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Vectors Magnitude + Direction I. Scalars and Vectors A. Scalars – Number and unit. No direction. Ex: Mass, time, temperature B. Vectors – Quantities that have a direction associated with them. Ex. Velocity, force, momentum 1. Each separate vector is called a COMPONENT. 2. The single vector that produces the same result as the combined components is called the RESULTANT. = V1 V2 V1 + V2 COMPONENTS RESULTANT Each vector has two ends TAIL HEAD; TIP Vectors can be used to represent the increasing speed of a ball in free fall. The distance the ball falls each second increases because the ball is accelerating. This is shown by drawing a longer vector arrow for each time interval. constant velocity (v) – motion in a straight line at a constant speed; vf = vo. Vectors can also be used to represent a ball rolling horizontally on a table at a constant velocity. . Each vector arrow is drawn the same length to represent the constant velocity. The velocity would remain constant but friction makes it slow down and eventually stop. II. Vector Addition and Subtraction A. Addition of COLINEAR Vectors – when vectors point along the same direction. Ex: If a person walks 10.0 meters, east, stops, and continues 5.0 meters, east. What is the resultant? Tail to Tip Method of Drawing Vectors: 10 m, E 5 m, E -Tail of the second arrow Vector A Vector B located At the head of the first arrow, the two lengths = add to give the length of the 10 m, E + 5 m E = 15 m, E total displacement. - Can only be done when the Vector R vectors point along the same direction. A+B=R Ex: If a person walks 10.0 meters, east, stops, and continues 5.0 meters, west. What is the resultant? Colinear Vector Addition in Opposite Directions: -Draw both vector components in 10 m, E 5 m, W their respective direction. Vector A Vector B - Find the difference between the two vectors and select the direction = of the vector with the greater 10 m, E - 5 m E = 5 m, E magnitude. Vector R A+B=R Ex: What is the velocity of a swimmer who is swimming 10 m/s upstream if the current downstream is 5 m/s? 5 m/s; 10 m/s up – 5 m/s; down 10 m/s; downstream = 5 m/s; up upstream = B. Addition of Vectors that are Neither Colinear Nor Perpendicular Graphical Technique – A diagram is constructed in which the arrow are drawn tail to tip. The lengths of the vector arrows are drawn to scale, and the angles are drawn accurately (with a protractor). Then, the length of the arrow representing the resultant vector is measured with a ruler. This length is converted into the magnitude of the resultant vector by using the scale factor with which the drawing was constructed. Ex: A car moves with a displacement A of 275 m, due west and then with a R displacement B of 125 m in a direction B 55.0° north of west. 55° A The two displacement vectors A and Vector A = 275 m; W B are neither colinear nor perpendicular but add to give the resultant vector R. Vector B = 125 m; @ 55° N of W Ex: I ran 10.0 km due east but was pushed 5.0 km, north by a crosswind. How far did I actually run? What was my angle from the start line? R 5.0 km; north Θ 10.0 km; east - The resultant, R can be found by using a ruler to measure the magnitude, and a protractor to measure, the angle Θ. C. Parallelogram Method to Vector Addition Two vectors are drawn to scale and joined at their tails. Parallel lines are drawn to complete the parallelogram. The resultant is drawn from where the tails touch to the opposite corner. Ex: A boat travels 2 km north, then 4 km north of east at an angle of 30°. Determine the magnitude and direction of the resultant. Two ways to solve for R - Use a ruler and a protractor R -Use vector resolution 2 km; n 4 km, e (math: Pythagorean’s Theorem; 30° Trig functions) D. Vector Component Method (Vector Resolution) 1. Resolve each vector into x and y components. 2. Assign positive or negative values to each component based on which quadrant it falls in. 3. Reduce the problem to the sum of the two vectors. Σ x = v1x + v2x + any others Σ y = v1y + v2y + any others 4. Determine the magnitude and direction of the resultant. Since Σ x and Σ y are at right angles, the Pythagorean Theorem can be used to determine the magnitude of the resultant. The definition of the tangent of the angle can be used to determine the direction of the resultant. tan Θ = vy Θ = tan-1 vx (v) v y x 4 km, e 4 km, e 28° 28° R R 2 km; n 2 km; n Vector 1 4 km, e Vector 2 (2 km; n) v2y (4 km; e @ 28°) 28° V1x = 0 km v2x V2x = 4 km V1y = 2 km V2y = sin θ = o/h V2x = cos θ = a/h V2y = 2 km V2y = o = sin θ (h) V2x = a = cos θ (h) V2y = sin (28°) ( 4 km) V2x = cos (28°) ( 4 km) V2y = 2 km V2x = 4 km v2x Σx v2y Σy R R v1y Σ x = v1x + v2x = 0 km + 4 km = 4 km Σ y = v1y + v2y = 2 km + 2 km = 4 km Σ x = 4 km Θ = tan-1 vy ( ) vx Σ y = 4 km R θ Θ = tan-1 Σ x ( Σy ) Θ = tan-1 4 km ( 4 km ) Θ = 45º; East of North sin Θ = o sin Θ = Σ x R h R= Σx R= 4 (sin Θ) (sin 37) R = 7 km Rule for Adding Vectors When adding vectors, it doesn’t matter which order they are added in. The resultant will always be the same. Ex: A man walks 100 ft at 30º north of east; 200 ft at 40º south of east; and 400 ft at 90º. ΣRx = V1x + V2x + V3x ΣRx = V1cosΘ + V2cosΘ + V3cosΘ R ΣRx = 100(cos30) + 200(cos40) + 0 V3 ΣRx = 240 ft V1 40º 30º V2 90º ΣRy = V1y + (-V2y) + V3y ΣRy = V1sinΘ - V2sinΘ + V3sinΘ ΣRy = 100(sin30) - 200(sin40) + 400 ΣRy = 321 ft Θ = tan-1 Ry ( ) Rx R Θ = tan-1 321 ft Ry = 321 ft ( 240 ft ) Θ Θ = 53º; North of East Rx = 240 ft sin Θ = Ry sin Θ = o R h R= 321 (sin 53) R = 402 ft III. Relative Motion/Velocity The concept of a frame of reference is important when solving for these problems. It is important to always keep in mind the relationship between each variable involved. A. Each velocity vector is labeled by two subscripts: First subscript refers to the object Second subscript refers to the reference frame in which it has a velocity Ex: Boat goes across a river VBW = velocity of the boat with respect to the water: how fast the boat travels in the water across the river. vws vBS VWS = velocity of the water with respect to the shore: vBW how fast the water in the river travels when compared to the shore. VBS = velocity of the boat with respect to the shore: how fast the boat (that’s in the water) is traveling Current when compared to the shore). vBS = vBW + vSW vBW = -vWB Ex: A passenger is walking towards the front of a moving train. The people sitting on the train see the passenger walking with a velocity of +2.0 m/s, where the plus sign denotes a direction to the right. Suppose the train is moving with a velocity of +9.0 m/s relative to an observer standing on the ground. What would the ground-based observer see the passenger moving with? vPT = velocity of the PASSENGER relative to the TRAIN = + 2.0 m/s vTG = velocity of the TRAIN relative to the GROUND = + 9.0 m/s vPG = vPT + vTG vPG = velocity of the PASSENGER relative to the GROUND = + 11 m/s = vPG vPT vTG Ex: The engine of a boat drives it across a river that is 1800 m wide. The velocity VBW of the boat relative to the water is 4.0 m/s, directed perpendicular to the current, as shown below. The velocity VWS of the water relative to the shore is 2.0 m/s. a.) What is the velocity VBS of the boat relative to the shore? b.) How long does it take for the boat to cross? a.) The velocity VBS of the boat relative to the shore is the vector sum of the velocity VBW of the boat relative to the water and the velocity VWS of the water relative to the shore: VBS = VBW + VWS. VBS = √(VBW 2 + VWS2) Θ = tan-1 VBW ( VWS ) VBS = √( (4.0 m/s)2 + (2.0 m/s)2 ) Θ = tan-1 4.0 = 4.5 m/s (2.0 ) = 63º b.) The component of VBS that is parallel to the width of the river determines how fast the boat is moving across the river; this parallel component is VBS (sin Θ) = VBW = 4.0 m/s. The time for the boat to cross the river is equal to the width of the river divided by the magnitude of this velocity component. t=x v t= width = 1800 m = 450 s VBS (sin Θ) 4.0 m/s IV. Projectile Motion The motion of an object that is projected into the air at an angle. Solve these problems in the same way as you would velocity problems. Use the kinematic formulas for constant acceleration to get formulas for projectile motion. Horizontal Motion Vertical Motion vfx = vix + axΔt vfy = viy + ayΔt Δx = xi + vixt + ½ axΔt2 Δy = yi + viyt + ½ ayΔt2 vfx2 = vix2 + 2axΔx vfy2 = viy2 + 2ayΔy Idea behind trajectory/projectile motion: Observe the motion of the ball as it falls focusing on the path followed (trajectory). In projectile motion, we are combining the motion of the ball in free fall with the motion of the ball rolling on the table at a constant velocity. This is seen when rolling the ball off of the table. The ball rolling on the table would continue forever in a straight line if gravity is ignored. The ball in free fall would continue to increase its speed if air resistance is ignored. Since the ball is moving at a constant velocity and in free fall at the same time, the horizontal and vertical vectors are added together during equal time intervals. This is done for each time interval until the ball hits the ground. The path the ball follows can be seen by connecting the intermittent resultant vectors. The two-dimensional motion of the spacecraft can be viewed as the combination of the separate x and y motions. Kinematic Variables y, x, ax, ay, vy, viy, vx, vix Ex: In the x direction, a spacecraft has an initial velocity component of vix = +22 m/s and an acceleration component ax = +24 m/s2. In the y direction, the analogous quantities are viy = +14 m/s and ay = +12 m/s2. The directions to the right and upward have been chosen as the positive directions. After a time of 7.0 s, find the: a.) x and vx b.) y and vy c.) final velocity (magnitude and direction of the aircraft) Treat the motion in the x and y direction separately a.) data for motion in the x direction ax = +24 m/s2 x = vixt + ½ axt2 vix = +22 m/s x = (22 m/s)(7.0 s) + ½ (24 m/s2)(7.0 s) 2 t = 7.0 s x=? x = +740 m vx = ? vx = vix + axt vx = (22 m/s) + (24 m/s2)(7.0s) vx = +190 m/s b.) data for motion in the y direction ay = +12 m/s2 y = viyt + ½ ayt2 viy = +14 m/s y = (14 m/s)(7.0 s) + ½ (12 m/s2)(7.0 s) 2 t = 7.0 s y=? y = +390 m vy = ? vy = viy + ayt vy = (14 m/s) + (12 m/s2)(7.0s) vy = +98 m/s c.) final velocity vfy = 98 m/s Vf = √(Vfx2 + Vfy2) Vf = √( (190 m/s)2 + (98 m/s)2 ) vf Vf = 210 m/s θ vfx = 190 m/s Θ = tan-1 vy ( )vx Θ = tan 98 -1 (190) = 27° above the positive x axis Ex: The figure shows an airplane moving horizontally with a constant velocity of +115 m/s at an altitude of 1050 m. The directions to the right and upward have been chosen as the positive directions. The plane releases a “care package” that falls to the ground along a curved trajectory. Ignoring air resistance, determine the time required for the package to hit the ground. The time required for the package to hit the ground is the time it takes for it to move 1050 m. It moves both downward and to the right. These two parts of motion occur independently. The package is not moving in the y direction at the instant of release, so viy = 0 m/s. When the package hits the ground, the y component of its displacement is y = -1050 m as the drawing shows. The acceleration is due to gravity, so g = - 9.80 m/s2. (Note, if you make y = +, then make g = +. I left it as a negative number to demonstrate that its just a frame of reference) What we know: y = viyt + ½ ayt2 y = -1050 m ay = -9.80 m/s2 viy = 0 m/s t = √(2y / ay) t=? t = √( 2(-1050 m) / -9.80 m/s2) ) t = 14.6 s Note: the vertical speed of the package picks up over time, but the horizontal component stays the same, v0x = vx = 115 cm. This is because the package is traveling at the same speed as the plane in the x direction. Therefore, the package does not accelerate in the x direction. Two packages are dropped from the plane. Package A drops vertically and package B drops at an angle. They both hit the ground at the same time because they have the same y components of their velocities. However, package B will hit the ground with a greater speed due to the magnitude and direction of its x component. Ex: Find the speed of the package and the direction of the velocity vector just before the package in the previous problem hits the ground. Image upon impact vfx = 115 m/s vfx = 115 m/s vf ground vf ground vfy = ? vfy = -143 m/s What we know to solve for vy : So, vf = √( (115 m/s)2 + (-143 m/s) 2 ) ay = -9.80 m/s2 vf = 184 m/s viy = 0 m/s t = 14.6 s Θ = cos-1 vfx ( ) vf vfy = viy + ayt Θ = cos 115 -1 vfy = 0m/s + (-9.80 m/s2)(14.6 s) ( 184 ) vfy = -143 m/s Θ = 51.3° below the x axis Angled projectile motion: A projectile launched at an angle would continue in a straight line at a constant velocity if gravity is ignored. However, gravity makes the projectile accelerate to Earth. This shows us why a projectile launched at an angle follows a parabolic trajectory. Since the projectile is launched at an angle, it now has both horizontal and vertical velocities. When dealing with angled projectile problems, we neglect air resistance and say that gravity is g = - 9.80 m/s2 (downward) and also say that the vertical velocity of any object moving upwards will have a + velocity. This makes solving the problems a lot “cleaner.” The horizontal component of the velocity remains constant. The vertical component of the velocity changes as the projectile moves up or down. This is shown when looking at the vector components of the trajectory at regular time intervals. The up and right vectors represent the velocity given to the projectile when launched. The vertical vectors decrease in magnitude due to gravity. Eventually, the effects of gravity will reduce the upward velocity to zero. This occurs at the top of the parabolic trajectory where there is only horizontal motion. After gravity reduces the upward (vertical) speed to zero it begins to add a downward velocity. This velocity increases until the projectile return to the ground. When looking at each half of the trajectory (up and down) you can determine that the speed of the projectile going up is equal to the speed of the projectile coming down. (Provided air resistance is ignored.) The only difference is the direction of the motion. Problem Solving Projectile Motion: Notes Since the horizontal and vertical components of a projectile launched at any angle are still independent of one other, the equations pertaining to linear motion and free fall can still be used in problem solving. (As long as we look at them separately.) Before the projectile is launched, t = 0, the position is zero, y = 0. The other half of the equation is the part that helps us find time; when y at the end of the trajectory is zero all the time needed to complete the trajectory has passed. y = viyt + ½ gt2 x = vixt + ½ at2 When solving problems, the following procedures will be used: 1. Make a sketch of the situation and label all given and unknown quantities. 2. Determine what type of projectile motion is being observed – angled. 3. List all of the equations that apply to the problem. 4. Determine what variables are known and needed. 5. Substitute the given quantities into the equation using proper units. 6. Solve the equation, carrying units throughout the problem. 7. Check your answer to make sure it makes sense. 8. Box or circle your final answer(s). Ex: A baseball player hits a home run, and the ball lands in the left-field seats, 7.5 m above the point at which the ball was hit. The ball lands with a velocity of 36 m/s at an angle of 28º below the horizontal. Ignoring air resistance, find the initial velocity with which the ball leaves the bat. What we want to find: vi and θ To solve for these, we need to find viy and vix Vi = √(Vix2 + Viy2) and Θ = tan-1 Viy ( ) Vix Since air resistance is ignored, the horizontal component of the velocity, vx is constant. vix = vfx = (36 m/s) cos 28° = 32 m/s What we know about the y component: y = + 7.5 m ay = -9.80 m/s2 vfy = (-36 sin 28°) m/s viy = ? vfy2 = viy + 2ayy viy = √(vfy2 – 2ayy) viy = √( (-36 sin 28°) m/s)2 – 2(-9.80 m/s2)(7.5 m) ) viy = 21 m/s Therefore… Vi = √(Vix2 + Viy2) Vi = √( (32 m/s)2 + (21 m/s)2 ) Vi = 38 m/s Θ = tan-1 Viy ( ) Vix Θ = tan-1 21 ( ) 32 Θ = 33° above the x axis