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course phy 201
9/23 7:00 pm
A ball rolls down a 60 cm incline and off the end. As it drops to the floor it travels an additional 16 cm in the horizontal direction, and it is known that the drop to the floor requires 0.4 seconds. From this we conclude that the ball was moving at 40 cm/s at the end of the ramp.`q001. If the ball started from rest, and if the acceleration on the ramp was uniform, then what does the graph of velocity vs. clock time look like?
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A strait line, going up.
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What new quantity or quantities can we determine, based on our graph, and what is the value of each? (Quantities in which we might be interested include acceleration, average velocity, time interval, change in velocity and possibly others; which ones can be determined directly from the graph?).
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The initial velocity is where the line begins, which would generally be the origin (from rest). The final velocity is the y-value where the line ends. The average velocity is the y-value halfway between the two endpoints. The change in time is the x-value. We can find the change in velocity by putting the change in velocity (y-axis) over the change in time (x-axis), in other words, the slope.
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You are given that the ball traveled 16 cm along the incline and ended up with a velocity of 40 cm/s. So you can use the procedure you outlined to determine the values of your new quantities.
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`q002. From the given information we know that the average velocity of the ball is 20 cm/s. Having found this, how can we determine the time interval for the motion down the ramp?
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If the ramp is 60 cm long, and the average velocity was 20 cm/s, then it would take 60/20 seconds, or 3 seconds
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`q003. If for an interval on which the graph of v vs. t is a straight line we know two of the three quantities initial velocity, final velocity and average velocity, then we can find the third.
Suppose for such an interval we know that the initial velocity is 10 cm / sec and the final velocity is 30 cm/s. What is the average velocity?
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20 cm/s
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Suppose we know for a different interval that the initial velocity is 20 cm/s and the average velocity is 30 cm/s. What is the final velocity?
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Assuming that the acceleration is constant, than it would be 40 cm/s. If the average velocity (which lies halfway between initial and final) is 10 cm/s greater than the initial, it follows that the final velocity would be 10 cm/s greater than the average velocity.
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On such an interval where the final velocity is 50 cm/s and the average velocity is 30 cm/s, what is the initial velocity?
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Assuming acceleration is constant, then it would be 10 cm/s. The average velocity is 20 cm/s below the final velocity, so it would follow that the initial velocity was also 20 cm/s below the average
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Describe in words how you would get the final velocity for an interval on which the initial and average velocities are known.
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If you know the initial and average velocities, then you simply need to find the difference between the two. By virtue of being the average between two values, the average velocity will be the same difference from the initial and final velocities. Once you have the difference, simply add it to the average velocity.
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`q004. The average rate of change of A with respect to B is defined to be the change in A, divided by the change in B, where A and B represent specific quantities.
If the average velocity of an object on some interval is defined to be the average rate of change of position with respect to clock time on that interval, how then do we apply the definition of average rate of change to find the average velocity?
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We take the time that has passed, and multiply it by the average rate of change.
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That's not how you find the average velocity using the definition of average velocity.
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You haven't specified which average rate of change. However if you multiply ave rate of change of velocity by change in clock time you get change in velocity, and if you multiply ave rate of change of position by change in clock time you get change in position. In neither case do you get average velocity.
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If the average acceleration of an object on some interval is defined to be the average rate of change of velocity with respect to clock time on that interval, how then do we apply the definition of average rate of change to find the average velocity?
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The average rate of acceleration with respect to clock time will give us the average velocity.
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This isn't so. Check my previous note.
You need to start with the definition of the average rate. Every part of your calculation needs to be connected to that definition.
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What quantity could you find for an interval on which the position of an object changes by 40 cm while the clock time changes by 8 seconds?
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average velocity: 5 cm/s
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What quantity could you find for an interval on which the velocity of an object changes by 300 cm/sec while the clock time changes by 20 seconds?
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Average acceleration: 15 m/s^2
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If you know the average rate of change of A with respect to B, and you also know the change in B, how would you find the change in A?
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Apply the rate of change with respect to B, to B, and you will end up with the average rate of change for A.
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'Apply' could mean multiply, divide, add, subtract, take the square root of ...
Your answer here needs to be very specific. Be sure you connect your answer to the definition.
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If you know the average rate of change of A with respect to B, and you also know the change in A, how would you find the change in B?
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Apply the ratio of change in A with respect to B, to the change in A, and you will end up with the rate of change in B.
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Again your answer needs to be very specific.
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If you know the average velocity of an object on an interval, and know the change in position, how do you find the change in clock time?
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Take the average velocity, and divide it by the change in position. The result will be the change in time it took to cross that distance.
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You haven't connected this to the definition.
Your calculation will give you the reciprocal of the time interval.
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If you know the average acceleration of an object on an interval, and know the change in clock time, what other quantity could you find?
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You could find the average velocity. Multiply the change in clock time by the acceleration.
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If you multiply change in clock time by acceleration you don't get the average velocity.
You need to connect your answer to the definition of average rate.
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`q005. Let's return to situation we started with, the ball on the ramp. We know its initial velocity to be 0 and its final velocity to be 40 cm/s, and the length of the ramp to be 60 cm.
Summarize everything we can reason out from this data, assuming our v vs. t graph to be a straight line and using the definitions of average velocity and average acceleration.
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The average velocity would be 20 cm/s, halfway between the initial and final. The time it would take to travel, would be 3 seconds, as the average velocity as 20 cm/s and it travelled 60 cm. The acceleration would be roughly 6.67 cm/s^2. The average velocity with respect to time.
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There is no such quantity as the average velocity with respect to time.
The quantity average rate of change of velocity with respect to time is defined.
You need to use all the words, and be very specific in your application of the definition.
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`q006. For your first short rubber band, what was the color of this rubber band, and what lengths of this rubber band corresponded to what lengths of the rubber band chain?
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Yellow. 1st: 15 cm to 33.5 cm; 2nd: 16.5 cm to 43.5 cm; 3rd: 17.75 cm to 55.5 cm; 4th: 18.5 cm to 58 cm.
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Give the same information for your second short rubber band.
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Red. 1st: 12.5 cm to 38.5 cm; 2nd: 14 cm to 47.5 cm; 3rd: 15.5 cm to 54.5 cm; 4th: 16.5 cm to 61 cm.
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Sketch a graph of the length of the first short rubber band vs. the length of the chain. Note the convention that a graph of y vs. x has the y quantity on the vertical axis and the x quantity on the horizontal.
Do your points lie pretty close to a single straight line? Do you think there's a tendency of your data points to curve upward or downward?
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Yes, there honestly doesn’t seem to be much of a curve, though it could be up.
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Sketch the straight line you think lies closest, on the average, to the points of your graph. What is the slope of this line?
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roughly 0.122
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Repeat for your second short rubber band.
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roughly 0.138
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If your two short rubber bands were allowed to oppose one another, you would be able to take data for the length of one vs. the length of the other. Based on your graphs, what do you think would be the slope of a graph of the length of your first rubber band vs. the length of the second?
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roughly 2.5
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The slope of length vs. length of chain is very similar for both rubber bands.
This means that their force response is also very similar.
The slope of this graph should therefore be fairly close to 1. It wouldn't be 2.5.
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Based on your two graphs, construct a graph of the length of the first short rubber band vs. the length of the second. Describe your graph.
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The first rubber band increases at a slightly increased rate, causing a positive slope. It eventually rises to a significantly increased level compared to the second rubber band.
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`q007. Give your data for the the number of dominoes vs. the length of the rubber band chain.
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no dominos: 32.5 cm
2 dominos: 35 cm
4 dominos: 37.5 cm
6 dominos: 41.5 cm
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Sketch a graph of the number of dominoes vs. the length of the rubber band chain. Describe your graph.
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The graph seems to increase at a relatively steady rate, giving a mostly strait line. However, there is a slight curve upward.
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Sketch the straight line you think comes closest, on the average, to your data points. What is its slope?
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roughly 1.25 cm
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Technically this is correct. It would be better to express that as 1.25 cm / domino, just to remind yourself of what the number means.
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Relabel your graph using the following assumptions: Each cm marking on the ruler you used corresponds to an actual metric measurement of 0.006 meters, and each domino has a weight of 0.14 kilogram meters / second^2 (University Physics group) or 0.18 kilogram meters / second^2 (General College Physics group). In terms of your relabeling, what now is the slope of your straight line?
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If you express the slope as 1.25 cm / domino, then upon relabeling the slope would be
1.25 *(.006 meters) / (0.18 kg m / s^2) = .004 meter / (kg m/s^2) = .004 s^2 / kg.
However your graph was to be # of dominoes vs. length, the reverse of the order in which you appear to have graphed the data. So the slope of the requested graph would be the reciprocal of the slope you calculated, and the slope of the relabeled graph would be
1 / (.004 s^2 / kg) = 250 kg / s^2.
This turns out to be identical to 250 Newtons / meter.
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`q008. Give your data for the number of cycles in 10 seconds vs. the number of dominoes.
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oscillations in one minute:
-four dominos: 173 oscillations
-Six dominos: 121 oscillations
divide result by six to get the 10 second oscillations.
-four dominos: 28.83 oscillations
-six dominos: 20.16 oscillations
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Sketch a graph of your data. Is it plausible that your graph could be fit by a straight line?
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Not really. With only two points of data, a plausible line of best fit cannot be made.
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Figure out the time required for a single cycles, for each number of dominoes. Give your results.
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four dominoes: 10 seconds for 28.83 oscillations. Time for one oscillation: 0.347 seconds;
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Graph the square of the time required for a single cycle vs. the number of dominoes. Is it plausible that your graph could be well fit by a straight line? If so, give its slope.
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Once again, it is impossible to gauge an accurate line of best fit with only two data points.
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&#This looks good. See my notes. Let me know if you have any questions. &#
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You are more than welcome to submit revisions on any solutions on which you still have questions after reading my responses.
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