PH 105-003&4 Sample Exam 2 P. B. Visscher [Note that coverage is slightly different – our exam will also cover Ch. 10,11, and 13.] Instructions [The same as for Exam 1, you won’t need them except for reference.] This is a closed-book exam; please put books and notebooks on the floor. Mark your answers on the answer form given, using a No. 2 or softer pencil. Turn in this test booklet with your green answer form – there will be a copy on the website after it is graded so you can see which questions you got wrong. Before turning in the answer form, be sure to write your name on it and to fill in the appropriate circles under the letters of your name. To get your score from MyBama, you need to also put your CWID on the test. FORMAT: Some problems contain more than one question, and so use more than one of the numbered lines on the answer sheet. At the end of each problem, the question line numbers are given in braces. For example, if question says • How many planets orbit the sun? value: {9} This doesn’t mean there are 9 planets, it means put the numerical value of the answer on line 9. Thus line 9 should look like this: If the answer isn’t an integer, round to the nearest integer. Some of the numerical answers are not between 0 and 9, for example: How many cents are in $33,000? Value: {3}×10{4} This means you should put the answer in scientific notation, 3.3×106 , round the mantissa (3.3) to the nearest integer, and bubble 3 and 6 into lines 3 and 4: Ignore signs in this process, i.e. bubble 3 and 6 for -3.3×10-6 as well. To allow partial credit if you know how to do a problem but make a numerical error, some of the problems also ask which equation to use. Choose it from the list on the next page, and darken the circle by the appropriate LETTER (A .. J) as in the following example: • A 300 kg mass is subjected to a force of magnitude 0.18 N. Its acceleration, in m/s2, is Formula:{8} Value: {9}×10{10} ANSWER: Formula A gives F = ma, so a = F/m = 18×10-2 kg m s-2 / 3 × 102 kg = 6 ×10-4 m s-2, so you bubble in A,6,4, and your answer sheet looks like the figure at right: There will be a some conventional multiple-choice problems as well, where choices (A), (B), etc., are given and you need to bubble the letter of the best answer. Special cases: I try to avoid answers close to a half-integer, but if you get exactly a halfinteger (and it doesn't go away when you double-check), raise your hand and show me your work – I might have made a mistake. There are many ways to write a number in the form (mantissa)×10exponent – for example, 91 = 91×100 = 9.1×101 = 0.91×102 = 0.091×103. To decide which to use, round each mantissa to an integer: … 91×100 ~ 9×101 ~ 1×102 ~ 0×103 , . . . The first isn’t going to work because the mantissa must be in 0…9. Of the others, choose the one with greatest accuracy, 9×101. If the answer is exactly 0, put 0 for both the mantissa and the exponent. To encourage you to have some familiarity with the formulas, there are a few wild cards (indicated by “?”) in the formula list. , each wild card represents an omitted number, letter, or phrase. You should probably begin by filling in the wild cards you know, to make the list easier to read. Not all formulas are included, but the list should include at least one that is needed for each problem where a formula is asked for. If two formulas seem to be equally important, pick the most recently covered -- never bubble two answers on the same line. You might want to tear out this page, for easier reference. Densities in kg/m3 Aluminum Water Iron 2700 1000 7900 CONVERSION FACTORS 1 meter = 3.28 feet = 1.09 yards 1 km = 0.61 mi g = 9.8 m/s 2 Constants Formulas (A) F = ma (B) (C) (D) (E) (F) (G) (H) (I) (J) v = v0 + a ? P=Fv a = v2/r m = ? * Volume KE + PE = constant a = dv/dt x2+ y2= ?2 (Pythag. Thm.) vAC = vAB + vBC (m1 + m2 + …) xcm = m1x1 + m2x2 + … Torque = F d p = mv Δ(KE) = work on ? α = dω/dt ► A skateboarder starts at rest and slides frictionlessly for a distance 4.5 meters along a ramp inclined at 45˚ to the horizontal. His velocity (in m/s) at the bottom is Formula:{1} Value: {2} (i.e., exponent not required here) ► A car of mass 1173 kg is moving at 15 m/s at the bottom of a hill. It coasts (with no engine power) up the hill to a height of 8 m, where it comes to a stop. What work was done against friction? Formula:{3} Value: {4}×10{5} Joules ► A block of mass 1.07 kg is at rest on a board inclined at 40˚ to the horizontal, with a static friction coefficient of 0.9. The normal force exerted on the block by the board is Value: {6} Newtons ► A 1500 kg Jeep on a safari has its center of mass 3 feet above the ground. A 1000 kg elephant is then tied to its roof rack, so that the elephant’s center of mass is 8 feet above the ground. How high above the ground is the center of mass of the Jeep + elephant, in feet? Formula:{7} Value: {8}×10{9} feet ► A 0.9 kg pileated woodpecker traveling east at 5.5 m/s is struck by a 2 kg inebriated war eagle traveling north at 3.5 m/s, and their talons become entangled. The total momentum vector [in the form (vEast,vNorth)] of the two birds is Formula:{10} Value: ({11},{12}) kg m/s (i.e., give 2 components, each should be < 10) During the collision their talons become entangled. After the collision they are moving with a common speed of Formula:{13} Value: {14} m/s. ► A 1 kg bead slides without friction on a circular wire (radius 1 m) in a vertical plane. It starts at the top with a very small velocity (say 0.001 m/s; it has to be nonzero or it would never slide down). What is the acceleration of the bead at the bottom, in g’s? (That is, give the value of a/g). Value: {15} What is N/mg at the bottom, i.e. the normal force in units of the weight? Value: {16} ► How much power does a motor have to produce to lift a 1500 kg elevator at a constant speed of 3.5 m/s? Formula:{17} Value: {18}×10{19} watts If the elevator accelerates to this speed from zero in 1 second, what is the instantaneous power produced by the motor at the end of this acceleration period (i.e. when the velocity has just reached 3.5 m/s, but the elevator is still accelerating)? {20}×10{21} watts ► A 0.63 kg mass hangs from a knot supported by two light strings as shown. The tension in the horizontal string is {22} Newtons And the tension in the diagonal string is {23} Newtons 45˚ 0.63 kg -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= The following questions have one answer each, in the traditional way: {24} To use Newton’s second law, you should first draw a free body diagram, showing all of the A) Forces exerted on a specific body. B) Accelerations acting on a specific body. C) Forces exerted by a specific body. D) Bodies that are involved in the problem. E) Conservative forces on a specific body. {25} In an elastic collision in one dimension between particles with (x-components of) velocities v1 and v2, a quantity that changes its sign but not its magnitude is A) v1 + v2 B) v1 C) v1 - v2 D) ½ m1v12 + ½ m2v22 E) ½ m1v12 - ½ m2v22 F) m1v1 + m2v2 {26} The potential energy stored by a spring that obeys Hooke’s law with a spring constant k is C) -mgh A) mgh B) ½ kx2 D) v2/r E) kx F) - ½ kx2 {27} The impulse of a force (i.e. the momentum transferred by the force) is the integral of A) F dx B) p dx C) F dt D) E dt E) kx dx F) ½ kx2

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