3. KINEMATICS IN TWO DIMENSIONS; VECTORS.

3. KINEMATICS IN TWO DIMENSIONS; VECTORS. Key words: Motion in Two Dimensions, Scalars, Vectors, Addition of Vectors by Graphical Methods, Tail to Tip Method, Parallelogram Method, Negative Vector, Vector Subtraction, Multiplication of Vector by Scalar, Adding and Subtracting Vectors by Components, Projectile Motion, We studied in details the motion of an object along straight line onedimensional motion. Now we will try to consider more general case the motion in two dimensions. But to reach this goal we need additional mathematical technique. We do not need this technique to work with physical quantities that have no relation to the direction in space. These physical quantities are called scalars. They can be characterized only by magnitude (number with unit). Examples of these quantities are as following: time, mass, density etc. We can use just arithmetic s to operate with them. Other situation is with the physical quantities that have a direction in space. When object is moving along straight line there are only two directions: in term of coordinate axis these are directions in the positive or negative direction of X-axis. In the two-dimensional case, we have two mutually perpendicular axes: horizontal X-axis and vertical Y-axis. Object can move now in any direction laying in the XOY plane, so the physical quantities that have direction in space displacement, velocity, acceleration, and others now can be directed in any direction. The physical quantities that characterized by magnitudes (some number with unit) but also by the direction in space are called vectors. Vector is a quantity that incorporates magnitude and direction. To operate with them we could not use just arithmetic. To add, subtract, multiply vectors we need to become familiar with some basic concepts of the special branch of mathematics vector algebra. ADDITION OF VECTORS BY GRAPHICAL METHODS. Each vector we represent as an arrow directed in space. The length of this arrow in some scale represents the magnitude of a vector. A vector quantity we will designate by corresponding letter with small arrow above. For example, the vector of velocity is represented by designation v r, the magnitude of this vector designated just v. The vector quantity can be

designated by the bold font also like v. The vector of acceleration is represented by designation a r or a, the magnitude of this vector is designated as a. For the vector of displacement we have correspondingly D r or D, and D. Due the spatial character of vectors it is reasonable to use graphical methods to operate with them. Graphical Methods. First of all we identify equality of two vectors. Two vectors supposed to be equal when their magnitudes are equal and these vectors are directed in the same direction. This definition allows us to displace vector without changing its magnitude and direction in any other position in space. This is the key point in the graphical methods techniques. The first of them is called tail to tip method of adding vectors. In this method, the tail of the added vector is brought to the tip of initial vector. Vector that results from this operation (it is called resultant) directed from the tail of the initial vector to the tip of an added vector. This operation is shown in Fig. 3.1a. Object was displaced from the point 1 to the point 2 (displacement vector D1), and then from the point 2 to the point 3 (displacement vector D2). Vector DR results from this vector addition (vector that results from vector addition sometimes is called resultant). By this method we can add more than two vectors (in this case, this type of vector addition is called polygon method. Addition of three vectors is shown in Fig. 3.1b as an example of application of the polygon method. Other wide used method of vector addition is called parallelogram method. In this method their tales brings two vectors together. The figure is completed to the parallelogram. Resultant vector is a diagonal of this parallelogram (see Fig. 3.1c). Comparison of Fig.3.1a and Fig.3.1c shows that tail to tip method and parallelogram method bring absolutely identical results. Before considering the vector subtraction we introduce the concept of negative vector. The negative vector has magnitude equals to the magnitude of initial vector but directed in opposite direction. Now we can consider vector subtraction as following: A B = A + (-- B) We just add to the vector A vector (-- B).

Fig. 3.1 Vector addition by the graphical methods: (a) Tail to Tip method; (b) Polygon Method; (c) Parallelogram method.

The result of the multiplication of vector by positive scalar is the vector directed in the same direction as initial vector but with magnitude equals to the magnitude of initial vector multiplied by scalar. If scalar is negative, the direction will be opposite to the direction of initial vector. Summarizing the consideration of the graphical methods of vector operations, we can notice that these methods are very clear and helpful to understand spatial nature of vector operations. At the same time they have serious disadvantageous. These methods are not precise. Their precision is limited by accuracy of instruments used (meter sticks and protractors). Most important, graphical methods are not applicable to the general three-dimensional case. To perform vector operations, we will use much more convenient and powerful method Method of Components. ADDING AND SUBTRACTING VECTORS BY COMPONENTS. Key point in consideration is as following: we bring tail of vector at the origin of coordinate axes (see Fig. 3.2). y A θ A x A y x Fig. 3.2. Resolution of a vector into components. The length of the arrow in some scale represents the magnitude A of the vector A. The direction of the vector is determined by an angle θa between the vector and the positive direction of the X-axis. What are the components? The projection of vector on the X-axis Ax is called x- component of the vector A. The projection of vector on the Y-axis Ay is called y-component of the vector A. Actually we have a right triangle situated at the origin of coordinates in which A plays role of diagonal. Side adjacent to the angle θa is Ax. Side opposite to the angle θa equals to the component Ay. The following trigonometric functions can be defined for this triangle: sin θa = Ay/A (3.1a)

cos θa = Ax/A (3.1b) tan θa = Ay/Ax (3.1c) Using these relationships we can get following useful formulas for operations that is called resolution vector into components. This operation allows finding components of vector when its magnitude and direction are given: Ax = A cos θa (3.2) Ay = A sin θa (3.3) If we are given by components and need to find magnitude of a vector and its direction (angle between the vector and the positive direction of the X- axis) can use the theorem of Pythagoras : V = Ax² + Ay² (3.4) θa = tan ¹(Ay//Ax) (3.5) Component situated on the positive part of the coordinate axis is positive, situated on the negative part of the coordinate axis is negative. Vector addition by components can be performed as follows. For example, we are given magnitudes and directions of two vectors A and B. It means that we are given their magnitudes A and B, and their angles with the positive direction of X-axis θa and θb. Suppose we are asked to find vector C. Actually we need to solve vector equation C = A + B (3.6) We could not solve it just substituting vectors by their magnitudes. This equation is only symbolical representation of operation. To really solve this equation we need to resolve vectors A and B into components using formulas (3.2) and (3.3). Than we should find components of resultant vector C according following formulas Cx = Ax + Bx (3.7) Cy = Ay + By (3.8)

Then we can find the magnitude of vector C and its direction (angle with respect the positive direction of X-axis) using formulas (3.4) and (3.5): C = Cx² + Cy² (3.9) θa = tan ¹(Cy/Cx) (3.10) If we need to subtract one vector from another, for example, to find vector D D = A B (3.11) In this case, we can find components of vector D as follows Dx = Ax Bx (3.12) Dy = Ay By (3.13) Then the magnitude and direction of vector D then can be found by the same way as for the vector C (see formulas (3.9) and (3.10)). EXAMPLE 3.1. Each of the displacement vectors A and B (Fig. 3.3) has a magnitude of 5.00 m. Their directions relative to the positive direction of the X- axis are 30.0º for the vector A and 45.0º for the vector B. Find: (a) x- and y-components of the vectors A and B; (b) x- and y- components of the vector D = B A; (c) the magnitude and direction of the vector D.

Fig. 3.3. Example 3.1. The same approach we can use to solve more complicated vector equation. For example, equation E = F + G H + I (3.14) We should to resolve each of the given vectors into components using formulas (3.7) and (3.8), and then to find components of the vector E as following Ex = Fx + Gx Hx + Ix (3.15) Ey = Fy + Gy Hy + Iy (3.16)

When the components of the vector E become known, its magnitude and direction can be found using in the formulas (3.9) and (3.10) components Ex and Ey. PROJECTILE MOTION. We will use the technique of the vector addition to describe one type of 2D motion that is called Projectile Motion (Fig. 3.4). y v 0 θ h O X P x Fig. 3.4. Projectile motion (general case). Projectile can be any object launched into space. We are not interested to know how and who launched the projectile into space. We only need to know the initial velocity v0 of the projectile (the magnitude of the initial velocity v0, and its direction -- the angle θ0 of the velocity with respect of the positive direction of the X-axis). To simplify the consideration, we will neglect the air resistance and assume that only the gravity influences the motion of the projectile. Nevertheless, the situation remains very complicated because the direction of the velocity is changing during all time of projectile motion. The solution of this complicated problem was found by Galileo Galilei. He performed the experiment in which to objects simultaneously start to move from the same height. One of them was falling vertically down experiencing free fall motion. Another object was simultaneously launched in the projectile motion. Both of them simultaneously reached the ground. Galilei deduced that motions of the object in vertical direction and in horizontal direction are absolutely independent. In accordance with our today s knowledge, we can say, that we can independently consider the x- and y-components of the projectile

motion. The gravity influences the motion of the projectile along vertical (ycomponent). We studied before this type of motion. This is the motion with constant acceleration (acceleration due to gravity) or free-fall motion. We studied before this type of motion, but now this approach is applied only to the description of the y-component of the position, velocity and acceleration of the projectile. ay = -- g = const (3.17) vy = vy0 -- g t (3.18) y= y0 + vy0 t -- (1/2) g t^2 (3.19) vy ² = vy0² -- 2 g (y y0) (3.20) Where vy0 is the y-component of the initial velocity vy0 = v0 sin θ0 (3.21) If the air resistance can be ignored, no forces influence the motion of the projectile along the horizontal. Therefore the projection of the projectile motion on the horizontal (x-component of the projectile motion) is the motion with the constant velocity. We also studied this type of motion, but now it is applicable only to the x-component of projectile motion: ax = = const = 0 (3.22) v = vx0 = const (3.23) Where vx0 is the x-component of the initial velocity x= x0 + vx0 t (3.24) vx0 = v0 cos θ0 (3.21) Solving problems, we should remember that these are only components of the motion. Real motion is occurred along the trajectory some curve in space so in the any instant of time, the time t in equations is the same for both of components. There is some times additional implicit information in

conditions of problems related to the projectile motion: when object is situated at ground level, it means that y = 0; when object reaches its highest elevation, its y-component of velocity is zero; x-component of velocity always is the same. How to apply this information in problems we can see for the following example. EXAMPLE 3.2. Projectile launched horizontally. A rescue plane is flying horizontally at a speed of 120 m/s and drops package at an elevation of 2000 m (Fig. 3.5). (a) How much time is required for the package to reach the ground? (b) How far does it travel horizontally (in other words, what is the largest distance range covered by the projectile before striking the ground)? (c) What is the x-component and y-component of the velocity of the package just before it reaches the ground? What is the magnitude of the velocity of the package just before it reaches the ground? O v 0 x h y Fig. 3.5. Example 3.2. Projectile launched horizontally. x0 = 0 y0 = 2000 m v0 = 120 m/s θ0 = 0º (a) tr? (b) xr? (c) vstr?

(a) When package reaches the ground t = tr, y = 0. The equation that relates these two physical quantities is (3.19). Solving this equation for unknown tr, we will get tr, = 20.2 s. (b) At the same time, when t = tr, x = xr. The equation that relates these two physical quantities is (3.24). Solving this equation for unknown tr, we will get xr = 2424 m. (c) At the same time, when t = tr, v = vstr. To find magnitude of the vector vstr, we need to find its components. The x-component of this vector is all the time the same vxstr = 120 m/s. To find vystr at t = tr, we will use Eq. (3.18) that relates these two physical quantities. As a result we will get vystr = -- 198 m/s. Now the magnitude of the vector vstr can be found using formula. Finally, we will get vstr = 231 m. EXAMPLE 3.3. Projectile launched at nonzero angle to the horizon. The projectile is launched with initial velocity v0 at some nonzero angle θ0 with the respect to the horizon. (Fig. 3.6). Find: (a) time needed to reach the range; (b) largest distance along X-axis (range) covered by the projectile before striking the ground; (c) time th needed for the projectile to reach highest elevation yh; (d) highest elevation. x0 = 0 y0 = 0 v0 = v0 θ0 = θ0 ---------------- (a) tr? (b) xr? (c) th? (d) yh?

y v 0 O θ some nonzero angle. x R Fig. 3.6. Example 3.3. Projectile launched with First two questions can be considered by the same way as it was done in the Example 3.2. (a) When projectiles reaches the range, t = tr, y = 0. The equation that relates these two physical quantities is (3.19). Solving this equation for unknown tr, we will get tr = (2 v0 sin θ0)/g (3.22) (b) At the same time, when t = tr, x = xr. The equation that relates these two physical quantities is (3.24). Solving this equation for unknown xr, we will get xr = ( v0² sin 2θ0)/g (3.23) The maximal value of the expression (3.23) will be when θ0 = 45º. Therefore the largest range will be occurred when projectile will be launched at the angle of 45º with the respect of the horizon. (c) At the highest elevation (t = th, y = yh) the y-component of the velocity equals zero. Using Eq. (3.18) related t and vy for this specific instant of time we will get th = (v0 sin θ0)/g (3.24) Comparing Eqs. (3.22) and (3.24), we come to conclusion that tr = 2 th. It means that the motion of the projectile is absolutely symmetrical: how many time is needed to reach the highest elevation, so many time then needed to reach the ground. (d) When t = th, y = yh. Using the equation that relates these two physical quantities (3.19), we can find

yh = ( v0² sin² θ0)/(2 g) (3.23) Discussing this beautiful Galileo s solution of the projectile problem, we need to remind that it is developed for the situations in which we can neglect the air resistance. Actually it means that results are applicable only to the case of comparatively small velocities of projectiles. Otherwise we should apply ballistics the scientific study of the fast projectiles (shells and rockets), their ejection and flight through the air.