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'. **
3.01, 2.98
2.50
.32
** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
28.74, 28.74, 28.68, 28.68, 28.45
28.66, .1201
First, I fixed the paper to the floor and marked its position, then I measured the distance from the edge of the paper to the edge of the table. Once I ran the trials, I marked the center of each of the marks left by the ball. I measured the distance from the mark to the edge of the paper. The sum of the distances from the mark and the table to the edge of the paper is the range of the ball.
** 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. **
39.36, 41.33, 41.43, 41.92, 41.75
15.72, 15.94, 16.10, 16.33, 15.84
41.16, 1.033
15.99, .2374
For each trial, I marked the center of each of the marks left by the ball and numbered them. Once I had done all five trials, I measured the distance from each mark to the edge of the paper. The sum of the distances from the mark and the table to the edge of the paper is the range of the ball.
** Vertical distance fallen, time required to fall. **
106.5 cm
.11 sec
The distance of fall is the sum of the table's height above the floor and the tee's height above the table. I am assuming that the force of air resistance is negligible, as is the difference between the heights of the bottoms of the balls.
** 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. **
2.60, 1.45, 3.74
2.62, 2.59
1.54, 1.35
3.84, 3.65
** 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. **
m1 * 2.6 m/sec
m1 * 1.45 m/sec
m2 * 3.74 m/sec
m1 * 2.6 m/sec
(m1 * 1.45) + (m2 * 3.74)
m1 * 2.6 = (m1 * 1.45) + (m2 * 3.74)
** 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(2.6 m/sec - 1.45 m/sec) = m2 * 3.74 m/sec
m1 = 3.25 m2
m1 / m2 = 3.25
The for each unit of mass that m2 contains, m1 contains 3.25 units of mass. If we had a mass value for either ball, we could find the mass of the other ball.
** Diameters of the 2 balls; volumes of both. **
2.4, 1.75
7.23, 2.81
** How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
If the center of the first ball is higher than the center of the second at collision, the horizontal component of the force the small exerts on the large one will be smaller than if the centers were level. The vertical component of this force only exists when the centers are not level.
The magnitude of the velocities of the balls will be the same, but the directions will change.
The velocity vector of the large ball now points slightly upward, and its horizontal component is smaller.
The velocity vector of the small ball now point downwards instead of being horizontal. Its horizontal component is also smaller.
This is assuming that at the instant just after collision, the vertical velocity of the balls is negligible if their centers collide.
** Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second: **
The smaller ball will have an initial downward velocity, so its range will be shorter.
The larger ball will have a small initial upward velocity, so its range will be longer.
The difference in the range of the smaller ball will probably be greater than the difference in the range of the larger ball.
** 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: **
m1 / m2 = 3.65
I substituted the given values into the equation for momentum: m1*(2.59-1.54) = m2 (3.84) and solved for the ratio as before.
** What percent uncertainty in mass ratio is suggested by this result? **
7 %
** What combination of before-and after-collision velocities gives you the maximum, and what combination gives you the minimum result for the mass ratio? **
Maximum mass ratio is 3.65:
m1 before = 2.59, m1 after = 1.54, m2 = 3.84
Minimum mass ratio is 2.87:
m1 before = 2.62, m1 after = 1.35, m2 = 3.65
** 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? **
m1 / m2 = (u2) / (v1 - u1)
** Derivative of expression for m1/m2 with respect to v1. **
-(u2) / (v1 - u1)^2
3.49/ m/sec
This is the change in mass ratio with respect to initial velocity. This would also be the error in mass ratio per m/sec error in calculated 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? **
.02 m/sec
-.075
I multiplied the difference in velocity by the rate of change in mass ratio with respect to velocity. This would be the error in mass ratio that corresponds to the uncertainty, based on standard deviation of the ranges I measured.
** Complete summary and comparison with previous results, with second ball 2 mm lower than before. **
The range of the first ball has a mean of 19.94 cm +/- .3720
This corresponds to a velocity of 1.81 m/sec with a maximum of 1.85 m/sec and a minimum of 1.78 m/sec.
The range of the target ball has a mean of 39.24 cm +/- 3.429
This corresponds to a velocity of 3.57 m/sec with a maximum of 3.88 m/sec and a minimum of 3.26 m/sec.
This correlates to a mass ratio of 4.41 / m/sec, a 26% increase in mass ratio. The masses of the balls did not change and the initial speed of the first ball did not change, but the velocity of the first ball relative to the target ball did change when the position of the target ball changed.
** 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? **
108.98, 39.24, .0097
82.79 cm/sec
89.99, 75.59
383.6, 364.8
291.4 cm/sec
There is a significant difference between the velocities of the level and lowered target balls.
** Your report comparing first-ball velocities from the two setups: **
109.0, 19.91, 0
42.08
43.99, 41.3
147.5, 143.2
103.3 cm/sec
There is a significant difference between the velocities of the ball when it strikes the target's center and when it strikes above the target's center.
** Uncertainty in relative heights, in mm: **
.32 mm
I used the singly reduced paper ruler. I estimated .5 mm and the conversion factor is 1.543 for that ruler.
** 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. **
Uncertainty in relative heights of the balls was a significant factor in the first setup. Changing the center of the smaller ball by 2mm resulted in a threefold increase in its calculated velocity. My uncertainty is 16 percent of that change. That is probably enough to introduce a significant error.
** How long did it take you to complete this experiment? **
6 hours
** Optional additional comments and/or questions: **
I had a hard time connecting some of the ideas in this lab, it was very helpful to work through them.
Excellent work. This lab does tie together a number of ideas; let me know if you have any questions.