conservation of momentum

Your work on conservation of momentum has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

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Distances from edge of the paper to the two marks made in adjusting the 'tee'.

Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process:

25.5, 25.6, 25.8, 25.9, 25.9

25.74, .1817

The horizontal range was measured from the point at which the ball lost contact with the ramp (the edge of the table) to the mark mad on the paper by the ball hitting the ground.

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.

32.9, 34.1, 32.5, 33.2, 33.2

23.7, 25.2, 21.5, 23.1, 23.2

33.18, .5891

23.34, 1.328

I made these measurements the same way as I did before. I used the same origin (the end of the table.)

Vertical distance fallen, time required to fall.

74.4cm

.359sec

I measured the distance from the edge of the table to the floor to determine the vertical fall distance. I dropped the bell from the edge of the table and used the timer program to find how long it took for the ball to hit the ground.

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.

71.7cm/sec, 65cm/sec, 92.4cm/sec

72.1cm/sec, 71.3cm/sec

68.8cm/sec, 61.3cm/sec

93.9cm/sec, 90.8cm/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.

m1 * 71.7cm/sec

m1 * 65cm/sec

m2 * 92.4cm/sec

(m1 * 71.7cm/sec) + (m2 * 0cm/sec)

(m1 * 65cm/sec) + (m2 * 92.4cm/sec)

(m1 * 71.7cm/sec) + (m2 * 0cm/sec) = (m1 * 65cm/sec) + (m2 * 92.4cm/sec)

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 * 71.7cm/sec) - (m1 * 65cm/sec) = (m2 * 92.4cm/sec)

m1 = (m2 * 92.4cm/sec) / 6.7cm/sec

m1/m2 = 92.4 / 6.7

m1/m2 = 13.8

This means that the mass of the first ball is 13.8 times as heavy as the second ball.

Diameters of the 2 balls; volumes of both.

2.7cm, 2cm

10.3cm^3

4.1cm^3

How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher?

The first ball would travel farther and faster. The speed of the first ball would be less if the centers are at the same height. It would affect the direction of the after collision velocity, the balls would travel at a different angle.

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 longer if the center of the first ball is higher than the center of the second at collision. Under the same circumstances, the second ball would have a shorter horizontal range.

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:

9.08

I plugged 71.3 in for v1 (before collision), 61.3 for v1 (after collision), and 90.8 for v2 and solved the equation as before.

What percent uncertainty in mass ratio is suggested by this result?

34%

What combination of before-and after-collision velocities gives you the maximum, and what combination gives you the minimum result for the mass ratio?

The maximum velocity for the second ball with the mimimum before collision velocity of the first with the maximum for the after collision velocity for the first ball will give me 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?

m1/m2 = u2 / (v1 - u1)

This equation would be solved in the manner.

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.

Mean and SD of the horizontal range of the first ball: 23.3, .5821

Mean and SD of the horizontal range of the second ball: 33.8, .1817

(m1*71.7cm/s) + (m2*0cm/s) = (m1*64.9cm/s) + (m2*94.2cm/sec)

m1/m2 = 13.9

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?

74.4, 38.3, .008

I could not get a velocity from the program because it asks for number of dominos rather than slope.

Your report comparing first-ball velocities from the two setups:

I do not know how to answer this question because of the data program.

Uncertainty in relative heights, in mm:

1mm

My data seems to be very close to the expected outcome

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 relative heights of the balls is significant if you make extremely accurate measurements, however, if you dont, it will not significantly change the results.

How long did it take you to complete this experiment?

4 hours