#$&*
PHY 121
Your 'ball down ramp' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
#$&* Your general comment **
September 27, 2011 at 11:26 p.m.
#$&* Will a steeper ramp give greater or lesser time? **
• Follow the instructions, fill in your data and the results of your analysis in the given format.
• Any answer you given should be accompanied by a concise explanation of how it was obtained.
• To avoid losing your work, regularly save your document to your computer.
• When you have completed your work:
Copy the document into a text editor (e.g., Notepad; but NOT into a word processor or html editor, e.g., NOT into Word or FrontPage).
Highlight the contents of the text editor, and copy and paste those contents into the indicated box at the end of this form.
Click the Submit button and save your form confirmation.
A ball is timed as it rolls from rest to the end of a ramp. The slope of the ramp is varied. Preliminary conclusions are drawn about the motion and the effect of ramp slope. A subsequent lab exercise uses the data from this lab to reach additional conclusions.
Most students report completion times between 45 minutes and 75 minutes hour, with a few reporting times as short as 25 minutes or as long as 2 hours. Median time of completion is around 1 hour.
Timing Ball down Ramp
The picture below shows a ball near the end of a grooved steel track (this steel track is a piece of 'shelf standard'); the shelf standard is supported by a stack of two dominoes. Your lab materials package contains two pieces of shelf standard; the shelf standard shown in the figure is white, but the one in your kit might be colored black, gold, silver or any of a variety of other colors.
If a ball rolls from an initial state of rest down three ramps with different slopes, the same distance along the ramp each time, do you think the time required to roll the length of the ramp will be greatest or least for the steepest ramp, or will the interval on the steepest ramp be neither the greatest nor the least? Explain why you think you have correctly predicted the behavior of the system.
Your answer (start in the next line):
I think that the steeper the ramp, the faster the ball cover the distance. I think that the steepness of the ramp will affect the speed in that the steeper the ramp, the faster the ball will accelerate.
#$&*
If we write down the slopes from least to greatest, next to the time intervals observed for those slopes, would you expect the time intervals to be increasing or decreasing, or do you think there would be no clear pattern? Explain why you think you have correctly described the behavior of the numbers in the table.
Your answer (start in the next line):
I would expect the time intervals to decrease in correlation with the increase in the ramps’ slopes. I think this will hold true for the same reason addressed above, that the steeper ramp will provide greater acceleration.
#$&*
Set up the shelf standard ramp on a reasonably level table, using a piece of 30-cm shelf standard and a single domino under the high end of the ramp. Position the dominoes so that the last .5 cm of the ramp extends beyond the point where the ramp contacts the domino,.and do the same in all subsequent setups.
Set the bracket on the table, touching the lower end of the ramp so that a ball rolling down the ramp will strike the bracket..
Mark a point about 3 cm below the top end of the ramp. Place a domino on the ramp to its high end is at this point, and place the ball just above the domino, so the domino is holding it back. Quickly pull the domino away from the ball so the ball begins to roll freely down the ramp. Allow the ball to roll until it strikes the bracket.
The bracket will probably move a little bit. Reset it at the end of the ramp.
Determine how far the ball rolled from release until it struck the bracket.
Now repeat, but this time use the TIMER. The first click will occur at the instant you release the ball, the second at the instant the ball strikes the bracket. Practice until you are as sure as you can be that you are clicking and pulling back the domino at the same instant, and that your second click is simultaneous with the ball striking the bracket.
When you are ready, do 5 trials 'for real' and record your time intervals.
Then reverse the system--without otherwise changing the position of the ramp, place the domino under the left end and position the bracket at the right end.
Time 5 trials with the ramp in this position.
In the space below, give the time interval for each trial, rounded to the nearest .001 second. Give 1 trial on each line, so that you will have a total of 10 lines, the first 5 lines for the first system, then 5 lines for the second system.
Beginning in 11th line give a short narrative description of what your data means and how it was collected.
Also describe what you were thinking, relevant to physics and the experiment, during the process of setting up the system and performing the trials.
Your answer (start in the next line):
2.017
2.000
2.109
2.102
2.047
1.902
1.922
1.922
1.910
1.938
It seem the ball travels at a faster rate of speed when traveling from the high end being on the right than on the left, but not by much. They were both generally the same time though, so I assume the discrepancy lies in the levelness of my dining room since the I have changed tables (that’s plural) that I perform the experiments on since the last one with the car.
#$&*
Now place two dominoes under the right end and repeat the process, obtaining the time interval for each of 5 trials.
Then place the two dominoes under the left end and repeat once more.
Enter your 10 time intervals using the same format as before.
Your answer (start in the next line):
1.281
1.225
1.250
1.221
1.266
1.359
1.359
1.313
1.531
1.388
Here the same as mentioned above, one side still seems a little faster than the other, but I still believe that is due to the evenness of the surface. My original belief, however, that the steeper the ramp the faster the descent is holding true.
#$&*
Repeat the preceding using 3 dominoes instead of 2. Enter your 10 time intervals using the same format as before.
Your answer (start in the next line):
1.078
1.078
1.080
1.014
1.156
1.269
1.241
1.209
1.231
1.266
Here I am seeing the same results as before as far as the discrepancies are concerned and also the same results as far the slope v. speed.
#$&*
Repeat the preceding again, still using the 3 domino setup, but this time place a CD or a DVD disk (or something of roughly similar thickness) on the 'low' end of the ramp. You need time only 5 intervals, but if you prefer you may use 10. Enter your 5 (or 10) time intervals using the same format as before.
Your answer (start in the next line):
1.273
1.259
1.213
1.244
1.231
Here I used a ruler, plastic, about the same width as a cd, just easier to stand up and I got similar results as above.
#$&*
Repeat the preceding one last time, still using the 3 domino setup, but remove the disk and replace it with a piece of paper. You need time only 5 intervals, but if you prefer you may use 10. Enter your 5 (or 10) time intervals using the same format as before.
Your answer (start in the next line):
1.250
1.233
1.300
1.304
1.256
Here the results are again similar to above, a couple of times were longer, but that may be because of the lack of sound when the ball hits the paper and my response in stopping the timer.
#$&*
Do your results support or fail to support the hypotheses you stated in the first two questions, regarding the relationship between time intervals and slopes? Explain.
Your answer (start in the next line):
I think my results support my original hypothesis, because not all of the results were exactly the same in each step, they all had something in common with its predecessor. The higher the slope, the faster the time, this proved true all the way throughout. The barrier which it hit may have caused slight differences, but the true difference maker in regards to the time is the height of the slope.
#$&*
How do you think the average velocity of the ball is related to the slope of the ramp? Explain in as much detail as possible.
Your answer (start in the next line):
Average Velocity
1 Domino
13.14 cm/s
14.197 cm/s
2 Domino
21.583 cm/s
19.992 cm/s
3 Domino
25.356 cm/s
21.792 cm/s
The average velocity was calculated by taking the average time (midpoint between highest and lowest time) in each set of 5 intervals measured for the first 3 sets of domino stacks taken, and dividing the distance traveled by the average time interval. Here you can see that the average velocity increased as the slope increased.
#$&*
Speculate on what it is that causes the average velocity on these ramps to change with slope.
Your answer (start in the next line):
The steeper the ramp the quicker the object accelerates, resulting in a faster average velocity.
#$&*
How might you verify whether your speculations are indeed valid explanations?
Your answer (start in the next line):
I think I verified this information as contained in the answer above.
#$&*
Do your data conclusively show that the disk made a difference?
Your answer (start in the next line):
Now I will do the same as above for the average of the disc and the paper.
Disc
21.722
Paper
21.285
Regular
21.792
Here, you can see there is no significant change between the original setup and the setup involving the disc.
#$&*
Do your data conclusively show that the piece of paper made a difference?
Your answer (start in the next line):
Using the above data, you can see there is no significant change between the original setup and the setup involving the paper.
#$&*
Imagine that someone is placing different objects below the 'low' end of the ramp, and you are timing the ball. Assume that somehow the object placed below the 'low' end is hidden from you in a way that does not interfere with the timing process. Compared to the thickness of the DVD, how thin would the object have to be before you would be unable, using the TIMER, to observe a difference in times down the ramp?
Answer this question in the first line below. Express your answer in multiples or fractions of the thickness of a disk.
Starting in the second line, explain how you came to your conclusion, based on the results you obtained in this experiment. Also discuss how you could modify or refine the experiment, still using the TIMER, to distinguish the effect of the thinnest possible object placed under the 'low end.
Your answer (start in the next line):
If you were solely relying on sound then a number of things could affect your timing. However, I think if you are watching the object travel, and you accurately judge when the ball touches the object, unless the object hangs over onto the ramp or is away from the ramp, it would hit at the same time.
#$&*
Had you placed the disk below the 'low' end of the ramp in a 1-domino setup, do you think the difference in times would have been greater or less? Do you think you would be better able distinguish the presence of a thinner object using the 1-domino setup, or the 3-domino setup? Explain your reasoning below:
Your answer (start in the next line):
Here, again as above, I think with accurate observation, if your setup is the same each time, you will get similar reults.
#$&*
Does the ball's velocity change more or less quickly with the 3-domino setup or the 1-domino setup? Explain as best you can how you could use your results to support your answer.
Your answer (start in the next line):
The velocity changes a greater rate the higher the domino setup. And this is shown by the faster times. If you were to divide the velocities by the average times used to calculate the velocity, since the object began at rest, you could clearly see the higher the domino stack, the quicker the velocity the greater the average increase in acceleration.
#$&*
Your instructor is trying to gauge the typical time spent by students on these experiments. Please answer the following question as accurately as you can, understanding that your answer will be used only for the stated purpose and has no bearing on your grades:
• Approximately how long did it take you to complete this experiment?
Your answer (start in the next line):
1 hour 45 minutes
#$&*
*#&!
&#This looks very good. Let me know if you have any questions. &#
If it then coasts back down the incline, accelerating at .5 m/s^2, how long will it take to travel from the position of your friend to your position?
****
vo=0ms
vf=?
ds=16m
dt=?
a=.5m/s^2
16m=(vf^2-0)/1m/s^2)=16m^2/s^2=vf^2=4m/s
4m/s=0m/s+.5dt=(4m/s)/.5m/s^2=8s
@& `q001. The meter, the kilogram and the second are the basic SI units (SI stands for 'standard international').
Velocity is the rate of change of position with respect to clock time. What therefore are the SI units of velocity?
****
velocity is rate of change of position with respect to clock time = (change in position) / (change in clock time).
position has units of meters so change in position has units of m
clock time has units of seconds so change in clock time has unit of s
Therefore velocity has units of m / s.
#$&*
Acceleration is the rate of change of velocity with respect to clock time. What therefore are the SI units of acceleration?
****
acceleration is rate of change of velocity with respect to clock time = (change in vel) / (change in clock time).
change in vel has units of m/s
change in clock time has unit of s
so acceleration has units of (m/s)/s = m/s^2.
#$&*
A Newton is the SI unit of force. Force = mass * acceleration. In terms of kg, meters and seconds, what therefore are the units of of force? The resulting units are the units of a Newton.
****
mass is in kg, acceleration in m/s^2, so force is in kg * m/s^2
#$&*
A Joule is the SI unit of work. Work = force * displacement. In terms of kg, meters and seconds, what therefore are the units of work? The resulting units are the units of a Joule.
****
Force is measured in kg m/s^2, displacement in meters, so work is in kg m/s^2 * m = kg m^2 / s^2.
#$&*
Power is measured in watts. A watt is a Joule per second. What therefore are the fundamental units equivalent to a watt?
****
work or energy is measured in kg m^2 / s^2 (i.e., Joules), time in s, so power is in units of Joules / sec = (kg m^2 / s^2) / s = kg m^2 / s^3.
#$&*
What are the fundamental units equivalent to Newtons * meters? What quantity would be expressed in these units?
****
A Newton is a kg m/s^2, so N * m is kg m/s^2 * m = kg m^2 / s^2.
This was seen earlier to be the fundamental units of the Joule.
#$&*
What are the units of kinetic energy?
****
ke=1/2kg*v^2
KE = 1/2 m v^2
m is in kg, v in m/s
So KE is in kg * (m/s)^2 = kg * m^2 / s^2.
Note that this was earlier shown to be equivalent to the Joule.
#$&*
`q002. Sketch an object on inclines of 10, 20 and 30 degrees; you will have three sketches. For each incline, sketch the arrow indicating the force exerted on the object by gravity (for each incline we'll refer to this as the 'first arrow'). For each sketch, construct the arrow which represents the component of the gravitational force parallel to the incline (you do this by projecting the arrow at a right angle onto a line parallel to the incline, as we did in class).
For each incline, estimate the length of this arrow as a percent of the first. Give your three estimated percents:
****
An accurate sketch would show that the percents are about 17%, 34% and 50%.
Discrepancies could be due to errors in estimating the angle of the incline, projecting at an angle other than perpendicular to the line of the incline, or simply errors in estimating one vector as a percent of the other.
#$&*
`q003. My truck has a mass of about 1400 kg. It reaches the part of the road which has a nearly constant incline, in front of VHCC, at a speed of 5 meters / second in the direction down the incline. In 12 seconds its speed has increased to 10 meters / second.
What is my acceleration on that incline?
****
acceleration is ave roc of vel wrt clock time = change in vel / change in clock time = (vf - v0) / `dt = (10m/s-5m/s)/12s=5m/s)/12s=.417m/s^2
#$&*
What is the net force on the truck as it coasts down that incline?
****
F_net = m a = 1400kg*.417m/s^2=583.33kg*m/s^2.
Note that kg m/s^2 is the fundamental unit of the Newton, so the net force is about 580 N.
#$&*
What is the change in its kinetic energy as it coasts down the incline?
****
KE_0 = 1/2 m v0^2 = 1/2 1400*(5m/s)^2= 700kg*25m^2/s^2=17500kg m^2/s^2
KE_f = 1/2 m vf^2 = 1/2 1400*(10m/s)^2= 700kg*100m^2/s^2=70000kg m^2/s^2
`dKE = KE_f - KE_0 = 70000kg m^2/s^2-17500kg m^2/s^2=52500kg m^2/s^2
#$&*
If friction exerts a force opposite to my direction of motion, with the magnitude of the frictional force equal to 2% of the force exerted by gravity on the truck, then during this interval how much work is done on the truck by the frictional force?
****
The acceleration of gravity is 9.8 m/s^2 so the force exerted by gravity on the truck is
F_grav = m a_grav = 1400 kg * 9.8 m/s^2 = 13 700 kg m/s^2.
Note that this is the same as 13 700 Newtons.
Now, 2% of this gravitational force is about 270 Newtons.
The work by the frictional force can be denoted `dW_frict. Using this notation we conclude that `dW_frict = F_frict * `ds.
We need to find `ds for this motion.
The average velocity is (5 m/s + 10 m/s) / 2 = 7.5 m/s, and the interval lasts 12 seconds, so the truck moves 7.5 m/s * 12 s = 90 meters.
So F_frict * `ds = 270 N * 90 m = 2500 N * m = 2500 kg m^2 / s^2.
This is actually not quite right. The frictional force is in the direction opposite the displacement. The signs of F_frict and `ds are therefore opposite, and we conclude that the work done by friction is
`dW_frict = -2500 kg m^2 / s^2, or -2500 Joules.
#$&*
The net force on the truck is the combination of the frictional force, and the component of the gravitational force which acts in the direction down the incline. What therefore is that component of the gravitational force, and what percent is this of the total gravitational force pulling the truck toward the center of the Earth?
****
The net force on the truck was found earlier to be about 580 Newtons; since the truck is speeding up this net force is in the direction of motion.
The frictional force was just found to be about 270 Newtons, and is in the direction opposite motion.
The net force is the sum of the frictional force and the 'parallel component' of the gravitational force, so
F_net = F_parallel + F_frict.
It follows that the parallel component of the gravitational force is F_net - F_frict.
Choosing the direction of motion as positive, we have
F_parallel = 580 Newtons - (-280 Newtons) = 580 N + 280 N = 960 N.
In fundamental units this is 960 kg m^2 / s^2.
#$&*
Make a reasonably accurate sketch depicting the incline, the truck, the gravitational force and its component along the incline.
****
#$&*
`q004. In the preceding, suppose that the direction down the incline is positive. The 5 meter / second initial velocity is therefore positive, and we would write v_0 = + 5 m/s. If the force of air resistance on the car was 2 kg m/s^2, then since that force is directed up the incline, it would be represented as -2 kg m/s^2.
Give each of the following, including its sign, its numerical value and its units:
The final velocity of the truck.
****
The final velocity is in the direction down the incline so is positive.
Thus vf = +10 m/s.
#$&*
The change in the kinetic energy of the truck.
****
As seen before, the change in KE is KE_f - KE_0. This was shown earlier to be about 50 000 kg m^2 / s^2, or 50 000 Joules.
#$&*
The acceleration of the truck.
****
The acceleration is in the direction of motion, since the truck speeds up.
Thus, as found earlier, a = +.417m/s^2
#$&*
The force of friction on the truck.
****
The force of friction was found earlier to have magnitude about 280 N.
This force is in the direction opposite motion.
So the frictional force is -280 N.
#$&*
The component of the gravitational force parallel to the direction of the incline.
****
This was found earlier to be about 960 Newtons.
This force has to be greater than the net force, since friction contributes its negative share to the net force.
#$&*
The net force on the truck.
****
F_net = m a = 1400 kg * (+.417 m/s^2) = +580 N, or +580 kg m/s^2.
#$&*
`q005. Suppose the truck coasts up the incline, starting at a velocity of 15 m/s, and continues until its velocity has decreased to 10 m/s. The frictional force still has a magnitude equal to 2% of the total gravitational force on the truck.
Let the direction down the incline be positive.
Give each of the following, including its sign, its numerical value and its units:
The final velocity of the truck.
****
vf is still + 10 m/s.
#$&*
The change in the kinetic energy of the truck.
****
KE_0 = 1/2 m v0^2 = 1/2 1400*(10m/s)^2=700kg*100m^2/s^2=70000kg*m^2/s^2
KE_f = 1/2 m vf^2 = 1/2 1400*(15m/s)^2=700kg*225m^2/s^2=157500kg*m^2/s^2
So
`dKE = KE_f - KE_0 = 70000kg*m^2/s^2-157500kg*m^2/s^2=-87500kg*m^2/s^2
#$&*
The force of friction on the truck.
****
The force of friction is, as before, about -280 N.
#$&*
The component of the gravitational force parallel to the direction of the incline.
****
This is as before about +960 N, or +960 kg m/s^2.
#$&*
The net force on the truck.
****
The net force on the truck is unchanged, provided we continue to neglect air resistance.
#$&*
The acceleration of the truck.
****
The net force on the truck is unchanged so its acceleration is unchanged. Still + .417 m/s^2.
#$&*
`q006. This problem requires that you use the four equations of uniformly accelerated motion. If the truck reaches an incline on which its acceleration is .5 m/s^2, with velocity 10 m/s (both velocity and acceleration in the same direction), then how long will it take to reach a point 200 meters down the incline and how fast will it be moving at that point? You will need to carefully identify which of the quantities v0, vf, ds, dt and a are given. Then you should jot down the four equations of uniformly accelerated motion and select the one that most easily gives you additional information, and proceed from that point.
****
.5 m/s^2 is the acceleration a.
10 m/s is its initial velocity v0.
200 meters is its displacement `ds.
If downward is the positive direction then all these quantities are positive.
Given a, v0 and `ds we can use the third or fourth equation to obtain additional information.
Using the fourth equation
vf^2 = v0^2 + 2 a `ds
we find that
vf = +- sqrt( v0^2 + 2 a `ds) = +- sqrt( (10 m/s)^2 + 2 * .5 m/s^2 * 200 m) = +- sqrt( 300 m^2 / s^2) = +- 17.3 m/s.
This velocity is downward, so we discard the negative solution and conclude that
vf = 17.3 m/s.
Its average velocity is therefore
vAve = (10 m/s + 17.3 m/s) / 2 = 13.7 m/s.
It travels the 200 meters in time interval `dt, such that vAve = `ds / `dt. Thus
`dt = `ds / vAve = 200 m / (13.7 m/s) = 15 sec, very roughly.
#$&*
`q007. This problem is fairly challenging. I expect all University Physics students to get it, and hope that at least some General College Physics students will also get it despite the fact that it's probably at least a little bit too challenging for this point of the course.
Suppose the truck is moving up the incline, on which its acceleration is .7 m/s^2 down the incline. It passes you moving at 12 meters / second, and then passes your friend, who is standing 16 meters up the incline from you.
How long does it take the truck to travel the intervening distance?
****
We can choose either up or down the incline as positive.
Let's choose up as positive.
Then the acceleration, being down the incline, is
-.7 m/s^2.
The car is initially moving up the incline so its initial velocity is
v0 = +12 m/s.
The displacement is also up the incline so
`ds = + 16 m.
Writing down the four equations of motion we find that we can most easily get additional information using the fourth equation
vf^2 = v0^2 + 2 a `ds
which yields
vf = +- sqrt( v0^2 + 2 a `ds) = +- sqrt( (12 m/s)^2 + 2 * (-.7 m/s) * 16 m) = +- sqrt(144 m^2 / s^2 - 22 m^2 / s^2) = +- sqrt(122 m^2 / s^2) = +-11 m/s, approximately.
The truck is still moving up the hill so
vf = + 11 m/s.
Its average velocity on the interval is
vAve = (vf + v0) / 2 = (12 m/s + 11 m/s) / 2 = 11.5 m/s, so the 16 meter displacement requires time
`dt = `ds / vAve = 16 m / (11.5 m/s) = 1.4 s, approx.
#$&*
How far up the incline does it go before coming to rest?
****
For the interval on which it comes to rest, starting at your position, we have vf = 0.
v0 is still 12 m/s and a is still -.7 m/s^2.
We can easily reason this out:
vAve = (vf + v0) / 2 = (12 m/s + 11.5 m/s) / 2 = 11.75 m/s.
`dv = vf - v0 = 0 - 12 m/s = - 12 m/s.
Since a = `dv / `dt, we have `dt = `dv / a = -12 m/s / (-.7 m/s^2) = 17 sec, approximately.
At average velocity 11.75 m/s, in 17 s the car travels about
`ds = vAve * `dt = 11.75 m/s * 17 s = 190 m.
#$&*
If it then coasts back down the incline, accelerating at .5 m/s^2, how long will it take to travel from the position of your friend to your position?
****
The car has 190 meters to travel back to you, starting from rest.
For the corresponding interval, still using up as positive, we have
`ds = -190 m
a = -.5 m/s^2
v0 = 0
so that
vf = +- sqrt( v0^2 - 2 a `ds) = ... = +- sqrt(190 m^2 / s^2) = +- 13.7 m/s, very approximately.
The car reaches you with a downward velocity, so
vf = - 13.7 m/s.
To analyze the motion between your friend's position and yours we set up a new interval. The initial event is the car reaching your friend, the final event is the car reaching you.
For this interval, still using upward as positive
vf = -13.7 m/s
a = -.5 m/s^2
`ds = -16 meters.
The fourth equation again works:
vf^2 = v0^2 + 2 a `ds
Solving for v0 we get
v0^2 = vf^2 - 2 a `ds
so that
v0 = +- sqrt(vf^2 - 2 a `ds ) = +- sqrt((13.7 m/2)^2 - 2 * (-.5 m/s^2) * (-16 m) ) = +- sqrt( 174 m^2 / s^2) = +- 13.3 m/s, approx..
This velocity is downward, so it's negative and
v0 = -13.3 m/s.
The car reaches your friend at 13.3 m/s in the downward direction, and you at 13.7 m/s in the same direction, therefore averaging 13.5 m/s for a displacement of 16 meters (both down the incline).
The time required is therefore
`dt = `ds / vAve = 16 m / (11.5 m/s) = 1.4 sec, approx.
However if calculated more accurately, there is a difference between the time required for the car to travel up the incline past you and your friend, and the time required to travel the same distance coming down.
*@
@& Check the appended document (above this note) for full solutions and discussion.
I've inserted notes on some of your work.
You're doing well overall. But be sure you're aware of all the details.*@