Chapter 3 Homework Packet. Conceptual Questions

Chapter 3 Homework Packet Conceptual Questions 1) Which one of the following is an example of a vector quantity? A) mass B) area C) distance D) velocity A vector quantity has both magnitude and direction. Mass, area and distance only have magnitude, while velocity has both magnitude and direction. Answer D 2) Which of the following operations will not change a vector? A) Rotate it. B) Multiply it by a constant factor. C) Translate it parallel to itself. D) Add a constant vector to it. To change a vector means to either change its direction or its magnitude, or both. If you rotate a vector this will change its direction. If you multiply it by a constant factor you will change its magnitude. If you add a constant vector to another vector this will at least change its magnitude but could also change the direction. Only translating a vector parallel to itself will preserve both the direction and magnitude of the vector. Answer C 3) Which of the following is an accurate statement? A) A vector cannot have zero magnitude if one of its components is not zero. B) The magnitude of a vector can be less than the magnitude of one of its components. C) If the magnitude of vector A is less than the magnitude of vector B, then the x-component of A is less than the x- component of B. D) The magnitude of a vector can be positive or negative. (A) True if the magnitude of the x or y component of a vector is a non-zero value, the vector itself will automatically at least have a magnitude in that direction; (B) False the least the magnitude of a vector can ever be compared to the magnitude of its components is when the vector is completely either in the y or completely in the x direction. In this case, the magnitude of the vector will equal the value of the magnitude of the component. If a vector has components in both the x and y direction, its magnitude (its length) will necessarily lie between the value of the magnitude of the shortest component, and the sum of the magnitudes of the components. (C) False while the magnitude of vector A could be less than the magnitude of vector B, this could occur if the y component of A is much less than the y component of B. For example, even though shorter, all of A could be in the x direction, while all of B could be in the Y direction. In this case, the x component of B would be less than that of A, even though A is the shorter vector. (D) False positive and negative only indicate the direction of a vector. If you applied a negative sign to a scalar quantity this would automatically give it a direction. Answer A 4) The resultant of two vectors is the smallest when the angle between them is A) 90. B) 180. C) 0. D) 45. 1

To answer this question we must consider this from the perspective of both vectors acting at the same point rather than two vectors action in a head to tail fashion. In this case, for an angle of 0 degrees, both vectors would be acting in the same direction and would sum together. On the other hand, if they were acting at an angle at 180 o they would add together in opposite directions where the resultant would have the least magnitude. If they were at some intermediate angle, the result would take on an intermediate value. Answer B 5) Two displacement vectors have magnitudes of 5.0 m and 7.0 m, respectively. When these two vectors are added, the magnitude of the sum A) is 2.0 m. B) is 12 m. C) is larger than 12 m. D) could be as small as 2.0 m, or as large as 12 m. The magnitude of the vector sum will depend on the direction in which the two vectors are added. If they are added together in the same direction the sum will be 12.0. If they are added together in opposite directions the magnitude will be the difference in the magnitudes of the two vectors meaning the magnitude of the sum would be 2.0 m. Answer D 6) Two vectors, of magnitudes 20 and 50, are added. Which one of the following is a possible answer for the magnitude of the resultant? A) 10 B) 20 C) 40 D) 80 The magnitude of the vector sum will depend on the direction in which the two vectors are added. If they are added together in the same direction the sum will be 70. If they are added together in opposite directions the magnitude will be the difference in the magnitudes of the two vectors meaning the magnitude of the sum would be 30. Therefore, the answer must lie between 20 and 50. The only answer that fits this criterion is 40. Answer C 7) Three forces, each having a magnitude of 30 N, pull on an object in directions that are 120 apart from each other. Make a statement concerning the resultant force. A) The resultant force is equal to 30 N. B) The resultant force is less than 30 N. C) The resultant force is zero. D) The resultant force is greater than 30 N. Because the direction of three forces of equal magnitude would be equidistant from each other, the three force would cancel each other out the resultant would be 0. Answer C 8) In the diagram shown, the unknown vector is A) B). C) D) Diagram 1 In this diagram the best approach is to ask, what vector is being 2

added to what vector to get a resultant. You can see that? is being added to A!" to get B!". This means that you could write the relationship: A!" +? = B!" Then, isolate? to get? = B!" A!" Answer C 9) Ignoring air resistance, the horizontal component of a projectile's velocity A) continuously increases. B) continuously decreases. C) is zero. D) remains constant. For projectile motion, velocity in the x direction (horizontal component), or v x, is constant. Answer D 10) A ball is thrown with a velocity of 20 m/s at an angle of 60 above the horizontal. What is the horizontal component of its instantaneous velocity at the exact top of its trajectory? A) 10 m/s B) 17 m/s C) 20 m/s D) zero The term instantaneous velocity refers to the velocity of the object at the instant the object is at the top of its flight. At that point recognize that there is only velocity in the x (horizontal) direction velocity in the y direction at that instant is 0 m/s. So, as velocity in the horizontal direction is constant, the horizontal component of the velocity at this point will be the same as the horizontal velocity at the start of the motion, which can be determined by finding the horizontal component of the velocity. Remember that we do this by multiplying the velocity at which the projectile is launched times the cos (because the horizontal component is the adjacent side in this case) of the angle at which it was launched. Further, you should know that the cos 60 0 is ½ so the horizontal velocity will be ½ (20 m/s) = 10 m/s. Answer A 11) Ignoring air resistance, the horizontal component of a projectile's acceleration A) is zero. B) remains a non-zero constant. C) continuously increases. D) continuously decreases. In projectile motion the horizontal component of the velocity is constant. Therefore, the horizontal acceleration is 0. Answer A 12) A soccer ball is kicked with a velocity of 25 m/s at an angle of 45 above the horizontal. What is the vertical component of its acceleration as it travels along its trajectory? A) 9.80 m/s2 downward B) (9.80 m/s2) sin (45 ) upward C) (9.80 m/s2) sin (45 ) downward D) (9.80 m/s2) upward For projectile motion you know that motion in the x (horizontal) direction is constant and so acceleration in the x direction is always 0. In the y (vertical) direction, however, the change in velocity is related to the acceleration due to gravity. Throughout the course of the motion, no matter where in the flight the object is, the acceleration due to gravity does not change it is always -9.8m/s 2. As far as the answer go, this would be 9.80 m/s 2 in the downward direction. Answer A 3

13) If the acceleration vector of an object is directed anti-parallel to the velocity vector, A) the object is turning. B) the object is speeding up. C) the object is slowing down. D) the object is moving in the negative x-direction. Anti-parallel is in a direction opposite to or at an angle of 180 0 to a given vector. You are aware by now that if acceleration directly opposes the velocity vector, the velocity in the direction it is moving will be decreasing. The positive or negative direction of the velocity is not specified in the given information so the answer will have to suffice for either. You can visualize both situations with a simple diagram. In the top diagram there is a decreasing positive velocity which is slowing down because of an opposing acceleration. In the bottom diagram there is a decreasing negative velocity which is also slowing down because of an opposing acceleration. Therefore, in both cases the result of an acceleration vector directed anti-parallel to a velocity vector is the slowing down of the object. Answer C 14) If the acceleration of an object is always directed perpendicular to its velocity, A) the object is speeding up. B) the object is slowing down. C) the object is turning. D) this situation would not be physically possible. The tendency of any object is to travel in a straight line. If a force causing an acceleration acts perpendicular to the direction of the velocity, if the acceleration is applied and then stopped, the object will then continue to move in a straight line in the slightly altered new direction after the acceleration is withdrawn. However, if the acceleration is continued, always at a direction perpendicular to the velocity, the object will be pushed into a path that is circular so, the object is turning. The following diagram may be useful. Answer C 15) At what angle should a water-gun be aimed in order for the water to land with the greatest horizontal range? A) 0 B) 30 C) 45 D) 60 4

According to the range formula, R = v2 sin(2θ), as the value of sin is greatest at an angle of a 90 0 (this is where sin = 1), if 2θ is 90 0, the range will accordingly be the greatest. So, if 2θ is 90 0, θ must be 45 0 Answer C 16) An Olympic athlete throws a javelin at four different angles above the horizontal, each with the same speed: 30, 40, 60, and 80. Which two throws cause the javelin to land the same distance away? A) 30 and 80 B) 40 and 60 C) 40 and 80 D) 30 and 60 The range formula is: R = v2 sin(2θ). Remember that sin is positive in both the first and a second quadrants. It increases from 0 to 1 in the first quadrant, and decreases from 1 to 0 in the second quadrant. Therefore, from 0 to 180 degrees there will be two angles where the value 2θ will be equal, and these values for 2θ will be caused by values of θ that are complements to each other. For example, if one angle is 17 0, sin 2θ = sin 34 0 =.559. The complement is 73 0 sin of sin 2θ = sin 146 0 =.559. So, the two throws that will cause the javelin to land the same distance away must be compliments that is, 30 0 and 60 0. Answer D 17) You are throwing a ball for the second time. If the ball leaves your hand with twice the velocity it had on your first throw, its horizontal range R (compared to your first serve) would be A) 1.4 times as much. B) half as much. C) twice as much. D) four times as much. Don t waste time trying to reason through this in your head write the range formula down, placing a 2 in front of velocity to indicate you are doubling the velocity. You can then easily see that if you multiply the velocity by 2, this velocity will be squared having the effect of 4 time the range of the first throw. Answer D 2 v 2 sin(2θ) a = 4R 18) A ball is thrown at an original speed of 8.0 m/s at an angle of 35 above the horizontal. What is the speed of the ball when it returns to the same horizontal level? A) 4.0 m/s B) 8.0 m/s C) 16 m/s D) 9.8 m/s We are being asked about speed here so we do not have to worry about directions being positive or negative. Also, we are not being asked about speed in the x or the y direction, just the overall speed of the ball as it reaches its original horizontal level. As we ignore air resistance, as the ball leaves the hand, while its x velocity will remain constant throughout the flight its y velocity will decrease until it reaches maximum height. Then, as the ball descends, it gain speed in the same fashion that it originally lost speed when it was ascending. As the acceleration due to gravity remains constant, when the ball reaches the same horizontal level, it should have exactly the same speed it had when it left the hand. Answer B 5

19) When a football in a field goal attempt reaches its maximum height, how does its speed compare to its initial speed? A) It is zero. B) It is less than its initial speed. C) It is equal to its initial speed. D) It is greater than its initial speed. We are not being asked specifically about the y or the x velocity, we are only being asked about the overall speed of the ball. As the ball leaves the foot, while its x velocity will remain constant throughout the flight, its y velocity will decrease until it reaches maximum height. At this point the y velocity will be 0 m/s. Therefore, as the overall speed is a result of the sum of the x and y velocities, the overall speed at maximum height will be less than its initial speed. Answer B 20) A stone is thrown horizontally from the top of a tower at the same instant a ball is dropped vertically. Which object is traveling faster when it hits the level ground below? A) It is impossible to tell from the information given. B) the stone C) the ball D) Neither, since both are traveling at the same speed. Be careful the tendency here is to think this will be a question about the time it takes for both objects to hit the ground, which would be the same time. However, this question is asking about which object is traveling faster when it hits the ground. As the ball is dropped vertically, all it will do is accelerate at the acceleration due to gravity and its final velocity will only have this y component of velocity. However, the stone that is thrown from the tower will not only have the same y component of velocity, it will also have whatever x component of velocity that was added to it when it was thrown. Therefore, from the perspective of overall speed, the stone s speed will be greater because it has the same y component of velocity the ball had, plus its x component of velocity. Answer B 21) A bullet is fired horizontally, and at the same instant a second bullet is dropped from the same height. Ignore air resistance. Compare the times of fall of the two bullets. A) The fired bullet hits first. B) The dropped bullet hits first. C) They hit at the same time. D) cannot tell without knowing the masses When comparing the fall of two objects, if we ignore air resistance, both objects will be under the influence of the same acceleration due to gravity. Therefore, both objects will fall at the same rate the motion in the x direction of bullet fired horizontally has no impact on the vertical motion of that bullet. The bullets will hit the ground at the same time. Answer C 22) A plane flying horizontally at a speed of 50.0 m/s and at an elevation of 160 m drops a package. Two seconds later it drops a second package. How far apart will the two packages land on the ground? A) 100 m B) 162 m C) 177 m D) 283 m As each package is moving at a velocity of 50.0 m/s because it is on an airplane moving at 50.0 m/s, at the instant each package is dropped, it still has a velocity in the x direction of 50.0 m/s. Each package will take the same amount of time to fall. As the second package was released two seconds after the first package, it was released (2)(50.0m) or 100 m past when the first 6

package was released. Therefore, it will also hit 100 m past where the first package hit. Answer A 23) A package of supplies is dropped from a plane, and one second later a second package is dropped. Neglecting air resistance, the distance between the falling packages will A) be constant. B) decrease. C) increase. The confusing part about this question is that you know by now that acceleration for both packages is the same they will both accelerate at the same rate so, the tempting thing to do would be to immediately say that as time progresses, the distance between the packages must remain constant. Trying to reason this out simply visualize what must be happening to each package as each second passes. After 1 second, the first package has dropped a certain amount but the second package has not yet moved. After 2 seconds, the first package has dropped an amount greater than it dropped the first second, while the second package has now dropped the same amount as the first package did the first second. You should quickly realize that at each second, the second package has dropped to the point where the first package had dropped to the preceding second, and the distance between the two must be increasing. You can also draw a very quick diagram that helps you visualize this by labeling each package with subscripts to show where they are at each second (see diagram) this only takes a few seconds and saves a lot of time trying to reason. Answer C 24) A pilot drops a bomb from a plane flying horizontally at a constant speed. Neglecting air resistance, when the bomb hits the ground the horizontal location of the plane will A) be behind the bomb. B) be over the bomb. C) be in front of the bomb. D) depend on the speed of the plane when the bomb was released. As the bomb is on the plane, it has a velocity in the x direction that is equal to the velocity of the plane. The instant it is released, it still has that velocity. If we ignore air resistance, the bomb will continue to travel in the x direction at the same velocity as the airplane. Therefore, when the bomb hits the ground, the plane will be directly over the bomb. Answer B 25) The acceleration of gravity on the Moon is only one-sixth of that on Earth. If you hit a baseball on the Moon with the same effort (and at the speed and angle) that you would on Earth, the ball would land A) the same distance away. B) one-sixth as far. C) 6 times as far. D) 36 times as far. You know that the distance at which the ball lands is a function of displacement in the x direction. You know the formula that determines displacement in the x direction is x = x 0 + v x t. You are told that speed and angle at which you hit the ball on the moon will be the same as on the earth so there will be no difference in the two situations related to v x. Therefore, if there is any difference between the two situations, it will be that the ball is in flight for a different amount of time on the moon than it is on the earth. Determining the amount of time the ball is in flight depends on how long it takes to reach maximum height and then return, which in turn, is controlled by gravity. It is intuitive to understand that in the weaker gravity of the moon the ball will take longer to 7

reach maximum height and then return the question is how much. Remember again, that according to our formula for x displacement, whatever we multiply time by, we will also multiply the distance by. So, we need to now figure out how time is affected by the change in gravity. We have two formulas related to motion in the y direction we could use to determine what will happen to time in the weaker gravity, and both of them will work to help us out. One of our y direction formulas is V y = V y0 + at. If we recognize that V y will equal the opposite of V y0, we can substitute V y0 in for V y and isolate t to get 2V y0 = t. If we multiply a the acceleration by 1/6, the effect on the time will be to increase the amount of time by six times. My practice is to write the formula out in this fashion, highlighting whichever variable we modified, and highlighting the resulting change in the desired variable (see below) v y = v y0 + at 2v y0 = t 2v y0 a 1 a = 6 t 6 We could also use the other formula for motion in the y direction to help us out. Because we are assuming that y and y 0 are the same value, we can manipulate the y displacement formula in the following way: y = y 0 t + 1 at 2 v 2 y0 t = 1 at 2 2v y0 = t At this point you should recognize that 2 a this is actually the same formula you already used, so even though you started with two different formulas, they accomplish the same thing. In either case, we find that the fact that the moon s gravity is 1/6 that of Earth s means that the time of flight is 6 times that of the flight on earth. According to our displacement formula then, the ball will travel six times as far. Answer C 26) You are traveling at 55 mi/h in the +x axis relative to a straight, level road and pass a car traveling at 45 mi/h. The relative velocity of your car to the other car is A) -10 mi/h. B) 10 mi/h. C) 65 mi/h. D) 35 mi/h. These can be tricky questions. Both cars have velocities that are relative to the road. For your car you could create a velocity vector labeled V CAR = 55 mi/h (meaning velocity of the car A relative to the road). For the other car, you could create a velocity vector labeled V CBR = 45 mi/h and place it directly under the vector for the your car. Then notice the gap between the tip of your car s vector and the other car s vector. If we are considering the velocity of your car relative to the velocity of the other car we can think, we need to add a vector of +10 m/s for the velocity of the other car to equal the velocity of your car. We can then draw a vector in from the tip of the other car s vector to the tip of your car s vector that represents that positive 10 m/s. Again, from your perspective, this is what would have to be added to the other car to make it equal to your car. We would label this vector V AB = 10 m/s To show the work for such a problem (although you don t need to for a multiple choice 8

question), you could say: Although we aren t asked to, you should also consider the other possibility here that we could have been asked to state what the velocity of your car is relative to the other car. In this case you would still create the same to vectors for the cars their motion hasn t changed. Then notice the gap between the tip of your car s vector and the other car s vector to consider the other car s velocity relative to your car you would have to think, we need to add a velocity of -10m/s for the velocity of the your car to equal the velocity of the other car. We can then draw a vector in from the tip of the your car s vector to the tip of the car s vector that represents that negative 10 m/s. Again, from the other car s perspective, this is what would have to be added to the your car to make it equal to the other car. We would label this vector V AB = -10 m/s. To show the work for such a problem (although you don t need to for a multiple choice question), you could say: Answer B 27) You are trying to cross a river that flows due south with a strong current. You start out in your motorboat on the east bank desiring to reach the west bank directly west from your starting point. You should head your motorboat A) due west. B) due north. C) in a southwesterly direction. D) in a northwesterly direction. The motion that you are trying to achieve, and the motion that a person standing on the shore would see, is a motion directly across the river. If you headed directly across the river (due west), you would be carried downstream to a point south of where you want to be. If you head the boat due north, you would travel slowly up stream but never reach the western bank. If you headed the boat in a southwesterly direction, you would eventually reach the western bank but you would be far downstream. Only if you headed the boat in a northwesterly direction a direction both against the current and with motion toward the western bank could you reach the western shore directly across from where you were. Answer D 28) Your motorboat can move at 30 km/h in still water. How much time will it take you to move 12 km downstream, in a river flowing at 6.0 km/h? A) 20 min B) 22 min 9

C) 24 min D) 30 min You want time in a situation of constant velocity, so you can use the displacement with constant velocity formula to help you calculate this start by writing down this formula. x = x 0 + vt t = x v and isolating t. You know that you have a given displacement of 12 km. If your boat can move 30 km/h in still water, if it does this in a river already flowing 6 km/h, the velocity you would be looking for would be the vector sum of the two velocities. As both velocities would be in the same direction. You know you would just need to sum the magnitudes to get the resultant: 30km/h + 6 km/h = 36 km/h. Adding this to the formula you can easily see without a calculator that the answer is 12/36 or 1/3 of an hour which would be 20 min. Answer A Quantitative Problems 1) Vector A has magnitude 8.0 m at an angle of 30 degrees below the +x axis.!" The y component of A is A diagram could help you see that the relationship between A!" and the y component of A is that A!" is the hypotenuse and the y component is the opposite side. Therefore, to find the y component of A!" you need to multiply the magnitude of A!" times the sin of θ. Also, because this is in the negative direction you need to apply a negative sign to the calculation and answer. I don t need to see the diagram but I do need to see the information in the box. 2) If a ball is thrown with a velocity of 25 m/s at an angle of 37 above the horizontal, what is the vertical component of the velocity? Let the velocity be represented by a vector, v! = 25m / s A diagram could help you see that the relationship between v! and the vertical component, v! x, is that v! is the hypotenuse and the vertical or y component is the opposite side. Therefore, to find the vertical component you need to multiply the magnitude of v! times the sin of θ. I don t need to see the diagram but I do need to see the information in the box. 10

3) If you walk 6.0 km in a straight line in a direction north of east and you end up 2.0 km north and several kilometers east. How many degrees north of east have you walked? Be careful here make sure you understand that the straight line you have walked for 6.0 km is the resultant of the problem. The 2.0 km you have walked north, and the distance you have walked east are the displacement vectors added together to get the resultant 6.0 km. A diagram that illustrates this is shown below. The resultant vector you would probably label D = 6.0 km, and the vertical vector you are given you would probably label D 2 = 2.0 km. At this point, you realize that you have an opposite side and a hypotenuse of a right triangle, and so can find the angle north of east by using the inverse sin. The diagram also helps you create the correct formula by providing labels. θ = sin 1 o h = sin 1 D 2 D = sin 1 2.0 6.0 = 190 (2SF) 4) A butterfly moves with a speed of 12.0 m/s. The x component of its velocity is 8.00 m/s. The angle between the direction of its motion and the x axis must be The butterfly must be moving in a direction that is some angle above the x axis, and this is the direction of the 12.0 m/s. Draw a vector at some angle above the x axis and label it V = 12.0 m/s. Picture this motion as being the result (that is, the vector sum ) of flying 8.00 m/s in a direction along the x axis (because we are told the x component is 8.00 m/s), plus, flying in a direction of an unknown number of m/s in the y direction (this would be the y component). Draw an x component vector labeled V x = 8.00 m/s and a y component vector labeled V y =? that add together to give the resultant vector, as shown below. This helps you realize that you have an adjacent side and a hypotenuse to the angle you want, so you can use the inverse cos to solve for the angle. The diagram also helps you create the correct formula by providing labels. θ = cos 1 a h = cos 1 V x V = cos 1 8.00 12.0 = 48.20 (3SF) 5) A 400-m tall tower casts a 600-m long shadow over a level ground. At what angle is the Sun elevated above the horizon? The length of the of the shadow will depend on the angle at which the rays of the sun s light is hitting the tower. To draw a diagram, the length of the shadow will form the adjacent side of a right triangle you could draw the line indicating this length and label it D 1 = 600 m. The length of the tower will form the opposite side. You could draw the line indicating this and label it D 2 = 400 m. The distance from the tip of the tower to the end of the shadow will form the hypotenuse of the triangle and the angle between this hypotenuse and the adjacent side is the angle we are seeking. After drawing the diagram you can easily see that you have an adjacent and an opposite side and can find the angle using the inverse tangent. 11

6) Two vectors A and Bhave components (0, 1) and (-1, 3), respectively. What are the components of the sum of these two vectors? Be aware that we can use x and y coordinates to represent the x and y components of a vector. The x coordinate represents the length along the x axis of the vector, and the y coordinate represents the length along the y axis of the vector. Just as we do when we add components of vectors together that do not form a right triangle, we can combine the x components of these vectors to get an overall x component, and we can combine the y components of these vectors together to get an overall y component. Therefore, the components of the sum of these two vectors are: x = A!" x + B!" x = 0 + ( 1) = 1 y = A!" y + B!" y = 1+ 3 = 4 vector sum = ( 1,4) Although it is not necessary, you could also represent this sum on an x-y coordinate graph as: Notice again that the coordinates of the added vector are simply the sums of the x and y components. You could actually sum as many vectors as you wanted in this fashion. 7) Two vectors Aand Bhave components (0, 1) and (-1, 3), respectively. What is magnitude of the sum of these two vectors? You should remember from math classes that on an x-y coordinate system we can determine the distance between any two points on the graph by finding the square root of the sum of the square of the difference in x components and the square of the difference in y components. If we consider the beginning point of the vector to be the origin, the difference in x components will simply be the x value of the resultant vector (-1), and the difference in y components will be the y value of the resultant vector (4). Therefore, as the length of the vector represents the magnitude of the vector, the magnitude of the sum of these two vectors will be x 2 + y 2. magnitude = x 2 + y 2 = ( 1) 2 + 4 2 = 4.1(2 SF) 8) Vector A= (1, 3). Vector B= (3, 0). Vector C= A+ B. What is the magnitude of C? Be aware that we can use x and y coordinates to represent the x and y components of a vector. The x coordinate represents the length along the x axis of the vector, and the y coordinate 12

represents the length along the y axis of the vector. Just as we do when we add components of vectors together that do not form a right triangle, we can combine the x components of these vectors to get an overall x component, and we can combine the y components of these vectors together to get an overall y component. Therefore, the components of the sum of these two vectors are: x = A!" x + B!" x = 1+ 3 = 4 A!"! + B!" = C!" = (4,3) y = A!" y + B!" y = 3 + 0 = 3 To find the magnitude, you should remember from math classes that on an x-y coordinate system we can determine the distance between any two points on the graph by finding the square root of the sum of the square of the difference in x components and the square of the difference in y components. If we consider the beginning point of the vector to be the origin, the difference in x components will simply be the x value of the resultant vector (4), and the difference in y components will be the y value of the resultant vector (3). Therefore, as the length of the vector represents the magnitude of the vector, the magnitude of the sum of these two vectors will be x 2 + y 2. magnitude = x 2 + y 2 = 4 2 + 3 2 = 5 9) A car travels 20 km west, then 20 km south. What is the magnitude of its displacement? This is a simple implementation of the Pythagorean theorem. To draw a diagram you would draw a velocity vector to the west labeled D 1 = 20 km, and then you would draw a vector from the tip of the first vector to the south of the same length and label this D 2 = 20 km. You would then draw a resultant vector from the beginning of the motion to the end of the motion and label it D =? From the diagram you could then easily see how to create a formula to solve for the resultant. You know from the Pythagorean theorem that the resultant, which is the hypotenuse of the triangle in this case, is equal to the square root of the sum of the squares of the lengths of the sides. Creating the diagram also then provides you with labels to use in formulas. D = D 1 + D 2 = 20 2 + 20 2 = 28.3km 10) Three vectors, expressed in Cartesian coordinates, are x comp y comp _ S - 3.5 +4.5 _ T 0-6.5 _ U +5.5-2.5 13

What is the magnitude of the resultant vector S+ T + U? We are not limited to adding two vectors together. We can add as many as we want to. To find the sum of the vectors, as with just using two vectors, all we have to do is sum the x components to get the overall x component and sum the y components to get the overall y component. x = S!" x + T!" x +U x!!"!"!!"!" = 3.5 + 0 + 5.5 = 2.0 S + T + U = (2.0, 4.5) y = S!" y + T!" y + U!" y = 4.5+ ( 6.5) + ( 2.5) = 4.5 To find the magnitude, you should remember from math classes that on an x-y coordinate system we can determine the distance between any two points on the graph by finding the square root of the sum of the square of the difference in x components and the square of the difference in y components. If we consider the beginning point of the vector to be the origin, the difference in x components will simply be the x value of the resultant vector (4), and the difference in y components will be the y value of the resultant vector (3). Therefore, as the length of the vector represents the magnitude of the vector, the magnitude of the sum of these two vectors will be x 2 + y 2. magnitude = x 2 + y 2 = 2.0 2 + ( 4.5) 2 = 4.9 11) Three vectors, expressed in Cartesian coordinates, are x comp y comp _ S - 3.5 +4.5 _ T 0-6.5 _ U +5.5-2.5 What is the angle of the resultant vector S+ T + Umeasured from the positive x axis? You have already found the components of this resultant vector: S!"! + T!" + U!" = (2.0, 4.5) The quickest way to figure out how find the angle (or direction) of this vector is to make a quick sketch to visualize what the direction of the vector is. The problem asks you to measure the resultant from the positive x axis. You can see this angle is in the fourth quadrant so you know that once you find the angle with reference to the negative y-axis you will have to add 270 0 to it. You have an opposite side (2) and an adjacent side (4.5 note, we only want the angle with reference to the 90 0 space that the angle is in, so use the absolute value of -4.5) so to find the angle with reference to the negative y-axis use the inverse 14

tangent: θ = 270 0 + tan 1 o a below the positive x axis). ( ) = 270 0 + tan 1 2 ( 4.5) = 294 0 (You could also have said 66 0 12) If vector A= (-3.0, -4.0) and vector B= (+3.0, -8.0), what is the magnitude of vector C= A- B?!" Be careful in this problem we are subtracting B!"!" from A, not adding them. So, to find the resultant, we must subtract the x component of B!"!" from A, not add them. Likewise, we must subtract the y component of B!" from A, not add them. x = A!" x B!" x = 3.0 3.0 = 6.0 C!"! = ( 6.0,4.0) y = A!" y B!" y = 4.0 ( 8.0) = 4.0 To find the magnitude, you should remember from math classes that on an x-y coordinate system we can determine the distance between any two points on the graph by finding the square root of the sum of the square of the difference in x components and the square of the difference in y components. If we consider the beginning point of the vector to be the origin, the difference in x components will simply be the x value of the resultant vector (0), and the difference in y components will be the y value of the resultant vector (-12). Therefore, as the length of the vector represents the magnitude of the vector, the magnitude of the sum of these two vectors will be x 2 + y 2. magnitude = x 2 + y 2 = 6.0 2 + 4.0 2 = 7.2 13) A runner runs halfway around a circular path of radius 10 m. What is the displacement of the jogger? Remember that the displacement is not the same thing as the distance. If the radius of the circle is 10 m, and the runner runs halfway around, he will be at a position 2r or 20 m away. As this is a quantitative problem you would need to show work for this: D = 2r = 2(10m) = 20 m Note: if you were finding the distance rather than displacement remember that the circumference around half a circle would be πr = π(10m) = 31.4m. FIGURE 3-1 14) Two forces are acting on an object as shown in Fig. 3-1. What is the magnitude of the resultant force? 15

As these forces are not acting at right angles to each other, you know you are going to have to sum the x components of the forces to get a combined x component, and sum the y components of the forces to get a combined y component. As you know that the magnitude will then equal the square root of the sum of the squares of the x and y components, you should write this formula down first and let it lead you through the rest of the problem. (see below). As this formula tells you you need F x, which will be the sum of the x components (F 1x + (-F 2x )), and F y, which will be the sum of the y components (F 1y + F 2y ), write down formulas for each of these next. Notice that we have subtracted F 2x rather than added this is because it is in the direction negative of F 1x ) Then, as the x components (the horizontal components) will be the adjacent side of each triangle we know we need to multiply the given force vectors times cosθ. As the y components (vertical components) will be the opposite side of each triangle we know we need to multiply the given force vectors times sinθ. Plug the values in and solve for the x and y components. Then plug these values into the original formula to get the magnitude: F = F x 2 + F y 2 = 39.3 2 +181 2 = 190 (2 SF) F x = F 1x + F 2x = (F 1 )(cosθ) + ( F 2 )(cosθ) = (120N )(cos60) (80N )(cos75) = 39.3N F y = F 1y + F 2 y = (F 1 )(sinθ) + (F 2 )(sinθ) = (120N )(sin60) + (80N )(sin75) = 181N 15) Two forces are acting on an object as shown in Fig. 3-1. What is the direction of the resultant force? If you were to make a quick sketch of the x and y components you just found in the previous problem you would see the proper relationship of the x and y component and know that you could solve for the angle using the inverse tangent. The diagram also helps you create a proper formula. Make sure you state what the 78 0 is with reference to. θ = tan 1 F y F x = tan 1 181 39.3 = 780 above x axis (2 SF) 16) Vector Ais 5.5 cm long and points along the x axis. Vector Bis 7.5 cm long and points at +30 to the negative x axis. (a) Determine the x and y components of Vector A. As you are asked for the x and y component of vector A!" simply write down the formulas for the components, plug in the values given and solve. Note, although you can easily see that!" because there is no y component of A, the x component will be 5.5 cm and the y component will be 0 cm, you still need to show this.!" Ax = (!" A )(cosθ) = (5.5cm)(cos0 0 ) = 5.5cm!" Ay = (!" A )(sinθ) = (5.5cm)(sin0 0 ) = 0 cm 16

(b) Determine the x and y components of Vector B. A sketch would probably be helpful. As you are asked for the x and y component of vector B!" simply write down the formulas for the components, plug in the values given and solve. Note, this x component is in the direction!" negative to the A vector so you will have to make the x component negative.!" Bx = (B!" )(cosθ) = (7.5cm)(cos30 0 ) = 6.5cm!" B y = (B!" )(sinθ) = (7.5cm)(sin30 0 ) = 3.8 cm (c) Determine the sum of these two vectors in terms of components. As always, to get the x component, sum the x components; to get the y component, sum the y components. x = A x + B x = 5.5cm + ( 6.5cm) = 1.0 cm y = A y + B y = 0cm + 3.8 cm = 3.8 cm (d) Determine the sum of these two vectors in terms of magnitude and direction. As always, to determine the magnitude, use the Pythagorean theorem.!"!" A + B = x 2 + y 2 = 1.0 2 + 3.8 2 = 3.9 cm A quick sketch will help you to see the proper relationship between the x and y components. you have an opposite (y) and adjacent (x) side to an angle so inverse tangent will allow you to find this angle. Make sure to state what the angle is with reference to. θ = tan 1 y x = 3.8 tan 1 1.0 = 750 above negative x axis 17) Vector Ais 75.0 cm long and points at 30 above the positive x axis. Vector Bis 25.0 cm long and points along the negative x axis. Vector Cis 40.0 cm long and points at 45 below the negative x axis. (a) Determine the x and y components of Vector A. (b) Determine the x and y components of Vector B. (c) Determine the x and y components of Vector C. (d) Determine the sum of these three vectors in terms of components. (e) Determine the sum of these three vectors in terms of magnitude and direction. As this problem involves so many vectors it would probably be useful to sketch the relationship of all three at the beginning. (a) As you are asked for the x and y component of vector A!" simply write down the formulas for the components, plug in the values given and solve.!" Ax = (!" A )(cosθ) = (75.0cm)(cos30 0 ) = 65cm (2SF)!" Ay = (!" A )(sinθ) = (75.0cm)(sin30 0 ) = 38cm 17

(b) As you are asked for the x and y component of vector B!" simply write down the formulas for the components, plug in the values given and solve. Note, although you can easily see that!" because there is no y component of B, the x component will be -25.0 cm and the y component will be 0 cm, you still need to show this. Also, remember the x component is in the negative direction.!" Bx = (B!" )(cosθ) = (25.0cm)(cos0 0 ) = 25cm (2 SF)!" B y = (B!" )(sinθ) = (25.0cm)(sin0 0 ) = 0cm (c) As you are asked for the x and y component of vector C!" simply write down the formulas for the components, plug in the values given and solve. Note, both the x and y components are in the negative direction so you will have to make the these components negative. C!" x = (C!" )(cosθ) = (40.0cm)(cos45 0 ) = 28cm (2 SF) C!" y = (C!" )(sinθ) = (40.0cm)(sin45 0 ) = 28cm (d) As always, to get the x component, sum the x components; to get the y component, sum the y components. x = A x + B x = 65cm + ( 25cm) + ( 28cm) = 12 cm y = A y + B y = 38cm + 0 cm + ( 28cm) = 10 cm (e) As always, to determine the magnitude, use the Pythagorean theorem.!"!" A + B = x 2 + y 2 = 10 2 + 12 2 = 16 cm A quick sketch will help you to see the proper relationship between the x and y components. you have an opposite (y) and adjacent (x) side to an angle so inverse tangent will allow you to find this angle. Make sure to state what the angle is with reference to. θ = tan 1 y x = 10 tan 1 12 = 400 above positive x axis 18) A stone is thrown horizontally with an initial speed of 10 m/s from the edge of a cliff. A stop watch measures the stone's trajectory time from the top of the cliff to the bottom to be 4.3 s. What is the height of the cliff? In projectile motion you know that to analyze the motion you have to break the motion into x and y components and analyze each of these separately. You know that motion in the x direction is constant that is, v x is constant. So to analyze motion in the x direction, you can only use the formula that is used for constant velocity: x = x 0 + v x t (although under certain circumstances you can use the range formula to find x displacement). You know that motion in the y direction is accelerated according to the acceleration due to gravity. Therefore, to analyze motion in the y direction you can use either v y = v y0 + at (from 18

the perspective of the initial and final y velocities) or y = y 0 t + 1 2 at 2 (from the perspective of displacement in the y direction). In this case, you are told that this is a horizontal launch problem, for which you immediately realize that initial y velocity is 0 m/s. You are asked to find the height of the cliff which is a y direction calculation, so you immediately know you are going to need to use: y = y 0 t + 1 2 at 2. Remember that in this case, it is not y 0 that equals 0 m, it is y. Therefore, you will want to isolate y 0 before you start plugging in values. Write out the y displacement formula and then isolate y 0 (see below) or, if you can write out the formula with y 0 already isolated just do that. Again, remember that the v y0 term drops out. Also notice that the negative sign after isolating y 0 cancels the negative sign of acceleration due to gravity. y = y 0 t + 1 2 at 2 y 0 = 1 2 at 2 = 1 2 ( 9.8m / s2 )(4.3) 2 = 91m (2 SF) 19) A girl throws a rock horizontally, with a velocity of 10 m/s, from a bridge. It falls 20 m to the water below. How far does the rock travel horizontally before striking the water? In projectile motion you know that to analyze the motion you have to break the motion into x and y components and analyze each of these separately. You know that motion in the x direction is constant that is, v x is constant. So to analyze motion in the x direction, you can only use the formula that is used for constant velocity: x = x 0 + v x t (although under certain circumstances you can use the range formula to find x displacement). You know that motion in the y direction is accelerated according to the acceleration due to gravity. Therefore, to analyze motion in the y direction you can use either v y = v y0 + at (from the perspective of the initial and final y velocities) or y = y 0 t + 1 2 at 2 (from the perspective of displacement in the y direction). In this case, you are told that this is a horizontal launch problem, for which you immediately realize that initial y velocity is 0 m/s. You are asked to determine how far the rock travels horizontally, which is an x direction calculation, so you immediately know you are going to need to use x = x 0 + v x t. NOTE-THIS IS IMPORTANT in the vast majority of problems, if you need to determine an x direction value (either x displacement or x velocity) you will almost always need to find time. The only way you will be able to do this is by using y direction calculations. MAKE THIS AN INTUITIVE PART OF YOUR PROBLEM SOLVING SKILLS FOR PROJECTILE MOTION. There are several possibilities: -for horizontal launch, you will almost always need to use y = y 0 t + 1 2 at 2 to find time (unless by some chance you are given the final velocity in the y direction). In this case v y0 will always be 0 m/s and the v y0 t term will drop out, and isolating t will give: t = 2y 0 a 19

-for angled launch (1) you will be able to use the range formula if you are given the velocity and angle: R = v2 sin(2θ) a -for angled launch (2) you will be able to use v y = v y0 + at if the object is launched and lands at the same height. In this circumstance, v y equals the opposite of v y0. Isolating t gives t = 2v y0 a. -for angled launch (3) if the launch height and landing height are not the same, you will have to use y = y 0 t + 1 2 at 2, but the v y0 t term will not drop out, and you will have to use the quadratic formula to solve for t. In this case, to solve for time, as this is a horizontal launch you must use y = y 0 t + 1 2 at 2. After isolating and solving for t, plug v x and t into x = x 0 + v x t and solve for x. 20) A ball thrown horizontally from a point 24 m above the ground, strikes the ground after traveling horizontally a distance of 18 m. With what speed was it thrown? In projectile motion you know that to analyze the motion you have to break the motion into x and y components and analyze each of these separately. You know that motion in the x direction is constant that is, v x is constant. So to analyze motion in the x direction, you can only use the formula that is used for constant velocity: x = x 0 + v x t (although under certain circumstances you can use the range formula to find x displacement). You know that motion in the y direction is accelerated according to the acceleration due to gravity. Therefore, to analyze motion in the y direction you can use either v y = v y0 + at (from the perspective of the initial and final y velocities) or y = y 0 t + 1 2 at 2 (from the perspective of displacement in the y direction). In this case, you are told that this is a horizontal launch problem, for which you immediately realize that initial y velocity is 0 m/s. You are asked to determine with what velocity the rock was initially thrown, meaning what was its horizontal velocity or v x this is an x direction calculation, so you immediately know you are going to need to use x = x 0 + v x t, isolating for v x. After isolating for v x you recognize that you don t have time. Following your strategy, (see THIS IS IMPORTANT discussion for problem 19), as this is a horizontal launch problem you 20

will need to use y = y 0 t + 1 2 at 2 to find time. Isolate and solve for t, and then plug this back into your original x direction formula to solve for v x : 21) A jumper in the long-jump goes into the jump with a speed of 12 m/s at an angle of 20 above the horizontal. How long is the jumper in the air before returning to the Earth? In projectile motion you know that to analyze the motion you have to break the motion into x and y components and analyze each of these separately. You know that motion in the x direction is constant that is, v x is constant. So to analyze motion in the x direction, you can only use the formula that is used for constant velocity: x = x 0 + v x t (although under certain circumstances you can use the range formula to find x displacement). You know that motion in the y direction is accelerated according to the acceleration due to gravity. Therefore, to analyze motion in the y direction you can use either v y = v y0 + at (from the perspective of the initial and final y velocities) or y = y 0 t + 1 2 at 2 (from the perspective of displacement in the y direction). In this case you are told that this is an angled launch and you know you are likely to have to calculate the v x and v y0 to solve the problem, so I would normally go ahead and do this first thing so I have this information available to me (you remember without hesitation that v x = (v)(cosθ ) and v y0 = (v)(sinθ ). (see below for calculations). Then, you are asked to find time. You quickly realize you would not be able to use the x direction calculation isolated for time because you do not have x displacement you will have to use a y direction calculation. Because the launch and landing heights are the same, (see THIS IS IMPORTANT discussion for problem 19), you can use v y = v y0 + at. As v y equals the opposite of v y0, isolating t gives t = 2v y0. Plug in the values for initial y velocity (which a you have already calculated) and a and solve for t. (Note it appears that we calculated v x for no apparent reason. You will almost always need this both velocities for an angled launch problem, so even though we did not here, it s always worth the few seconds of extra time it takes to solve for both at the beginning.) v x = (v)(cosθ) = (12m / s)(cos20) = 11.3m / s v y0 = (v)(sinθ) = (12m / s)(sin20) = 4.10m / s t = 2v y0 a = 2(4.10m / s) 9.8 m / s =.84s 21