Two-dimensional Motion Part 1

Two-dimensional motion An Object Accelerating on a Ramp Learning Goal: Understand that the acceleration vector is in the direction of the change of the velocity vector. In one dimensional (straight line) motion, acceleration is accompanied by a change in speed, and the acceleration is always parallel (or antiparallel) to the velocity. When motion can occur in two dimensions (e.g. is confined to a tabletop but can lie anywhere in the x-y plane), the definition of acceleration is in the limit . In picturing this vector derivative you can think of the derivative of a vector as an instantaneous quantity by thinking of the velocity of the tip of the arrow as the vector changes in time. Alternatively, you can (for small ) approximate the acceleration as . Obviously the difference between and is another vector that can lie in any direction. If it is longer but in the same direction, will be parallel to . On the other hand, if has the same magnitude as but is in a slightly different direction, then will be perpendicular to . In general, can differ from in both magnitude and direction, hence can have any direction relative to . This problem contains several examples of this.Consider an object sliding on a frictionless ramp as depicted here. The object is already moving along the ramp toward position 2 when it is at position 1. The following questions concern the direction of the object's acceleration vector, . In this problem, you should find the direction of the acceleration vector by drawing the velocity vector at two points near to the position you are asked about. Note that since the object moves along the track, its velocity vector at a point will be tangent to the track at that point. The acceleration vector will point in the same direction as the vector difference of the two velocities. (This is a result of the equation given above.) Part A Which direction best approximates the direction of when the object is at position 1? Hint 1. Consider the change in velocity At this point, the object's velocity vector is not changing direction; rather, it is increasing in magnitude. Therefore, the object's acceleration is nearly parallel to its velocity. ANSWER: straight up downward to the left downward to the right straight down Correct Part B Which direction best approximates the direction of when the object is at position 2? Hint 1. Consider the change in velocity At this point, the speed has a local maximum; thus the magnitude of is not changing. Therefore, no component of the acceleration vector is parallel to the velocity vector. However, since the direction of is changing there is an acceleration. ANSWER: straight up upward to the right straight down downward to the left Correct Even though the acceleration is directed straight up, this does not mean that the object is moving straight up. Part C Which direction best approximates the direction of Hint 1. Consider the change in velocity when the object is at position 3? At this point, the speed has a local minimum; thus the magnitude of is not changing. Therefore, no component of the acceleration vector is parallel to the velocity vector. However, since the direction of is changing there is an acceleration. ANSWER: upward to the right to the right straight down downward to the right Correct Problem 3.03 A dragonfly flies from point Part A to point along the path shown in the figure in 2.20 . Find the and components of its position vector at point . Express your answers in meters to two significant figures separated by a comma. ANSWER: , = 2.0,1.0 , Correct Part B What is the magnitude of its position vector at ? Express your answer in meters to two significant figures. ANSWER: = 2.20 Correct Part C What is the direction of its position vector at ? Express your answer in degrees to three significant figures. ANSWER: = 26.6 counterwise from axis. Correct Part D Find the and components of the dragonfly's average velocity between and . Express your answers in meter per second to two significant figures separated by a comma. ANSWER: , = 3.6,2.3 , Correct Part E What is the magnitude of its average velocity between these two points? Express your answers in meter per second to two significant figures. ANSWER: = 4.3 Correct Part F What is the direction of its average velocity between these two points? Express your answers in degrees as an integer. ANSWER: = 32 counterwise from axis. Correct Introduction to Projectile Motion Learning Goal: To understand the basic concepts of projectile motion. Projectile motion may seem rather complex at first. However, by breaking it down into components, you will find that it is really little different from the one-dimensional motions that you may already have studied. One of the most often-used techniques in physics is to divide two- and three-dimensional quantities into components. For instance, in projectile motion, a particle has some initial velocity . In general, this velocity can point in any direction in the xy plane and can have any magnitude. To make a problem more manageable, it is common to break up such a quantity into its x component and its y component . Consider a particle with initial velocity above the negative x axis. that has magnitude 12.0 and is directed 60.0 Part A What is the x component of ? Express your answer in meters per second. Hint 1. Determine which trignometric function to use Suppose that were directed at 60.0 above the positive x axis. Which of the following would give the correct relationship between (the magnitude of ) and (the magnitude of )? ANSWER: Correct The fact that is actually directed at 60 above the negative x axis does not affect the magnitude of , but think about how it affects the sign of . ANSWER: = -6.00 Correct Part B What is the y component of ? Express your answer in meters per second. Hint 1. Determine which trignometric function to use Which of the following gives the correct relationship between ) and (the magnitude of )? ANSWER: Correct (the magnitude of ANSWER: = 10.4 Correct Breaking up the velocities into components is particularly useful when the components do not affect each other. Eventually, you will learn about situations in which the components of velocity do affect one another, but for now you will only be looking at problems where they do not. So, if there is acceleration in the x direction but not in the y direction, then the x component of the velocity will change, but the y component of the velocity will not. Part C Click on the image below to launch the video: Projectile Motion. Once you have watched the entire video, answer the graded follow-up questions. You can watch the video again at any point. The motion diagram for a projectile is displayed, as are the motion diagrams for each component. The x component motion diagram is what you would get if you shined a spotlight down on the particle as it moved and recorded the motion of its shadow. Similarly, if you shined a spotlight to the left and recorded the particle's shadow, you would get the motion diagram for its y component. How would you describe the two motion diagrams for the components? ANSWER: Both the vertical and horizontal components exhibit motion with constant nonzero acceleration. The vertical component exhibits motion with constant nonzero acceleration, whereas the horizontal component exhibits constant-velocity motion. The vertical component exhibits constant-velocity motion, whereas the horizontal component exhibits motion with constant nonzero acceleration. Both the vertical and horizontal components exhibit motion with constant velocity. Correct As you can see, the two components of the motion obey their own independent kinematic laws. For the vertical component, there is an acceleration downward with magnitude . Thus, you can calculate the vertical position of the particle at any time using the standard kinematic equation . Similarly, there is no acceleration in the horizontal direction, so the horizontal position of the particle is given by the standard kinematic equation . Now, consider the case in which two balls are simultaneously dropped from a height of 5.0 . Click on the image below to launch the second video: Projectile Motion. This video shows the situation described in the problem. Once you have watched the entire video, answer the graded follow-up questions. You can watch the video again at any point. Part D How long does it take for the balls to reach the ground? Use 10 magnitude of the acceleration due to gravity. for the Express your answer in seconds to one significant figure. Hint 1. How to approach the problem The balls are released from rest at a height of at time . Using these numbers and the kinematic equation , you can determine the amount of time it takes for the balls to reach the ground. ANSWER: = 1.0 Correct This situation, which you may have dealt with before (motion under the constant acceleration of gravity), is actually a special case of projectile motion. Think of this as projectile motion in which the horizontal component of the initial velocity is zero. Part E Imagine that the ball on the left is given a nonzero initial velocity in the horizontal direction, while the ball on the right continues to fall with zero initial velocity. What horizontal speed must the ball on the left start with so that it hits the ground at the same position as the ball on the right? Express your answer in meters per second to two significant figures. Hint 1. How to approach the problem Recall from Part B that the horizontal component of velocity does not change during projectile motion. Therefore, you need to find the horizontal component of velocity such that in time , the ball will move horizontally 3.0 . You can assume that its initial x coordinate is . ANSWER: = 3.0 Correct