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.
** Your optional message or comment: **
** Distances from edge of the paper to the two marks made in adjusting the 'tee'. **
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: 0.35 cm
Height of the top of the 'tee' above the tabletop: 1.5 cm
Uncertainty: I tried to make the distance between where the ball and the collision of balls hit the paper minimal, which I believe to be fairly accurate, at a distance of 0.35 cm. I conducted this set of trial runs for the set up of the experiment a few different times to try to make it more accurate. The 1.5 height of the straw corresponds to this measurement, as the height of the straw on which the little ball is propped up helps to dictate whether the two marks are close enough or not.
** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
Horizontal ranges: trial 1 = 20.2 cm, trial 2 = 20 cm, trial 3 = 20.3 cm, trial 4 = 20.8 cm, trial 5 = 20.8 cm.
Their mean is 102.1/ 5 = 20.42 cm.
The standard deviation is = .3633 cm.
I got the measurements for the horizontal ranges listed here by conducting the experiment as the big ball rolled along the ramp and horzontal ramp and hit the carbon paper on the floor. I did this 5 times to get the results above. The measurements of the spots marked for each trial were found by meauring the mark's distance from the edge of the paper I had set up. This is taped down to make sure the measurements are consistent. I obtained the mean and standard deviation by plugging in the numbers into the data program. The mean is the average of all the trials and the standard deviation is how far the mark of the ball's distance tends to deviate from the mean.
** Five horizontal ranges observed for the second ball; corresponding first-ball ranges; mean and standard deviation of the second-ball ranges; mean and standard deviation of ranges for the first ball. **
Five horizontal ranges for the big ball:
trial 1 = 14.3 cm, trial 2 = 15.2 cm, trial 3 = 14.8 cm, trial 4 = 14.9 cm, trial 5 = 13.34 cm.
FIve horixontal ranges for the small ball (or target): trial 1 = 24.7 cm, trial 2 = 25 cm, trial 3 = 24 cm, trial 4 = 23.6 cm, trial 5 = 25.6 cm.
Big ball mean and standard deviation:
Their mean is 72.54/ 5 = 14.51 cm.
The standard deviation is = .7289 cm.
Small ball (target) mean and standard deviation:
Their mean is 122.9/ 5 = 24.58 cm.
The standard deviation is = .7949 cm.
I found these measurements the same way I did in the last set of five trials. By placing the target ball on the straw and having the big ball come down the ramp, causing a head-on collision with the target ball, I obtained two marks, one from each ball. I measured these marks by using the same reference as the edge of the paper I used for the other set of five earlier in the experiment. I got the mean and standard deviation from the data program.
** Vertical distance fallen, time required to fall. **
The vertical distance through which the two balls fell = 70.9 cm
Time required to fall from rest for big ball = .2135 sec
Time required to fall from rest for small ball = .2813 sec
I found the vertical distance through which the two balls fell by measuring the top of the table the balls fell from to the bottom of the floor where they hit. I found the time for each ball to fall from rest by placing each on the edge of the table and letting them fall from rest. My mom helped me with this so that I could time each of the balls to get an accurate read. Thus, I found the time intervals for each. I took 3 free falls for each and then got the average for each ball's time to drop using the data program. I assumed that my timing may have been subject to error (because of human error); thus, I used three trials for each to try to alleviate some of that variability.
You can't accurately time interval this short--nobody can, our neural equipment isn't up to the task.
The time of fall is the same for both balls, assuming both start with zero vertical velocity and accelerate downward at 980 cm/s^2. Using v0 = 0, a = 980 cm/s^2, `ds = 70 cm you find that the time will be a little under .4 second.
Big Ball
.1875
.21875
.234375
Average = .2135 sec
Small Ball
.265625
.3125
.265625
Average = .2813 sec
** Velocity of the first ball immediately before collision, the velocity of the first ball after collision and the velocity of the second ball after collision; before-collision velocities of the first ball based on (mean + standard deviation) and (mean - standard deviation) of its uninterrupted ranges; same for the first ball after collision; same for the second ball after collision. **
In order to calculate the velocity before collision, I need to measure the distances of the sloped and level tracks and the time to traverse each.
Time for Big Ball Before Collision on ramp (just sloped ramp)
.984375
.96875
.984375
Average = .9792 sec
Standard deviation = .009
Time for Big Ball on non sloped ramp & sloped ramp
1.21875
1.21875
1.234375
Average = 1.224 sec
Big ball nonsloped ramp = 1.224 sec - .9792 sec = .2448 sec
Velocity on sloped ramp = 54.23 cm / .9792 sec = 55.38 cm/sec
Velocity on nonsloped ramp = 14 cm / .2448 sec = 57.19 cm/sec
Total velocity of the big ball before collision = 55.38 cm/sec + 57.19 cm/sec = 112.57 cm/sec; Average = 56.29 cm/sec
The velocity of the first ball immediately before collision = 56.29 cm/sec
The velocity of the first ball immediately after collision = 67.96 cm/sec
The velocity of the second ball immediately after collision = 87.38 cm/sec
I got each of these velocities by taking the change in horizontal distance over the change in time. For the before collision of the big ball, I split the two ramps into the sloped ramp and horizontal ramp and found their average velocities separately. I then found the average of these two to use as the total average velocity for the big ball before the collision. The other two velocities (after the collision) were determined by measuring the horizontal distance from the collision to the point that landed on the ground divided by the time from collision to the time it hit the ground.
The big ball before collision's range was calculated by doing upper time = 1.224 sec + .009 sec = 1.233 sec, lower time = 1.224 sec - .009 sec = 1.215 sec. The velocity upper then is calculated by 68.23 cm / 1.233 sec = 55.34 cm/sec (the distance is the total traveled from both ramps, both sloped and nonsloped, as calculated earlier). The velocity lower = 68.23 cm / 1.215 sec = 56.16 cm/sec.
Big ball after collision:
To find the upper bound of the big ball, we use the mean + st dev of the time = .2135 sec + .024 = .2375 sec. The lower bound is mean - st dev = .2135 sec - .024 sec = .1895 sec. The small ball upper bound (same calculations) is .3083 sec and lower bound is .2543 sec. After getting the upper and lower, we need to divide each by its respective upper and lower times. This gives us an upper of 14.51 cm / .1895 sec = 76.57 cm/sec. and a lower of 14.51 cm/ .2375 sec = 61.09 cm/sec.
Small Ball after collision:
For the small ball, the upper is 24.58 cm / .3083 sec = 79.73 cm/sec and the lower is 24.58 cm / .2543 sec = 96.66 cm/sec.
** First ball momentum before collision; after collision; second ball after collision; total momentum before; total momentum after; momentum conservation equation. All in terms of m1 and m2. **
Momentum of ball 1 (big ball) before collision: p = m1v1 = m1 * 56.29 cm/s
Momentum of ball 1 (big ball) after collision: p = m1v2 = m1 * 67.96 cm/s
The first ball isn't going faster after the collision. The second ball does slow it. The uninterruped large ball goes further than it does when it collides with the second ball.
Momentum of second ball (small ball) after collision: p = m2v2 = m2 * 87.38 cm/s
Total momentum of the 2 balls before collision: p = m1v1 + m2v1 = m1 * 56.29 cm/s + m2 * 0 cm/s
Total momentum after the two balls collide: p = m1v2 + m2v2 = m1 * 67.96 cm/s + m2 * 87.38 cm/s
Law of conservation of momentum: m1v1 + m2v1 = m1v2 + m2v2 so that m1 * 56.29 cm/s + m2 * 0 cm/s = m1 * 67.96 cm/s + m2 * 87.38 cm/s
** Equation with all terms containing m1 on the left-hand side and all terms containing m2 on the right; equation rearranged so m1 appears by itself on the left-hand side; preceding the equation divided by m2; simplified equation for m1 / m2. **
m1 * 56.29 cm/sec - m1 * 67.96 cm/sec = m2 * 87.38 cm/sec
m1 = (m2 * 87.38 cm/sec) / (56.29 cmm/sec - 67.96 cm/sec)
By dividing both sides by m2, we effectively get the equation asked for in the 4th line: m1 / m2 = 87.38 cm/sec / (56.29 cm/sec - 67.96 cm/sec) = 87.38 cm/sec / -11.61 cm/sec = -7.5, so m1 / m2 = -7.5
The meaning of the ratio m1 / m2 = -7.5 says that, when simplified, the ratio of the first mass to the second mass is -7.5.
** Diameters of the 2 balls; volumes of both. **
Big ball diameter = 2.5 cm, Small ball diameter = 1.25 cm
Big ball volume = 8.18 cm^3, Small ball volume = 1.02 cm^3
To get the volume of a sphere, we use (4/3) pi * r^3 (where pi = about 3.14).
For each, we must first find r, which is 1/2 of the diameter, so big ball will be r = 1.25cm and small ball r= .625 cm.
Then, we plug in the r values to get big ball volume = (4/3) pi * (1.25 cm)^3 = 8.18 cm^3 and small ball volume of (4/3) * pi * (.625 cm)^3 = 1.02 cm^3.
** How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
For the first ball (big ball): If the center of the first ball is higher than the center of the second, then the magnitude and velocity should be greater than if the centers were aligned. This is probably true because it is not a head-on collision in this case. I actually observed this happen when adjusting the set up of the experiment. The speed would be synonymous with the velocity, so it should also increase moreso than it would be if the centers were at the same height. The direction will also differ, as the angle from which the small ball takes off should be lesser and the angle from which the big ball leaves the table should be greater.
The small ball's speed/velocity will be increased as a result of the big ball not being aligned with its center. The direction will therefore have a smaller angle and the magnitude will be greater.
** Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second: **
The horizontal range of the first ball will then be greater, whereas the horizontal range of the second, or small, ball will be lesser. This is due to where the balls will land on the carbon paper on the ground as a result of the misalignment of the balls' centers.
** ratio of masses using minimum before-collision velocity for the first ball, maximum after-collision velocity for the first ball, minimum after-collision velocity of the second: **
Ratio of masses if I used the minimum before-collision velocity in the time interval for the first ball: m1 / m2 is found by replugging in the v1 for the equation to get:
m1 * 56.16 cm/sec = m1 * 67.96 cm/sec + m2 * 87.38 cm/sec and then simplifying to get m1/m2 = 87.38 cm/sec / (56.16 cm/sec - 67.96 cmm/sec) = -7.41
Ratio of the masses if I used the maximum after-collision velocity for the first ball: We use the same method above to restructure the equation we used before by plugging in a new v2 for m1, to get:
m1 * 56.29 cm/sec = m1 * 76.57 cm/sec + m2 * 87.38 cm/sec and getting an equation of m1/m2 = 87.38 cm/sec / (56.29 cm/sec - 76.57 cm/sec) = -4.31
Ratio of the masses if I used the minimum after-collision velocity of the second ball: To get this m1/m2 ratio, we use the same method as the other two answers above. Thus, we use the equation:
m1 * 56.29 cm/s = m1 * 67.96 cm/sec + m2 * 96.66 cm/sec and restructure to get m1/m2 = 96.66 cm/sec / (56.29 cm/sec - 67.96 cm/sec) = -8.28
** What percent uncertainty in mass ratio is suggested by this result? **
My original result of the mass ration of m1 to m2 was -7.5. When I found these other three answers:
1st ball before collision min used = -7.41
1st ball after collision max used = -4.31
2nd ball after collision min used = -8.28
I found that the respective percent uncertainty for each should be:
1st ball before collision min used = (-7.41-(-7.5)) / -7.5 * 100 = 1.2%
1st ball after collision max used = (-4.31-(-7.5)) / -7.5 * 100 = 42.53%
2nd ball after collision min used = (-8.28-(-7.5) / -7.5 * 100 = 10.4%
** What combination of before-and after-collision velocities gives you the maximum, and what combination gives you the minimum result for the mass ratio? **
** 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? **
** Derivative of expression for m1/m2 with respect to v1. **
** 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? If v1 changes by this amount, then by how much would the predicted mass ratio change? **
** Complete summary and comparison with previous results, with second ball 2 mm lower than before. **
** Vertical drop of the second ball, its mean horizontal range and the slope of the line segment connecting the two centers; the velocity given by the program based on mean; velocity interval for 2-mm decrease in 2d-ball height; velocity interval from the original run at equal heights; difference in the mean-based velocities; is new velocity significantly different than original? **
** Your report comparing first-ball velocities from the two setups: **
** Uncertainty in relative heights, in mm: **
** 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 a significant factor in the first setup. **
** How long did it take you to complete this experiment? **
This experiment took me about 15 hours to do, over a period of 3 days.
** Optional additional comments and/or questions: **
Your data were good, as was most of your analysis. There were a few errors in the analysis--see my notes.