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course phy 201
12/10 2:30
Ramp & Ball
You allowed a ball to roll from rest down a ramp on a tabletop. The lower end of the ramp was positioned so that the ball could roll continuously off the ramp and into free fall. You observed the horizontal range of the falling ball and the distance of fall.
Insert a copy of your data here, along with any previously submitted work you wish to include:
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Ramp 1 Ramp 2 Ramp 3
60cm 60cm 60cm
2.5s 3.5s 6s
5cm rise 2.5cm rise .7cm rise
16.2cm 10.5cm 5.7cm distance travel horizontally after leaving ramp
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You did trials for three ramps.
Analysis based on projectile motion:
Phy 201 students may assume that the initial projectile velocity is horizontal. University Physics students can do the same as a first approximation, but are expected to also then solve assuming that the initial projectile velocity is parallel to the ramp.
What was the horizontal velocity of the ball as it fell to the floor?
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At this point it appears you've got a good analysis of motion on each ramp, based on the timing.
You haven't done the analysis based on projectile motion. After leaving the end of the ramp, the ball fell about 90 cm to the floor while it moved at a constant horizontal velocity (the measured distance to the floor should have been included with your data, but 90 cm is close enough for this analysis).
The vertical velocity of the ball at the end of the ramp is for the purposes of this analysis regarded as negligible.
So the ball's vertical motion started with velocity 0, the vertical acceleration was 980 cm/s^2 downward and the vertical displacement was 90 cm downward.
How long did it take the ball to fall?
What do you conclude, based on your observed horizontal ranges, were the ball's horizontal velocities for the three trials?
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Ramp 1 Ramp 2 Ramp 3
slope 5cm/60cm=0.08 2.5cm/60cm=.04 .7cm/60cm=.01
Vave 60cm/2.5s=24cm/s 60cm/3.5s=17.14cm/s 60cm/6s=10cm/s
Vf 2(24cm/s)-0=48cm/s 2(14.14cm/s)-0=34cm/s 2(10cm/s)-0=20cm/s
a 48(cm/s)-0/2.5s=19.2cm/s^2 (34(cm/s)-0)/3.5s=9.7cm/s^2 20(cm/s)-0/6s=3.3cm/s^2
VERTICAL MOTION
Ramp 1 Ramp 2 Ramp 3
`ds 160cm 160cm 160cm
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160 cm is a little bit over your head, about the height of my forehead.
The ball did not travel 160 cm in the vertical direction. More like 90 cm.
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a 9.8m/s^2 or 980cm/s^2 980c/s^2 980cm/s^2
V0 48cm/s 34cm/s^2 20cm/s^2
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The ball's velocity at the beginning of the fall was almost all in the horizontal direction. The vertical velocity was not much different than 0. So we assume, without too much error, that the initial vertical velocity was 0.
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Vf^2 (48cm/s)^2+2(980cm/s^2)(160cm) (34cm/s)^2+2(980cm/s^2)(160cm) (20cm/s)^2+2(980cm/s^2)(160cm)
Vf =sqrt315904(cm^2/s^2)=562cm/s =sqrt341756(cm^2/s^2)=561cm/s =sqrt31400=177cm/s
`dt (562cm/s-48cm/s)/980(cm/s^2)=.52s (561cm/s-34cm/s)/980(cm/s^2)=.54s (177cm/s-20cm/s)/980(cm/s^2)=.16s
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For a fall of 160 cm your results, about 560 cm/s, are in the right ballpark, as is the .52 s time of fall. The 177 cm/s result is incorrect; it's not completely clear what steps you took to get this result.
However if you use initial velocity 0 for each trial, and 90 cm for the distance of fall, you will get the same result for all three trials (around .4 seconds in each case).
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HORIZONTAL MOTION
`ds 16.2cm 10.5cm 5.7cm
a 0 0 0
`dt .52s .54s .16s
Vave 16.2cm/.52s=31cm/s 10.5cm/.54s=19.8cm/s 5.7cm/.16s=35.6cm/s
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What was the final velocity for each ramp slope?
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Ramp 1 Ramp2 Ramp 3
Vf 2(24cm/s)-0=48cm/s 2(14.14cm/s)-0=34cm/s 2(10cm/s)-0=20cm/s
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Good. These results are consistent with your timing data and distance down the ramp. They will also be consistent with your horizontal velocities, as determined by the projectile motion.
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Based on your result and the length of the ramp, what would have been the acceleration of the ball on the ramp?
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Ramp 1 Ramp 2 Ramp 3
a 48(cm/s)-0/2.5s=19.2cm/s^2 (34(cm/s)-0)/3.5s=9.7cm/s^2 20(cm/s)-0/6s=3.3cm/s^2
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Good.
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Graph acceleration vs. ramp slope and find the slope of this graph.
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Ramp 1 Ramp 2 Ramp 3
slope 5cm/60cm=0.08 2.5cm/60cm=.04 .7cm/60cm=.01
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You found the accelerations in the preceding.
You've found the ramp slopes here.
You need to construct the graph of acceleration vs. ramp slope, and report the slope of that graph.
The graph will have three points. You will sketch the straight line you think best fits these points (the best line, the one that comes the closest to the points on the average, won't go through any of the points), find two points on the line and use them to find the slope of the graph.
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What do you think is the percent uncertainty in your measurement of the horizontal distance traveled by the ball from the end of the ramp to the floor?
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.2% uncertainty
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What do you think is the percent uncertainty in your measurement of the vertical distance traveled by the ball from the end of the ramp to the floor?
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01% uncertainty
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What do you think is the percent uncertainty in your measurement of the distance traveled by the ball from release to the end of the ramp?
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.1% uncertainty
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What therefore do you think is the percent uncertainty in each of the following quantities, as you have calculated them?
· the time required to fall to the floor .1% uncertainty
· the horizontal velocity of the ball during its fall .3% uncertainty
· the acceleration of the ball on the ramp .1% uncertainty
· the slope of your graph of acceleration vs. ramp slope .1% uncertainty
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Analysis based on timing:
Based on the time down the ramp and the distance the ball traveled from rest as it accelerated down the ramp, what was the acceleration on each ramp, and what is the slope of the graph of acceleration vs. ramp slope?
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Ramp 1 Ramp 2 Ramp 3
accel v/s slope (19.2cm/s^2)/.08 (9.7cm/s^2)/.04 (3.3cm/s^2)/.01
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What do you think is the percent uncertainty in your measurement of the time required to travel down the ramp?
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.1% uncertainty
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For each ramp:
What was the change in the gravitational PE of the ball? You may assume a 70 gram mass.
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k=mg/L
K for all ramps= (70g*980cm/s^2)/160cm=728.8g Cm^2/s^2
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k = m g / L applies to the motion of a pendulum. It's not relevant in the context of this experiment.
PE change is equal and opposite to the work done by gravity. What is the gravitational force on the ball, how far did it move in the direction of that force as it rolled down the ramp, and what therefore was the change in PE?
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Ramp 1 Ramp 2 Ramp 3
PE (1/2)428.8(16.2)^2=56267g s^2 (1/2)428.8(10.5cm)^2=23638g s^2 (1/2)428.8(5.7cm)^2=6966g s ^2
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What velocity would have been attained if all the lost PE went into translational KE?
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KE=(1/2)mv^2
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Your analysis of motion on the ramps was good. So you've got half the analysis done.
You haven't yet done the projectile analysis correctly.
Please revise according to my notes.
&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you. &#
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