course Phy 121 ??????}??????M?Student Name:
......!!!!!!!!...................................
13:25:57 `q001. Note that this assignment contains 5 questions. . If you move 3 miles directly east then 4 miles directly north, how far do end up from your starting point and what angle would a straight line from your starting point to your ending point make relative to the eastward direction?
......!!!!!!!!...................................
RESPONSE --> The distances traveled east and north represent the legs of a right triangle on a graph. Therefore, the distance from the starting point can be found with the Pythagorean Theorem. c^2 = a^2 + b^2 c^2 = (3)^2 + (4)^2 = 25 c = 5 miles Now we can find the angle of the hypotenuse with the x-axis or eastward direction by the formula: `theta = tan-1 (`dy/`dx) `theta = tan-1(4mi/3mi) = 53.13 degrees
.................................................
......!!!!!!!!...................................
13:27:38 If we identify the easterly direction with the positive x axis then these movements correspond to the displacement vector with x component 3 miles and y component 4 miles. This vector will have length, determined by the Pythagorean Theorem, equal to `sqrt( (3 mi)^2 + (4 mi)^2 ) = 5 miles. The angle made by this vector with the positive x axis is arctan (4 miles/(three miles)) = 53 degrees.
......!!!!!!!!...................................
RESPONSE --> I understand this problem.
.................................................
......!!!!!!!!...................................
13:34:28 `q002. When analyzing the force is acting on an automobile as it coasts down the five degree incline, we use and x-y coordinate system with the x axis directed up the incline and the y axis perpendicular to the incline. On this 'tilted' set of axes, the 15,000 Newton weight of the car is represented by a vector acting straight downward. This puts the weight vector at an angle of 265 degrees as measured counterclockwise from the positive x axis. What are the x and y components of this weight vector? Question by student and instructor response:I have my tilted set of axes, but I cannot figure out how to graph the 15,000 N weight straight downward. Is it straight down the negative y-axis or straight down the incline? ** Neither. It is vertically downward. Since the x axis 'tilts' 5 degrees the angle between the x axis and the y axis is only 85 degrees, not 90 degrees. If we start with the x and y axes in the usual horizontal and vertical positions and rotate the axes in the counterclockwise direction until the x axis is 'tilted' 5 degrees above horizontal, then the angle between the positive x axis and the downward vertical direction will decrease from 270 deg to 265 deg. ** It might help also to magine trying to hold back a car on a hill. The force tending to accelerated the car down the hill is the component of the gravitational force which is parallel to the hill. This is the force you're trying to hold back. If the hill isn't too steep you can manage it. You couldn't hold back the entire weight of the car but you can hold back this component.
......!!!!!!!!...................................
RESPONSE --> wt = 15000 N `theta = 265 degrees To find the vertical component, y, we use the following equation: y = L*sin(`theta) = 15000N * sin(265) = -14942 N To find the horizontal component, x, we use the following equation: x = L*cos(`theta) = 15000N * cos(265) = -1307 N
.................................................
......!!!!!!!!...................................
13:35:52 The x component of the weight vector is 15,000 Newtons * cos (265 degrees) = -1300 Newtons approximately. The y component of the weight vector is 15,000 Newtons * sin(265 degrees) = -14,900 Newtons. Note for future reference that it is the -1300 Newtons acting in the x direction, which is parallel to the incline, then tends to make the vehicle accelerate down the incline. The -14,900 Newtons in the y direction, which is perpendicular to the incline, tend to bend or compress the incline, which if the incline is sufficiently strong causes the incline to exert a force back in the opposite direction. This force supports the automobile on the incline.
......!!!!!!!!...................................
RESPONSE --> Ok
.................................................
......!!!!!!!!...................................
13:39:54 `q003. If in an attempt to move a heavy object resting on the origin of an x-y coordinate system I exert a force of 300 Newtons in the direction of the positive x axis while you exert a force of 400 Newtons in the direction of the negative y axis, then how much total force do we exert on the object and what is the direction of this force?
......!!!!!!!!...................................
RESPONSE --> Fx = 300 N Fy = -400 N We can say that these forces are graphically, the legs of a right triangle. So we use the Pythagorean Theorem to find the direction of the force which is the hypotenuse. c^2 = a^2 + b^2 = (300N)^2 + (-400)^2 = 250000 N c = 500 Newtons To find the direction of the force, we find the angle of the force from the +x-axis. Since y is negative we must add 360 to the angle. `theta = tan-1(y/x) + 360 = tan-1(-400N/300N) + 360 = -53.13 degrees + 360 = 306.87 degrees.
.................................................
......!!!!!!!!...................................
13:40:29 Force is a vector quantity, so we can determine the magnitude of the total force using the Pythagorean Theorem. The magnitude is `sqrt( (300 N)^2 + (-400 N)^2 ) = 500 N. The angle of this force has measured counterclockwise from the positive x axis is arctan ( -400 N / (300 N) ) = -53 deg, which we express as -53 degrees + 360 degrees = 307 degrees.
......!!!!!!!!...................................
RESPONSE --> I understand.
.................................................
......!!!!!!!!...................................
13:42:10 `q004. If I exert a force of 200 Newtons an angle of 30 degrees, and you exert a force of 300 Newtons at an angle of 150 degrees, then how great will be our total force and what will be its direction?
......!!!!!!!!...................................
RESPONSE --> F = 200 N `theta = 30 degrees y component = L*sin(`theta) = 200*sin(30) = 100 Newtons x component = L*cos(`theta) = 200*cos(30) = 173 Newtons F = 300 N `theta = 150 degrees y component = L*sin(`theta) = 300*sin(150) = 150 Newtons x component = L*cos(`theta) = 300*cos(150) = -260 Newtons Total y component = 100 + 150 = 250 Newtons Total x component = 173 - 260 = -87 Newtons These two components represent the legs of a right triangle. In order to find the magnitude of the force, we use the Pythagorean Theorem. c^2 = a^2 + b^2 = (250N)^2 + (-87N)^2 = 70069 N c = 265 Newtons The angle of direction from the +x-axis can be found using the following equation: `theta = tan-1(y/x) = tan-1(250N/-87N) = -71 degrees Since the x component is negative, we must add 180 degrees `theta = -71N + 180 = 109 degrees
.................................................
......!!!!!!!!...................................
13:44:27 My force has an x component of 200 Newtons * cosine (30 degrees) = 173 Newtons approximately, and a y component of 200 Newtons * sine (30 degrees) = 100 Newtons. This means that the action of my force is completely equivalent to the action of two forces, one of 173 Newtons in the x direction and one of 100 Newtons in the y direction. Your force has an x component of 300 Newtons * cosine (150 degrees) = -260 Newtons and a y component of 300 Newtons * sine (150 degrees) = 150 Newtons. This means that the action of your force is completely equivalent the action of two forces, one of -260 Newtons in the x direction and one of 150 Newtons in the y direction. In the x direction and we therefore have forces of 173 Newtons and -260 Newtons, which add up to a total x force of -87 Newtons. In the y direction we have forces of 100 Newtons and 150 Newtons, which add up to a total y force of 250 Newtons. The total force therefore has x component -87 Newtons and y component 250 Newtons. We easily find the magnitude and direction of this force using the Pythagorean Theorem and the arctan. The magnitude of the force is `sqrt( (-87 Newtons) ^ 2 + (250 Newtons) ^ 2) = 260 Newtons, approximately. The angle at which the force is directed, as measured counterclockwise from the positive x axis, is arctan (250 Newtons/(-87 Newtons) ) + 180 deg = -71 deg + 180 deg = 109 deg.
......!!!!!!!!...................................
RESPONSE --> Ok
.................................................
......!!!!!!!!...................................
13:46:37 `q005. Two objects, the first with a momentum of 120 kg meters/second directed at angle 60 degrees and the second with a momentum of 80 kg meters/second directed at an angle of 330 degrees, both angles measured counterclockwise from the positive x axis, collide. What is the total momentum of the two objects before the collision?
......!!!!!!!!...................................
RESPONSE --> p1 = 120 kg*m/s `theta1 = 60 degrees y component = L*sin(`theta) = (120 kg*m/s)(sin 60) = 104 N x component = L*cos(`theta) = (120 kg*m/s)(cos 60) = 60 N p2 = 80 kg*m/s `theta 2 = 330 degrees y component = L*sin(`theta) = (80 kg*m/s)(sin 330) = -40 Newtons x component = L*cos(`theta) = (80 kg*m/s)(cos 330) = 69 Newtons Total y component = 104 - 40 = 64 Newtons Total x component = 60 + 69 = 129 Newtons These two components represent the legs of a right triangle. In order to find the magnitude of the force, we use the Pythagorean Theorem. c^2 = a^2 + b^2 = (64N)^2 + (129N)^2 = 20737 N c = 144 Newtons The angle of direction from the +x-axis can be found using the following equation: `theta = tan-1(y/x) = tan-1(64N/129N) = 26 degrees
.................................................
......!!!!!!!!...................................
13:47:36 The momentum of the first object has x component 120 kg meters/second * cosine (60 degrees) = 60 kg meters/second and y component 120 kg meters/second * sine (60 degrees) = 103 kg meters/second. The momentum of the second object has x component 80 kg meters/second * cosine (330 degrees) = 70 kg meters/second and y component 80 kg meters/second * sine (330 degrees) = -40 kg meters/second. The total momentum therefore has x component 60 kg meters/second + 70 kg meters/second = 130 kg meters/second, and y component 103 kg meters/second + (-40 kg meters/second) = 63 kg meters/second. The magnitude of the total momentum is therefore `sqrt((130 kg meters/second) ^ 2 + (63 kg meters/second) ^ 2) = 145 kg meters/second, approximately. The direction of the total momentum makes angle arctan (63 kg meters/second / (130 kg meters/second)) = 27 degrees, approximately.
......!!!!!!!!...................................
RESPONSE --> Ok
.................................................
???????l????assignment #019 019. `query 19 Physics I 07-01-2006
......!!!!!!!!...................................
00:53:17 Query class notes #20 Explain how we calculate the components of a vector given its magnitude and its angle with the positive x axis.
......!!!!!!!!...................................
RESPONSE --> The magnitude of the vector represents the hypotenuse of a right triangle.The angle with the positive x-axis represents `theta. The vector components are the legs of the right triangle. The x component can be found by multiplying the magnitude by cos(`theta) and multiplying magnitude by sin(`theta) to get the y component.
.................................................
......!!!!!!!!...................................
00:53:40 ** STUDENT RESPONSE: x component of the vector = magnitude * cos of the angle y component of the vector = magnitude * sin of the angle To get the magnitude and angle from components: angle = arctan( y component / x component ); if the x component is less than 0 than we add 180 deg to the solution To get the magnitude we take the `sqrt of ( x component^2 + y component^2) **
......!!!!!!!!...................................
RESPONSE --> I understand how to find vector components, magnitude, and the angle with the positive x axis.
.................................................
......!!!!!!!!...................................
00:55:08 Explain what we mean when we say that the effect of a force is completely equivalent to the effect of two forces equal to its components.
......!!!!!!!!...................................
RESPONSE --> The best way to explain this is with a graphical representation. If we consider the effect of the force vector, we can see that it is equivalent to the effect of force in the horizontal direction plus the force in the vertical direction.
.................................................
......!!!!!!!!...................................
00:56:26 ** If one person pulls with the given force F in the given direction the effect is identical to what would happen if two people pulled, one in the x direction with force Fx and the other in the y direction with force Fy. **
......!!!!!!!!...................................
RESPONSE --> Ok
.................................................
......!!!!!!!!...................................
00:59:08 Explain how we can calculate the magnitude and direction of the velocity of a projectile at a given point.
......!!!!!!!!...................................
RESPONSE --> The magnitude of the velocity of a projectile can be found by finding the horizontal (x) and vertical (y) velocities with an equation of uniformly accelerated motion. The magnitude is then found by the Pythagorean Theorem. We would pluge the horizontal and vertical velocitie in as the legs of the right triangle. Magnitude of velocity = Sqrt [(X velocity)^2 + (Y velocity)^2] The direction of velocity is the inverse tangent of the y velocity component over the x velocity component. `theta = tan-1(Y velocity/X velocity)
.................................................
......!!!!!!!!...................................
00:59:49 ** Using initial conditions and the equations of motion we can determine the x and y velocities vx and vy at a given point, using the usual procedures for projectiles. The magnitude of the velocity is sqrt(vx^2 + vy^2) and the angle with the pos x axis is arctan(vy / vx), plus 180 deg if x is negative. **
......!!!!!!!!...................................
RESPONSE --> Ok
.................................................
......!!!!!!!!...................................
01:02:06 Explain how we can calculate the initial velocities of a projectile in the horizontal and vertical directions given the magnitude and direction of the initial velocity.
......!!!!!!!!...................................
RESPONSE --> To find the x component we would multiply the magnitude by cos(theta). Theta is the direction of the initial velocity. This gives us the initial velocity in the horizontal direction. To find the y component we would multiply the magnitude by sin(theta). This gives us the initial velocity in the vertical direction.
.................................................
......!!!!!!!!...................................
01:02:43 ** Initial vel in the x direction is v cos(theta), where v and theta are the magnitude and the angle with respect to the positive x axis. Initial vel in the y direction is v sin(theta). **
......!!!!!!!!...................................
RESPONSE --> I understand.
.................................................
......!!!!!!!!...................................
01:02:59 Univ. 8.58 (8.56 10th edition). 40 g, dropped from 2.00 m, rebounds to 1.60 m, .200 ms contact. Impulse? Ave. force?
......!!!!!!!!...................................
RESPONSE --> I am not required to answer this question.
.................................................
......!!!!!!!!...................................
01:03:03 ** You have to find the momentum of the ball immediately before and immediately after the encounter with the floor. This allows you to find change in momentum. Using downward as positive direction throughout: Dropped from 2 m the ball will attain velocity of about 6.3 m/s by the time it hits the floor (v0=0, a = 9.8 m/s^2, `ds = 2 m, etc.). It rebounds with a velocity v0 such that `ds = -1.6 m, a = 9.8 m/s^2, vf = 0. This gives rebound velocity v0 = -5.6 m/s approx. Change in velocity is -5.6 m/s - 6.3 m/s = -11.9 m/s, approx. So change in momentum is about .04 kg * -11.9 m/s = -.48 kg m/s. In .2 millliseconds of contact we have F `dt = `dp or F = `dp / `dt = -.48 kg m/s / (.002 s) = -240 Newtons, approx. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
01:03:19 Query Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE -->
.................................................
"