conservation of momentum

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'.*********************************** **

22mm,22mm

16mm

Pretty close, maybe off a mm

** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process:*********************************** **

38cm,38cm,38cm,38cm,38cm

used carbon paper and they all contributed to one slightly smudged dot so I used the measurement to the center of the dot

** 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.*********************************** **

30cm,30cm,30cm,30cm,30cm

68cm,68cm,68cm,68cm,68cm

30cm,0

68cm,0

These are the ranges for the largeand then the small ball and then the mean and std dev of each. tried to use carbon paper which is getting ragged. Very consistent marks -I've got everything taped down.

** Vertical distance fallen, time required to fall.*********************************** **

70cm

.38sec

measured the height of the desk from the floor then used the equation ds=1/2*a*t^2 to get time to hit floor

** 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.*********************************** **

1m/s,.8m/s,1.8m/s

1m/s,1m/s

.8m/s,

1.8m/s

** 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*1m/s

m1*.8m/s

m2*1.8m/s

m1*1m/s + m2*0

m1*.8m/s + m2*1.8m/s

m1*1m/s + m2*0=m1*.8m/s + m2*1.8m/s

** 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*1m/s)-(m1*.8m/s)=(m2*1.8m/s)-(m2*0)

m1*.2m/s = m2*1.8m/s

m1*.2m/s/m2 = 1.8 m/s

m1/m2 = 1.8m/s / .2 m/s = 9

The mass of the first ball is 9 times that of the second. Measuring the balls (1 dia and 1/2 dia) and calculating their mass gives a similar answer.

** Diameters of the 2 balls; volumes of both.*********************************** **

2.5cm,1.25cm

8.16 cubic cm,1.02 cubic cm

** How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher?*********************************** **

I think velocity would be the same but direction would be different as it would tend to lift the 2nd ball upward and push the first ball downward.

An initial upward velocity will give the rolling ball more time to fall, which will allow it to travel further than if the collision is head-on.

If you assume zero initial vertical velocity, you will therefore tend to underestimate the time of fall and to overestimate the after-collision horizontal velocity.

If the collision is not head-on the rolling ball will also tend to maintain more of its horizontal velocity. Depending on the angle of impact this might or might not tend to mitigate the above effect.

The second ball, on the other hand, will have an initial velocity in the downward direction and the assumption of zero initial vertical velocity will cause an overestimate of the time of fall, tending to result in an underestimate of the horizontal velocity of that ball.

** Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second:*********************************** **

I think it would drop the range of the first ball and possibly increase the range of the 2nd

** 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:*********************************** **

I reported all velocites the same with no deviation so there's no differences.

** What percent uncertainty in mass ratio is suggested by this result?*********************************** **

No change.

** What combination of before-and after-collision velocities gives you the maximum, and what combination gives you the minimum result for the mass ratio?*********************************** **

As the estimated velocity of either ball goes up the estimate of its mass goes down by the inverse of the v^2

** 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*v1=m1*u2-m1*v1+m2*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.*********************************** **

Had some trouble setting this up, but raising the smaller ball certainly changes the path and range of both balls.

**   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?*********************************** **

I think I made some mistake here. My original ramp was elevated by 5 cm and was 30cm long so slope was 5/30 or .17

Velocity and range of small ball decreased as alignment worsens. The velocity and range of the large ball,at the same time increases.

** Your report comparing first-ball velocities from the two setups: *********************************** **

The large (first) ball's velocity and range increased as the small ball was no longer perfectly aligned in its path of travel

** Uncertainty in relative heights, in mm:*********************************** **

I measured these initial heights carefully and believe I was within a 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.*********************************** **

My results were very consistent but even small variations in the height of the small ball have a major impact on its path and velocity

** How long did it take you to complete this experiment?*********************************** **

5 hours

** Optional additional comments and/or questions:*********************************** **

I think I got lost a little on the last part of this.

You did fine and got excellent results. See my notes.

Your mass ratio of 9:1 was very close to the volume ratio, which is 8:1. Assuming that the steel is of the same or very nearly the same density for both balls, we would expect the ratios to be the same. Given the limits of precision in this experiment, your ratio is consistent with very careful and accurate work.