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Ch0110 - Chapter 10 Constant Acceleration Problems in Two...

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Chapter 10 Constant Acceleration Problems in Two Dimensions 52 10 Constant Acceleration Problems in Two Dimensions In solving problems involving constant acceleration in two dimensions, the most common mistake is probably mixing the x and y motion. One should do an analysis of the x motion and a separate analysis of the y motion. The only variable common to both the x and the y motion is the time. Note that if the initial velocity is in a direction that is along neither axis, one must first break up the initial velocity into its components. In the last few chapters we have considered the motion of a particle that moves along a straight line with constant acceleration. In such a case, the velocity and the acceleration are always directed along one and the same line, the line on which the particle moves. Here we continue to restrict ourselves to cases involving constant acceleration (constant in both magnitude and direction) but lift the restriction that the velocity and the acceleration be directed along one and the same line. If the velocity of the particle at time zero is not collinear with the acceleration, then the velocity will never be collinear with the acceleration and the particle will move along a curved path. The curved path will be confined to the plane that contains both the initial velocity vector and the acceleration vector, and in that plane, the trajectory will be a parabola. (The trajectory is just the path of the particle.) You are going to be responsible for dealing with two classes of problems involving constant acceleration in two dimensions: (1) Problems involving the motion of a single particle. (2) Collision Type II problems in two dimensions We use sample problems to illustrate the concepts that you must understand in order to solve two-dimensional constant acceleration problems. Example 10-1 A horizontal square of edge length 1 . 20 m is situated on a Cartesian coordinate system such that one corner of the square is at the origin and the corner opposite that corner is at (1 . 20 m, 1 . 20 m). A particle is at the origin. The particle has an initial velocity of 2 . 20 m/s directed toward the corner of the square at (1 . 20 m, 1 . 20 m) and has a constant acceleration of 4 . 87 m/s 2 in the +x direction. Where does the particle hit the perimeter of the square?
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Chapter 10 Constant Acceleration Problems in Two Dimensions 53 Solution and Discussion Let’s start with a diagram. Now let’s make some conceptual observations on the motion of the particle. Recall that the square is horizontal so we are looking down on it from above. It is clear that the particle hits the right side of the square because: It starts out with a velocity directed toward the far right corner. That initial velocity has an x component and a y component. The y component never changes because there is no acceleration in the y direction. The x component, however, continually increases. The particle is going rightward faster and faster. Thus, it will take less time to get to
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