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.
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Distances from edge of the paper to the two marks made in adjusting the 'tee'.
1.7, 1.65
1.15
.05
Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process:
24.0, 24.8, 24.9, 25.1, 25.2
24.8, .4743
The ball was first dropped off the table from the position at the end of the ramp. Then the ball was placed at the top of the ramp set up and rolled down to hit the paper. The measurements were taken as the difference from the vertical drops and the drops down the ramp.
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.
34.1, 34.6, 34.7, 32.6, 34.9
19.2, 19.7, 20.0, 19.4, 19.5
34.18, .9311
19.56, .3049
The measurements were made with the starting point as the point in which the ball strikes if it is dropped vertically from the end of the ramp.
Vertical distance fallen, time required to fall.
104.8 cm
.46 sec
This distance was measured from the bottom of the ramp to the ground. Then `dt was found by using the equation `ds=v0`dt + .5a(`dt^2) where a =9.8, v0=0, and `ds=1.048.
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.
55.69, 43.72, 77.40
56.78, 54.59
45.83, 41.59
78.11, 76.69
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.
m1*55.69 cm/s
m1*43.72 cm/s
m2*77.40 cm/s
m1*55.69 cm/s + m2*0cm/s
m1*43.72 cm/s + m2*77.40 cm/s
m1*55.69 cm/s + m2*0cm/s = m1*43.72 cm/s + m2*77.40 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.
55.69cm/s (m1) - 43.72cm/s (m1)=77.4(m2)
m1=6.47 (m2)
m1/m2=6.47
m1/m2=6.47
The ratio represents the relationship between the two masses of the ball. This makes sense because ball 1 is heavier than ball 2. This has a direct correlation with the momentum of the system described earlier. The momentum before collision of the two masses is equal to the momentum after the collision of the two masses. The velocities obtained in the system after collision are dependent on the ratio of the mass of the balls.
Diameters of the 2 balls; volumes of both.
1.2, 1.6
7.24, 17.16
How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher?
The velocity in the horizontal direction will decrease, because the center of mass will be focused on the bottom of the heavier ball, and will push the ball in a more upward motion.
Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second:
It will be less. It will be more.
Note that the first ball will spend more time in its fall, due to its initial upward vertical velocity, which would tend to make it travel further, assuming equal horizontal velocities. If the horizontal velocity is less than in a direct collision, this could offset the effect of greater time.
The second ball will spend less time falling, but if it has a greater velocity than in the first case then this might offset the effect of the shorter time interval.
In a barely glancing collision the first ball will not lose much momentum (not much force in a glancing collision) so will travel further than if it had collided head-on, and if the upward velocity increases its time of fall it might possibly travel further than if the second ball was not there; whether or not this would actually be the case will be regarded for the moment as an open question.
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:
8.75=m1/m2
I found this by using m1b(v1b) + m2b(v2b)= m1a(v1a) + m2a(v2a)
What percent uncertainty in mass ratio is suggested by this result?
35%
What combination of before-and after-collision velocities gives you the maximum, and what combination gives you the minimum result for the mass ratio?
To obtain the maximum mass ratio, one must use the maximum velocity after collision for the second ball, the minimum for the velocity before collision(first ball), and the maximum velocity after collision of the first ball.
As for the minimum it is the opposite.
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?
u2/(v1-u1)
Derivative of expression for m1/m2 with respect to v1.
-u2/{(v1-u1)^2}
-.54 cm^-1/s^-1
the rate that the mass ratio changes
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?
1.09
-0.6213
This result was done by finding the derivative and multiplying by the standard deviation for v1. The derivative was found using the quotient rule. The derivative can find a relationship between the beginning speed of the ball which is dependent on the ramp slope and the mass of the two balls involved.
Complete summary and comparison with previous results, with second ball 2 mm lower than before.
The average distance recorded for the first ball was 14.1 +/- 1.4 cm and the average for the second ball (smaller ball) was 29.62 cm +/- 1.827 cm. Ball 1 before collision has a value of 55.69 cm/s and ball 2 has a value of 0 cm/s. The values for after collision are 36.15 cm/s and for ball 2 77.36. Therefore the prediction of mass ratio was m1/m2=3.96.
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?
79.5, 29.62, .7143
73.43
68.91, 77.96
76.69, 78.11
3.97
not significantly different, but there is an apparent difference
Your report comparing first-ball velocities from the two setups:
79.5, 14.1, .7143
34.98
31.51, 38.45
41.59, 45.83
8.74
in these results there is a more significant difference in data with the two different collision styles
Uncertainty in relative heights, in mm:
If we take the value of 34.18 horizontal average, and subtract the deviation then divide that quantity by the 34.18 we get about a 3% error. If we say that the total length of the straw is about 2 cm then 3% of this is about .06 cm so I would estimate .6 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.
The uncertainty I calculated should not be a big factor relative to the change with the 2mm experiment. If the 2mm experiment only yielded a slight difference, then .06 of a mm should not yield a big difference.
How long did it take you to complete this experiment?
8 hrs
Optional additional comments and/or questions:
Good work on this experiment. See my notes.
Note also that if the diameters are 1.6 and 1.2, then the ratio of radii is 4/3; the volumes would then be in proportion to the cube of this ratio, or (4/3)^2 = 64/27 = 2.4 or so. Discrepancies between this result and the mass ratios you observed in the experiment are probably due mostly to the difficulty of centering the two balls so they collide in a common horizontal plane.