Phy 231
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'. **
I'm not sure what exactly they want me to measure, but the ball's strike at the same height at the edge of the table.
1 cm is the height of the collision
** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
19.5, 19, 19.6, 19.2, 19.8
19.42,.3194
The horizontal distances were measured by measuring the distance between the point of impact and the edge of the table.
** 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. **
15.5, 16.8, 17.5, 17.5, 17.4
26.4, 23.5, 24, 22.4, 21
23.46, 2.007
16.94, .8562
** Vertical distance fallen, time required to fall. **
71 cm
0.38 s
The vertical distance was found by measuring the height of the table. The time was found by using the equation D = v0*t + 1/2 at^2. D,a, and v0 were known, so solving for t you get .38.
** 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. **
70.42, 56.82, 32.5
69.56, 71.28
55.96, 57.68
30.5, 34.5
I found these velocities assuming that the collisions lasted 0.1 seconds. Using this and the acceleration found by using the kinematic equations, I added/subtracted the dv's to the initial velocites for each ball.
** 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*70.42
m1*56.82
m2*32.5
m1*70.42 + m2*0
m1*56.82 + m2*32.5
m1*70.42 = m1*56.82 + m2*32.5
** 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*70.42 - m1*56.82 = m2*32.5
m1 = (m2*32.5)/(70.42 - 56.82)
m1/m2 = 32.5/(70.42 - 56.82)
m1/m2 = 2.39
This means that m1 is 2.39 times larger than m2.
** Diameters of the 2 balls; volumes of both. **
2.5, 1
8.18, 0.52
** How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
If struck above the center, the force of the velocity vector will be angled towards the ground. The speed will be less, because some of the velocity will be vertical, instead of all the velocity being horizontal. Also, this will introduce more friction into the system. If the ball was struck exactly center, then the whole of the velocity will be horizontal, and so it will have faster horizontal speed.
** Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second: **
Since this will reduce the horizontal speed of the second ball, its horizontal range will then be reduced as well. According to the conservation of momentum, that would mean the second ball has a less after-collision momentum, so that must mean the first ball's momentum is thus increased. This means that its velocity increases which means that it will then have a greater 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: **
m1/m2 = 2.57
Using the original equation found, I substituted the new (max/min) values for the old values.
** What percent uncertainty in mass ratio is suggested by this result? **
7.5%
** What combination of before-and after-collision velocities gives you the maximum, and what combination gives you the minimum result for the mass ratio? **
min/ max-min would give me the smallest ratio. This means the min after v of the second ball, the max before v of the first and the min after v of the first.
max/min-max would give me the biggest ratio. This means the max after v of the second, the min before v of the first, and the max after v of the first.
** 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
-.1757
I think the units are cancelled out since its velocity/velocity. I'm not sure as to what the derivative of the ratio indicates.
** 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? **
If the range changes by a std dev, then the new velocity would change by std dev/time. since v = d/t. The new v would be v = d-std dev/t. I'm not sure how the number I found above correlates to this.
** Complete summary and comparison with previous results, with second ball 2 mm lower than before. **
The avg distance of the second ball is now 38 cm. The avg distance for the first ball is about 24 cm. The velocity of the first ball before collision is 91 cm/s and 76.35 cm/s after collision. The velocity of the second ball after collision is 52.6 cm/s.
Since m1/m2 = u2/v1-u1, I found the mass ratio to be 3.59.
** 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? **
71, 38, .038
109.1
106.5, 111.7
61.15, 65.15
36
The velocity looks significantly different in the 2mm increase.
** Your report comparing first-ball velocities from the two setups: **
The first ball also increased velocity from 45 to about 63 cm/s. I assume this is because the ramp was raised 2mm, I don't know how much the raising of the second ball had much to do with this.
** Uncertainty in relative heights, in mm: **
I didn't understand this measurement at the beginning of the experiment so I did not take it.
** 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. **
Since the uncertainty was about the same for the 0 to the 2 mm setup, I don't think it had much affect on this experiment. You could still see the difference it data from the different heights of the second ball.
** How long did it take you to complete this experiment? **
The angle of the slope was very low at about 2 degrees. Since the mangitude of the after-collision velocity was found to be 52.6 cm/s, the vertical component of this would be 52.6*sin2 = 1.84 cm/s.
** Optional additional comments and/or questions: **
If the before collisions v-momentum was 0, then the after collision momentum must also be 0. That would mean I have to offset the 1.84 cm/s, which would mean that I would have a negative number? I'm not sure how this would work, how a negative momentum would work.
** **
The hv of the first ball before collision is 70.42 cm/s. The hv of the second ballafter is 32.5 cm/s. Based on the ratio, the hv of the first ball after would be about 52 cm/s.
** **
Since the velocity found was the velocity directly AFTER the collision and not the average velocity that it would attain, I don't know if the velocity found * time would give me an accurate distance of the first ball. It would be 52*.38 = 19.76.
** **
I don't think the spinning o fthe first ball would affect the path of the second ball. I'm not sure of the physics behind this, but I assumed this from watching billiards. The players would put various spins on the cue ball, but it wouldn't affect the path of the ball being struck. I think this is because the impact only lasts for a split second and there is not enough spin velocity to cause a trasnfer of the spin.
** **
0.095
98.84
I gave the program a slope of .095, a height of 71, and a range of 38.
** **
3 hours
** **
You report that the velocity of the second ball is less than the after-collision velocity of the first. Unless the first ball somehow passes through the second, it doesn't seem possible that the first ball would travel further. If the second ball travels further than the first, then its after-collision velocity would be greater.
The ball velocities are determined by their horizontal ranges and their time of fall. Your .39 s time of fall looks right for the distance of the fall. With that time of fall, a horizontal range of, say, 20 cm would imply a velocity of about 50 cm/s.
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