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course PHY 231
3:35 pm 9/14/10
More notes related to today's class will follow, and when they are complete they will be posted along with this part of the document. You should submit this using the Submit Work Form:Report your counts for the five trials with the toy car going in the first direction in the first line below, separated by commas. Report you counts for the five trials with car going in the opposite direction in the second line below, separated by commas.
5, 49, 4 38, 4, 38, 5 ,48, 6, 56
4, 30, 4, 44, 4, 35, 4, 40, 4 40
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Report the five resulting accelerations for the first direction in the first line below, separated by commas, and use the same format to report in the second line the five for the second direction. Starting in the third line show the details of how you found one of your accelerations.
-3.92, -4.75, -4.75, -3.84, -3.11
-3.75, -5.5, -5, -4.38, -5
Here I used x=v0*t+ 1/2*a*t^2, and vf=v0+a*t. I know vf=0, so a= -v0/t, and x= 1/2*v0*t. I substitute in for x and t to solve first for v0 and then a, e.g. 49= 1/2*vo*(5)---->v0=19.6 cm/c---->a = -19.6/5= -3.92 cm/c^2.
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How far from the lower end of the ramps did you have to position the two balls in order to synchronize their consecutive time intervals with the time interval for the third ball released from rest at the top of the third ramp?
I ended up at 8 cm.
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You didn't time the intervals for either trial. Suppose that the ball on the third ramp required 3 seconds to travel the 30-cm length of that ramp. What was the acceleration of that ball? Report the acceleration in the first line. Explain how you found it starting in the second line.
a= 6.667 cm/s^2. I used ∆x/∆t= 30/6= 10 cm/s, v0=0 ----> vf= 2*10= 20 cm/s---> 20=a*t---->a= 20/3= 6.667 cm/s^2.
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Assuming the 3-second interval for the ball on the third ramp, what was the time interval for ball released on the first ramp, and what was the resulting acceleration? Report the acceleration in the first line, and explain how you calculated it starting in the second line.
This is confusing. My ramps were labelled A, B, and C where A is the 30 cm, C is released with A, and B released when A 'stopped'. So I'm not sure which ramp you mean, B or C. Assuming C, released together, I take this to mean that however long it took A to traverse 8 cm is also how long it took C, and I found that to be t=1.55 seconds. So C went 8 cm/ 1.55 seconds-----> v_Ave= 5.16 cm/s, then vf= 2*5.16= 10.3, a= 10.3/ 1.55= 6.66 cm/s^2. So this means that it took B, released last, 1.45 s to traverse 8 cm, 8/1.45= 5.52 cm/s---> vf= 11 cm/s--> a= 11/ 1.45= 7.59 cm/s^2.
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Are your two results for the accelerations reasonably consistent? Why would you or would you not expect them to be so?
A and B accelerations seem to be reasonably close but C is very suspect. For very similar balls on very similar ramps and conditions I would expect these to be very close to one another. I've looked at this some more and it should be released at 7.5 cm for all the accelerations to be the same with 30 cm in 3 seconds for the A ball. Though I do realize that 8 cm might be correct for my actual scenario if I had timed the A ball.
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Are your results in any way relevant to the problem of ordering v0, vf, `dv, v_mid_x, v_mid_t and vAve? If so, what conclusions can you draw?
The ordering here seems to be close to saying that v_mid_t= v_Ave for constant acceleration, something I asserted in the last work document. Here we also know that v0 = 0 and vf = 2* v_Ave and ∆v= vf. v_mid_x doesn't seem to popup here, at least as far as I can see.
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This problem is optional for Phy 201 students. University Physics students should attempt to solve this problem: Use your knowledge of uniformly accelerated motion to answer the following: If the accelerations are uniform on all three ramps, with all three accelerations being equal, then at what position would the balls on the first two ramps need to be placed in order to achieve the desired result?
Based on my observations the C ball, short distance at same relase time, should be released from the distance that the A ball, longer distanced ball, has traveled when t= ∆t/2. In a general symbolic form this works out to mean that if x_A= 1/2*a*t^2 then x_C=1/2*a*(t/2)^2, the latter simplifies to: x_C= 1/4( 1/2*a*t^2). Substituting the x_A in for (1/2*a*t^2) gives: x_C= 1/4 x_A. I'm not sure if that is all I need to do to show what I mean and it definitely doesn't seem enough to 'prove' for all cases that 1/4 x_A will always work but I am pretty confident that there is something to this. The assumption of t/2 seems to be the part that is still iffy for me but if it is true then my assertion will hold that it will be 1/4 the distance.
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Brief notes
Force and energy:
1 Newton is the net force required to accelerate a 1 kg mass at 1 meter / second^2.
1 Joule is the energy required to exert a force of 1 Newton on an object which moves a distance of 1 meter in the direction of the force.
Newton's Second Law: F_net = m * a, where m is the mass, a its acceleration and F_net the net force acting on the mass.
Definition of work: The work `dW performed by a force F_net acting in the same direction of as the displacement `dx is `dW = F_net * `dx. (For University Physics students: The force doesn't have to be in the direction of the displacement. For any force and displacement `dW = F_net dot `dx).
We have a lot of details to work through in order to understand forces, work and energy. However the following 'story line' might provide a useful reference point as we continue:
When I brought the toy car (and its magnet) closer to the fixed magnet, I had to exert an increasing force through a displacement. So I had to do work. That work obviously came from the Cheerios I ate for breakfast. Naturally I had to breathe more air, in order to oxidize more Cheerios and give off the energy I required to do work. This is why I got so out of breath.
Once the car was in position near the fixed magnet, all I had to do was release it and it quickly accelerated away (that acceleration was nonuniform, falling off as the car moved further away from the magnet). I did work on the car, most of which was transformed into the potential energy of the system as it awaited release. Upon release that potential energy became kinetic energy (energy of motion), which was gradually dissipated by friction as the car rolled across the tabletop and eventually came to rest."
&#This looks good. Let me know if you have any questions. &#
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