course Phy 121 {ܺyUՋؾOC挆Student Name:
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17:26:58 `q001. Note that this assignment contains 7 questions. . The Pythagorean Theorem: the hypotenuse of a right triangle has a length c such that c^2 = a^2 + b^2, where a and b are the lengths of the legs of the triangle. Sketch a right triangle on a set of coordinate axes by first locating the point (7, 13). Then sketch a straight line from the origin of the coordinate system to this point to form the hypotenuse of the triangle. Continue by sketching a line straight down from (7, 13) to the x axis to form one leg of the triangle, then form the other leg by continuing straight along the x axis back to the origin. How long are these two legs? How long therefore is the hypotenuse?
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RESPONSE --> The length of the vertical leg is 13. The length of the horizontal leg is 7. To find the length of the hypotenuse, we must use the Pythagorean Theorem. a^2 + b^2 = c^2 c^2 = (7)^2 + (13)^2 = 218 Take square root c = 14.76
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17:28:14 The leg from (7, 13) to the x axis drops from the point (7, 13) to the point (7,0) and so has length 13. The second leg runs from (7,0) to the origin, a distance of 7 units. The legs therefore have lengths a = 13 and b = 7, so that the hypotenuse c satisfies c^2 = a^2 + b^2 and we have c = `sqrt(a^2 + b^2) = `sqrt( 13^2 + 7^2 ) = `sqrt( 216) = 14.7, approximately.
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RESPONSE --> I understand this problem.
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17:32:22 `q002. Sketch a right triangle on a set of coordinate axes whose leg along the x axis has length 12 and whose hypotenuse has length 15. What must be the length of the second leg of the triangle?
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RESPONSE --> c^2 = a^2 + b^2 (15)^2 = (12)^2 + b^2 b^2 = 225 - 144 = 81 Take square root b = 9
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17:32:29 Let c stand for the length of the hypotenuse and a for the length of the known side, with b standing for the length of the unknown side. Then since c^2 = a^2 + b^2, b^2 = c^2 - a^2 and b = `sqrt(c^2 - a^2) = `sqrt( 15^2 - 12^2) = `sqrt(225 - 144) = `sqrt(81) = 9.
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RESPONSE --> I understand.
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17:35:02 `q003. If a line of length L is directed from the origin of an x-y coordinate system at an angle `theta with the positive x axis, then the x and y coordinates of the point where the line ends will be y = L * sin(`theta) and x = L * cos(`theta). Sketch a line of length 10, directed from the origin at an angle of 37 degrees with the positive x axis. Without doing any calculations estimate the x and y coordinates of this point. Give your results and explain how you obtained your estimates.
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RESPONSE --> Since the angle with the positive x-axis is less than 45 it will be closer to that axis than the y-axis. This will make the run of the graph longer than the rise. Therefore, the x coordinate will be higher than the y coordinate. If the hypontenuse length is 10, then the legs will both by smaller. The x coordinate should be closer to 10 than the y coordinate.
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17:40:45 The line will run closer to the x axis then to the y axis, since the line is directed at an angle below 45 degrees. It won't run a whole lot closer but it will run significantly closer, which will make the x coordinate greater than the y coordinate. Since the line itself has length 10, it will run less than 10 units along either the x or the y axis. It turns out that the x coordinate is very close to 8 and the y coordinate is very close to 6. Your estimates should have been reasonably close to these values.
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RESPONSE --> I understand this problem.
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17:42:31 `q004. Using your calculator you can calculate sin(37 deg) and cos(37 deg). First be sure your calculator is in degree mode (this is the default mode for most calculators so if you don't know what mode your calculator is in, it is probably in degree mode). Then using the sin and cos buttons on your calculator you can find sin(37 deg) and cos(37 deg). What are these values and what should therefore be the x and y coordinates of the line directed from the origin at 37 degrees from the x axis?
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RESPONSE --> cos(`theta) = cos(37) = 0.799 sin(`theta) = sin(37) = 0.602 x: (length)(cos 37) = 10(0.799) = 7.99 = about 8 y: (length)(sin 37) = 10(0.602) = 6.02 = about 6 Coordinates of line: (8, 6)
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17:43:30 sin(37 deg) should give you a result very close but not exactly equal to .6. cos(37 deg) should give you a result very close but not exactly equal to .8. Since the x coordinate is L cos(`theta), then for L = 10 and `theta = 37 deg we have x coordinate 10 * cos(37 deg) = 10 * .8 = 8 (your result should be slightly different than this approximate value). Since the y coordinate is L sin(`theta), then for L = 10 and `theta = 37 deg we have y coordinate 10 * sin(37 deg) = 10 * .6 = 6 (your result should be slightly different than this approximate value).
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RESPONSE --> Ok
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17:50:07 `q005. Show that the x and y coordinates you obtained in the preceding problem in fact give the legs of a triangle whose hypotenuse is 10.
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RESPONSE --> To show that the legs of the triangle are actually 8 and 6, we can use the Pythagorean Theorem to check. If they are correct, the hypontenuse, c, will be 10. c^2 = a^2 + b^2 c^2 = (8)^2 + (6)^2 = 64 + 36 = 100 Take square root c = 10
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17:52:06 If c stands for the hypotenuse of the triangle, then if a = 8 and b = 6 are its legs we have c = `sqrt(a^2 + b^2) = `sqrt(8^2 + 6^2) = `sqrt(100) = 10. The same will hold for the more precise values you obtained in the preceding problem.
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RESPONSE --> I understand.
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17:53:24 `q006. A vector drawn on a coordinate system is generally depicted as a line segment of a specified length directed from the origin at a specified angle with the positive x axis. The vector is traditionally drawn with an arrow on the end away from the origin. In the preceding series of problems the line segment has length 10 and was directed at 37 degrees from the positive x axis. That line segment could have been drawn with an arrow on its end, pointing away from the origin. The components of a vector consist of a vector called the x component drawn from the origin along the x axis from the origin to x coordinate L cos(`theta), and a vector called the y component drawn from the origin along the y axis to y coordinate L sin(`theta). What are the x and y components of a vector directed at an angle of 120 degrees, as measured counterclockwise from the positive x axis, and having length 30 units?
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RESPONSE --> L = 30 `theta = 120 x: 30 * cos(120) = -15 y: 30 * sin(120) = 26
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17:54:38 The x component of this vector is vector along the x axis, from the origin to x = 30 cos(120 degrees) = -15. The y component is a vector along the y axis, from the origin to y = 30 sin(120 degrees) = 26, approx.. Note that the x component is to the left and the y component upward. This is appropriate since the 120 degree angle, has measured counterclockwise from the positive x axis, takes the vector into the second quadrant, which directs it to the left and upward.
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RESPONSE --> Ok
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18:01:17 `q007. The angle of a vector as measured counterclockwise from the positive x axis is easily determined if the components of the vector are known. The angle is simply arctan( y component / x component ) provided the x component is positive. If the x component is negative the angle is arctan( y component / x component ) + 180 deg. If the x component is positive and the y component negative, arctan( y component / x component ) will be a negative angle, and in this case we generally add 360 degrees in order to obtain an angle between 0 and 360 degrees. The arctan, or inverse tangent tan^-1, is usually on a calculator button marked tan^-1. Find the angle and length of each of the following vectors as measured counterclockwise from the positive x axis: A vector with x component 8.7 and y component 5. A vector with x component -2.5 and y component 4.3. A vector with x component 10 and y component -17.3.
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RESPONSE --> For -x, +y: tan-1(y/x) + 180 = `theta For +x, -y: tan-1(y/x) + 360 = `theta (8.7, 5) `theta = tan-1(5/8.7) = 30 degrees L = sqrt((8.7^2)+(5^2)) = 10.03 (4.3, -2.5) `theta = tan-1(4.3/-2.5) + 180 = 120 degrees L = sqrt((-2.5^2)+(4.3^2)) = 4.97 = 5 (10, -17.3) `theta = tan-1(-17.3/10) + 360 = 300 degrees L = sqrt((10^2)+(-17.3^2)) = 20
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18:01:41 A vector with x component 8.7 and y component 5 makes an angle of arctan (5/8.7) = 30 degrees with the x axis. Since the x component is positive, this angle need not be modified. The length of this vector is found by the Pythagorean Theorem to be length = `sqrt(8.7^2 + 5^2) = 10. A vector with x component -2.5 and y component 4.3 makes an angle of arctan (4.3 / -2.5) + 180 deg = -60 deg + 180 deg = 120 degrees with the x axis. Since the x component is negative, we had to add the 180 degrees. The length of this vector is found by the Pythagorean Theorem to be length = `sqrt((-2.5)^2 + 4.3^2) = 5. A vector with x component 10 and y component -17.3 makes an angle of arctan (-17.3 / 10) = -60 degrees with the x axis. Since the angle is negative, we add 360 deg to get 300 deg. The length of this vector is found by the Pythagorean Theorem to be length = `sqrt(10^2 + (-17.3)^2) = 20.
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RESPONSE --> I understand.
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Y럵}Ƀ}zdz|˓ assignment #018 018. `query 18 Physics I 06-29-2006
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22:00:34 Query intro problem sets Explain how we determine the horizontal range of a projectile given its initial horizontal and vertical velocities and the vertical displacement.
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RESPONSE --> Vertically: We know the initial velocity and displacement. We also know that the vertical acceleration is the acceleration of gravity, 9.8m/s. Given these three variables, we can solve vf^2 = vo^2 + 2a*`ds. Then we use the final and initial velocity to find the average velocity. We can then find the time interval by dividing the displacement by the average velocity. Horizontally: We use the time interval from our vertical calculations in horizontal calculations as well because they are equal. We also know the initial horizontal velocity. For horizontal movement the acceleration is 0m/s^2, therefore the velocity is constant. So we know that the initial velocity is equal to the final velocity and the average velocity. We can then find the horizontal displacement or range by multiplying the horizontal velocity by the time interval.
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22:01:38 ** We treat the vertical and horizontal quantities independently. We are given vertical displacement and initial velocity and we know that the vertical acceleration is the acceleration of gravity. So we solve the vertical motion first, which will give us a `dt with which to solve the horizontal motion. We first determine the final vertical velocity using the equation vf^2 = v0^2 + 2a'ds, then average the result with the initial vertical velocity. We divide this into the vertical displacement to find the elapsed time. We are given the initial horizontal velocity, and the fact that for an ideal projectile the only force acting on it is vertical tells us that the acceleration in the horizontal direction is zero. Knowing `dt from the analysis of the vertical motion we can now solve the horizontal motion for `ds. This comes down to multiplying the constant horizontal velocity by the time interval `dt. **
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RESPONSE --> I understand this process.
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22:05:43 Query class notes #17 Why do we expect that in a collision of two objects the momentum change of each object must be equal and opposite to that of the other?
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RESPONSE --> The forces exerted in the collision are equal and opposite, while the time interval is the same. This means that the impulses will be equal and opposite. Therefore, the momentum changes will be equal and opposite. Since the total momentum is the sum of the two momenta, there will be no change in total momentum of the objects during the collision.
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22:06:13 **COMMON ERROR AND INSTRUCTION CORRECTION: This is because the KE change is going to be equal to the PE change. Momentum has nothing directly to do with energy. Two colliding object exert equal and opposite forces on one another, resulting in equal and opposite impulses. i.e., F1 `dt = - F2 `dt. The result is that the change in momentum are equal and opposite: `dp1 = -`dp2. So the net momentum change is `dp1 + `dp2 = `dp1 +(-`dp1) = 0. **
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RESPONSE --> I understand this concept.
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22:07:08 What are the six quantities in terms of which we analyze the momentum involved in a collision of two objects which, before and after collision, are both moving along a common straight line? How are these quantities related?
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RESPONSE --> Six quantities we use to analyze the momentum of two objects before and after collision: mass 1 & 2 = m1, m2 velocity 1 & 2 = v1, v2 Total momentum 1 & 2 = (m1v1) + (m2v2) Relationships can be expressed in this equation: BEFORE(m1v1 + m2v2) = AFTER(m1v1 + m2v2)
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22:07:30 ** We analyze the momentum for such a collision in terms of the masses m1 and m2, the before-collision velocities v1 and v2 and the after-collision velocities v1' and v2'. Total momentum before collision is m1 v1 + m2 v2. Total momentum after collision is m1 v1' + m2 v2'. Conservation of momentum, which follows from the impulse-momentum theorem, gives us m1 v1 + m2 v2 = m1 v1' + m2 v2'. **
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RESPONSE --> I understand this concept.
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22:08:27 prin phy and gen phy 6.47. RR cars mass 7650 kg at 95 km/hr in opposite directions collide and come to rest. How much thermal energy is produced in the collision?
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RESPONSE --> v0 = 95km/hr, we must convert this to the standard units for velocity. v0 = (95km/hr)(1000m/km)(1hr/3600s) = 26.39m/s m = 7650kg We know that in this situation, the kinetic energy and thermal energy would be almost equal since there is no `dPE. Therefore, we can use the equation, 1/2mv0^2 = KE0 = Thermal energy. 1/2(7650kg)(26.39m/s)^2 = 2660000 Joules We know this is the case for both automobiles, so the total kinetic energy for the collision is the sum of KE of both cars: 2660000J + 2660000J = 5330000 Joules Therefore, the kinetic energy is almost equal to the thermal energy produced because it is almost completely converted during impact.
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22:08:37 There is no change in PE. All the initial KE of the cars will be lost to nonconservative forces, with nearly all of this energy converted to thermal energy. The initial speed are 95 km/hr * 1000 m/km * 1 hr / 3600 s = 26.4 m/s, so each car has initial KE of .5 m v^2 = .5 * 7650 kg * (26.4 m/s)^2 = 265,000 Joules, so that their total KE is 2 * 265,000 J = 530,000 J. This KE is practially all converted to thermal energy.
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RESPONSE --> Ok
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22:08:58 Query* gen phy roller coaster 1.7 m/s at point 1, ave frict 1/5 wt, reaches poin 28 m below at what vel (`ds = 45 m along the track)?
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RESPONSE --> I am not required to answer this question.
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22:09:04 **GOOD STUDENT SOLUTION WITH ERROR IN ONE DETAIL, WITH INSTRUCTOR CORRECTION: Until just now I did not think I could work that one, because I did not know the mass, but I retried it. Conservation of energy tells us that `dKE + `dPE + `dWnoncons = 0. PE is all gravitational so that `dPE = (y2 - y1). The only other force acting in the direction of motion is friction. Thus .5 M vf^2 - .5 M v0^2 + M g (y2 - y1) + f * `ds = 0 and .5 M vf^2 - .5M(1.7m/s)^2 + M(9.8m/s^2)*(-28 m - 0) + .2 M g (45m) It looks like the M's cancel so I don't need to know mass. .5v2^2 - 1.445 m^2/s^2 - 274 m^2/s^2 + 88 m^2/s^2 = 0 so v2 = +- sqrt( 375 m^2/s^2 ) = 19.3 m/s.
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RESPONSE -->
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22:09:15 Univ. 7.74 (7.62 in 10th edition). 2 kg pckg, 53.1 deg incline, coeff kin frict .20, 4 m from spring with const 120 N/m. Speed just before hitting spring? How far compressed? How close to init pos on rebound?
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RESPONSE --> I am not required to answer this question.
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22:09:19 ** The forces acting on the package while sliding down the incline, include gravitiational force, normal force and friction; and after encountering the spring the tension force of the spring also acts on the package. The normal force is Fnormal = 2 kg * 9.8 m/s^2 * cos(53.1 deg) = 11.7 N, approx.. This force is equal and opposite to the component of the weight which is perpendicular to the incline. The frictional force is f = .2 * normal force = .2 * 11.7 N = 2.3 N, approx.. The component of the gravitational force down the incline is Fparallel = 2 kg * 9.8 m/s^2 * sin(53.1 deg) = 15.7 N, approx.. Friction acts in the direction opposite motion, up the incline in this case. If we choose the downward direction as positive the net force on the package is therefore 15.7 N - 2.3 N = 13.4 N. So in traveling 4 meters down the incline the work done on the system by the net force is 13.4 N * 4 m = 54 Joules approx. Just before hitting the spring we therefore have .5 m v^2 = KE so v = +-sqrt(2 * KE / m) = +-sqrt(2 * 54 J / (2 kg) ) = +- 7.4 m/s. If we ignore the gravitational and frictional forces on the object while the spring is compressed, which we really don't want to do, we would conclude the spring would be compressed until its elastic PE was equal to the 54 J of KE which is lost when the object comes to rest. The result we would get here by setting .5 k x^2 equal to the KE loss is x = sqrt(2 * KE / k) = .9 meters, approx.. However we need to include gravitational and frictional forces. So we let x stand for the distance the spring is compressed. As the object moves the distance x its KE decreases by 54 Joules, its gravitational PE decreases by Fparallel * x, the work done against friction is f * x (where f is the force of friction), and the PE gained by the spring is .5 k x^2. So we have `dKE + `dPE + `dWnoncons = 0 so -54 J - 15.7N * x + .5 * 120 N/m * x^2 + 2.3 N * x = 0 which gives us the quadratic equation 60 N/m * x^2 - 13.4 N * x - 54 N m = 0. (note that if x is in meters every term has units N * m). Suppressing the units and solving for x using the quadratic formula we have x = ( - (-13.4) +- sqrt(13.4^2 - 4 * 60 (-54) ) / ( 2 * 60) = 1.03 or -.8 meaning 1.07 m or -.8 m (see previous note on units). We reject the negative result since the object will continue to move in the direction down the incline, and conclude that the spring would compress over 1 m as opposed to the .9 m obtained if gravitational and frictional forces are not accounted for during the compression. This makes sense because we expect the weight of the object (more precisely the weight component parallel to the incline) to compress the spring more than it would otherwise compress. Another way of seeing this is that the additional gravitational PE loss as well as the KE loss converts into elastic PE. If the object then rebounds the spring PE will be lost to gravitational PE and to work against friction. If the object goes distance xMax back up the incline from the spring's compressed position we will have`dPE = -.5 k x^2 + Fparallel * xMax, `dKE = 0 (velocity is zero at max compression as well as as max displacement up the incline) and `dWnoncons = f * xMax. We obtain `dPE + `dKE + `dWnoncons = 0 so -.5 k x^2 + Fparallel * xMax + 0 + 33 N * xMax = 0 or -.5 * 120 N/m * (1.07 m)^2 + 15.7 N * xMax + 2.3 N * xMax = 0 We obtain 18 N * xMax = 72 N m, approx., so that xMax = 72 N m / (18 N) = 4 meters, approx.. This is only 2.93 meters beyond the position of the object when the spring was compressed. Note that the object started out 4 meters beyond this position. **
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RESPONSE -->
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