#$&* course PHY 201 Question: `q001. The accepted value of the acceleration of gravity is approximately 980 cm/s^2 or 9.8 m/s^2. This will be the acceleration, accurate at most places within 1 cm/s^2, of any object which falls freely, that is without the interference of any other force, near the surface of the Earth. If you were to step off of a table and were to fall 1 meter without hitting anything, you would very nearly approximate a freely falling object. How fast would you be traveling when you reached the ground? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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Given Solution: You would have an initial vertical velocity of 0, and would accelerate at 9.8 m/s^2 in the same direction as your 1 meter vertical displacement. You would also have a slight horizontal velocity (you don't step off of a table without moving a bit in the horizontal direction, and you would very likely maintain a small horizontal velocity as you fell), but this would have no effect on your vertical motion. So your vertical velocity is a uniform acceleration with v0 = 0, `ds = 1 meter and a = 9.8 m/s^2. The equation vf^2 = v0^2 + 2 a `ds contains the three known variables and can therefore be used to find the desired final velocity. We obtain vf = +- `sqrt( v0^2 + 2 a `ds) = +- `sqrt ( 0^2 + 2 * 9.8 m/s^2 * 1 m)= +- `sqrt ( 19.6 m^2 / s^2) = +- 4.4 m/s, approx. Since the acceleration and displacement were in the direction chosen as positive, we conclude that the final velocity will be in the same direction and we choose the solution vf = +4.4 m/s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& self-critique (if necessary): OK Self-critique rating OK ********************************************* Question: `q002. If you jump vertically upward, leaving the ground with a vertical velocity of 3 m/s, how high will you be at the highest point of your jump? Note that as soon as you leave the ground, you are under the influence of only the gravitational force. All the forces that you exerted with your legs and other parts of your body to attain the 3 m/s velocity have done their work and are no longer acting on you. All you have to show for it is that 3 m/s velocity. So as soon as you leave the ground, you begin experiencing an acceleration of 9.8 m/s^2 in the downward direction. Now again, how high will you be at the highest point of your jump? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `dt = `dv/a `dt = (-3m/s)/-9.8m/s^2 = .31 seconds. average velocity =1.5m/s multiplied by .31 seconds=.465 meters. confidence rating #$&* 2
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Given Solution: From the instant the leave the ground until the instant you reach your highest point, you have an acceleration of 9.8 m/s^2 in the downward direction. Since you are jumping upward, and since we can take our choice of whether upward or downward is the positive direction, we choose the upward direction as positive. You might have chosen the downward direction, and we will see in a moment how you should have proceeded after doing so. For now, using the upward direction as positive, we see that you have an initial velocity of v0 = + 3 m/s and an acceleration of a = -9.8 m/s^2. In order to use any of the equations of motion, each of which involves four variables, you should have the values of three variables. So far you only have two, v0 and a. {}What other variable might you know? If you think about it, you will notice that when objects tossed in the air reach their highest point they stop for an instant before falling back down. That is precisely what will happen to you. At the highest point your velocity will be 0. Since the highest point is the last point we are considering, we see that for your motion from the ground to the highest point, vf = 0. Therefore we are modeling a uniform acceleration situation with v0 = +3 m/s, a = -9.8 m/s^2 and vf = 0. We wish to find the displacement `ds. Unfortunately none of the equations of uniformly accelerated motion contain the four variables v0, a, vf and `ds. This situation can be easily reasoned out from an understanding of the basic quantities. We can find the change in velocity to be -3 meters/second; since the acceleration is equal to the change in velocity divided by the time interval we quickly determine that the time interval is equal to the change in velocity divided by the acceleration, which is `dt = -3 m/s / (-9.8 m/s^2) = .3 sec, approx.; then we multiply the .3 second time interval by the 1.5 m/s average velocity to obtain `ds = .45 meters. However if we wish to use the equations, we can begin with the equation vf = v0 + a `dt and solve to find `dt = (vf - v0) / a = (0 - 3 m/s) / (-9.8 m/s^2) = .3 sec. We can then use the equation `ds = (vf + v0) / 2 * `dt = (3 m/s + 0 m/s) / 2 * .3 sec = .45 m. This solution closely parallels and is completely equivalent to the direct reasoning process, and shows that and initial velocity of 3 meters/second should carry a jumper to a vertical height of .45 meters, approximately 18 inches. This is a fairly average vertical jump. If the negative direction had been chosen as positive then we would have a = +9.8 m/s^2, v0 = -3 m/s^2 (v0 is be in the direction opposite the acceleration so if acceleration is positive then initial velocity is negative) and again vf = 0 m/s (0 m/s is the same whether going up or down). The steps of the solution will be the same and the same result will be obtained, except that `ds will be -.45 m--a negative displacement, but where the positive direction is down. That is we move .45 m in the direction opposite to positive, meaning we move .45 meters upward. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique OK ********************************************* Question: `q003. If you roll a ball along a horizontal table so that it rolls off the edge of the table at a velocity of 3 m/s, the ball will continue traveling in the horizontal direction without changing its velocity appreciably, and at the same time will fall to the floor in the same time as it would had it been simply dropped from the edge of the table. If the vertical distance from the edge of the table to the floor is .9 meters, then how far will the ball tavel in the horizontal direction as it falls? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: vf^2 = v0^2 + 2a * `ds vf^2 = 0^2m/s + 2(9.8m/s^2) * .9m vf^2 = 19.6m/s^2* .9m vf^2 = 17.64m/s^2 vf 4.2m/s -4.2m/s divided by -9.8m/s^2=.43 seconds for the ball to hit the ground. The ball travels 1.29 meters horizontally confidence rating #$&* 2
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Given Solution: A ball dropped from rest at a height of .9 meters will fall to the ground with a uniform vertical acceleration of 9.8 m/s^2 downward. Selecting the downward direction as positive we have `ds = .9 meters, a = 9.8 m/s^2 and v0 = 0. Using the equation `ds = v0 `dt + .5 a `dt^2 we see that v0 = 0 simplifies the equation to `ds = .5 a `dt^2, so `dt = `sqrt( 2 `ds / a) = `sqrt(2 * .9 m / (9.8 m/s^2) ) = .42 sec, approx.. Since the ball rolls off the edge of the table with only a horizontal velocity, its initial vertical velocity is still zero and it still falls to the floor in .42 seconds. Since its horizontal velocity remains at 3 m/s, it travels through a displacement of 3 m/s * .42 sec = 1.26 meters in this time. STUDENT COMMENT: I don’t see why you don’t use the formula that I used in my answer. INSTRUCTOR RESPONSE: In your solution you chose to solve the third equation ds = v0 * dt + 0.5 * a * dt^2 for `dt. Since v0 = 0 for the vertical motion, the equation simplifies and is easy to solve for `dt. However I avoid using the third equation to solve for `dt because it is quadratic in `dt, and is therefore very confusing to most students. It is less confusing to use the fourth equation to find vf, after which we can easily reason out `dt. Of course in the present case v0 = 0 and the equation becomes every bit as simple as the fourth equation; in fact when v0 = 0 it's simpler to use the equation you used. However most students have problems with special conditions and special cases, so I choose not to confuse the issue, and consistently use the fourth equation in my solutions. My convention of using the fourth equation in this situation does no harm to students who understand how to solve the quadratic, and who know how consider and apply special conditions. You and other students who are sufficiently comfortable with the mathematics should always use the most ppropriate option. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating OK ********************************************* Question: `q004. A ball starts out with a horizontal velocity of 4 m/s and vertical velocity 0 m/s. It falls freely for 2 seconds. During this 2 seconds how far does it travel in the vertical direction, and how far in the horizontal direction? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x v0 = 4 m/s 'dt = 2 seconds a = 0 'ds = v_0*'dt + a/2 * 'dt^2 = 4m/s * (2s) + 0 = 8m Y 'ds = v_0*'dt + a/2 * 'dt^2 = 0 + (9.8 m/s^2) / 2 * (4 s^2) = 19.6 m Confidence rating: 2 ********************************************* Question: `q005. A ball starts out with a horizontal velocity of 4 m/s and vertical velocity 0 m/s. It falls freely to the ground 20 meters below. During its fall how far does it travel in the vertical direction, and how far in the horizontal direction? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 'ds = v_0 * 'dt + a/2 * 'dt^2 20 m = 0 + 4.9 m/s^2 * 'dt^2 'dt = sqrt(4.08 s^2) = 2.02 s 'ds = 4 m/s * 2.02 s + 0 = 1.97 m Confidence rating: 2 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
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Given Solution: If the pieces attract, then the tape at point A is pulled toward point B. • The vectors AB_v and AB_u point from A to B. • Of these the vector AB_u is the unit vector. • So the tape at A experiences a force in the direction of the vector AB_u. If the pieces repel, then the tape at point B is pushed away from point A. • The direction of the force is therefore from A towards B. • The direction is therefore that of the vector AB_u. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):good ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qUsing the notation of the preceding question, which you should have noted on paper (keep brief running notes as you do qa's and queries so you can answer 'followup questions' like this), how does the magnitude of the vector AB_v depend on the magnitude of BA_v, and how does the magnitude of each vector compare with the distance between A and B? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The magnitude of the two vectors should be equal but opposite, the distances between the points should be the sum of the lengths of the vectors between them. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The distance from A to B is the same as the distance from B to A. So the vectors AB_v and BA_v have the same length. Each vector therefore has magnitude equal to the distance between A and B. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): good ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qUsing the notation of the preceding question, how is the force experienced by the two pieces of tape influenced by the magnitude of AB_v or BA_v? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The pieces of tape do not exhibit characteristics of point charges when brought together fail to exhibit the inverse square force. Each individual charge experiences the inverse square force from any other charged particle surrounding it. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The expected answer is that the force exerted by two charges on one another is inversely proportional to the square of the distance between them. So as the magnitudes of the vectors, which are equal to the distance between A and B (i.e., to the separation between the two pieces of tape), increases the force decreases with the square of the distance (similarly if the distance decreases the force increases in the same proportion). This answer isn't exactly correct, since the two pieces of tape are not point charges. Since some parts of tape A are closer to B than other parts of tape A, and vice versa, the inverse square relationship applies only in approximation for actual pieces of tape. STUDENT COMMENT: Still not sure how to answer this question. I dont know what sort of difference there is between the magnitude vector and the unit vector. INSTRUCTOR RESPONSE: As noted in the preceding solution, AB_v stands for the vector whose initial point is A and whose terminal point is B, and AB_u stands for a vector of magnitude 1 whose direction is the same as that of AB_u. To get a unit vector in the direction of a given vector, you divide that vector by its magnitude. So AB_u = AB_v / | AB_v |. The purpose of a unit vector is to represent a direction. If you multiply a unit vector by a number, you get a vector in the same direction as the unit vector, whose magnitude is the number by which you multiplied it. The problem does not at this point ask you to actually calculate these vectors. However, as an example: Suppose point A is (4, 5) and point B is (7, 3). Then AB_v is the vector whose initial point is A and whose terminal point is B, so that AB_v = <3, -2>, the vector with x component 3 and y component -2. The magnitude of this vector is sqrt( 3^2 + (-2)^2 ) = sqrt(11). Therefore AB_u = <3, -2> / sqrt(11) = (3 / sqrt(11), -2 / sqrt(11) ). That is, AB_u is the vector with x component 3 / sqrt(11) and y component -2 / sqrt(11). (Note that these components should be written with rationalized denominators as 3 sqrt(11) / 11 and -2 sqrt(11) / 11, so that AB_u = <3 sqrt(11) / 11, -2 sqrt(11) / 11 > ). The vector AB_u is a unit vector: it has magnitude 1. The unit vector has the same direction as AB_v. If we want a vector of magnitude, say, 20 in the direction of this vector, we simply multiply the unit vector AB_u by 20 (we would obtain 20 * <3 sqrt(11) / 11, -2 sqrt(11) / 11 > = <60 sqrt(11) / 11, -40 sqrt(11) / 11 >). You don't need to understand this example at this point, but if you wish to understand it you should probably sketch this situation and identify all these quantities in your sketch. STUDENT RESPONSE AB_v/(AB_v)absolute value So AB_u would either be 1 or -1. INSTRUCTOR COMMENT AB_u and BA_v are vectors, so they have both magnitude and direction. Your answer is correct if everything is in one dimension (e.g., if all charges are on the x axis and all forces directed along the x axis). In one dimension direction can be specified by + and - signs. In two dimensions + and - signs are not sufficient. In two dimensions the direction of a vector is generally specified by its angle as measured counterclockwise from the positive x axis. In two dimensions, the magnitude of AB_u would be the same as that of BA_u, but the vectors would be in the opposite direction. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):good ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `qQuery introductory set #1, 1-5 Explain how we calculate the magnitude and direction of the electrostatic force on a given charge at a given point of the x-y plane point due to a given point charge at the origin. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using coulombs’ law of kqq/r^2. The direction is found by using trigonometry and adjusting for a negative solution. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The magnitude of the force on a test charge Q is F = k q1 * Q / r^2, where q1 is the charge at the origin. The force is one of repulsion if q1 and Q are of like sign, attraction if the signs are unlike. The force is therefore directly away from the origin (if q1 and Q are of like sign) or directly toward the origin (if q1 and Q are of unlike sign). To find the direction of this displacement vector we find arctan(y / x), adding 180 deg if x is negative. If the q1 and Q are like charges then this is the direction of the field. If q1 and Q are unlike then the direction of the field is opposite this direction. The angle of the field would therefore be 180 degrees greater or less than this angle.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):good ------------------------------------------------ Self-critique Rating:3