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'. **
2.00, 2.10
1.45
.025
** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
33.1, 32.8, 32.7, 32.5, 32.4
32.7, 0.2739
The position of 5 drops was first averaged to determine the 'start' point for measurement. The ball was started at the same position on the ramp for 5 trials, marking on carbon paper each landing position. Measurements were made of each landing position, recorded above, and analyzed to determine mean and standard deviation. All positions were measured from the center of the mark made by the ball on the landing surface.
** 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. **
61.0, 59.2, 58.5, 57.6, 58.8
31.6, 32.3, 31.9, 32.0, 31.8
59.02, 1.254
31.92, 0.2588
The straw was 1.3 cm from the end of the ramp. Each trial resulted in two marks on the landing surface, one for the little ball (#2), and one for the big ball (#1). Each mark was measured from the center to the 'drop' position, which is the zero reference for landing position.
** Vertical distance fallen, time required to fall. **
92.8 cm
0.425 seconds
Assumptions: the balls both fell from a distance equal to the distance from the landing surface to the top of the straw. Actually, the big one did not fall as far because its diameter is larger.
`ds = v0`dt + 0.5(a)(`dt^2)
`dt^2 = `ds / (0.5 a)
`dt^2 = 92.8 cm / (490 cm/s/s)
`dt^2 = 0.189 sec^2
`dt = 0.435 seconds
** 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. **
76.9, 75.1, 139
77.6, 76.3
75.7, 74.5
142, 136
** 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. **
76.9*m1 g*cm/s
75.1*m1 g*cm/s
139*m2 g*cm/s
76.9*m1 g*cm/s + 0
75.1*m1 g*cm/s + 139*m2 g*cm/s
76.9*m1 g*cm/s + 0 = 75.1*m1 g*cm/s + 139*m2 g*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. **
76.9*m1 g*cm/s + 0 = 75.1*m1 g*cm/s + 139*m2 g*cm/s
m1 g = m2(139) g*cm/s / 1.8 cm/s
m1 = 77 * m2 g
m1 / m2 = 77
The m1 (larger) ball has mass 77 times as large as the m2 (smaller) ball.
** Diameters of the 2 balls; volumes of both. **
2.54, 1.27
8.58 cm^3, 1.07 cm^3
** 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 second, the first ball will have a trajectory directed toward the floor. The speed will be greater than if centers were at the same height because less energy would transfer in the collision.
** 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 would be reduced, and the horizontal range of the second ball increased.
** 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: **
42.5
m1v01 + m2v02 = m1vf1 + m2vf2
m1(76.3) g*cm/s = m1(79.5) g*cm/s + m2(136) g*cm/s
m1 / m2 = 136 / 3.2
m1 / m2 = 42.5
** What percent uncertainty in mass ratio is suggested by this result? **
(77-42.5) / 77
45%
** 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 = min first before, min first after, max 2nd after.
minimum mass ratio = max first before, max first after, min 2nd after.
** 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? **
m1v1 + m2v2 = m1u1 + m2u2
m1(v1-u1) = m2(u2-v2)
m1 / m2 = (u2 - v2) / (v1 - u1)
** 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. **
Ball 1, 5 trials: 31.4, 31.7, 31.8, 31.8, 31.6
Ball 1 mean, sd: 31.66, 0.1673
Ball 2, 5 trials: 59.1, 56.9, 56.0, 55.7, 56.1
Ball 2 mean, sd: 56.76, 1.381
Mass ratio: m1 / m2 = 56
** 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? **
92.5, 56.76, 0.11
136.5
130, 137
136, 142
139, 136.5
In the second run the velocities appear to be lower than the first run, both in mean and interval.
** Your report comparing first-ball velocities from the two setups: **
ball 1 same height:
74.5, 75.1, 75.7
ball 1 higher than 2:
74.1, 74.5, 74.9
In the second run the velocities appear to be lower than in the first run, both in mean and in interval.
** Uncertainty in relative heights, in mm: **
0.5 mm because the dots for vertical checks were repeatable and were all within 0.5 mm of each other.
** 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. **
The uncertainty of relative heights of the balls was not significant because I had error that far outweighed the differences in range caused by as much as a 2mm change in setup. My calculated mass ratio was 49(best case) and should have been 8. If I had the rest of the measurements locked in, the relative heights in the first setup would have been significant.
** How long did it take you to complete this experiment? **
3.25 hours
** Optional additional comments and/or questions: **
Good work. Your analysis follows correctly from your data and you clearly understand everything that's going on.
The first ball will almost certainly have lost more momentum than indicated by your data. After colliding with the 'target' ball, if the collision is center-to-center and there is no interference from the tee, the first ball should lose more range than your data indicate. I suspect interference from the 'tee' gave the ball a slightly upward trajectory and resulted in the excess horizontal range; it is difficult with the setup used here to control for this type of interference.