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: **
resubmit v.3--
Got stuck using the momentum conservation theory.
** Distances from edge of the paper to the two marks made in adjusting the 'tee'. **
.35 cm, .4cm
.4 cm
+-.05
** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
12.9, 13.2, 13.4, 13.6, 13.62,
13.34, 0.3011
to obtain my results i rolled the big ball down the incline and allowed it to hit the floor. I marked where it hit the floor each time.
** 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. **
9.9, 10.3, 10.5, 11.1, 11.5 ( big ball )
12.2, 12.3, 12.6, 12.8, 13.1 ( small ball )
12.60, 0.3674 ( small ball )
10.66, 0.6387 ( big ball )
** Vertical distance fallen, time required to fall. **
9.9, 10.3, 10.5, 11.1, 11.5 ( big ball )
12.2, 12.3, 12.6, 12.8, 13.1 ( small ball )
12.60, 0.3674 ( small ball )
10.66, 0.6387 ( big ball )
** Vertical distance fallen, time required to fall. **
62.42 cm - ball 1, 62.78 cm - ball 2
.357 s, .358 s
To determine just how far the balls feel I used Pyth Theory. The distance from the table to the floor was 61.5 cm (A^2). I then took the avg distance (B^2). I added A^2 + B ^2 and received C^2. I took the squareroot of C^2 and got the total distance.
61.5 ^ 2 + 10.66 ^ 2 = square root of ( 3782.25 + 113.64 ) = 62.42 cm
61.5 ^ 2 + 12.60 ^ 2 = square root of ( 3782.25 + 158.76 ) = 62.78 cm
To determine the time I used an equation for uniform acceleration:
x = x 0 + v 0 t + 1 / 2 a t ^ 2
62.42 = 0 + 0 + 1/2 * 980 cm / s ^2 * t ^ 2
.357 = t
x = x 0 + v 0 t + 1 / 2 a t ^ 2
62.78 = 0 + 0 + 1/2 * 980 cm / s ^2 * t ^ 2
.358 = t
Note: you told me that Pyth Theory would not work here, but I could keep this stat just because it is close enough to the exact value.
** 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. **
*****LINE 1
34.5 cm / 1.09 s = 31.65 cm / s
4 cm / .164 s = 24.39 cm / s
2 cm / .164 s = 12.20 cm / s
*****LINE 2
13.34 cm +- 0.3011
13.37 cm / .36 s = 37.14 cm / s
13.31 cm / .36 s = 36.97 cm / s
*****LINE 3
12.60 cm +- .3674
12.9674 cm / .164 s = 29.07 cm / s
12.2326 cm / .164 s = 74.59 cm / s
******LINE 4
10.66 cm +-.06387
10.72 cm / .164 s = 65.39 cm / s
10.59 cm / .164 s = 64.61 cm / s
In the above it isn't clear where 1.09 sec and .164 sec came from.
However to calculate the velocity of a ball at the 'top' of an initially horizontal projectile path, you divide its horizontal range by the time required to fall.
Every ball starts its fall with a horizontal path, and you have measure the horizontal range of the first ball when it doesn't collide and when it does, and that of the second ball when it collides. You need to use the horizontal ranges to find the velocities of the balls before and after their collisions. The velocity of the second ball before collision is, of course, zero.
** 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 = big ball
m2 = small ball
u signifies vector velocity before the collision
v signifies vector velocity after the collision.
m1u1 + m2u2 = m1v1 + m2v2
m1 * 31.65 m/s + m2 * 0 = m1 * 24.39 + m2 * 12.20 m/s
m1 * 31.65 m/s = m1 * 24.39 + m2 * 12.20 m/s
The above equation is very plausible, except that I believe your units should be cm/s (see my previous note to be sure you have calculated those velocities correctly; they aren't implausible but as noted above I can't tell how they were obtained).
However the equation below does not follow. The 31.65 m/s cannot just disappear from the equation.
If you subtrace m1 * 24.39 cm/s from both sides you will get something like m1 * 7.3 cm/s = m2 * 12.2 cm/s.
If you then divide both sides by 7.3 cm/s, and then divide both sides by m2, you will get m1 / m2 on the left-hand side and a meaningful ratio on the right.
m1 = m1 * 24.39 + m2 * 12.20 m/s
m1/m1 = 24.39 + m2 * 12.20 m/s
1 = 24.39 + m2 * 12.20 m/s
-23.39 = m2 * 12.20 m/s
-1.917 = m2
m1 u1 + m2 u2 = m1 v1 + m2 v2
m1 * 31.65 m/s + (-1.917 * 0) = m1 * 24.39 + (-1.917)12.20 m/s
m1*31.65 = m1 * 24.39 - 23.39 m / s
Desperately need some help here, I am sure I messed up somewhere. I should be over in Abingdon tomorrow to meet with the tutor.
** 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. **
** Diameters of the 2 balls; volumes of both. **
** How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
** Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second: **
** 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: **
** What percent uncertainty in mass ratio is suggested by this result? **
** 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 1/2 hours
** Optional additional comments and/or questions: **
You aren't far off in your procedures. You got the right type of info into the equation, then the algebra got in your way. Your velocities are questionable but shouldn't be hard to recalculate.
Let me know if you have questions.