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: **
I need some help with these labs, is there anyway at all I can come to abingdon and you can have a tutor assist me?
** 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 )
10.66, 0.6387 ( big ball )
12.60, 0.3674 ( small 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
Your time of fall is close enough that it doesn't need to be corrected. However be sure you understand two things:
1. The acceleration 980 cm/s^2 is only applicable to the vertical motion, and the vertical displacement is 61.5 cm.
2. While the Pythagorean Theorem was used correctly, and does give you the displacement from the end of the ramp to the floor, that displacement is greater than the vertical displacement and wouldn't be used in this equation.
Since the Pythagorean displacement isn't relevant to the calculation of the time interval, this is not relevant to this particular analysis, but it's also worth noting that the path traveled by the ball is parabolic, so the distance traveled by the ball is a bit greater than the straight-line distance calculated by the Pythagorean Theorem.
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
** 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. **
distance of ramp / time to travel ramp = 49.5 cm / 1.668 s = 29.68 cm / s
62.78 cm / .357 s = 175.85 cm / s
62.42 cm / .358 s = 174.36 cm / s
distance of fall / time to travel fall = 62.42 cm / .357 s = 174.85 m / s
distance of fall / time to travel fall = 62.78 cm / .358 s = 175.36 m / s
Immediately after collision the balls are still traveling in the horizontal direction; they haven't had time to build any vertical velocity.
A basic principle of 'ideal' projectile motion is that the horizontal acceleration is zero, so that the horizontal velocity is unchanging. The steel ball in this experiment behaves in pretty much the ideal way, with very negligible horizontal acceleration.
So the velocity of the projectile is determined only by the projectile's horizontal displacement and time of fall.
What therefore are your before- and after-collision velocities?
** 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. **
** 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? **
3 hours
** Optional additional comments and/or questions: **
See if you can answer the question posed about horizontal velocities. Give me your best answer, or additional questions.