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'. **
45.7cm
0.75cm
I first want to clarify that the 45.7cm is the horizontal distance that the ball travels before it collides with the other ball. The height of the tee is 0.75cm. There is a minimal uncertainty for the vertical height of the tee(the straw does not shrink). The horizontal distance, after running a few tests, does not vary by more than about 0.25cm. The track is on a ramp, and there is slight movement before and after collisions.) There is no bump, I have prevented that.
** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
90.9cm, 93.0cm, 90.4cm, 89.6cm, 93.9cm
91.56cm, 1.185cm
This is the horizontal range of every test when the ball ran. I included the initial distance of 45.7cm that the ball travels horizontally on the track(so the ball traveled 45.2cm after colliding with the other ball, but it already traveled 45.7cm before that in the first trial) In the second line is the mean and standard deviation of this test.
** 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. **
48.0cm, 47.0cm, 49.`cm, 50.`cm, 47.7cm
90.9cm, 93.0cm, 90.4cm, 89.6cm, 93.9cm
48.2cm, 1.304cm
91.56cm, 1.185cm
On the first line is the horizontal range the second ball(the one being hit) traveled. The third line is the mean and standard deviation of that horizontal range. The second and fourth lines are the range for the first ball. I want to note that the reason the horizontal range is much larger is because I included the horizontal range the ball traveled before colliding with the second ball. If I subtract every value by 45.7cm, then it would be how far it traveled after colliding. Both balls traveled very similar horizontal ranges after colliding if I note only the horizontal range after collision.
** Vertical distance fallen, time required to fall. **
73cm
1.3 This was done from the top of a tabletop. The vertical distance from the ground to the top of the table top is 73cm. On average it took about 1.3 seconds to hit the ground.
It doesn't take the ball 1.3 seconds to fall 73 cm. It takes significantly under half the second. You will probably need to refigure this and modify your subsequent results accordingly.
** 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. **
35.153cm/sec, 56.154cm/sec, 56.154cm/sec
36.122cm/sec, 34.887cm/sec
57.969cm/sec, 55.185cm/sec
57.123cm/sec, 55.185cm/sec
The first ball does not speed up as a result of the collision. However this is what your results appeared to indicate.
The 47.5 cm traveled by the ball along the ramps is not relevant to the analysis of the following ball. The horizontal accelerations are different for the projectile than for either of the two ramps. In this part of the experiment you need to analyze the projectile motion, which will reveal the end-of-ramp velocity of each ball on each trial.
** 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. **
m1v1 = 35.153cm/sec * m1
m1v2 = 56.154cm/sec* m1
56.154m2
m1* delta v(35.153 - 0cm/sec)=m2v2
m2*delta v
m1v1=m2v2
** 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. **
m1v1=m2v2
There is only one mass moving before collision, so there would be only one term in the before-collision expression for the momentum of the system. However after the collision there are two masses moving, and you therefore have nonzero contributions to the total momentum from both masses.
m1 = m2v2/v1
m1/m2 = v2/v1
v2 = 56.154cm/sec and v1 = 35.153cm/sec
so v2/v1 = 1.579cm/sec
The units of v2 / v1 are not cm/s.
So m1/m2 = 1.579
This means that the mass of the first ball is 1.579 times greater than the second ball. Hypothetically, if the second marble weighed 1kg, then the first would weigh 1.579cm/sec.
** Diameters of the 2 balls; volumes of both. **
2.2cm, 1.5cm
5.57cm^3, 1.77cm^3
** How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
The direction will differ because instead of the first ball hitting the second one directly, it hits it at an angle. This would cause a less drastic change in momentum between two points of mass. Momentum is conserved, but what is lost by one side is gained by the other. If less of the ball makes contact, then it will not lose as much momentum, so the other ball will gain less momentum. The direction(because it is still being hit from the same direction) will not change, it will still go forward. The ball originally not moving will not go as fast.
** Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second: **
The first ball will travel farther if it makes less of an impact with the second ball. As a result, the second ball will not travel as far.
** 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: **
Keeping everything proportional, The ratio of the masses is still 1.579.
This ratio was calculated several boxes above. When I add and subtract uncertainty, the ratio remains the same. They differ slightly by about 2%
** What percent uncertainty in mass ratio is suggested by this result? **
2%
** 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 minimum velocity gives the maximum mass ratio. The maximum velocity gives the minimum 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? **
m1v1 = m2v2
m1/m2 = v1/v2
so m1/m2 = v1/delta (u1 and u2)
** Derivative of expression for m1/m2 with respect to v1. **
First I need to assume that my answer previously was correct( v1/ delta u)
Using the quotient rule, the derivative would be delta u / v^2.
Taking the derivative of a ratio of masses and velocities can lead to a change in position.
** 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? **
v1 changes proportionally
The predicted mass ratio would also change proportionally
Every change here is proportional to another. Since we treated 'u's as constants, using the quotient rule was sufficient for above.
** Complete summary and comparison with previous results, with second ball 2 mm lower than before. **
My assumption was that there would less change in momentum. I mean that a gain from one would be smaller so a loss from another would be smaller(as compared to the first experiment) this held true to be valid. The mass ratio was about the same(1.62). There was an average change yes, but still close. The mass never changes, so the ratio should not change.
** 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? **
70cm, 45.354cm, .34
50.399cm/sec
49.987cm/sec-53.34cm/sec, 1.35cm/sec
56.154cm/sec, 1.54cm/sec
5.755cm/sec
There is a gap of 5.755cm/sec. This shows there is am obvious change when the tee is lowered.
** Your report comparing first-ball velocities from the two setups: **
When the tee was lowered, the difference of velocities was almost 6cm/sec slower. This validates my assumption stated earlier.
** Uncertainty in relative heights, in mm: **
0.05cm
I was careful to measure heights precisely. There will always be some uncertainty, however.
** 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. **
If there is more of a height difference, there will be more of a significant factor in velocities even though the momentum will be conserved.
** How long did it take you to complete this experiment? **
200 minutes
** Optional additional comments and/or questions: **
You data report is mostly clear, but it is unclear how you used your data in the calculation of the velocities of the two balls before and after collision. A couple of other points require clarification as well.
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