#$&*
PHY 201
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 **
** **
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experiment has been revised since those videos were recorded, and any specific instructions given on the videos are superceded by the instructions given here, and by more recent practices in observing horizontal range.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):
2.1 cm
0.9 cm
91.1 cm
accuracy within 0.1 cm.
#$&*
Run your first set of trials:
Now you will remove the 'tee' and release the ball from the rest at the high end of the sloped track. You will use the same procedures as in previous experiments for observing the horizontal range of the ball as it falls to the floor.
Be sure the ramps remain well aligned, and if necessary 'shim' the end of the inclined ramp to ensure that there is no 'bump' when the ball moves from one ramp to the next.
Conduct 5 trials, and in the first line give 5 horizontal ranges; in the second line give the mean and standard deviation of the range of the ball. Starting in the third line explain in detail how you got your results.
-------->>>>>>>> 5 ranges uninterrupted, mean & sdev, explanation
Your answer (start in the next line):
20.3, 22.9, 21.6, 22.9,22.9
22.12, 1.163
I put a measuring stick on the floor and measured the horizontal distance from the table that the ball hit the ground. All distances are in cm.
#$&*
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):
27.9, 30.5, 30.5, 33.0, 30.5
15.2, 16.5, 16.5, 17.8, 16.5
30.48, 1.803
16.5, 0.9192
All measurements are in cm. Did five trials where i watched where the second ball hit on the floor while I got some help from my wife who watched where the first ball hit on the floor.
#$&*
Do not disassemble the system until you are sure you are done with it. General College Physics and University Physics students will use the system again in subsequent activities, and should leave it as it is.
Analysis of Results from First Setup:
Give in the first line below the vertical distance through which the two balls fell after collision, and in the second line the time required to fall this distance from rest. Starting in the third line, explain precisely how you determined these distances, how you determined the time of fall and what assumptions you made in determining the time of fall:
-------->>>>>>>> vertical fall, time to fall, explanation
Your answer (start in the next line):
91.1 cm
0.4 seconds
Measured the total distance between where the ball set on the tee down to the floor. Then dropped a ball from rest at this height to measure the total time to hit the floor.
#$&*
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):
60.85 cm/s, 21.65 cm/s, 152.4 cm/s
60.76, 0.8295, 22.12, 1.163
30.48, 1.803
16.5, 0.9192
#$&*
The masses of both balls are unknown. Using momentum conservation, you will determine the ratio of their masses:
Let m1 stand for the mass of the large ball and m2 the mass of the small ball. In terms of m1 and m2 write expressions for each of the following:
The momentum of the first ball immediately before collision, using the velocity you reported above (the velocity based on the mean range and distance of fall). Be sure to use both the numerical value of the velocity and its units. This will be reported in the first line below.
The momentum of the first ball immediately after collision, using the velocity you reported above. This will be reported in the second line below.
The momentum of the second ball immediately after collision, using the velocity you reported above. This will be reported in the third line below.
The total momentum of the two balls immediately before collision. This will be reported in the fourth line below.
The total momentum of the two balls immediately after collision. This will be reported in the fifth line below.
The total momentum immediately before collision is equal to the total momentum immediately after collision. Set the two expressions equal to obtain an equation. Report this equation in the sixth line below.
-------->>>>>>>> equation for momentum conservation
Your answer (start in the next line):
m1(60.85 cm/s)
m1(21.65 cm/s)
m2(152.4 cm/s)
m1(60.85 cm/s) + m2(0 cm/s)
m1(21.65 cm/s) + m2(152.4 cm/s)
m1(60.85 cm/s) + m2(0 cm/s) = m1(21.65 cm/s) + m2(152.4 cm/s)
#$&*
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):
m1(60.85 cm/s) - m1(21.65 cm/s) = m2(152.4 cm/s) - m2(0 cm/s)
m1 = (m2(152.4 cm/s)) / 39.2 cm/s = m2(3.89)
m1 / m2 = 3.89
The ratio m1 / m2 is the ratio of the mass of the first ball versus the mass of the second ball.
#$&*
Measure and report the diameter of ball 1 and the diameter of ball 2, in comma-delimited format in the first line below.
Calculate the volumes of the two balls and report them in the second line.
-------->>>>>>>> diameters, volumes
Your answer (start in the next line):
2.3, 2.2
6.37, 5.58
Volume = (4/3)*'pi*r^3. All measurements are in cm.
#$&*
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):
The direction would be straight but down into the direction of the tee. The velocity would be less than when hitting it in center of mass. It would affect the magnitude and direction of the velocity of the first ball immediately after collision, possible causing the first ball to move upward a little. The speeds would be greater if the centers are the same height. If the ball hits the other ball higher, the second ball will experience general plane motion at impact versus straight translation movement.
#$&*
How do you think this will affect the horizontal range of the first ball? How will it affect the horizontal range of the second?
-------->>>>>>>> effect on horizontal ranges
Your answer (start in the next line):
The horizontal range of the first ball may go a little further because it would get a little boost from the second ball because it will jump off the top part. The second ball would not go as far.
#$&*
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):
m1 / m2 = 5.03
#$&*
What percent uncertainty in mass ratio is suggested by comparing this result to your original result?
-------->>>>>>>> % uncertainty suggested by previous
Your answer (start in the next line):
22.7%
#$&*
#$&*
Your instructor is trying to gauge the typical time spent by students on these experiments. Please answer the following question as accurately as you can, understanding that your answer will be used only for the stated purpose and has no bearing on your grades:
Approximately how long did it take you to complete this experiment?
-------->>>>>>>>
Your answer (start in the next line):
2 hr
#$&*
*#&!
&#Your work on this lab exercise looks good. Let me know if you have any questions. &#