IDS 101 Fall 2006 Study Guide For Exam 2. Primary ideas:

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1 IDS 101 Fall 2006 Study Guide For Exam 2 Primary ideas: 1) You should be able to construct a position-time graph given some data and be able to interpret the meaning of the slope of the line. If you are given a graph of position versus time for some object you should be able to describe the motion of the object. 2) You should be able to identify and discuss what a negative slope of the trendline means on a position-time graph. You should be able to predict what the position vs. time graphs would look like for two objects that have the same speed but opposite directions. You should be able to find the velocity and/or speed of an object from a position vs. time graph. 3) You should be able to determine the velocity and discharge of a stream given some data. 4) You should be able to evaluate what is needed to produce balance in a given situation and determine the turning effect of varying masses of objects. 5) You should be able to determine the time it would take to accelerate a specific mass to given velocity (and the variations such as the effect of adding more mass to the object and the velocity of an object if given the rate of acceleration and other variations on this theme). 6) You should be able to relate mass, density and volume. You should be able to compare the masses of objects given varying information about their density and volume. You should be able to compare the volumes of objects given varying information about their density and mass. 7) You should be able to predict floating/sinking behavior if given information on density, volume, and mass. 8) Use isostasy to explain the variations in the elevation of the Earth s surface. Some practice questions from previous exams follow on the extra pages. Note than in addition to the questions below, you should consider all of your past end of modules questions (since the last exam) as practice questions for this exam.

2 1. In ancient Mediterranean cultures, coins were made out of pure metal and the value of a coin was determined by the mass of the metal used to make the coin. Once upon a time, three merchants from different cultures met to buy and sell goods. Merchant #1 had a pocket full of Nara coins, Merchant #2 had a pocket full of Filsa coins, and Merchant #3 had a pocketful of Mac coins. The first thing they needed to do was figure out how many Nara in a Filsa, how many Filsa in a Mac, and so on. a) Explain in words (and pictures if you like) how the three merchants could have determined the relative values of their coins using only a few sticks, some cloth, and some string. Any description of the creation of a balance and use of the balance to measure mass is okay here. One could hang a long stick by a string tied around the middle and then hang cloth pouches off of each end to make a two "pan" balance. b) It turns out that four Filsas have the same mass as one Mac. Three Macs have the same mass as two Naras. i) How many Filsa would have the same value as 1 Nara? (Show your work) 4 Filsas = 1 Mac 3 Macs = 2 Naras So 1 Nara would be the same as one and a half (1½) Macs. One Mac is 4 Filsas and one-half of a Macs is 2 Filsa, so 1½ Macs is (4 Filsa + 2 Filsa) or 6 Filsa. ii) Hoping that Merchant #3 didn't pass IDS 101, Merchant #2 decides to try and fool her. He quickly demonstrates that he can balance one Mac with one Filsa. How did he do it and where did he have to hang the two coins to pull of his trick? Since the Mac has more mass than the Filsa, Merchant #2 has to cleverly hang the Filsa farther from the center of the balance than the Mac. Since the Mac has FOUR TIMES as much mass as the Filsa, Merchant #2 has to hang the Filsa FOUR TIMES FARTHER from the center than the Mac. 2) The following questions all refer to a set of experiments with the balance shown below. A meter stick is balanced on a pivot point. Four pieces of aluminum metal are then placed 22 cm from the balance point. Each piece of aluminum has dimensions of 1 cm x 1 cm by 1 cm.

3 22 cm a. If you remove just one of the four pieces of aluminum, where will you have to place it on the other side to balance the remaining three pieces? Show your work or explain your reasoning. If we remove one block, that leaves three behind. In order to balance those three with the one in our hand, we have to place that one three times as far from the fulcrum. Three times as far away would be 66 cm from the center, so 66 cm away on the right. b. Next you are going to try balancing all four pieces of aluminum (in the original position shown in the diagram) with one piece of lead. Lead is 4 times as dense as aluminum. If you use a 1 cm x 1cm x 1 cm piece of lead, where will you have to place the lead to balance the four pieces of aluminum? Show your work or explain your reasoning. If lead is four times as dense as aluminum, then one block of lead will have the same density as four blocks of aluminum of the same size. Ah ha! We have four blocks on aluminum of the same size on the balance already. We only need one block of lead to have the same mass, and since they have the same mass we can balance them by putting them the same distance away from the fulcrum. So place this block 22 cm away on the right. If you use three of pieces of lead, where will they need to be placed in order to balance the four pieces of aluminum (in their original position)? Show your work or explain your reasoning. Three of these blocks of lead will have three times as much mass as one block of lead which in turn has the same mass as four blocks of aluminum. Thus based on the reasoning above, three blocks of lead have three times as much mass as four blocks of aluminum. Since they have more mass they need to be placed closer to the fulcrum to get the lever to balance. They have three times as much mass so they need to be placed one third of the distance away, or 7.3 cm away on the right

4 Finally, imagine repeating the experiment with a 2 cm x 2 cm x 2 cm piece of lead. Where should this piece of lead be placed in order to balance the four pieces of aluminum? Show your work or explain your reasoning. 2 cm x 2 cm x 2 cm = 8 cm 3 of lead. Since lead is four times as dense as aluminum, this would have the same mass as 8 x 4 = 32 of the original aluminum blocks. This is eight times as many blocks as were on the lever on the left-hand side, so they need to be placed 1/8 of the distance away. This turns out to be 2.75 cm away from the fulcrum on the righthand side. 3. (To make this problem easier to complete we have downsized the Earth to a sphere that is cm in radius and the mantle/core boundary is 30 cm from the surface (these dimensions are times smaller than the real Earth). The volume of a sphere is 4/3π r 3 crust and mantle core 30 cm a) If our downsized Earth has a mass of 5.98 X 10 6 g. What is the average density of the Earth? cm Using the formula for the volume of a sphere, we get 1.08 x 10 6 cubic cm x10 g x10 cm 3 = 5.5g / cm 3 b) We have simplified the layers inside the Earth to two main layers, the combination of the crust and mantle and the core. From direct observations of crust and mantle rocks, we think that the density of the crust and mantle together is about 4.2 g/cm 3. We have never drilled to the Earth s core, but let s imagine that we have three models for the composition of the core of the Earth.

5 Model Composition Density (g/cm 3 ) 1 Aluminum Iron 13 3 Gold 19.3 Which of the models for the composition of the core best matches the correct density of the Earth you calculated in part a above? Explain your logic and show your work. First one must determine the volume of the core. To determine the radius of the core we subtract cm-30cm= cm The volume of the core is 1.60 x 10 5 cubic cm. The next question is what percentage of the volume of the whole Earth is the core: x10 cm x10 cm 3 = or 14.8 percent The average density of the Earth will be the sum of the densities of the two parts (the crust/mantle and the core). X below is the density of the core. 5.5 g/cc= 0.148(x) (4.2 g/cc) x = 13.0 g/cc, so iron is the best model for the composition of the Earth s core. 4. We have two blocks of wood. We measured the volume and mass of each block and the results are below: Block mass (g) volume (cubic cm) If we have a cubic foot of block 1 and block 2, which one will weigh more? a) block 1 wood b) block 2 wood c) they would be the same The density of block 2 is greater, therefore a cubic foot of block 2 will weight more. If we have 50 pounds of wood, which will occupy the largest volume? a) block 1 wood b) block 2 wood c) they would be the same

6 block1 has the lowest density so if the weight of the two blocks is the same, block 1 will occupy a greater volume. If fresh water has a density of 1 g/cubic cm, which of the following is most accurate: a) block 1 will float on the water b) block 2 will float on the water c) both blocks will float on the water answer a is correct. The other two answers are incorrect. Block 2 has a density greater than water, so it will sink yes, there is a type of wood that has a density higher than water and it sinks! 5) Consider the lever shown below. If there are no weights on the lever it will be perfectly balanced on the fulcrum shown in the center. A B a) Block B has the same width and thickness as block A, but block B is twice as tall. The lever balances when blocks A and B are the same distance from the center. How does the density of block B compare to the density of block A? (Be quantitative and explain your reasoning.) Keep in mind that many answers would be acceptable here. Since the lever is balanced with the two blocks equidistant from the balance point, the two blocks have the same mass. The cross-sections are the same, but block B is twice as tall, so block B has twice the volume of block A. Since the masses are the same and block B has twice the volume, the density of block B is half as great as the density of block A. b) The two blocks labeled C in the diagram below have masses of 30 g each. The lever is balanced when one block C is 20 cm from the center and the other block C rests on block D 5 cm from the center. What is the mass of block D? (Be quantitative and explain your reasoning and/or show your work.) C C D

7 The mass on the left is four times as far from the balance point as the masses on the right. This means the combined mass of the two masses on the right must be four times as great as the mass on the left. The combined mass on the right must be equal to (4 x 30 g) or 120 g. If the mass of B plus the mass of C is 120 g, then the mass of D must be 120 g 30 g = 90 g. c) Two identical 12 g blocks are placed on the lever, 20 cm and 25 cm from the center. Where would one have to place a single 60 g block to balance the lever? (Be quantitative and explain your reasoning and/or show your work 60 g 12 g 12 g? Again, more than one solution is possible. Going directly to the concept of turning effect is one way to approach this. The turning effect on the left is (12 g x 20 cm) + (12 g x 25 cm) = 540 g cm In order to make the lever balance, we also need the turning effect on the right side to be 540 g cm. The position where we need to place the 60 g block is then: ( ) ( 60 g) 540 g cm = 9 cm Notice that if we keep track of units we will be guided to the answer by simply knowing that the answer has to be some number of cm. SIDE NOTE: The unit g cm may seem strange, but that s just what we get in this case. It turns out to be a reasonable unit of measure for turning effect. You may have thought that you didn t have an interpretation for the concept of g cm but now you do.

8 6) A marshmallow is through in the river at point 0 and floats 4 miles downstream as shown on the map below The following graph of the change in distance downstream as a function of time was recorded from some marshmallow measurements: Marshmallow position as a function of time Distance downstream (miles) Minutes a) In which section was the marshmallow going the highest speed? How did you determine this? What was the speed of the marshmallow in feet per second? The section of the graph from 0-1 mile. If we connect the 0 and 1 data points, the slope of the line is steeper than other sections. b) What segment had the lowest speed in feet per second? What was the speed in this segment? The segment from 3-4miles has the slowest speed because the slope of the line between those two points has the lowest slope. c) What was the average speed of the marshmallow (in ft/sec) over the whole distance? 4miles 1min 5280 ft x x = 1.7 ft / sec 210 min utes 60sec 1mile