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'. **
1.8cm
1.3cm
The uncertainty would be about +-.05cm since the smallest markings on the ruler are .1cm.
** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
20.1 cm, 19.8 cm, 20.2 cm, 19.6 cm, 20.6 cm
20.06 cm +- .3847
These are the horizontal ranges of the larger ball if you remove the tee. It hit the carbon paper taped to a string of 8x11 pieces of computer paper taped to the floor. The mean and sd were obtained from the data program.
** 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. **
13.9 cm, 14.9 cm, 14.5 cm, 14.3 cm, 14.2 cm
23.9 cm, 24.6 cm, 23.9 cm, 24.7 cm, 25.1 cm
14.36 +- .3715
24.44 +- .5273
I just set it up as per instructions, with the head on collision of the smaller ball and the larger ball, and got their positions from their point of impact on the carbon paper taped to the computer paper taped to the floor. Since the computer paper is taped to the floor and the table isn't going anywhere, the position of the paper is controlled.
** Vertical distance fallen, time required to fall. **
68cm
.373s
I just used ds = v0(dt) + .5(a)(dt^2). Since v0 = 0, a = 9.8m/s^2, ds = .68m, .68m = (.5)(9.8m/s^2)(dt^2), and t=~ .373s
** 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. **
0cm/s, 65.5cm/s, 38.5cm/s
52.75cm/s, 54.81 cm/s (second_before), 53.78 cm/s (average of those)
64.1cm/s, 66.94 cm/s (first_after)
37.5cm/s, 39.5cm/s (second_after)
For the horizontal velocity of the first ball before and after collision, and the horizontal velocity of the second ball after collision, I just divided the horizontal ds by dt. This is supposed to be the total initial velocity after collision since it's (hopefully) setup to be horizontal, and horizontal velocity is constant.
For the before-collision velocity of the first(second?) ball, I used the mean +- sd, which was 20.06 cm +- .3847, so (19.6753cm/.373s =~ 52.75 cm/s), and (20.4447m/.373s) =~ 54.81 cm/s. I'm a bit confused by the phrasing of the question, though, since the second ball would be the only one with an uninterrupted range, not the first. Or maybe (probably) I've mixed them up.
Line 3 and 4 uses the same calculation for the predicted velocity of the first ball after collision and the second ball after collision using, respectively horizontal ds's of 24.44 +- .5273 cm and 14.36 +- .3715 cm.
** 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. **
0cm/s
m1(65.5cm/s)
m2(38.5cm/s)
m2(53.78cm/s) + 0
m1(65cm/s) + m2(38.5cm/s)
First, I set up the m1v1_before + m2v2_before = m1v1_after + m2v2_after equation with m1 as the first (small) ball and m2 as the second (big) ball, so the bigger ball should have a greater mass, but let's check. So m1(0cm/s) + m2(53.78cm/s) = m1(65.5cm/s) + m2(38.5cm/s), and then you rearrange for whatever the question is asking, or just pick out whatever term the question wants.
** 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(65.5cm/s) = m2(53.78cm/s) - m2(38.5cm/s)
m1 = m2(.2333)
m1/m2 = .2333
m1/m2 = .2333
It's definitely supposed to be a fraction since the big ball (m2) is supposed to have a greater mass, and this ratio implies that the smaller ball is about 23% the mass of the bigger ball. Even if that's incorrect, at least it came out smaller and the general principle worked. I'll take what I can get.
** Diameters of the 2 balls; volumes of both. **
1.6cm, 2.5cm
2.14cm^3, 8.18cm^3
V = (4/3)pi(r^3), so I just plugged in half my diameter. Actually, it works out pretty well, since 23.3% of 8.18cm^3 is about 1.91cm^3, which is in the general vicinity of 2.14cm^3. Assuming the density is the same (and it appears to be the same material), then that means my m1/m2 ratio is fairly decent.
Your results worked out very well.
** How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
I'm not entirely sure about this one, but it seems like if the big ball is higher than the smaller ball, then it will increase the speed of the smaller ball, because the force of impact will have a component in the same direction as gravity, and it would be opposite for the big ball since the big ball will have a component opposing gravity. I'm imagining the point of impact as the origin of a coordinate plane (with the x-axis parallel to the table), and the smaller ball would have a force acting on it toward the third quadrant and the bigger ball would have a force acting on it toward the first quadrant. So these forces would change the respective momentum (impulse). The direction for the smaller ball would definitely be angled downward more sharply (say, from 180 deg to 240 or something like that depending on how much higher the bigger ball is). Then, for the bigger ball, obviously it's net force is still to the left (or away from the ramp to the floor), but the force directed into the first quadrant would combine with that horizontal force (so the direction might change from 180 deg to, say, 160 deg or something). That's how I'm imagining it, anyway. I could have totally made that up, though.
That's very good. The small ball will get to the floor more quickly. The big ball will have an initial upward velocity so will get to the floor more slowly. Horizontal velocities will be affected slightly, but the main effect will be related to the times of fall.
** 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 small ball will be less, I think, since there's a y component acting along with gravity (greater acceleration downward), which means there would be less time for the horizontal component to act. Then, the big ball would be the opposite. That force toward the first quadrant would have a component opposite to gravity, which would cut into gravity's acceleration, which would make the dt longer, which would give the horizontal velocity longer to act.
** 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: **
.198, .27
So if we use the minimum before-collision and the maximum after-collision, we get m1(0cm/s) + m2(52.75cm/s) = m1(66.94cm/s) + m2(39.5cm/s), and the ratio will be m1/m2 = .198, and max before collision, min after, would be m1(0cm/s) + m2(54.81cm/s) = m1(64.1cm/s) + m2(37.5cm/s), and we get m1/m2 = .270. So if we look at the volume ratio (which is about .261), which should correspond to the mass ratio assuming identical density, it's within this range. Score.
** What percent uncertainty in mass ratio is suggested by this result? **
15.1%, 15.7%
So the original result is .2333, then .0353/.2333 = 15.1%, and .0367/.2333 = 15.7%, so around 15 or 16 percent error. The .0353 and .0367 are just the absolute values of the difference in original result and max-min results.
** What combination of before-and after-collision velocities gives you the maximum, and what combination gives you the minimum result for the mass ratio? **
Well, as we just saw, using the maximum after-collision and minimum before-collision would give you the minimum m1/m2 result, and using the minimum after collision and maximum before-collision would give you the maximum m1/m2 result.
** 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-v2)/(-u1)
If m1v1 + m2v2 = m1u1 + m2u2, then m1(v1-u1) = m2(u2 - v2), and m1/m2 = (u2-v2)/(v1-u1). I guess, though, if we're just talking about this particular example of a collision, then v1 will always be 0cm/s since the smaller ball is stationary prior to collision. Not sure if we're supposed to keep it all in symbols or not, but in that case, it'd be m1/m2 = (u2-v2)/-u1). My m1 is the target ball, though, and my m2 is the big ball.
** Derivative of expression for m1/m2 with respect to v1. **
So if the expression is (u2-v2)/(-u1), and u1 and u2 are constants, then f(v) = -u2/u1 + (1/u1)(v2), and the derivative is 1/u1 with u1 being the velocity of the small ball after collision. I guess it would be the change in the rate with respect to v1. So it would help us predict how much the rate changes with errors in v1. However, that makes my rate constant, which is wrong, and my v1 was always stationary (0cm/s), anyway, so I'll try it assuming ball 1 is the bigger and ball 2 is the smaller, so in that case the expression would be -u2/(u1-v1), or -u2(u1-v1)^-1, and the derivative would be u2(u1-v1)^-2. So we'll try that. (Since the exponent on v1 is 1, I don't believe I have to do anything else for the chain rule, but I could be forgetting something here). Or if you don't take out the 0cm/s v2, then the expression would be m1/m2 = (v2 - u2)/(u1 - v1), in which case the derivative would be ((u1-v1)(1) - (v2-u2)(-1))/(u1-v1)^2 = 1/(u1-v1) + (v2-u2)/(u1-v1)^2 or it would be = (u1 - v1 - u2)/(u1 - v1)^2.
** 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? **
1.03cm/s
So, if v1 was predicted to be 53.78 cm/s, and (as we predicted before), if the horizontal range changes by the sd, then v1 changes by +- 1.03 cm/s. Since I had ball 1 and ball 2 switched around, then u2 would now be the velocity_after of the small ball, and u1 would be velocity_after of the large ball, and v1 would be the velocity_before of the large ball, but those are all constant except for v1, anyway. So, we've got u2(u1+-1.03m/s)^-2 = u2/(u1^2 +- 1.03u1 +- 1.03u1 +1.06) = u2/(u1^2 +-1.06ul + 1.06). I have a suspicious feeling I've done this wrong.
You've done most of this right. See the discussion in the link given at the end of this document.
** Complete summary and comparison with previous results, with second ball 2 mm lower than before. **
13.8 cm, 14.1 cm, 13.7 cm, 14.4 cm, 14.0 cm
24.6 cm, 25.7 cm, 25.1 cm, 25.7 cm, 25.2 cm
14 +- .2739
25.26 +- .4615
Your previous results were
14.36 +- .3715
24.44 +- .5273
The first ball would take longer to reach the floor and would hence be expected to travel a bit further than in the previous. The second would reach the floor more quickly and would be expected not to travel as far.
This is consistent with the change in your mean distances. However the standard deviations are high enough that this conclusion can't be regarded as reliable.
.51
I did 5 trials as in the first experiment, and the values were fairly similar. Using m1v1 + m2v2 = m1u2 + m2u2, and the new horizontal ds values and the old uninterrupted value for v1, then m1(53.78 cm/s) + m2(0cm/s) = m1(37.53cm/s) + m2(67.72cm/s), and the ratio will be m1/m2 = 1.96, which is greater than one this time since m1 is the larger mass, and I think I'd had them backwards before. So, you could say it that way too, though, and smaller/larger would be about .51. This implies that the smaller ball is about half the mass of the larger instead of a quarter.
** 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? **
68cm, 25.26 cm, .1
69.55 cm/s
68.25 cm/s, 70.85 cm/s
64.1cm/s, 66.94 cm/s
4.03cm/s
It's about 5 or 6% off, so I guess you could argue it either way. That 5 or 6% could be used as evidence of velocity difference or it could just be error. I think I'd have to do the experiment a few more times before I'd say that a 5-6% difference was conclusive.
It dropped 68 cm, the mean horizontal range was 25.26 +- .4615cm, the vertical rise between balls was .2 cm, the hypotenuse was radius1+radius2 = 2.05cm, and the horizontal ds was 2.04cm, so the slope was .2cm/2.04cm =~ .1. The horizontal ds mean sd was +- .4615, so I used 25.26+-.4615 in the data program, and the data program did the work for the next couple of lines (oh, and dominoes were 3.333, or .1/.03).
** Your report comparing first-ball velocities from the two setups: **
68cm, 14 cm, .1
38.20 cm/s
37.45 cm/s, 38.96 cm/s
37.5cm/s, 39.5cm/s
0.3cm/s difference
About 0.8% off, so probably not that significant.
** Uncertainty in relative heights, in mm: **
Well, I would say +-.05cm, since that's half the smallest mark on the ruler, but since we did the experiment with the carbon paper in order to get it to match up exactly, it would just be the limits of human perception, which I'm really not sure how to calculate. The two dots looked like they eclipsed each other exactly, and I actually think a +-.05cm difference would be really visible, so I'm going to say about +-.1mm would still probably look like it's eclipsing the previous dot, and +-.1mm would be smaller than the size of the dot, so it could be off that much and still be overlapping it, which might just look to the naked eye like a bigger dot. Greater differences might still be overlapping, but would be noticeable (kind of like a venn diagram), so you could see where that second circle deviated from the other.
** 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. **
I would argue that the uncertainty in the relative heights of the balls was not a significant factor in the first setup since the size of the dot made on the carbon paper was fairly small, and any significant deviation from that tiny dot (well, splotch, anyway) would be noticed and corrected (and was). So if the dots ranged from about 2-5 millimeters (on that part of the experiment, anyway--strangely the dots made by the carbon paper were larger on the horizontal ds) on all of those test runs, they're in the same ballpark and are noticeable when the height is changed. If you had the bigger ball making a 5mm mark on one side and the smaller ball making a 4mm mark on the other side, it's easier to see that the 4mm mark strays out of the bounds of the 5mm mark. If you had a really tiny ball that made a .00001mm mark, then you could pretty much put it anywhere within the 5mm mark and a difference in height of 5mm would go unnoticed since the tiny circle would still be contained within the 5mm. When the 4mm circle was not at the right height, there was noticeable overlap (I actually used thin tracing paper so I could transpose the pieces of paper).
** How long did it take you to complete this experiment? **
So if the slope was .1 for the second ball, then for every unit x, there's .1 unit y, and its velocity was 69.55 cm/s. So 69.55cm/s = x + .1y, and x^2 + y^2 = 69.55^2, so we substitute and 69.55cm/s = +-sqrt(69.55^2 - y^2) + .1y, and 69.55cm/s - .1y = +-sqrt(69.55^2 - y^2), and (69.55cm/s - .1y)^2 = (69.55cm/s)^2 - y^2, and y =~ 13.9, so, by the p-theorem, x=~68.1cm/s.
Or it probably would be easier to get the angle of the hit, or tan^-1(.2/2.04) = 5.6 deg, so 69.55cm/s(sin(5.6deg)) =~ 6.8cm/s. Hmm, I'm gonna trust that one more.
** Optional additional comments and/or questions: **
So, m1/m2 =~ 1/.2333 or 4.29 (I had had the balls reversed), and m1_verticalv1 = -m2_verticalv2, so m1_vertv1 = -m2(6.8cm/s), and m1/m2_vertv1 = -6.8cm/s, and (4.29)_vertv = -6.8, and vertv =~ -1.6cm/s or 1.6cm/s in the downward direction
** **
According to our prior tests, the horizontal velocity of the first ball is about 53.78 cm/s before collision, and the horizontal velocity of the second ball after collision from 2mm up can be obtained from (69.55cm/s)^2 = (6.8cm/s)^2 + b^2, with b =~ 69.2cm/s.
m1_horizu1 = m1_horizv1 + m2_horizv2, or m1(53.78cm/s) = m1_hv1 + m2_(69.2cm/s), and m1(53.78cm/s - hv1) = m2(69.2cm/s), and m1 = m2(69.2cm/s)/(53.78cm/s - hv1), and 4.29 = 69.2cm/s/(53.78cm/s -hv1), and 230.72 - 4.29(hv1) = 69.2cm/s, and 161.52cm/s = 4.29(hv1), and hv1 =~ 37.6cm/s
** **
So ds = v0(dt) + .5(a)(dt^2), and 68cm = 1.6cm/s(dt) + .5(980cm/s^2)(dt^2), and dt =~ .374s, so dt(vAve) = ds, and vAve for horizontal v is just v0_horizontal since a=0, and (.374s)(37.6cm/s) = 14.0624cm, which is actually pretty close to what I got.
** **
So the friction would be the normal force * mu, and I feel like I should know how to calculate the spinning rate, but I just can't remember. I guess you could use something like aCent = v^2/r, but I'm a little iffy on this one since you don't have aCent or v of the spin. I guess you could also count how many times it spins on its way down the incline and over the horizontal ramp and also time it, but that seems a bit tedious. I would guess that the spin would affect the path of the ball by picking up air resistance and slowing it down, but, again, I'm not sure about this one.
** **
69.55cm/s
I think it would be the same as before. So, the line segment between centers is .8cm+1.25cm = 2.05cm, and the run is 2.04cm with p-theorem, and rise/run is .2cm/2.01cm =~ .1, then that would be equivalent to 3.333etc dominoes, 25.26cm is the average for the horizontal ds for ball 2 for the second setup, and for drop height 68 cm, horizontal range 25.26 cm and 3.3333 dominoes in the stack, the velocity of the ball should be 69.55 cm/s.
** **
7 hr
You've done outstanding work here. See my notes, and check discussion at the link given below, especially about the meaning of the derivative function.
You did have the two objects reversed in your original analysis, but you eventually became aware of the reversal and worked everything out nicely.
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