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'.

0.9 cm, 1.1 cm

1.0 cm

+-0.1 cm

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

24.0, 23.9, 23.5, 24.0, 24.2

23.92, 0.2588

To measure the horizontal range of the ball as it falls to the floor, I measure the distance from the edge of the table to the point where the ball strikes the paper on the floor in centimeters with a ruler. This is the horizontal distance that the ball moved during its fall. The ball started at the top of the 5cm inclined ramp and then rolled onto the horizontal ramp and off the table. We do not count the horizontal ramp as part of the horizontal range of the ball because it is not falling at that point.

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.

32.0, 32.5, 32.7, 33.2, 33.7 (I am assuming that the “second” ball is the target ball)

14.0, 14.7, 15.3, 15.5, 15.7

32.82, 0.6535

15.04, 0.6914

I laid two pieces of paper end to end at the edge of the table on the floor. When the ball hit the floor I numbered the marks they made in order to keep track of which horizontal range of the first ball corresponded to that of the second ball. The first ball was the large ball that was initially positioned at the top of the 5cm incline. The second ball was the target ball that was positioned on the tee at the edge of the ramp. To obtain the horizontal distances, I measured the distances in centimeters of the marks from the edge of the paper which was at the edge of the table. The target ball obviously traveled farther than the first ball.

Vertical distance fallen, time required to fall.

48cm

0.390625 seconds

The vertical distance is the height of the table above the floor. The ball flew off the table and fell 48 centimeters to the floor. I used the TIMER program to determine the time required to fall 48 cm from the edge of the table to the floor. In determining this time, I assumed that the ball dropped straight to the ground without moving horizontally.

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.

61.2352, 38.50, 84.02

60.57, 61.90

36.73, 40.27

82.35, 85.69

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.

61.2352cm/s * m1

38.50cm/s * m1

84.02cm/s * m2

61.2352cm/s * m1 + 0cm/s*m2

38.50cm/s * m1 + 84.02cm/s * m2

61.2352cm/s * m1 = 38.50cm/s * m1 + 84.02cm/s * m2

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.

22.7352cm/s * m1 = 84.02cm/s * m2

m1 = (84.02cm/s) / (22.7352cm/s) * m2

m1/m2 = 3.6956

Diameters of the 2 balls; volumes of both.

2.4cm, 1.8cm

7.2382 cm^3, 3.0536 cm^3

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

The magnitude of the velocity will decrease. However, the velocity will be greater at this position than if the centers were at the same height. The direction after collision for the first ball will be higher than it would be if the centers were at the same height. I think the speed would be greater than if the centers were at the same height because the path of the balls is distorted after collision.

The magnitude of the velocity for the second ball will be less than it would be if the balls were centered. The direction of the velocity will be lower because the larger ball is making contact with the ball at a higher point, sending it downward. The speed of the ball will be less than if the balls were centered.

QUESTION: I don’t really know how to be sure about this information. I was just trying to visualize the path of the ball. How can I accurately answer these questions?

This is something we might discuss more extensively. However for the present your answers and speculations are very good.

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 be greater. The horizontal range of the second ball is be less.

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 = 4.0567

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

[(4.0567 – 3.6956) / 3.6956 ] * 100 = 9.77%

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?

5 hours