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course PHY 231
9/12 5:49 pm
If you weren't in class to do this:We measured the landing positions of the ball after rolling down three ramps, one supported by a domino lying flat on its side (least steep), one supported by the domino lying on its long edge and one supported by the domino lying on its short edge (steepest). You have some dominoes, a ball and a ramp so if you didn't get to do this at the beginning of class, you should be able to do it now.
Assuming that the ball fell to the floor in .4 seconds, after leaving the end of the ramp, and that after leaving the ramp its horizontal velocity remains constant:
How fast was it traveling in the horizontal direction when the domino was lying flat on its side?
5 cm/ 0.4 s = 12.5 cm/s
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How fast was it traveling in the horizontal direction when the domino was lying on its long edge?
16.5 cm/ 0.4s = 41.25 cm/s
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How fast was it traveling in the horizontal direction when the domino was lying on its short edge?
27 cm/0.4 s = 67.5 cm/s
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Pendulum count:
What was the length of the pendulum you counted, and how many counts did you get in 30 seconds?
L= 11.5 cm and 43 counts / 30 seconds.
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What therefore is the period of motion of that pendulum?
T= 30/43 = 0.682
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How does your result compare with the formula given on the board, T = .2 sqrt(L) where T is period of oscillation in seconds and L the length in centimeters?
My T= 0.682 s and formula T= 0.678 s.
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How well did the freely oscillating pendulum synchronize with the bouncing pendulum of the same length? Which was 'quicker'?
For about the first 4 to 6 bounces it was as close as I could tell but then the bouncing one deteriorated very quickly and it was hard to tell. It seems though that bouncing one was more sensitive to very small differences in release height.
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Ball drop
From what height did the drop of the ball synchronize with the second 'hit' of the pendulum, and what was the length of the pendulum?
My pendulum here was still 11.5 cm and my ball drop height was 221 cm.
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How long should it have taken the pendulum between release and the second 'hit'? On what do you base this answer?
I saw that two hits was equal to one oscillation in the swinging pendulum and the formula gave 0.678s for that.
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Given you answer to the preceding, you know the time required for the ball to fall from rest to the floor, and you know how far it fell. What therefore was its acceleration?
2.21 cm/0.678 s= 3.26 m/s =v_Ave ---> 2*3.26= 6.52=v_f ---> 6.52/0.678= 9.61 m/s^2= a_Ave. So I found my ball to be accelerating at 9.61 m/s^2 which is pretty close to the actual value of g.
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Ball down long ramp
How would you design an experiment to measure the velocities v0, v_mid_x, v_mid_t and v_f for different values of v0?
Well there are all kinds of way to do this but I will stick to the methods we have been using thus far. To be able to set v0 on the second ramp to whatever we want we can make measurements of the first ramp to find v_f at whatever x0 so that we will obtain our desired value of v0 for the second ramp. Though it seems that you want us to assume that we don't know v0 and I'm having some difficulty imagining a way to find anything but ∆x, ∆t, and v_Ave without already knowing v0 by either my first method above or some device to measure v0 from the start. I do know that to find v_mid_x we just time the ball to the middle of the ramp, which is 1/2 ∆x and for v_mid_t we measure the distance that the ball travels in 1/2 ∆t time. However, without knowing v0 these would only be average values.
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How would you design an experiment to measure v0 and `dv for different values of v0?
Here again this has the same trouble for me as above since without v0 I don't see how you find ∆v.
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Rotating strap
For the strap rotating about the threaded rod, give your data indicating through how many degrees it rotated, how long it took and the average number of degrees per second. Report one trial per line, with a line containing three numbers, the number of degrees, the number of seconds, and the average number of degrees per second, separated by commas.
385º, 5 s, 77º/s
1320º, 10 s, 132º/s
1325º, 5s, 265º/s
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To do with the materials you took home:
Using the TIMER program (link to Java version: Timer_b) with the materials you took home:
Bracket pendulum:
Shim the bracket pendulum until the 'strikes' appear to occur with a constant interval. Click when you release the bead, then click for alternate 'strikes' of the ball on the bracket pendulum (that is, click on release, on the second 'strike', on the fourth 'strike', etc., until the pendulum stops striking the bracket). Practice until you think you think your clicks are synchronized with the 'strikes'. Report the length of the pendulum in the first line, then in the second line report the corresponding time intervals below, separated by commas:
I have a hard time with this as I can't clearly make out the hits past 4 and even with practice I still get erratic results so my 'best' times are what I'm reporting.
L= 11.5 cm
0.689, 0.624, 0.608, 0.689, 0.663
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Using the same length, set the pendulum so it swings freely back and forth. Click each time the bead passes through the equilibrium position. Continue until you have recorded 11 'clicks'. Report the corresponding time intervals below in one line, separated by commas.
0.424, 0.384, 0.425, 0.439, 0.44, 0.384, 0.447, 0.441, 0.408, 0.383, 0.384
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For both sets of trials, how do your results compare with the prediction of the formula T = .2 sqrt(L)?
The bouncing pendulum was much closer than my swing counts as my T=0.2√(11.5) ≈ 0.678
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Ball down ramp:
Do your best to take measurements you can use to find vf, v_mid_x, v_mid_t and `dv using your ramp and ball, releasing the ball from rest. (You could use the TIMER to get decent data. If you wish you can use the fact that a ball falling off a typical table or countertop will reach the floor in about .4 seconds. Note: Don't let the ball fall on a tile or vinyl-covered floor. You don't want broken tile, and you don't want dents in your vinyl. You could put your book on a carpeted or otherwise protected floor and land the ball on the book.)
Briefly describe what you did and what your results were:
Okay, I'm confused. Why do we need to let the ball fall if we are measuring the balls ramp velocities? there's a very good answer to this question but I'm probably going to let people think about it for awhile longer.
Since the problem doesn't specify, I am going to use two dominos. I used the TIMER here and I found that the ball traveled 30 cms in 1.4 seconds, so v_Ave= 30/1.4= 21.4 cm/s. Now since v0=0 ---> v_f= 2(v_Ave)= 42.8 cm/s. Next I used a ruler to the find mid_x, which here is 15 cm, then I found the time it took to travel that distance and it was 1 second. So the average value of v_mid_x is 15 cm/s and since v0= 0 ----> v_mid_x = 2(15)= 30 cm/s . I estimated v_mid_t to be about half the distance of mid_x and clicked the timer at that 7.5 cm mark and that did it. So the average value of v_mid_t is 7.5/0.7 = 10.7 cm/s and since v0= 0 ----> v_mid_t = 21.4 cm/s. This is interesting that the v_mid_t position is exactly half of mid_x
very good
we can use the equations to obtain the same results
Galileo did something very similar to get the ball rolling, so to speak, in the development of the science of physics.
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Rotating strap:
Let the strap rotate on the threaded rod, as before. Click the TIMER at the start, and then at 180 degree and/or 360 degree intervals (the latter if it's moving too fast to do the former). Copy the output of the TIMER program below:
1 13.464 13.464
2 14.271 0.807
3 15.102 0.831
4 16.454 1.352
5 18.19 1.736
(I counted 360º rotations and the first time is the time since the page reloaded and the last time is when the strap stopped rotating.)
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On the average through how many degrees per second was the strap rotating during each interval? Report in a single line, giving the numbers separated by commas. Starting in the second line explain how you did your calculations.
446, 433, 266, 207, (176)
I divided 360º by the time between clicks, and the last number I found by measuring how far the strap had rotated between the last click and when it stopped, which was 305º.
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The second column of the TIMER output shows the clock times. For a given interval the 'midpoint clock time' is the clock time in the middle of the interval. Report clock times at the beginning, middle and end of your first interval in the first line below. Do the same for your second interval, in the second line. Starting in the third line explain how you got your results.
13.464, 13.8675, 14.271
14.271, 14.6865, 15.102
Here I added the beginning and end times and divided by two.
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These should be submitted using the Submit Work Form. You can submit the entire document at once, or you can submit the document in parts.
Very Short Preliminary Activity with TIMER (should take 5 minutes or less once you get the TIMER loaded)
This exercise can be put off until you are near a computer. However it is best done before some of the problems that follow. If you can't do it before starting the problems, at least imagine doing it, actually doing the 8-counts and clicking an imaginary mouse, and making your best estimate of the time intervals.
Click the mouse as you start an 8-count, doing your best to count at the same rate you used in class. Complete four 8-counts and click the mouse again. Note the time interval required to complete your set of four 8-counts.
Repeat four more times.
Report your five time intervals in the first line below, separated by commas:
15.964, 15.735, 15.463, 16.187, 15.658
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Based on your results, how long does your typical 8-count last?
Typical would be about 15.7 seconds.
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Based on your result, what is the time interval of each of your counts?
3.93 seconds
that's time between 8=counts; time between individual counts in much shorter
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If you counted the motion of a ball down the ramp, completing two 8-counts and 1-2-3-4-5 of a third, how long would you conclude the ball spend moving down the ramp? Based on the TIMER data you reported above, what do you think is the percent uncertainty in your result?
It took about 10.3 seconds with about 1% uncertainty, so about ±0.1 second.
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Preliminary problems:
a. A ball travels down a ramp in 2 seconds, accelerating uniformly. Its initial velocity on the ramp is 20 cm/s and its final velocity is 40 cm/s.
• Reasoning from the definitions of velocity and acceleration, and assuming a linear v vs. t graph, how long is the ramp, and what is the ball's acceleration (i.e., rate of change of velocity with respect to clock time)?
• Ramp length is (40 + 20)/2 * 2 = 60 cm, and the acceleration is (40-20) / 2= 10 cm/s^2.
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• In the first line below list the quantities v0, vf, `ds, `dt and a for this motion and give the value of each (or as many as you were able to identify or reason out). In the second line identify which of the quantities were given, and which were reasoned out. In the reasoning process you would have found vAve and `dv; identify these quantities also and give their values.
• v0 = 20 cm/s, vf= 40 cm/s, ∆s= 60 cm, ∆t = 2 seconds, a = 10 cm/s^2.
• v0, vf, and ∆t were all given and ∆s and a were reasoned out using those. vAve= (v0+vf)/2 and ∆v= vf-v0 and they were 30 cm/s and 20 cm/s respectively.
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b. A ball travels down a ramp for 3 seconds, starting with velocity 20 cm/s and with its velocity changing with respect to clock time at 10 cm/s^2.
• Reasoning from the definitions of velocity and acceleration, and assuming a linear v vs. t graph, how far did the ball travel along the ramp, and what is the ball's velocity at the end of the 3 seconds?
• Final velocity is easily seen to be 20+10*3= 50 cm/s and distance is then (20+50)/2 *3 = 105 cm.
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• In the first line below list the quantities v0, vf, `ds, `dt and a for this motion and give the value of each (or as many as you were able to identify or reason out). In the second line identify which of the quantities were given, and which were reasoned out. In the reasoning process you would have found vAve and `dv; identify these quantities also and give their values.
• v0= 20 cm/s, vf = 50 cm/s, ∆s= 105 cm, ∆t = 3 seconds and a = 10 cm/s^2
• v0, ∆t, and a were all given and ∆s and vf were reasoned from those.
• vAve is defined above and is 35 cm/s and ∆v is also defined above and is 30 cm/s.
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c. A ball travels 30 cm down a ramp in 5 seconds, ending with a velocity of 20 cm/s.
• Identify, by giving the value of each, which of the quantities v0, vf, a, `ds and `dt are given.
• ∆s= 30cm, ∆t = 5 seconds, vf = 20 cm/s.
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• Identify which of the four equations of uniformly accelerated motion contain the three given quantities (identify all the equations that apply; there will be at least one such equation, and no more than two).
• ∆x= (v0+vf)/2 * ∆t is the only one that has all three.
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• For each of the equations you identified, identify the quantity that was not given, and do your best to solve that equation for that quantity.
• v0 was not given though I found it to be 8 cm/s. Symbolically we have v0= 2*(∆x/∆t)- vf (the problem with this is that this gives the negative of v0 and I'm not sure how to change that.)
If the object was speeding up 2 `dx/`dt would be greater than vf so you would get a positive result for v0
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d. A ball travels 30 cm down a ramp, accelerating at 10 cm/s^2 and ending with a velocity of 20 cm/s.
• Identify, by giving the value of each, which of the quantities v0, vf, a, `ds and `dt are given.
• Here we have ∆s, a, and vf, which are 30 cm, 10 cm/s^2, and 20 cm/s respectively.
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• Identify which of the four equations of uniformly accelerated motion contain the three given quantities (identify all the equations that apply; there will be at least one such equation, and no more than two).
• vf^2 = v0^2 + 2*a*∆s is the only with all three quantities.
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• For each of the equations you identified, identify the quantity that was not given, and do your best to solve that equation for that quantity.
• The unknown in the above equation is v0 and it is found using v0=√(vf^2-2*a*∆x) (This again presents a problem because the vf^2 is less than 2a∆x and then the square root is a complex number. I'm not sure why this is happening in this one and the above equation.
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There is no guarantee that a stated situation is physically possible. In this case the situation is not physically possible, so it's good that you couldn't get a solution.
All students should be able to make a good attempt at the questions in the Preliminary Questions, though it is expected that there will be questions on some of the details.
Problems
1. Each ramp used in constructing the series of ramps used in today's lab was 24 inches long. The 5-ramp series had a length of 10 feet, or about about 300 cm. Assume that the ball takes 10 seconds to travel the length of the ramp when released from rest. If this time interval is accurate, then what is the value of each of the following:
• The average velocity of the ball on the ramp.
• ∆x/∆t= 300/10 = 30 cm/s= v_Ave
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• The final velocity of the ball on the ramp.
• Since v0=0, vf = 2* v_Ave= 60 cm/s
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• The change in the velocity of the ball from start to finish.
• ∆v= vf-v0= 60-0= 60 cm/s
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• The average rate of change of the velocity with respect to clock time.
• ∆v/∆t= 60/10= 6 cm/s^2= a_Ave
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• The velocity of the ball at the midpoint of the ramp.
• mid_x= ∆x/2= 150 cm, 150=a/2 t^2= 6/2 * t^2, t=√(50)= 5√2≈7.1 seconds, v_mid__x=6*7.1= 42.6 cm/s.
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• The velocity of the ball at the clock time halfway between the start and the ball reaching the end of the ramp.
• ∆t/2= 5 seconds, v(t)= v0+a*t= 0+6*5= 30 cm/s
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Hint: sketch a trapezoid that you think represents the v vs. t behavior of the ball on the ramp (... time on each ... need equal-area divisions ... etc.)
It is expected that some phy 201 students will be able to make a good attempt on all the above questions, and all should be able to answer the first two and make a good attempt on the next two. University Physics students should be able to make a good attempt on all questions.
2. Based on your in-class counts and your timing of your counts, estimate as accurately as you can the time required for the ball to travel the length of this series of ramps, starting from rest.
• Find the average velocity of the ball, and based on this result find its final velocity.
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• Using your results, find its acceleration.
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• Find the velocities v_mid_x and v_mid_t.
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Everyone should be able to do the first. University Physics students should certainly be able to do the second, and General College Physics students should be able to make a good attempt.
3. If the ball was given an initial velocity of 20 cm/s, then given the acceleration you found in the preceding problem:
• How long would it take the ball to travel the length of the ramp, and what would be its final velocity?
I had 3 + 7/8 counts and based on my timing that is 15.2 s, so 300 cm/ 15.2 s= 19.7 cm/s and v0=0 so vf=2*19.7= 39.4 cm/s
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• Where would it be at the halfway clock time?
∆t/2= 7.6 s, ∆v/∆t= 39.4/15.2= 2.59 cm/s^2, x(7.6)= 1/2*a*t^2= 74.8 cm.
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• How fast would it be moving at the midpoint between the two ends of the ramp?
• mid_x= ∆x/2= 150 cm, 150 = 1/2 a t^2, t=√(116)≈10.8 s, v(10.8) = a*t= 28 cm/s.
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• What would be its velocity at the halfway clock time?
• v_mid_t= a*t= 2.59*7.6= 19.7 cm/s
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• What would be the change in its velocity from one end of the ramp to the other?
• ∆v= vf-v0= 39.4 - 0= 39.4 cm/s
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Everyone should be able to make a good attempt at some of these questions. University physics students should be able to make a good attempt at all.
4. A ball requires a count of 24 to accelerate from rest down a 60 cm ramp. It rolls from that ramp onto an identical ramp with an identical slope, and requires 13 counts from one end of the ramp to the other. Does it lose any speed in making the transition? If you simply answer 'yes' or 'no' without supporting your answer in detail, you haven't answered the question.
It did lose speed in the transition. Heres why (note c=counts). v_Ave_1= 60/24= 2.5 cm/c and v_Ave_1+2= 120/37= 3.2 cm/c, so vf_1= 2*2.5= 5 cm/c and v_f_1+2= 2*3.2 = 6.4 cm /c. So, for ramp two we should have v_f_2= 6.4 cm/s and v0_2= 5 cm/c, then v_Ave_2 should be 5.7 cm/c but it is in fact v_Ave_2 = 60/13= 4.6 cm/c. Since this last value is less than the expected value we have v0_2 < vf_1 and in fact the actual of v_0_2 is found by the following: 4.6= (6.4+v0_2)/2 -----> v0_2 = 2.8 cm/c and so the ball lost 5-2.8= 2.2 cm/c in speed during its transition. Whew, I'm by no means certain here but this is my instinctive reaction to this problem.
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This question is somewhat challenging. You should begin by figuring out everything you can from the given information. Then see how your information might be used to answer the question.
5. I just timed myself for five sets of counts, with four fast 8-counts in each set (similar to the preliminary exercise I asked you to do at the beginning of these problems). My times for the sets were all between 4.4 and 4.6 seconds. Starting with 1 at release and counting until the ball reached the end of the last ramp, I counted two sets of four 8-counts, plus a count of 1-2-3 at the end. On three additional repetitions I always got two sets of four 8-counts, and the counts at the end were always 1-2 or 1-2-3. Based on these figures:
• What is the best estimate of the time required for the ball to travel the entire distance?
• My estimate is based on an average of the count times, 4.5 s, then 2 sets of 4 8 counts= 9 seconds, and dividing 4.5/32 = 0.14 s/#. Here though I will subtract one from the count (fencepost: I did not take this into consideration before) and so on average I found you had 2 sets of 4 8 counts + 2/32 counts = 9.3 seconds.
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• What is the percent uncertainty in the time required for the ball to travel down the ramp, based on the given information and without making any extraneous assumptions?
• Probably around ±0.1 seconds so 1% uncertainty. The first part seems right but looking at just 1% seems a little low.
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4.5 sec +- .1 sec is about a +- 2% uncertainty.
• How long should the ball have spent on the first ramp, if the acceleration was indeed constant?
• About 6.6 seconds.
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• If I got to the 8 of the third set of 8-counts by the time the ball reached the end of the first ramp, what was the acceleration on that ramp?
• Here I found ∆t to be 3.2 seconds so v_Ave= 60/3.2= 18.75 cm/s, vf= 18.75*2= 37.5 cm/s=∆v, ∆v/∆t= 37.5/3.2= 11.7 cm/s^2.
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Everyone should be able to answer the first questions and make a good attempt on the others.
6. The ball is at the end of the first ramp when I reach the count 1 of the fourth set of 8-counts. Where will it be when I get to the 1 of the seventh set of 8-counts, assuming a constant acceleration throughout?
Here we have: v0= 37.5 cm/s, x0=60 cm, t0 = 3.4 s and tf= 6.8 seconds, thus ∆t = 3.4 seconds. x=x0+v0*t+1/2*a*t= 60 + 37.5*3.4+1/2*11.7*3.4^2= 255 cm.
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7. I give the ball a quick flick, starting from the low end of the ramp, starting my count at 1 at the instant the ball leaves the end of my finger. It traveled up the ramp through one count of 8, and came to rest for an instant as I counted 5 during the next count of 8. I continued my count as it rolled back down, getting to the end of my third count of 8 and reaching 1-2 of the next set of 8 before the ball reached its original point.
• Was the magnitude of the ball's acceleration the same going up as coming down?
• Not quite because by the fact that 8 + 5/8 -1=12 and 8*3+2-1= 25 so it took 12 to go up and 25-12= 13 to go down. (I subtract one again because of fencepost).
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• If not, what was the approximate percent difference in the accelerations?
• The difference between the two is about 0.14 seconds so I would say roughly 14%.
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8. If acceleration down a ramp is constant, then where will an object released from rest reach its average velocity? This will be at mid_t.
If the initial velocity is not zero, how will this affect the position at which the object reaches its average velocity? If the acceleration is the same for both cases, v0=0 and v0 ≠ 0, then it doesn't matter: v_Ave will always be at mid_t for the same acceleration.
"
Very good. See my notes.
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