course PHY 121
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15:55:46 1) When masses of 35, 70 and 105 grams are hung from a certain rubber band its respective lengths are observed to be 39, 46, and 53 cm. What are the x and y components of the tension of a rubber band of length 50.6 cm if the x component of its length if 14.02232? What horizontal force, when added to this force will result in a total force of magnitude 150 grams?
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RESPONSE --> x = 50.6 cm * cos(14.02232) x = 49.1cm y = 50.6 cm * sin(14.02232) y = 12.3 cm
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15:59:52 STUDENT SOLUTION WITH INSTRUCTOR COMMENTS INDICATED BY ** I am assuming that you plug in what you know into the equation of the line. Y = .2x + 32 ** If y represents the mass and x the length that approach would give you the right mass--good thinking, though the equation would be more like y = 5 x - 150. Your equation looks reasonably good if y is length and x is supported mass. However you do need to use the force, which will be the force exerted by gravity on the mass. Force and mass are not the same thing. In general you will use the graph to determine the force corresponding to a given length. If there is significant curvature to the graph the linear model won?t work very well. In that case you could just draw a smooth curve to fit the data and read your forces from the curve. ** When the length is 50.6 cm the corresponding weight is 93 grams and when the length is 14.02232 cm the weight is ?89.9 g, but I don?t think weight can be negative???? ** Plug the rubber band length into the force relationship or read the force from your graph. If your graph is of supported mass vs. length you need to find the force. In this case you have a length of 50.6 cm, which corresponds to a supported mass of about 93 grams. This corresponds to a force of about .093 kg * 9.8 m/s^2 = .91 Newtons. The direction of the rubber band is arctan(y/x). Using the Pythagorean Theorem with length 50.6 cm and x component 14 cm we get y component about 48.6 cm, so the direction of the rubber band is arctan(48.6 / 14) = 74 deg. The angle at which the tension acts is parallel to the rubber band, so the tension also acts at 74 deg. The x and y components of the force are therefore Fx = .91 N * cos(74 deg) = .25 N and Fy = .91 N * sin(74 deg) = .87 N. If a horizontal force Fhoriz is added then the x component becomes Fx = .25 N + Fhoriz and the magnitude of the resulting force is | F | = sqrt(Fx^2 + Fy^2) = sqrt( (.25 N + Fhoriz)^2 + (.87 N)^2). If this force is 150 grams (or, converting this to a force, 1.47 N) then we have sqrt( (.25 N + Fhoriz)^2 + (.87 N)^2) = 1.47 N. Squaring both sides we get .0625 N^2 + .5 N * Fhoriz + Fhoriz^2 + .76 N^2 = 2.16 N^2. This is quadratic in Fhorz and rearranges to Fhoriz^2 + .5 N * Fhoriz ? 2.1 N^2 = 0. Using the quadratic formula we get Fhoriz = (-.5 N +- sqrt( (.5 N)^2 ? 4 * 1 * (-2.1 N^2) ) / (2 * 1) = (-.5 N +- 2.8 N) / 2, giving us two forces: Fhoriz = (-.5 N + 2.8 N ) / 2 = 1.15 N and Fhoriz = (-.5 N ? 2.8 N) / 2 = -1.65 N.**
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RESPONSE --> I did not know how to solve this problem. I can follow your solution but do not know if I could solve it on my own.
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16:15:45 2) A ball of mass .3 kg is tossed vertically upward from altitude 4.1 meters and allowed to rise to a max altitude of 4.68 meters before falling to an altitude of 4.06 meters. * Determine the work done by gravity on the ball and the work done by the ball against gravity as it rises from initial to max altitude and determine its KE change during that displacement. * Using energy considerations determine its initial velocity. * Determine the work done by gravity on the ball and the work done by the ball against gravity from its initial all the way to the final position. ** Determine the change in KE between these positions and the use of energy considerations to determine its velocity at the final position. * How are the work done by the ball and its KE change related?
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RESPONSE --> Wgrav = F * 'ds Wgrav = (0.3 kg * 9.8 m/s^2) * (4.68 m - 4.1m) Wgrav = 2.94 N * 0.58 m Wgrav = 1.7 J 'dKE = 1.7 J KE0 = 1/2 m v0^2 v0^2 = (2 * KE0) / m v0 = 'sqrt(2 * 1.7 J / 0.3 kg) v0 = 3.4 m/s Wgrav = F * 'ds Wgrav = (0.3 kg * 9.8 m/s^2) * (4.68 m - 4.06 m) Wgrav = 2.94 N * 0.62 m Wgrav = 1.8 J 'dKE = 1.8 J KE0 = 1/2 m v0^2 v0^2 = (2 * KE0) / m v0 = 'sqrt(2 * 1.8 J / 0.3 kg) v0 = 3.5 m/s
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16:18:37 Intuitively the PE increases as the ball rises, so that KE decreases. Since we know that the KE of the rising ball is 0 at the top of the arc we can find the initial KE by finding the PE increase. As the ball falls, PE decreases and KE increases. Between initial and final position the ball has a small net downward displacement, hence a small net loss of PE, so it has a small net gain in KE. PE changes are easily found by multiplying the weight of the ball by the displacement. The detailed solution that follows is based on these concepts, but uses the work-energy theorem in a rigorous fashion: The work done by gravity on the ball is equal to the product of the force exerted by gravity, which is .3 kg * 9.8 m/s^2 = 2.94 N downward, and the displacement, which is 4.68 m ? 4.1 m = .58 m upward. This work is therefore -2.94 N * .58 m = -1.8 Joules, approx.. The work done by the ball against gravity is +1.8 J, equal and opposite to the work done by gravity on the ball. The work done against gravity is the change in the PE of the ball. Assuming no dissipative forces we have `dPE + `dKE = 0, so the KE change during this upward displacement is `dKE = -`dPE = -1.8 J. Since the final KE is 0, we have `dKE = KEf - KE0 = -1.8 J so that KE0 = 1.8 J + KEf = 1.8 J + 0 = 1.8 J. From initial to final position altitude changes by 4.06 m ? 4.1 m = -.04 m. Thus PE changes by -.04 m * 2.94 N = -.12 J. Since KE0 = 1.8 J and `dKE + `dPE = 0 we have `dKE = -`dPE = +.12 J so that KEf ? KE0 = .12 J and KEf = KE0 + 1.2 J = 1.8 J + .12 J = 1.92 J. The final velocity of the ball therefore satisfies .5 m vf^2 = KEf so that vf = +-sqrt(2 KE / m) = +- sqrt(1.92 J / (.3 kg) ) = +- sqrt( 6.4 m^2/s^2) = +- 2.5 m/s, approx.. Since final velocity is downward, our previous implicit choice of upward as the positive direction tells us that the final velocity is -2.5 m/s. The work done by the ball against gravity is equal and opposite to its change in KE.
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RESPONSE --> OK. My solution was on track with this one and I was able to follow this one.
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16:20:19 3) Give an example of a situation in which you are given v0, a, and ?ds, and reason out all possible conclusions that could be drawn from these three quantities assuming uniform acceleration.
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RESPONSE --> An object start at initial velocity 1 m/s and accelerates 0.1 m/s^2 through a displacement of 20 m. You could find the final velocity and time.
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16:20:35 ** Your solution should show the use of the fourth equation of motion vf^2 = v0^2 + 2 a `ds to find vf. You then have the initial and final velocities, which since acceleration is uniform can be averaged to give you the average velocity vAve and the change `dv in velocity.. Dividing displacement `ds by average velocity we obtain `dt.
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RESPONSE --> OK.
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16:30:59 4) A ball reaches a ramp of length 74 cm with an unknown initial velocity and accelerates uniformly along the ramp, reaching the other end in 5.2 seconds. Its velocity at the end of the ramp is 7.46154 cm/s. What is its acceleration on the ramp?
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RESPONSE --> 'ds = 74 cm 'dt = 5.2 s vf = 7.46154 cm/s 'ds = (vf + v0) / 2 * 'dt v0 = (2 'ds / 'dt) - vf v0 = (2 * 74 cm / 5.2 s) - 7.46154 cm/s v0 = 21 cm/s vf = v0 + a 'dt a = (vf - v0) / 'dt a = (7.46154 cm/s - 21 cm/s) / 5.2 s a = -2.6 cm/s^2
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16:31:13 STUDENT SOLUTION: I found the average velocity to be 74 cm / (5.2 s) = 14.23 cm/s. Since vAve = (vf + v0) / 2 we can easily solve for v0, obtaining v0 = 2 vAve ? vf = 2 * 14.23 cm/s ? 7.46 cm/s = 21 cm/s. Having vf and v0 we easily find that `dv = vf ? v0 = 7.46 cm/s ? 21 cm/s = -13.54 cm/s, and therefore a = `dv / `dt = -13.54 cm/s / (5.2 s) = ?2.6 cm/s^2
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RESPONSE --> OK.
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16:48:56 5) A ball starting from rest rolls 10 cm down an incline on which its acceleration is constant, requiring .83 seconds to cover the distance. It then rolls onto a second incline 41 cm long on which its acceleration is 14 cm/s^2. How much time does it spend on the second incline and what is its acceleration on the first?
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RESPONSE --> v0 = 0 cm/s 'ds = 10 cm 'dt = 0.83 s 'ds = v0 'dt + 1/2 a 'dt^2 a = 2 'ds / 'dt^2 a = 2 * 10 cm / (0.83 s)^2 a = 29 cm/s^2 'ds = (vf + v0) * 'dt vf = ('ds / 'dt) - v0 vf = (10 cm / 0.83 s) vf = 12 cm/s v0 = 12 cm/s 'ds = 41 cm a = 14 cm/s^2 'ds = v0 'dt + 1/2 a 'dt^2 1/2 a 'dt^2 + v0 'dt - 'ds = 0 (1/2 * 14 cm/s^2 * 'dt^2) + (12 cm/s * 'dt) - 41 cm (7 cm/s^2 * 'dt^2) + (12 cm/s * 'dt) - 41 cm 'dt = ?
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16:49:37 STUDENT SOLUTION: Ramp 1 I found the average velocity vAve = 10 cm / (.83 s) = 12 cm/s, approx.. Since v0 = 0 we have vf = 2 * vAve = 24 cm/s, approx. Acceleration on this incline is a = `dv / `dt = 24 cm/s / (.83 s) = 29 cm/s^2 approx. Ramp 2 I plugged in what I knew to find the unknowns, using v0 = 24 cm/s (the final velocity on the first incline is the init vel on the second), a = 14 cm/s^2 and `ds = 41 cm. I obtained vf = sqrt( v0^2 + 2 a `ds) = sqrt( (24 cm/s)^2 + 2 * 14 cm/s^2 * 41 cm) = 41 cm/s, approx.. Average velocity on the second incline is therefore (41 cm/s + 24 cm/s) / 2 = 32.5 cm/s, and the time required is `dt = `ds / vAve = 41 cm / (32.5 cm/s) = 1.3 sec, approx.
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RESPONSE --> ?
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17:01:50 6) A cart of mass 1.7 kg coasts 30 cm down an incline at 4 degrees with horizontal. Assuming Ffr is .042 times the normal force and other nongravitational forces parallel to the incline are negligible. * What is the component of the carts weight parallel to the incline? * How much work does this force do as the cart rolls down the incline? * How much work does the net force do as the cart rolls down the incline? * Using the definition of KE, determine the velocity of the cart after coasting the 30 cm assuming it initial velocity is zero.
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RESPONSE --> (1.7 kg * 9.8 m/s^2) / 4 = 4 N 'dW = Fnet * 'ds 'dW = (4 N - 0.042 N) * 30 cm 'dW = 119 N cm
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17:03:08 The weight of the cart is 1.7 kg * 9.8 m/s^2 = 16.7 N, approx.. The incline makes an angle of 4 degrees with horizontal, so if the positive x axis is directed up the incline the weight will be at angle 270 deg ? 4 deg = 266 deg with the positive x axis and the x and y components of the weight will be x comp of weight = 16.7 N * cos(266 deg) = -1.2 N, approx., and y comp of weight = 16.7 N * sin(266 deg) = -16.6 N approx. The x component is parallel and the y component perpendicular to the incline. The normal force exerted by the incline is the elastic reaction to the y component of the weight and is +16.6 N. The parallel component of the gravitational force is in the direction of the displacement down the incline so the work done by this component on the cart is positive. We get work by parallel component of weight = 1.2 N * .30 m = .36 Joules, approx.. The frictional force is .042 times the normal force, or frictional force = .042 * 16.6 N = .7 N, approx., directed opposite to the displacement (up the incline). So the net force is net force = parallel component of gravitational force + frictional force = -1.2 N + .7 N = -.5 N, or .5 N down the incline. The work done by this force is .5 N * .3 m = .15 J. The KE of the cart after coasting down the incline from rest is by the work-energy theorem equal to the work done by the net force. So we have .5 m vf^2 = KE, so that vf = +- sqrt(2 KE / m) = +- sqrt( 2 * .15 J / (1.7 kg) ) = +- sqrt( .16 m^2/s^2) = +- .4 m/s, approx. Using the sign conventions specified by our choice of the upward direction as the positive x direction, the velocity will be -.4 m/s.
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RESPONSE --> I do not understand where the 270 degrees comes from.
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17:11:59 7) A ball mass 8 kg rolls off the edge of a ramp with a horizontal speed of 70 cm/s. * What is its KE as it rolls off the ramp? * How much work does gravity do on the ball as it falls 22 cm? * What will be its KE after falling 22 cm? *How much of this KE is accounted for by its horizontal velocity and how much by its vertical velocity? * What then is its vertical velocity at this point?
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RESPONSE --> KE = 1/2 m v^2 KE = 1/2 * 8 kg * (0.7 m/s)^2 KE = 1.96 J 'dWgrav = F * 'ds 'dWgrav = (8 kg * 9.8 m/s^2) * 0.22 m 'dWgrav = 17 J KEtot = 1.96 J + 17 J KEtot = 18.96 J KE = 1/2 m v^2 v = 'sqrt(2 KE / m) v = 'sqrt(2 * 17 J / 8 kg) v = 2 m/s
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17:12:59 We have KE = .5 m v^2 = .5 * 8 kg * (.70 m/s)^2 = 2 Joules, approx.. As the ball falls 22 cm gravity does work `dWgrav = 8 kg * 9.8 m/s^2 * .22 m = 17 Joules, approx.. So after the 22 cm fall we will have KE = 2 J + 17 J = 19 J. There is no net force in the horizontal direction so the horizontal velocity will be unchanged and its horizontal KE will be 2 J, leaving 17 J of vertical KE. The vertical velocity will therefore be vertical velocity = +-sqrt(2 * vertical KE / m) = +- sqrt( 2 * 17 J / (8 kg)) = +-2.1 m/s.
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RESPONSE --> OK. I left off the +- sign but understand why that is needed.
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ןŰXƅؒI Student Name: assignment #020
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14:24:01 `q001. Note that this assignment contains 3 questions. . A 5 kg block rests on a tabletop. A string runs horizontally from the block over a pulley of negligible mass and with negligible friction at the edge of the table. There is a 2 kg block hanging from the string. If there is no friction between the block in the tabletop, what should be the acceleration of the system after its release?
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RESPONSE --> F = 9.8 m/s^2 * 2 kg F = 19.6 N F = -9.8 m/s^2 * 5 kg F = -49 N a = Fnet / m a = (-29.4 N) / 7 kg a = -4.2 m/s
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14:26:04 Gravity exerts a force of 5 kg * 9.8 meters/second = 49 Newtons on the block, but presumably the tabletop is strong enough to support the block and so exerts exactly enough force, 49 Newtons upward, to support the block. The total of this supporting force and the gravitational force is zero. The gravitational force of 2 kg * 9.8 meters/second = 19.6 Newtons is not balanced by any force acting on the two mass system, so we have a system of total mass 7 kg subject to a net force of 19.6 Newtons. The acceleration of this system will therefore be 19.6 Newtons/(7 kg) = 2.8 meters/second ^ 2.
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RESPONSE --> F = 9.8 m/s^2 * 2 kg F = 19.6 N a = Fnet / m a = 19.6 N / 7 kg a = 2.8 m/s^2 I factored in the Force on the stationary object but understand now why this is not included.
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14:28:46 `q002. Answer the same question as that of the previous problem, except this time take into account friction between the block in the tabletop, which exerts a force opposed to motion which is .10 times as great as the force between the tabletop and the block. Assume that the system slides in the direction in which it is accelerated by gravity.
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RESPONSE --> F = 9.8 m/s^2 * 2 kg F = 19.6 N F = (-9.8 m/s^2 * 5 kg) * 0.10 F = -49 N * 0.10 F = -4.9 N a = Fnet / m a = (19.6 N - 4.9 N) / 7 kg a = 2.1 m/s^2
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14:29:04 Again the weight of the object is exactly balance by the upward force of the table on the block. This force has a magnitude of 49 Newtons. Thus friction exerts a force of .10 * 49 Newtons = 4.9 Newtons. This force will act in the direction opposite that of the motion of the system. It will therefore be opposed to the 19.6 Newton force exerted by gravity on the 2 kg object. The net force on the system is therefore 19.6 Newtons -4.9 Newtons = 14.7 Newtons. The system will therefore accelerate at rate a = 14.7 Newtons/(7 kg) = 2.1 meters/second ^ 2.
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RESPONSE --> OK.
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14:31:33 `q003. Answer the same question as that of the preceding problem, but this time assume that the 5 kg object is not on a level tabletop but on an incline at an angle of 12 degrees, and with the incline descending in the direction of the pulley.
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RESPONSE --> I am not sure how to obtain the Force for the 5 kg object.
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14:39:06 In this case you should have drawn the incline with the x axis pointing down the incline and the y axis perpendicular to the incline. Thus the x axis is directed 12 degrees below horizontal. As a result the weight vector, rather than being directed along the negative y axis, lies in the fourth quadrant of the coordinate system at an angle of 12 degrees with the negative y axis. So the weight vector makes an angle of 270 degrees + 12 degrees = 282 degrees with the positive x axis. The weight vector, which has magnitude 5 kg * 9.8 meters/second ^ 2 = 49 Newtons, therefore has x component 49 Newtons * cosine (282 degrees) = 10 Newtons approximately. Its y component is 49 Newtons * sine (282 degrees) = -48 Newtons, approximately. The incline exerts sufficient force that the net y component of the force on the block is zero. The incline therefore exerts a force of + 48 Newtons. Friction exerts a force which is .10 of this force, or .10 * 48 Newtons = 4.8 Newtons opposed to the direction of motion. Assuming that the direction of motion is down the incline, frictional force is therefore -4.8 Newtons in the x direction. The weight component in the x direction and the frictional force in this direction therefore total 10 Newtons + (-4.8 Newtons) = + 5.2 Newtons. This force tends to accelerate the system in the same direction as does the weight of the 2 kg mass. This results in a net force of 5.2 Newtons + 19.6 Newtons = 24.8 Newtons on the 7 kg system. The system therefore accelerates at rate {} a = (24.8 Newtons) / (7 kg) = 3.5 meters/second ^ 2, approximately.
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RESPONSE --> I do not follow where you came up with the 270 degrees + 12 degrees. I follow the rest of the solution.
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wwT Student Name: assignment #021
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21:27:52 `q001. Note that this assignment contains 3 questions. . A projectile has an initial velocity of 12 meters/second, entirely in the horizontal direction. After falling to a level floor three meters lower than the initial position, what will be the magnitude and direction of the projectile's velocity vector?
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RESPONSE --> x = 12 m/s * cos(3) x = 12 m/s y = 12 m/s * sin(3) y = 0.6 m/s c^2 = a^2 + b^2 c = 'sqrt[(12 m/s)^2 + (0.6 m/s)^2] c = 12 m/s I am not sure if this is right, but I wanted to at least try.
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21:30:56 To answer this question we must first determine the horizontal and vertical velocities of the projectile at the instant it first encounters the floor. The horizontal velocity will remain at 12 meters/second. The vertical velocity will be the velocity attained by a falling object which is released from rest and allowed to fall three meters under the influence of gravity. Thus the vertical motion will be characterized by initial velocity v0 = 0, displacement `ds = 3 meters and acceleration a = 9.8 meters/second ^ 2. The fourth equation of motion, vf^2 = v0^2 + 2 a `ds, yields final vel in y direction: vf = +-`sqrt( 0^2 + 2 * 9.8 meters/second ^ 2 * 3 meters) = +-7.7 meters/second. Since we took the acceleration to be in the positive direction the final velocity will be + 7.7 meters/second. This final velocity is in the downward direction. On a standard x-y coordinate system, this velocity will be directed along the negative y axis and the final velocity will have y coordinate -7.7 m/s and x coordinate 12 meters/second. The magnitude of the final velocity is therefore `sqrt((12 meters/second) ^ 2 + (-7.7 meters/second) ^ 2 ) = 14.2 meters/second, approximately. The direction of the final velocity will therefore be arctan ( (-7.7 meters/second) / (12 meters/second) ) = -35 degrees, very approximately, as measured in the counterclockwise direction from the positive x axis. The direction of the projectile at this instant is therefore 35 degrees below horizontal. This angle is more commonly expressed as 360 degrees + (-35 degrees) = 325 degrees.
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RESPONSE --> I did not know to solve this the way you did, but I do understand your solution.
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21:40:36 `q002. A projectile is given an initial velocity of 20 meters/second at an angle of 30 degrees above horizontal, at an altitude of 12 meters above a level surface. How long does it take for the projectile to reach the level surface?
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RESPONSE --> x = 20 m/s * cos(30) x = 17 m/s y = 20 m/s * sin(30) y = 10 m/s
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21:43:54 To determine the time required to reach the level surface we need only analyze the vertical motion of the projectile. The acceleration in the vertical direction will be 9.8 meters/second ^ 2 in the downward direction, and the displacement will be 12 meters in the downward direction. Taking the initial velocity to be upward into the right, we situate our x-y coordinate system with the y direction vertically upward and the x direction toward the right. Thus the initial velocity in the vertical direction will be equal to the y component of the initial velocity, which is v0y = 20 meters/second * sine (30 degrees) = 10 meters/second. Characterizing the vertical motion by v0 = 10 meters/second, `ds = -12 meters (`ds is downward while the initial velocity is upward, so a positive initial velocity implies a negative displacement), and a = -9.8 meters/second ^ 2, we see that we can find the time `dt required to reach the level surface using either the third equation of motion `ds = v0 `dt + .5 a `dt^2, or we can use the fourth equation vf^2 = v0^2 + 2 a `ds to find vf after which we can easily find `dt. To avoid having to solve a quadratic in `dt we choose to start with the fourth equation. We obtain vf = +-`sqrt ( (10 meters/second) ^ 2 + 2 * (-9.8 meters/second ^ 2) * (-12 meters) ) = +-18.3 meters/second, approximately. Since we know that the final velocity will be in the downward direction, we choose vf = -18.3 meters/second. We can now find the average velocity in the y direction. Averaging the initial 10 meters/second with the final -18.3 meters/second, we see that the average vertical velocity is -4.2 meters/second. Thus the time required for the -12 meters displacement is `dt = `ds / vAve = -12 meters/(-4.2 meters/second) = 2.7 seconds.
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RESPONSE --> I understand what you did. I just don't always understand when to use what formula. After I see your solution I usually understand.
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21:46:41 `q003. What will be the horizontal distance traveled by the projectile in the preceding exercise, from the initial instant to the instant the projectile strikes the flat surface.
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RESPONSE --> 'dt = 2.7 s x = 20 m/s * cos(30) x = 17 m/s 'ds = vAve * 'dt 'ds = 17 m/s * 2.7 s 'ds = 46 m
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21:47:11 The horizontal velocity of the projectile will not change so if we can find this horizontal velocity, knowing that the projectile travels for 2.7 seconds we can easily find the horizontal range. The horizontal velocity of the projectile is simply the x component of the velocity: horizontal velocity = 20 meters/second * cosine (30 degrees) = 17.3 meters/second. Moving at this rate for 2.7 seconds the projectile travels distance 17.3 meters/second * 2.7 seconds = 46 meters, approximately.
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RESPONSE --> OK. I got this one right.
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