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course Phy 241
8/29 9 pm
Explain your reasoning as you answer each question. Not all questions are easy, but do your best on each:If a ball requires 2 seconds to accelerate from rest along 3 dominoLengths of track, what was its average speed during that time?
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Since the ball moved 3 dominoLengths (d) in 2 seconds (s), its average speed is 3 d/ 2 s, or 1.5 d/s.
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Answer the same question if the time interval for the 3 dominoLength distance is 1 second, and again if the time is 1.5 second.
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Using the same process as the previous question, I can tell that if the ball moved 3 d in 1 s, its average speed was 3 d/s. If it moved 3 d in 1.5 seconds, the average speed is 3 d/ 1.5 s, or 2 d/s.
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Does the average speed change as much between the 1 second and 1.5 second time intervals as between the 1.5 and 2 second intervals?
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The average speed changes more between the 1 second and 1.5 second intervals.
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For the case where the ball required 1.5 seconds to coast from rest through 3 dominoLengths, the initial speed of the ball was zero. You have calculated its average speed. What final speed, averaged with the initial speed, would result in the same average speed?
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When the ball took 1.5 seconds to travel 3 dominoLengths, the average speed was 2. This means when you average the final velocity (Vf) and initial velocity (Vi) together, you will get 2. The initial velocity was 0, so:
½ (Vf + Vi) = 2
(Vf + 0) = 4
Vf = 4 d/s
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Assuming the final speed is the one you just calculated, how quickly was the speed changing during the 1.5 seconds? Your final answer to this question should be one single quantity.
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The speed was changing at a rate of (4d/s)/ 1.5s, or approximately 2.66 d/ s^2
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For the 1 second and 2 second intervals, how quickly was the speed changing?
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The final speed for the 1 second interval was 6 d/s, so the speed was changing at a rate of 6 d/ s^2.
The final speed for the 2 second interval was 3 d/s, so the speed was changing at a rate of (3 d/s)/2 s, or 1.5 d/ s^2.
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If you had four dominoes with thicknesses 10 mm, 11 mm, 12 mm and 13 mm, and repeatedly scrambled and stacked them into stacks of two, it would be possible to get stacks of equal height and stacks of unequal height.
How many different heights are possible?
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There are 5 different height possibilities: 21mm, 22 mm, 23 mm, 24 mm, and 25 mm.
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What height differences are possible between the two stacks (0 would be regarded as a height difference)?
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The possible height differences are 4 mm, 2 mm, and 0.
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What would be the probability of each of the possible height differences?
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Since there are only three possible height differences and all three are equally likely, the probability of each is 1/3.
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Have you really examined your assumption that all three differences are equally likely?
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If you had six dominoes with heights 10, 10.5, 11, 11.5, 12 and 12.5 cm, what height differences are possible when comparing stacks of three, and what is the probability of each on a random stacking?
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The possible height differences are: 4.5 cm, 3.5 cm, 2.5 cm, 1.5 cm, and 0.5 cm.
There are seven possible ways to combine the dominoes, so the probability of each difference is:
Height difference Probability
4.5 cm 1/7
3.5 cm 1/7
2.5 cm 2/7
1.5 cm 1/7
0.5 cm 2/7
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If each domino pip is a hemisphere and the diameter of a pip is 1/4 of the domino's width and 1/2 of its thickness, then what percent of the domino's volume would need to be removed to make a single pip?
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I will use the thickness of the domino (T) as the unit of measurement. If the diameter of the pip is half the thickness of the domino (1/2 T), then the radius would be 1/4 T. Since a pip is a hemisphere, I will use the following expression:
½ * (4/3 * pi * r^3)
2/3 * pi * (1/4)^3
2/3 *pi * (1/64)
pi / 96 = 0.032 T approx.
Given the width of the domino is twice the thickness and the length is about twice the width, I can find the volume of the domino to be (1 * 2 * 4) T, or 8 T. To find out the percentage value of one pip, we divide its volume by the total volume.
0.032/ 8 = 0.004, or 0.4%.
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How much would the presence of pips add to the percent uncertainty in the volume of a domino?
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Most dominos have between 0 and 12 pips, so if each pip is worth 0.4% of the volume, we multiply 0.4% by 12, and that will give us the percentage variation.
0.4% * 12 = 4.8%
However, this percentage is the difference between one extreme and the other. We want the percentage to apply to the average value, so we divide 4.8 by 2, which gives us 2.4%
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If the thicknesses of a set of dominoes varies by as much as 5%, then how much uncertainty would you expect in the actual weight of an object which was found to be equal to that of 4 randomly selected dominoes?
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If there 5% uncertainty of the thickness of a domino, assuming the length and width are constant that would cause a 5% uncertainty of the volume of the domino. Take into account that the pips can cause as much as a 2.4% variation in the dominoes. Together that makes a total possible uncertainty percentage of 7.4%, though this is the percentage difference between the two extremes and is unlikely to occur. Since this is a percentage, whether the number of dominoes used is 1 or 100 the answer will not change.
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Very well done. Do check my one note.
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