conservation of momentum

Phy 121

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'. **

2cm, 5cm

1cm

.05cm

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

11cm,10cm,10cm,10cm,11cm

10.4cm, .5477

I allowed the ball to roll off the ramp onto the floor and I measured the horizontal range from the table to where the ball hit the floor.

** 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. **

16.5cm,17cm,16cm,16.5cm,17cm

13.5cm,14cm,13cm,13.5cm,14cm

16.6cm, .4183

13.6cm, .4183

I allowed the larger ball(first ball)to roll down the ramp and hit the smaller ball(second ball) and both fell to the floor, I then noted where they first hit on the floor and measured that horizontal distance from the table.

first-ball range after collision won't be less than for the uniterrupted ball

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

67cm

.66 seconds

I measured the distance from where the ball began to fall to the floor wher it hit. I timed the ball falling from rest to hitting the floor to determine the time it took the ball to reach the floor. This was assuming the ball fell flat off the table.

** 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. **

17.3712 m/s,17.3712 m/s, 21.34cm/s

17.9cm/s,16.82cm/s,

17.9cm/s, 16.82cm/s

21.9cm/s, 20.8cm/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. **

p=mv 66.9n-cm/s=(3.98N)(16.82cm/s)

p=mv 66.9n-cm/s=(3.98N)(16.82cm/s)

p=mv 67.01n-cm/s=(3.06N)(21.9cm/s)

133.91N-cm/s

133.91N-cm/s

m1v1=m2v2

It's not clear where you are getting 3.98 N and 3.06 N, and I don't think these forces of 3.98 N and 3.06 N are relevant to this situation. However it is very good that you are including units in your calculations.

The masses of the two balls are regarded as unknown. See the link at the end for more information.

** 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. **

m=pv 3.98N=66.9N-cm/s(16.82cm/s)

m=pv 3.06N=67.01cm/s(21.9cm/s)

1=479.6cm/s

3.98/3.06=1.3

The meaning of the ration letss you see your inaccuracy of your measurements.

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

7cm,3cm

179.6,14.13

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

The first ball may not follow in a direct downward pattern.

Less, because when the ball hits the top it loses some momentum it has built up.

Yes, with the first ball not hitting directly in the middle of the second, it will cause the balls to have different velocities.

The second ball will not go as far on the horizontal range because it will not have the extra push from ball 1.

If the centers are at the same height, the speed will be greater than if they were at different heights, this way the first ball reacts on the second one, sending its momentum to the secon ball.

Yes because the second ball will not receive as much power from the first ball if the balls do not hit directly in the center.

** 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 may actualy have a larger horizontal range due to the fact that it has been able to keep all of its momentum, sincie it did not hit directly in the center of the smaller ball.

The second ball will not go as far on the horizontal range due to the fact that it did not get the momentum from the first ball.

** 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: **

3.98/3.06=1.3

3.06/3.98=.767

These are the ratios of the masses of the balls.

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

5%

** 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? **

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

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2 hours

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