (start in the next line):
Uncertainty = Change in result/original result *100% = .2/3.0 *100% = 6.7%
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General College Physics students may stop here. University Physics students will continue with additional error analysis, then with an investigation of how errors in the relative heights of the two balls at collision might affect results.
Suppose you can choose either the maximum or minimum of the velocities in each of the reported velocity intervals. What combination of before-and after-collision velocities gives you the maximum, and what combination gives you the minimum result for the mass ratio?
-------->>>>>>>> combination of three obs velocities to get max, min mass ratio
Your answer (start in the next line):
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In symbols, what mass ratio is indicated if the before-collision velocity of ball 1 is v1, its after-collision velocity u1 and the after-collision velocity of the 'target' ball is u2? You will solve the same equation in the same manner as before, but use the symbols v1, u1 and u2 instead of the numerical results you used earlier:
-------->>>>>>>> mass ratio using symbols
Your answer (start in the next line):
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You just obtained an expression for m1 / m2 in terms of v1, u1 and u2. Treating v1 as a variable and u1 and u2 as constants, take the derivative of this expression with respect to v1. Give your expression in the first line of the space below. In the second line give the value of this expression for your velocities v1, u1 and u2, as you reported them earlier based on the mean value of each. Include units in your result. Starting in the third line explain what you think this quantity might mean and how it might be related to error analysis.
-------->>>>>>>> derivative wrt v1;value for observed v1, u1, u2 incl units; interpret
Your answer (start in the next line):
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The derivative you just reported is the rate at which the predicted mass ratio changes with respect to v1; that is, your derivative is the approximate value of
approximate derivative = (change in mass ratio) / (change in v1)
Again, this is an approximation, for small values of the change in v1.
It follows that the change in the mass ratio as predicted by this experiment is equal to the value of the derivative, multiplied by the change in v1:
approximate change in predicted mass ratio = approximate derivative * change in v1
v1 is the before-collision velocity of the first ball.
Answer the following:
If the range of the uninterrupted first ball changes by an amount equal to the standard deviation, then how much does the predicted value of v1 change? Answer in the first line below.
If v1 changes by this amount, then based on the numerical value you reported for above the derivative with respect to v1, by how much would the predicted mass ratio change? Answer in the second line below.
Starting in the third line explain how you got your results and what they mean in terms of this experiment. You might also address how this chain of reasoning illustrates the chain rule for derivatives. You might also consider how this analysis illustrates the meaning of the differential of a function. These applications of calculus to the process of error analysis are subtle but very important.
-------->>>>>>>> change in v1 if range changes by std dev ; resulting change in mass ratio; discussion
Your answer (start in the next line):
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Repeat the Experiment with Second Setup (University Physics Students Only):
The second setup will change the system so that the target ball is 2 mm higher than before. The same observations will be made. Analysis will take account of the non-horizontal direction velocities of the balls immediately after collision.
Repeat the preceding experiment with the target ball 2 mm higher than before:
Set up the system with the second ball 2 millimeters higher than before.
Repeat the preceding experiment with this new setup.
Analyze as before, including the previous assumption that the two balls both have negligible vertical velocity after collision. This assumption is no longer completely valid, and we will soon correct for this.
Give a summary of your data and your prediction of mass ratio, and compare to your previous prediction of mass ratio
-------->>>>>>>> summary expt repeated 2 mm higher
Your answer (start in the next line):
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Make a detailed sketch of the two balls at collision:
Sketch a horizontal line on a sheet of paper.
Place a dot on that horizontal line, representing the position of the center of the first ball at the instant of collision.
Sketch a circle representing the first ball, making the circle about the same size as the ball.
Sketch a dot which represents the center of the second ball at collision, and a circle representing that ball, again more or less actual size. The sketch should be more or less to scale, with the second dot 2 mm higher than the horizontal line.
The velocity of the second ball immediately after collision will be in the direction of the line connecting the centers of the two balls. Find the slope of this line:
Sketch a right triangle whose hypotenuse runs from the center of the first ball to the center of the second, with one vertical and one horizontal leg. The vertical leg of the triangle will consist of the 2 mm vertical segment from the line to the center of the second ball.
You know the length of the hypotenuse and the vertical leg of the triangle. What therefore is the length of the horizontal leg? Is this length significantly different than the hypotenuse?
The vertical and horizontal legs represent the rise and the run between the centers of the two balls. What therefore is the slope of the line segment connecting the centers?
Recall that the data analysis program will give you the velocity of a projectile if you give it the horizontal range and initial slope of its path. This feature was applied in a previous experiment to a ball rolling off an incline with given slope. It can also be applied to this situation.
Run the data analysis program.
Click on 'Experiment-Specific Calculations'
Enter the number for the Ball and Ramp Projectile Experiment.
When prompted, enter the vertical drop of the second ball.
When prompted, enter the mean horizontal range observed for the second ball.
Rather than the slope, you will be asked for the 'number of dominoes'. The program uses the number of dominoes to calculate the slope, based on a 30 cm ramp; for small slopes each domino changes the slope of a 30 cm ramp by approximately .03 so if you divide the slope by .03 you get the equivalent number of dominoes (e.g., if the slope is .16 the equivalent number of dominoes is .16 / .03 = 5.33). However note the following qualification:
If the slope is not small then the number of dominoes can be calculated by dividing the slope by .03, as before, then dividing the result by sqrt( 1 + m^2), where m is the slope.
You obtained a standard deviation for the observed ranges of the second ball.
Repeat the calculations, this time using the horizontal velocity predicted by mean + std deviation of the horizontal ranges of the second ball.
Repeat once more using the velocity predicted by mean - std deviation of horizontal ranges.
In the space below:
Report the vertical drop of the second ball, its mean horizontal range and the slope of the line segment connecting the two centers.
In the second line enter the velocity given by the program, based on the mean horizontal range.
In the third line enter the velocity interval obtained this run of the experiment, giving the second-ball velocities based on mean horizontal range + standard deviation and on mean horiz range - std dev. Give these quantities in cm/sec as two numbers in comma-delimited format.
In the fourth line enter the velocity interval obtained in the original run of the experiment, giving the second-ball velocities based on mean horizontal range + standard deviation and on mean horiz range - std dev. Give as two numbers in comma-delimited format.
In the fifth line give the difference in the mean-based 2d-ball velocity for the original run, and the mean-based velocity for the current run.
Starting in the sixth line state, based on the information given above, whether the velocity resulting from the 2 mm increase in 2d-ball altitude appears to be significantly different than that obtained when the centers were at the same level.
-------->>>>>>>> target 2 mm higher 2d ball vert drop mean range slope of segment connecting centers; velocity based on mean horiz range; vel interval, vel interval 1st run, difference in vel between runs, is difference significant
Your answer (start in the next line):
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In the space below give a similar report comparing the first-ball velocity obtained with the second ball 2 mm higher, with the first-ball velocity obtained with the centers at the same height.
-------->>>>>>>> same info 1st ball
Your answer (start in the next line):
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The first run was set up so the two centers would be at the same altitude. You took a certain amount of care to ensure that this was so, within the limits of precision possible with available apparatus. What do you think were the limits of your precision?
Report in the following manner.
Since they were ideally at the same altitude, the relative height of the second ball relative to the first was ideally 0; its actual altitude is therefore 0 +- uncertainty. If you were careful at the beginning of the experiment, measuring as precisely as possible the distances of the dots made by the carbon paper when the balls collided, your uncertainty will probably be less than 1 millimeter.
In the first line, give the uncertainty in millimeters (e.g., .5 mm, .8 mm, .03 mm), and state beginning in the second line the reasons for the number you give.
-------->>>>>>>> uncertainty in relative ball hts, reasoning
Your answer (start in the next line):
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Based on the results you have obtained to this point, argue for or against the hypothesis that the uncertainty in the relative heights of the balls was not a significant factor in the first setup.
-------->>>>>>>> argue for and against hypothesis
Your answer (start in the next line):
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Based on the slope of the initial after-collision path of the second ball and its velocity, as determined in the second setup, what is the vertical component of the second ball's immediate after-collision velocity?
-------->>>>>>>> init vert comp of 2d ball vel
Your answer (start in the next line):
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Based on this result and on the mass ratio determined in the first setup, what is the vertical velocity of the first ball immediately after collision? Note that vertical momentum is conserved, and that since immediately before collision nothing was moving in the vertical direction, the total vertical momentum is zero immediately before collision.
-------->>>>>>>> infer init vert vel of 1st ball
Your answer (start in the next line):
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What is the horizontal velocity of the first ball before collision and the horizontal velocity of the second ball after collision? Based on the mass ratio obtained in the first setup, what therefore is the horizontal velocity of the first ball after collision?
-------->>>>>>>> horiz vel of 1st before and of 2d after; iner horiz of 1st after
Your answer (start in the next line):
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Using these horizontal and vertical velocities, what should be the horizontal range of the first ball after collision?
-------->>>>>>>> predicted horz range of 1st
Your answer (start in the next line):
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Before collision, the first ball was actually spinning fairly rapidly. When the balls collide, the coefficients of static and kinetic friction are those between two smooth steel surfaces. Can you make a reasonable estimate of the relative spinning rates of the two balls after collision? Would the fact that the first ball is spinning affect the path of the second ball, or just its spin?
-------->>>>>>>> estimate rel spin rates after collision, effect on path of 2d ball
Your answer (start in the next line):
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What is the slope of the line connecting the centers of the two balls? Hint: The rise is 2 mm. The run is very nearly equal to the length of the line segment between the centers. You will report this slope in the first line below.
Using this slope and the mean of the observed second-ball ranges, what does the program give you for the after-collision speed of the second ball? Report in the second line.
Starting in the third line report exactly what numbers you gave the program, and how you determined each of these numbers.
-------->>>>>>>> slope of segment connecting centers, after coll speed of 2d based on this slope and on ball range, numbers into program and how determined
Your answer (start in the next line):
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Your instructor is trying to gauge the typical time spent by students on these experiments. Please answer the following question as accurately as you can, understanding that your answer will be used only for the stated purpose and has no bearing on your grades:
Approximately how long did it take you to complete this experiment?
-------->>>>>>>>
Your answer (start in the next line):
2.5 hours
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Please copy your completed document into the box below and submit.
Author information goes here.
Copyright © 1999 [OrganizationName]. All rights reserved.
Revised: 06 Aug 2012 00:06:38 -0400
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You are encouraged but not required to fill out the form (I'll never know in any
case who has and has not filled it out). Your feedback will be very
valuable to me in the ongoing process of improving your course.
I will be including this message at the end of my responses on most days for the
next week, so you will probably see it more than once. However, it will
only be possible for you to fill out the form once. So you can ignore
redundant requests.
Click on this link to participate in the survey:
https://www.surveymonkey.com/s/Distance_Learning_Student_Survey