PHU3300 Session 1: Vectors and Scalars PDF
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This document is lecture notes from The Open University of Sri Lanka about vectors and scalars, and motion. It covers the basic concepts of vectors and scalars, motion in one, two, and three dimensions, including vector notation and operations.
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PHU3300 – Unit 1 Session 1: VECTORS AND SCALARS, MOTION IN ONE, TWO AND THREE DIMENSIONS Session 1 VECTORS AND SCALARS, MOTION IN ONE, TWO AND THREE DIMENSIONS Contents 1.1 INTRODUCTION 1.2 VECTORS, SCALARS AND UNIT VECTORS 1.3 MOTION IN ONE DIMENSIONS: REVISION 1.4...
PHU3300 – Unit 1 Session 1: VECTORS AND SCALARS, MOTION IN ONE, TWO AND THREE DIMENSIONS Session 1 VECTORS AND SCALARS, MOTION IN ONE, TWO AND THREE DIMENSIONS Contents 1.1 INTRODUCTION 1.2 VECTORS, SCALARS AND UNIT VECTORS 1.3 MOTION IN ONE DIMENSIONS: REVISION 1.4 MOTION IN TWO AND THREE DIMENSIONS Summary Learning Outcomes Self-Assessment Questions and Answers 1.1 INTRODUCTION The session one is intend student to learn some basic concepts in motion. Two areas have been identified with this regard as follows; (1) vectors and scalars (2) Motion in one, two and three dimensions. The first part of the session one aims at introducing the student to the concepts of vectors and scalars in relation to physical quantities. The composition and resolution of vectors dealt with more or less as a revision of what the student has already has learned at the G.C.E. (A/L) or Foundation levels. The multiplication of vectors is next discussed, and the ideas of scalar (or dot) and vector (or cross) products are explained. These are clarified as far as possible by actual physical examples. Unit vectors are introduced, and the student is introduced to basic vector notation. Through examples and self-assessment questions certain vector operations are described. The scalar and vector products of two vectors, expressed in terms of their component magnitudes are discussed in some detail. Further applications of vector operations, including a more detailed knowledge of vector analysis, will be dealt with in subsequent sessions, so far as their applications become necessary in the study of concrete physical problems. The second part of the session deals with motion in one, two and three dimensions. In physics, we can categorize motion into three types: translational, rotational, and vibrational. In this and the next few sessions, we are concerned only with translational motion. Later in the book we shall discuss rotational motion. However, the vibrational motion will be discussed in a separate course in Level 3. 1 Copyright © 2017, The Open University of Sri Lanka (OUSL) PHU3300 – Unit 1 Session 1: VECTORS AND SCALARS, MOTION IN ONE, TWO AND THREE DIMENSIONS The motion in one dimension is also more or less a revision of what the student has already has learned at the G.C.E. (A/L) or Foundation levels. In two and three dimensional motion we shall explore the study of kinematics of a particle moving in two and three dimensions. By knowing the basics of two and three dimensional motions the student will eventually prepare him/her self to examine a variety of situations, ranging from the motion of satellites in orbit to the motion of electrons in a uniform electric field. Further this will allow the student to study in greater detail the vector nature of position, velocity, and acceleration. The student will further learn about the projectile motion and uniform circular motion as special cases of motion in two dimensions during this course. 1.2 VECTORS, SCALARS AND UNIT VECTORS 1.2.1 Vectors and Scalars When a particle changes its position, it is said to have undergone a displacement. If a particle moves from a point A to a point B (Figure 1.la), its displacement is represented by a straight line drawn from A to B. The arrowhead at B gives the direction of displacement in this instance, from A to B. This straight line does not necessarily represent the actual motion of the particle from A to B. It only shows the net effect of the· motion. Figure 1.1: Particle changes its position and undergoes a displacement (a) displacement is represented by a straight line (b) dotted line shows the actual path of the particle from A to B (c) when particle undergoes a subsequent displacement from B to C, the resultant displacement is A C In Figure 1.1b, for example, the dotted line shows the actual path of the particle from A to B. Thus, the actual path is not the same as the displacement, which is given by AB. Next consider a displacement such as A (Figure l.1a). Here is parallel to AB i.e. in the same direction, and equal in length to AB. No distinction is made between these two displacements. A displacement is thus characterized only by a length and a direction. 2 Copyright © 2017, The Open University of Sri Lanka (OUSL) PHU3300 – Unit 1 Session 1: VECTORS AND SCALARS, MOTION IN ONE, TWO AND THREE DIMENSIONS Consider Figure 1.1 (c). Suppose the particle undergoes a subsequent displacement from B to C. In effect, the displacements A B, and B C would be equivalent to a displacement A C. AC is thus known as the sum or resultant of the displacements AB and BC. In representing displacements in the above way, the lengths of the straight lines should be proportional to the magnitudes of the displacements; their directions would indicate the directions of the displacements. Quantities like displacement cannot be described by a magnitude alone. To describe them we must specify both a magnitude and a direction. Such physical quantities are known as vectors. Other examples of vectors are force, velocity, acceleration, weight, electric field strength, magnetic induction etc. There are other physical quantities, which can be completely described by a magnitude and unit only. They are known as scalars. Some examples of scalars are mass, time, density, energy, temperature, speed etc. Scalar quantities can be added or subtracted by the ordinary rules of algebra. For instance, the sum of two masses 2 g and 3 g would be equal to 2 g + 3 g = 5 g. Similarly two amounts of energy 5 J and 10 J would add up to 5 J + 10 J = 15 J. 1.2.2 Notation and Properties of Vectors Notation Vector quantities are generally denoted by drawing an arrow or a bar above them. Thus F represents a force, which is a vector, where the magnitude of a vector quantity is a positive number (with its units) that describes its size and denoted either by F or F. A scalar, such as temperature, can be directly represented by its magnitude alone, thus the scaler is denoted by its quantity T. Properties of Vectors Two vectors A and B may be defined to be equal if they have the same magnitude and if they point in the same direction. Thus A B only if the magnitude of vectors A and B are equal (i.e. A B ), and both vectors point in the same direction along parallel line. This property allows us to move a vector to a position parallel to itself in a diagram without affecting the vector. Since vectors involve a direction as well as a magnitude, they cannot be simply added by the rules of algebra. Conveniently a geometrical method is used for adding of vectors. That is, if we are to add vector B to vector A , first we draw vector A on a paper, with its magnitude represented by a convenient length scale, and then draw vector B to the same scale, with its tail starting from the tip of A (Figure 1.2). The resultant vector R A B is the vector drawn from the tail of A to the tip of B. Further, if two vectors are added, the sum is independent of the order of the addition. This means A B B A and is known as the commutative law of addition. 3 Copyright © 2017, The Open University of Sri Lanka (OUSL) PHU3300 – Unit 1 Session 1: VECTORS AND SCALARS, MOTION IN ONE, TWO AND THREE DIMENSIONS Figure 1.2: When vector B is added to vector A , the resultant R is the vector that runs from the tail of A to the tip of B Suppose if we have more than two vectors, the same geometric construction can be used to add those vectors as shown in Figure 1.3. In this case, the resultant vector R A B C D , is the vector that completes the polygon. A , B , C and A Figure 1.3: Geometric construction for summing four vectors, D The negative of the vector is a vector which has the same magnitude but point in opposite directions and it is represented as A A (Figure 1.4). Figure 1.4: Negative vector is a vector which has the same magnitude but point in opposite directions We can use the definitions of the operation of vector addition and the negative of a vector to define the operation of vector subtraction. Accordingly, we define the operation A B as vector B added to vector A. i.e. A B A B If vector A is multiplied by a positive scalar quantity m, the product mA is a vector that has the same direction as A and magnitude mA. If m is a negative scalar quantity then the magnitude is mA but the direction is opposite to A. 4 Copyright © 2017, The Open University of Sri Lanka (OUSL) PHU3300 – Unit 1 Session 1: VECTORS AND SCALARS, MOTION IN ONE, TWO AND THREE DIMENSIONS 1.2.3 Addition of Vectors Since vectors involve a direction as well as a magnitude, they cannot be simply added by the rules of algebra. We indicate below a geometrical method for the addition of vectors. Suppose we wish to find the sum or resultant of two velocities AB 20 m s1 and BC 10 m s1 (Figure 1.5). Their directions are obviously different. Straight lines AB, and BC are drawn proportional in length to 20 and 10 respectively, and parallel to the actual directions of the velocities. As we have shown earlier, the effective sum or resultant of the velocities AB and BC would be given by AC, where the direction and length of AC would represent this resultant, it provided that the length of AC is converted into a magnitude of velocity on the same scale as AB and BC. Figure 1.5: The sum or resultant of two velocities This triangle of vectors can be converted into a parallelogram of vectors. Consider the same two velocities AB 20 m s1 and BC 10 m s1 , suppose we draw a straight line AB (Figure1.6), equal in length to BC, and parallel to it, then AB would also represent the velocity10 m s1. Figure 1.6: Parallelogram of vectors From Figure 1.5, we found the resultant of the velocities to be AC. We now have a parallelogram ABCB, where AB and AB, represent the two velocities considered, and their resultant is given by the diagonal AC of the parallelogram. Suppose when two vectors a and b are added and the resultant is c , then by using the operation of vector addition we can write resultant c as c a b. Here, the addition is not algebraic. We can use the graphical technique mentioned earlier. 5 Copyright © 2017, The Open University of Sri Lanka (OUSL) PHU3300 – Unit 1 Session 1: VECTORS AND SCALARS, MOTION IN ONE, TWO AND THREE DIMENSIONS Activity 1.1 A car moves eastward for a distance of 50 km, then northward for 30 km and finally in a direction 30o east of north for 25 km. Draw the vector diagram and determine the total displacement of the car from its starting point. Solution The displacement of the car must be represented to scale on the vector diagram - say 1 cm represents 10 km (see Fig.1.7) Figure 1.7: Schematic diagram for Activity 1.1 (not drawn to scale) Draw the AB eastward with a length of 5 cm; BC northwards with a length of 3 cm; CD is drawn with a length of 2.5 cm; in a direction making an angle of 30o with. BC (i.e. with the S-N direction). Now AD is the total displacement of the car. Measure the length AD, and estimate the corresponding distance in km i.e. if AD is measured in cm, the distance is (AD x10) km. By measuring the angle the direction of the car’s resultant displacement can be found. Remember that in stating the resultant displacement as a vector, both its magnitude and direction must be indicated. 6 Copyright © 2017, The Open University of Sri Lanka (OUSL) PHU3300 – Unit 1 Session 1: VECTORS AND SCALARS, MOTION IN ONE, TWO AND THREE DIMENSIONS 1.2.4 Resolution of Vectors Suppose we wish to determine the “effect” of a vector OA in some, other given direction, OX, say, making an angle with OA. Figure 1.8: The vector OA and its respective components OX and OY which are mutually perpendicular to each other. We draw a rectangle, as shown in Figure 1.8, where OX, and OY are perpendicular to one another. XA is then, effectively the same as OY. The resultant of OX and XA is the original vector OA. The vectors OX and OY are then known as the components of OA , which are in the mutually perpendicular directions. The component OX is given by OA cos , and the perpendicular component OY by OA sin. We have resolved the vector OA thus in two mutually perpendicular directions. The relationship between OA and its two perpendicular components is; i.e. OA2 OX2 OY2 If we wish to resolve a vector OA in a three dimensional space, we must select three axes mutually perpendicular to one another. Let the components of the vector along these three axes be OX , OY and OZ respectively. The relationship between OA and its three perpendicular components is; OA2 OX2 OY2 OZ2 We should note that the system of axes as a whole can be rotated to fit in with any desired direction. Activity 1.2 Two vectors of lengths a and b make an angle with each other when placed tail to tail. Prove, by taking components along two perpendicular axes that the length of the resultant vector is r a2 b2 2abcos 7 Copyright © 2017, The Open University of Sri Lanka (OUSL) PHU3300 – Unit 1 Session 1: VECTORS AND SCALARS, MOTION IN ONE, TWO AND THREE DIMENSIONS Solution We select the axes OX and OY as perpendicular axes, where OX is the direction of the vector b. Figure 1.9: Schematic diagram for Activity 1.2 The sum of the components of vectors a and b along the x-axis is: a cos b Similarly the sum of the components along the y-axis is: a sin b cos900 a sin Thus, if r is the length of the resultant vector, then r2 a cos b asin a2 cos2 sin2 2abcos b2 a2 b2 2abcos 2 2 Hence, r a2 b2 2abcos. Activity 1.3 Consider the three vectors shown in Figure1.10, where A 10 units , B 30 units and C 20 units. Find (i) the resultant vector in x, y direction and (ii) the magnitude and direction of the resultant vector. Figure 1.10: Schematic diagram for Activity 1.3 8 Copyright © 2017, The Open University of Sri Lanka (OUSL) PHU3300 – Unit 1 Session 1: VECTORS AND SCALARS, MOTION IN ONE, TWO AND THREE DIMENSIONS Solution (i) The resultant vector along the x – axis: 30 20 Rx B cos 450 C cos 450 35.36 units 2 2 The resultant vector along the y – axis: 30 20 Ry A B sin 450 C sin 450 10 17.07 units 2 2 (ii) Magnitude of the resultant vector: R R2x Ry2 35.362 17.072 39.26 units Direction of the resultant vector: 17.07 tan 0.48 25.770 35.36 1.2.5 Multiplication of Vectors From what we have so far said, you would have seen that only scalar quantities of a similar kind could be added to one another. For instance, it would be meaningless to add together quantities such as mass and temperature - in fact it would be ridiculous to do so. Similarly only vector quantities of a like kind can be added together. It would be meaningless to add together acceleration and electric field strength. However, scalars and vectors of different kinds can be multiplied by one another, and this leads to quantities with new physical qualities or dimensions. Scalars can be multiplied by the ordinary rules of algebra. But this cannot be done with vectors, since they have direction as well as magnitude. New rules are necessary for vectors. Vector multiplication can have many forms, but we will define three: (a) multiplication of a vector by a scalar; (b) multiplication of two vectors resulting in a scalar; and (c) multiplication of two vectors resulting in another vector. When we multiply a vector a by a scalar m (i.e. ma ), we obtain a new vector whose magnitude is m times the magnitude of a. If m is positive, the new vector has the same direction as of a ; if m is negative, it has the opposite direction. Suppose, if we wish to divide a vector by a scalar, we simply multiply the vector by the reciprocal of the scalar. However, when we multiply two vectors, the resultant can have two forms; one is resulting as the scalar (or called dot) product and the other as a vector (or called cross) product. 9 Copyright © 2017, The Open University of Sri Lanka (OUSL) PHU3300 – Unit 1 Session 1: VECTORS AND SCALARS, MOTION IN ONE, TWO AND THREE DIMENSIONS Scalar product (dot product) Consider two vectors a and b with their tails at the same point, and making an angle between them as shown in the Figure 1.11. Figure 1.11: Two vectors a and b with their tails at the same point, and is the angle between them. The scalar product of two vectors a and b (written as a b ), is defined as: a b ab cos (1.1) Here, a and b are the magnitudes of the vectors a and b respectively, and is the angle between the two vectors. Now, remember that a and b are scalars, and cos is a pure number. Thus, the scalar product of two vectors is a scalar. Figure 1.12: Schematically showing one vector and the component of the other vector in the direction of the first. If we consider the Figure 1.12, you will see that the scalar product of two vectors can also be defined as the product of one vector and the component of the other vector in the direction of the first. Here are some examples of the scalar product of two vectors in physics: (a) If the point of application of a force F is displaced through a displacement x , the mechanical work done is F x Fx cos (see Figure 1.13). Figure 1.13: A force F is displaced through a displacement x. (b) The increase in the gravitational potential energy of a mass m, displaced upwards through a distance h is; mg h mghcos1800 mgh The negative sign arises from the fact that external work is being done on the body. Here g is the acceleration due to gravity. Please note that in later studies the significance of the scalar product of two vectors will become clearer. 10 Copyright © 2017, The Open University of Sri Lanka (OUSL) PHU3300 – Unit 1 Session 1: VECTORS AND SCALARS, MOTION IN ONE, TWO AND THREE DIMENSIONS Vector product (cross product) The vector product of two vectors a and b is written as a b. This is another vector c , where c a b. The magnitude of c is given by: c absin (1.2) where is the angle between a and b. The vector c is perpendicular to the plane of a and b. To sense the direction of vector c we can use the right-hand corkscrew. To sense the direction, consider Figure 1.14; Figure 1.14: Vector product a b showing the sense of resultant vector c. A right-handed corkscrew, whose axis is perpendicular to a and b , is rotated so as to turn it through the angle from a to b. Then, the sense of the vector product a b is given by the direction in which the corkscrew advances. The order of the vectors in a vector product is important. From the above corkscrew rule it is evident that b a a b. In fact a b b a Some examples of vector products, which we will discuss later, are torque, angular momentum and the force on a moving electric charge in a magnetic field. 1.2.6 Unit Vectors The component of a vector in any given direction may be represented by the product of a unit vector in that direction, and the component of the given vector. Consider the vector a. If u is a unit vector in the direction of a , the vector itself can be represented by: a u (Figure 1.15a). In a rectangular co-ordinate system (Figure 1.15b), the unit vectors along the x, y and z axes are usually denoted by iˆ , ˆj and kˆ respectively. 11 Copyright © 2017, The Open University of Sri Lanka (OUSL) PHU3300 – Unit 1 Session 1: VECTORS AND SCALARS, MOTION IN ONE, TWO AND THREE DIMENSIONS Figure 1.15: (a) Unit vector u in the direction of vector a (b) rectangular co-ordinate system showing the unit vectors iˆ , ˆj and kˆ along the x, y and z axes respectively. In such a rectangular system co-ordinates (or referred to as Cartesian co-ordinates system), which is in two dimensional, a vector a may be written as: a axiˆ ay ˆj (1.3) Where ax and ay are the scalar components of a along the x and y axes; iˆ and ˆj are the unit vectors in these directions. In a three-dimensional system of rectangular co-ordinates a may be represented by: a axiˆ ay ˆj az kˆ (1.4) Where az is the scalar components of a along the z axis; and kˆ is the unit vector in this direction. Activity 1.4 In a Cartesian (rectangular) co-ordinates system having the axes of x, y and z, you are given a vector a in the + x direction, a vector b in the +y direction, and a scalar quantity d. (a) What is the direction of a b ? (b) What is the direction of b a ? b (c) What is the direction of ? d (d) What is the direction of a b ? 12 Copyright © 2017, The Open University of Sri Lanka (OUSL) PHU3300 – Unit 1 Session 1: VECTORS AND SCALARS, MOTION IN ONE, TWO AND THREE DIMENSIONS Solutions (a) Using the corkscrew rule, where a is rotated to b , the corkscrew advances in the direction +z. Hence, the direction of a b is +z. (b) Using the same rule, but now rotating b to a , the direction of b a is –z. b (c) The direction of is the same as that of b , since multiplication of division d by scalar does not alter the direction or sense of a vector. Thus, the direction b of is +y. d (d) The angle between a and b is 900 , hence a b ab cos 900 0. Activity 1.5 Show that iˆ iˆ ˆj ˆj kˆ kˆ" 1 and iˆ ˆj' ˆj kˆ kˆ iˆ 0 Solution iˆ iˆ iˆ iˆ cos 00 i2 1 , Similarly ˆj ˆj j 2 1 and kˆ kˆ k 2 1 iˆ ˆj iˆ ˆj cos 900 0 , Similarly ˆj kˆ 0 and kˆ iˆ 0. Activity 1.6 Show that iˆ iˆ ĵ ĵ ' k̂ k̂ 0 and iˆ ˆj kˆ; k̂ íˆ ĵ; ˆj kˆ iˆ Solution Refer Figure 1.15(b); The magnitude of iˆ iˆ' iˆ iˆ sin 00 0 , similarly ĵ ĵ k̂ k̂ 0. Using the corkscrew rule (or right hand rule) the direction of iˆ ˆj is +z, its magnitude is iˆ ˆj 'sin 900 1. Since iˆ and ˆj are unit vectors, thus iˆ ˆj is a unit vector in the +z direction, i.e. iˆ ˆj kˆ. Similarly kˆ iˆ ˆj and ˆj kˆ iˆ. 13 Copyright © 2017, The Open University of Sri Lanka (OUSL) PHU3300 – Unit 1 Session 1: VECTORS AND SCALARS, MOTION IN ONE, TWO AND THREE DIMENSIONS Activity 1.7 If vectors a and b are express in terms of their components as a axiˆ ay ˆj az kˆ and b bxiˆ by ˆj bz kˆ , where ax , ay , az and bx ,by ,bz are the components of a and b respectively. Show that (i) a b axbx ayby azbz (ii) a b a y bz az by iˆ az bx ax bz ĵ ax by a y bx k̂ Solutions (i) a b ax iˆ a y ĵ az k̂ bx iˆ by ĵ bz k̂ a b ax bx iˆ iˆ a x by iˆ ĵ ax bz iˆ k̂ a y bx ĵ iˆ a y by ĵ ĵ a y bz ĵ kˆ a z bx kˆ iˆ + azby kˆ ˆj azbz kˆ kˆ From Activity 1.5, we know iˆ iˆ ˆj ˆj kˆ kˆ 1 and iˆ ˆj ˆj kˆ kˆ iˆ 0 Hence, we have a b axbx ayby azbz Thus the scalar product of two vectors is the sum of the products of their respective components. (ii) a b ax iˆ a y ĵ az k̂ bxiˆ by ĵ bz k̂ a b a x bx iˆ iˆ a x by iˆ ĵ ax bz iˆ k̂ a y bx ĵ iˆ a y by ĵ ĵ a y bz ĵ kˆ az bx k̂ iˆ + azby kˆ ˆj azbz kˆ kˆ Hence, by using the result of Activity 1.6, we have a b a y bz az by iˆ az bx ax bz ĵ ax by a y bx k̂ Note: The vector product can also be expressed in determinant form as iˆ ˆj kˆ a b ax ay az bx by bz 14 Copyright © 2017, The Open University of Sri Lanka (OUSL) PHU3300 – Unit 1 Session 1: VECTORS AND SCALARS, MOTION IN ONE, TWO AND THREE DIMENSIONS Activity 1.8 For the vectors a 2iˆ 2 ˆj kˆ and b 2iˆ 2 ˆj 3kˆ , determine (i) a b (ii) ab (iii) Angle between vector a and b Solutions (i) a b 2 2 22 13 4 4 3 3 iˆ ˆ j kˆ (ii) a b 2 2 1 2 3 1 2iˆ 1 2 2 3 ˆj 2 2 2 2kˆ 2 2 3 Hence a b 6 2iˆ 2 6 ˆj 4 4kˆ 8iˆ 4 ˆj 8kˆ a b (iii) a b a b cos cos , we have a b 3 and a b a 22 22 12 9 3 , b 22 2 32 17 2 3 Hence, cos 0.243 75.960 3 17 1.3 MOTION IN ONE DIMENSION: REVISION 1.3.1 Speed and Velocity If we say that a motorcar is travelling at 40 km/h due north, we are specifying both a magnitude (40km/h) and a direction (due north). We are here referring to the velocity of the motorcar. On the other hand, if we say that the motorcar is travelling at 40 km/h, we are specifying only a magnitude. This is the speed of the car. You will note that there is a difference between speed and velocity. Speed is a scalar quantity, whereas velocity is a vector, subject to the normal rules governing vectors, which we have already studied. 1.3.2 Definitions of Velocity and Acceleration If a particle make a displacement s in a given direction in a time t , its average s velocity can be defined as. However, the instantaneous velocity of a particle at t a given point has to be defined in the language of the calculus. If s is the 15 Copyright © 2017, The Open University of Sri Lanka (OUSL) PHU3300 – Unit 1 Session 1: VECTORS AND SCALARS, MOTION IN ONE, TWO AND THREE DIMENSIONS displacement of the particle in the neighbourhood of the point considered in a small interval of time t , its instantaneous velocity is defined as s v lim t 0 t You must be familiar with this notation. It indicates the ratio of displacement to time as the latter tends to infinitesimal values t 0 as we say. Suppose a particle moves in a given direction with a velocity v1 at a time t1 and suppose at a time t2 its velocity increases to v2. Then the average acceleration of the particle is v2 v1 (1.5) t2 t1 However, if we consider a very small increment of velocity v , occurring the small time interval t , the instantaneous acceleration of the particle is defined as a v a lim (1.6) t 0 t Graphical representations can be made of the variation of displacement with time or the variation of velocity with time. Revise your knowledge of these graphs, how they are drawn and what information can be derived from them. Velocity and acceleration can be either constant or variable. The displacement - time and velocity - time curves will change accordingly. If a particle moves with a constant acceleration a , and if its initial velocity is u , its velocity after a time interval t is v , and its displacement during this interval is s , the quantities s , u , v , t and a could be combined in three different ways, namely: v u at (1.7) 1 s ut at 2 (1.8) 2 v 2 u 2 2as (1.9) These are the equations of motion. To solve any given problem in motion you could select the appropriate equation depending on the data at your disposal. You should know the proofs of these equations. Revise carefully what you have already learned about relative velocity. Although the topics mentioned are not dealt with in any detail in this course, they are of fundamental importance. Therefore your tutorial assignments may include questions and problems from these areas. So, do not fail to refresh your mind about them. 1.4 MOTION IN TWO AND THREE DIMENSIONS In previous section (section 1.3) we were able to revise the motion of a particle along a straight line which you would have studied extensively in your GCE (A/L) course in physics. Generally in one dimensional analysis, we consider the motion along the x axis and the motion is completely known if its position is known as a function of time. Let us now extend this idea to describe the motion of a particle in 16 Copyright © 2017, The Open University of Sri Lanka (OUSL) PHU3300 – Unit 1 Session 1: VECTORS AND SCALARS, MOTION IN ONE, TWO AND THREE DIMENSIONS space. To describe the motion of a particle in space, we must first be able to describe the particle’s position. Figure 1.16: (a) The position vector r from the origin O to point P has components x, y, and z (b) particle’s position vectors r1 at position P1 and r2 at position P2. Consider a particle that is at a point P at a certain instant. The position vector of the particle at this instant is a vector that goes from the origin O of the coordinate system to the point P as shown in Figure 1.16(a). Here, x, y, and z are the components of the position vector r of the point P in a Cartesian co-ordinates x, y, and z. If we use the unit vectors to express the position vector r , we have r xiˆ' yˆj zkˆ (1.10) Suppose during a time interval t the particle moves from a position P1 , where its position vector is r1 at time t1 to another position P2 , where its position vector is r2 at time t2 as shown in Figure 1.16(b). Thus, change in position which is referred to as displacement during this time interval t is given as r r2 r1 x2 x1 iˆ' y2 y1 ˆj z2 z1 kˆ xiˆ' yˆj zkˆ (1.11) Where x, y and z are the displacements along direction of x, y and z axes. Then the average velocity vav of the particle during the time interval t t2 t1 is r2 r1 r (1.12) vav t2 t1 t r The instantaneous velocity v is defined as the limit of the average velocity as t the time interval approaches zero, and it equals the instantaneous rate of change of position with time. As such the instantaneous velocity v can be written as r dr v lim (1.13) t 0 t dt The magnitude of the vector v at any instant is the speed of the particle at that instant. Also the direction of the instantaneous velocity v at any instant is the same 17 Copyright © 2017, The Open University of Sri Lanka (OUSL) PHU3300 – Unit 1 Session 1: VECTORS AND SCALARS, MOTION IN ONE, TWO AND THREE DIMENSIONS as the direction in which the particle is moving at that instant. Further, we could realize that at every point along the path, the instantaneous velocity vector is tangent to the path at that point as shown in the Figure 1.17(a). Figure 1.17: (a) Instantaneous velocities v1 and v2 at the points P1 and P2 respectively (b) The two velocity components vx , vy for a motion in the xy-plane. Suppose if vx , vy and vz are the components of the instantaneous velocity v along the direction of x, y and z axes respectively, using the equations (1.11) and (1.12) we could write dx dy dz v , v , vz (1.14) x y dt dt dt Now you can easily identify that the components of the instantaneous velocity v , vx , vy and vz , are the time derivatives of the coordinates x, y, and z respectively. Hence, we can write ˆ ˆ dx ˆ dy ˆj dz kˆ v v iˆ v j v k i x y z (1.15) dt dt dt Using the Pythagorean relation we can obtain the magnitude of the instantaneous velocity v as v v vx2 vy2 vz2 (1.16) Now consider the two dimensional situation where a particle is moving only in the xy plane as shown in Figure 1.17(b). In this situation, the components in the direction of z-axis is zero (i.e. z 0 and vz 0 ). Hence, the magnitude and direction of the instantaneous velocity v is vy v v vx2 vy2 tan (1.17) v x Here, the direction of the instantaneous velocity v is express by the angle . As shown in Figure 1.17(a), let the instantaneous velocities at P1 at time t1 be v1 and P2 at time t2 be v2. If the particle move from the position P1 to the position P2 along the path during the time interval t , the average acceleration aav is define as v v v (1.18) a 2 1 av t t t 2 1 18 Copyright © 2017, The Open University of Sri Lanka (OUSL) PHU3300 – Unit 1 Session 1: VECTORS AND SCALARS, MOTION IN ONE, TWO AND THREE DIMENSIONS The instantaneous acceleration a is defined as the limit of the average acceleration v as the time interval approaches zero, and it equals the instantaneous rate of t change of velocity with time. As such the instantaneous acceleration a can be written as v dv a lim (1.19) t 0 t dt We have already seen that the direction of the instantaneous velocity v is tangent to the path of the particle. However, in the case of instantaneous acceleration a , it does not have to be tangent to the path. Suppose if the path is curved, it points toward the concave side of the path, this means toward the inside of any turn that the particle is making. We shall discuss in detail when we study the circular motion in session 4. You should note that the acceleration is tangent to the path only if the particle moves in a straight line. Suppose if ax , ay and az are the components of the instantaneous acceleration a , then each component of the instantaneous acceleration a is the derivative of the corresponding component of velocity. Thus dvx dvy dvz a , a , a , (1.20) x y z dt dt dt Therefore, in terms of unit vectors we can write the instantaneous acceleration a as dv dvy a a iˆ a ˆj a kˆ x iˆ ˆj dvz kˆ (1.21) x y z dt dt dt Since each component of velocity is the derivative of the corresponding coordinate, the instantaneous acceleration a also be written as d 2 x ˆ d 2 y ˆ d 2z ˆ (1.22) j a 2 i dt2 k dt dt2 Activity 1.9 Consider a particle moving in a horizontal xy plane. The particle starts from the origin of the Cartesian co-ordinate system at t 0 with an initial velocity v 20iˆ'15 ˆj m s1. Suppose, the particle experiences a constant acceleration in the x direction, given by ax 5 m s2 , determine (i) the velocity vector at time t. (ii) the magnitude and the direction of the particle at t 5 s. 19 Copyright © 2017, The Open University of Sri Lanka (OUSL) PHU3300 – Unit 1 Session 1: VECTORS AND SCALARS, MOTION IN ONE, TWO AND THREE DIMENSIONS Solutions (i) Since the acceleration is constant we can use the equation of motion. By using the equation (1.7) we have v u at 20iˆ 15 ˆj 5iˆt 20 5t iˆ 15 j (ii) At t 5 s the velocity is v 20 5 5iˆ 15 j 45iˆ 15 j m s1 The magnitude of the velocity; v v 452 15 2 47.43 m s1 The direction of the velocity of15 the particle; tan1 18.43 0 45 Activity 1.10 Consider a particle moving in a horizontal xy plane. The particle starts from the origin of the Cartesian co-ordinate system and has x- and y-coordinates that varies with time t given by the equations x 2 m 3 m s2 t 2 y 1 m s1 t 2 m s t3 3 (i) Determine the particle’s coordinates and distance from the origin at t 2 s. (ii) Determine the particle’s displacement and average velocity vectors for the interval t 0 to t 2 s. (iii) Find a general expression for the particle’s instantaneous velocity vector v (iv) Determine the instantaneous velocity v in component form and in terms of magnitude and direction at t 2 s. (v) Find the instantaneous acceleration at t 2 s. Solution 1.10 (i) At t 2 s , the particle’s coordinates; x 2 3 22 10 m , y 1 2+2 23 18 m r 10iˆ 18 ˆj The distance from the origin; r r 10 182 20.59 m 2 (ii) At t 0 ; r0 2i , at t 2 s ; r2 10iˆ 18 ˆj The displacement; r r2 r0 12iˆ 18 j , The average velocity; v r 12iˆ 18 ˆj 6iˆ 9 ˆj av t 2 (iii) rˆ 2 3t2 iˆ t 2t3 ˆj v dr 6tiˆ 1 6t 2 ˆj dt Particle’s instantaneous velocity vector v 6tiˆ 1 6t2 ˆj 20 Copyright © 2017, The Open University of Sri Lanka (OUSL) PHU3300 – Unit 1 Session 1: VECTORS AND SCALARS, MOTION IN ONE, TWO AND THREE DIMENSIONS (iv)At t 2 s ; vx 12 m s1, v y 25 m s1 , Magnitude of the velocity; v v 12 252 27.73 m s1 2 1 25 The direction; tan 64.36 0 12 dv (v) At t 2 s ; instantaneous acceleration; a 6iˆ 12 2 ˆj 6iˆ 24 ˆj dt t 2 Summary The first part of this session deals in greater detail with the two main types of physical quantities - namely, vectors and scalars. The student should be now able to easily identify any physical quantity as belonging to one or the other of these two categories. Where scalar quantities can be completely described by a magnitude - i.e. a number and a unit, vectors must be expressed in terms of both magnitude and direction. Scalar quantities can be added, subtracted, multiplied and divided by the usual rules of algebra. However, the compounding, resolution and multiplication of vectors requires different rules and techniques. After studying this session, the student should be familiar with these techniques. Vector notation has been indicated. In the multiplication of vectors the physical significance of scalar (dot) product, and vector (cross) product has been explained and illustrated by a few physical examples. The student will grow more familiar with these concepts by their application in subsequent sessions. The device of utilizing unit vectors iˆ , ˆj and kˆ , along the x, y and z axes of a rectangular (Cartesian) coordinate system has been described. Their use and the relations between them have also been discussed. Some of the self-assessment questions that follow will give further information in this area. Using the unit vectors iˆ , ˆj and kˆ , the position vector r of a point P in space has been identified and realize that its components are the coordinates x, y, and z. The average velocity vector vav and the average acceleration vector aav of a moving particle during the time interval t has been defined and comprehended that the instantaneous velocity vector v is the time derivative of the position vector r and its components are the time derivatives of x, y, and z. Similarly, we found that the instantaneous acceleration vector a is the time derivative of v and its components are the time derivatives of vx , vy and vz. The use of vector techniques – vector analysis - is a powerful tool in the solution of many problems in physics. It also helps to simplify and, shorten the analysis of many physical problems. However, vector analysis itself is outside the immediate scope of this course. The student will encounter this technique in some of the courses to follow. 21 Copyright © 2017, The Open University of Sri Lanka (OUSL) PHU3300 – Unit 1 Session 1: VECTORS AND SCALARS, MOTION IN ONE, TWO AND THREE DIMENSIONS Learning Outcome Now you should be able to; (a) Explain the concept of vectors and scalars. (b) Learn what vectors are and how they can be used to model real-world situations. (c) Realize the concept of unit vectors and its usage. (d) Perform various operations with vectors. (e) Recognize the use of vector techniques (vector analysis) as a powerful tool in the solution of many problems in physics. (f) Solve simple problems related to Motion in One, Two and Three dimensions. Self-Assessment Questions 1) Vector A has magnitude 6 and points in the negative y direction and vector B has magnitude 5 and points at an angle of 450 above the x-axis as shown in the Figure 1.18. Figure 1.18: Illustration for SAQ 1 (i) Determine the x and y components of vector A and B vector. (ii) If C A B , calculate the magnitude and direction of the vector C. 2) For the vectors; a 2iˆ ˆj kˆ, b 3iˆ 2 ˆj kˆ, c 3iˆ ˆj kˆ. Calculate (i) a b c (ii) a b c (iii) a b c (iv) a b c 3) Show that the area of the triangle contained between the vectors a and b 1 is a b. 2 4) Show that a b c is equal in magnitude to the volume of the parallelepiped formed on the three vectors a , b and c. 22 Copyright © 2017, The Open University of Sri Lanka (OUSL) PHU3300 – Unit 1 Session 1: VECTORS AND SCALARS, MOTION IN ONE, TWO AND THREE DIMENSIONS 5) A toy rocket is launched with an initial velocity of 10 m/s in the horizontal direction from the roof of a 40 m tall building. The rocket’s engine produces a horizontal acceleration of 3t in the same direction as the initial velocity (where t is the time), but in the vertical direction the acceleration due to gravity is g, downward. If air resistance is neglected, determine (i) the time taken for the rocket to reach the ground. (ii) the horizontal distance the rocket travel before it reaches the ground. NOTE: Although the solutions are given below, it is in your own interest to attempt the self-assessment questions on your own, as an indication of how far you have understood the material of the session. Solutions 1) (i) Ax 0, Ay 6 ; B 5cos 450 3.54, B 5sin 450 3.54 x y (ii) Cx 0 3.54 3.54, Cy 6 3.54 2.46 The magnitude; C C2x C2y 3.542 2.462 4.31 2.46 The direction; tan1 34.8 0 3.54 2) (i) a b c 2iˆ ˆj kˆ 3iˆ 2 ˆj kˆ 3iˆ ˆj kˆ 2iˆ 3kˆ iˆ ˆ j kˆ (ii) b c 3 2 1 iˆ2 1 ˆj 3 3 kˆ3 23 3iˆ 9kˆ 3 1 1 a b c 2iˆ ˆj kˆ 3iˆ 9kˆ 6 1 5 (iii) b c 3iˆ 2 ˆj kˆ 3iˆ ˆj kˆ 9 2 1 8 a b c 2iˆ ˆj kˆ 8 16iˆ 8 ˆj 8kˆ iˆ ˆ j kˆ (iv) a b c 2 1 1 iˆ9 0 ˆj 3 18 kˆ0 3 9iˆ 21 ˆj 3kˆ 3 0 9 3) From elementary geometry, the area of a triangle = ½ (base x the perpendicular drawn from the vertex to the base of the triangle). Consider the Figure 1.19. The vertical lines refer to the absolute or numerical values of the cross product a b. Figure 1.19: Illustration for SAQ 3 If the angle between the vector a and b is , the area of the triangle will 1 1 thus be ba sin 1 ab sin , which is the numerical value of a b. 2 2 2 23 Copyright © 2017, The Open University of Sri Lanka (OUSL) PHU3300 – Unit 1 Session 1: VECTORS AND SCALARS, MOTION IN ONE, TWO AND THREE DIMENSIONS 4) The base of the parallelepiped (Figure 1.20) formed by the vectors b and c 1 can be divided into two equal triangles, each of area of b c. 2 Thus, the area of the base is b c. The corresponding vector is b c perpendicular to the plane of b and c and directed upwards. From elementary geometry, we have; the volume of a parallelepiped = area of base perpendicular height Figure 1.20: Illustration for SAQ 4 The perpendicular height of the component of a multiplied by the area of the base. The area of the base in vector form is b c , and represented by a vector perpendicular to the base. Thus, the volume of the parallelepiped is b c a cos , where is the angle between b c and the vector a. In vector notation this can be written as a b c . 1 5) (i) Applying the equation of motion s ut at 2 direction 2 40 0 10t2 t 2 s (ii) In horizontal direction, the acceleration is not constant. Acceleration is a function of time. Therefore, dv v t 3 3t v t 2 10 dt dv 3tdt 10 0 2 dx 3 x 2 3 t3 2 i.e., t 2 10 dx t 2 10 dt x 10t dt 2 0 0 2 2 0 8 x 2 20 m 24 m Reference Physics for Scientists and Engineers with Modern Physics 7th/8th/9th Edition By Serway and Jewett University Physics 11th /12th /13th Edition by Young and Freedman Fundamentals of Physics by Halliday, Resnick and Walker 8th /9th/10th Edition 24 Copyright © 2017, The Open University of Sri Lanka (OUSL)
