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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 **
Submitted @ 9:44 PM on May 14, 2013..
I have not finished this lab but there are questions within the lab.. and if you could see how well I am doing with the lab and if everything makes sense so If not I could fix it or continue with my data and figuring out the velocities of the first and target ball..
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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experiment has been revised since those videos were recorded, and any specific instructions given on the videos are superceded by the instructions given here, and by more recent practices in observing horizontal range.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):
Collision pt 1 ball against vert
5.45 centimeters
Coll pt 2 balls
5.38 centimeters
Height of top of tee above tabletop
1.67 cm
Tabletop to floor
76.6 cm
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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):
18.51 cm
18.47 cm
18.65 cm
18.54 cm
18.42 cm
Mean
18.52 cm
Standard Deviation
0.08642
This is when the ball is rolled down the inclined ramp and there is nothing in front of it so it could hit the bottom of the floor..
Each measurement is how far the ball went from the beginning of the 30 cm inclined ramp....
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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):
Five ranges for Target ball
18.76 cm
18.56 cm
18.49 cm
18.69 cm
18.58 cm
Five ranges for First ball
13.69 cm
13.47 cm
13.58 cm
13.42 cm
13.87 cm
Mean and Standard Deviation for Target ball
Their mean is 93.08/ 5 = 18.62.
The standard deviation is therefore sqrt( 0.04652/( 5 - 1) ) = .1078.
Mean and Standard Deviation for First ball
Their mean is 68.03/ 5 = 13.61.
The standard deviation is therefore sqrt( .1305/( 5 - 1) ) = .1806.
What I did for this experiment was place the Target ball on the tee at the end of the ramp and then had the first ball roll down the incline in order for it to hit the Target ball..
Then I measured from the origin that I drew to the place where the Target ball and First ball landed on the floor...
I tried to have everything as accurate as possible in the measurements of where it landed and when I would place the Target ball on the tee...
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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):
The vertical distance that the two balls fell is
76.6 cm
The time it takes to reach the floor from rest is..
.568 seconds
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A ball will fall 76 cm from rest in less than .568 second.
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The time it takes to reach the floor from rest at the top of the inclined ramp (30 cm long) is..
1.958 seconds..
I determined the time it takes to reach the floor from rest by placing the ball where table top is and timing how long it takes for the ball to reach the floor and it took .568 seconds..
I also determined the time it takes to reach the floor from rest at the top of the inclined ramp then to the bottom of the floor to be 1.958 seconds.. but I am pretty sure we just need the time it takes it to reach the floor in the vertical direction, which is at rest where the table top is to the floor...
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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):
Velocity of first ball immediately before collision..
30 cm / 1.359 seconds = 22.075 cm/s
Velocity of the first ball immediately after collision..
Is the velocity for this one the distance it travelled? And the time that the ball took to land from the tabletop from rest to the floor?
Velocity of the second ball immediately after collision..
Is the velocity for this one the distance it travelled? And the time that the ball took to land from the tabletop from rest to the floor?
Before-collision velocities of the first ball based on mean and standard devition of its uninterrupted ranges....
18.62 cm /
Before-collision velocities of the first ball after collision....
Second ball after collision..
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You should have observed three quantities:
The horizontal distance traveled by the first ball after leaving the end of the ramp, when the second ball was not present, so that the first ball travels unimpeded down the ramp before falling unimpeded to the floor.
The horizontal distance traveled by the first ball after leaving the end of the ramp, when the second ball was present, so that the first ball collided with the second before falling to the floor. This distance will obviously be less than when the ball was unimpeded.
The horizontal distance traveled by the second ball after being struck by the first. This distance will be greater than either of the other two.
You will then calculate how fast each ball was traveling in each set of trials, by considering the horizontal distance traveled and the time of fall.
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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):
v
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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):
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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):
Diameter of Ball 1
2.34 cm
Volume of Ball 1
6.709 cm^3
Diameter of Ball 2
1.22 cm
Volume of Ball 2
.951 cm^3
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*#&!*#&!
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I'm not sure you have all the data, though some of it looks good.
Check my notes, then submit a revision (and/or additional questions if you have them).
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