Physics Chapter 3 Notes. Section 3-1: Introduction to Vectors (pages 80-83)

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1 Physics Chapter 3 Notes Section 3-1: Introduction to Vectors (pages 80-83) We can use vectors to indicate both the magnitude of a quantity, and the direction. Vectors are often used in 2- dimensional problems. Vectors are represented by arrows: 36m/s 14m Scalar: a quantity that ha magnitude but no direction (something like distance or speed) Vector: a quantity that has magnitude and direction (velocity, acceleration, or displacement). A resultant is the sum of two or more vectors. We can add vectors together by combining them tip- to- tail and finding the resultant Section 3-2: Vector Operations (pages 84-92) If the two vectors that we are using make a right angle triangle, we can calculate the resultant vector by using Pythagorean Theorem. For example: If Mary walks 12m to the east and then 7.0m to the north, what is Mary s displacement? First step: make and label a vector diagram 7.0m 12m Second step: Use Pythagorean Theorem to solve for the resultant a 2 + b 2 = c 2 (12m) 2 + (7.0m) 2 = c 2 193m 2 = c m = c 14m = c

2 Now that we have the resultant, we have an idea as to what the actual displacement was of Mary. However, without a direction of Mary s displacement, it is difficult to describe to someone which way she walked. To improve this, we can also calculate an angle by using a trigonometry function. Most times, our angle will be found using the inverse function of tan. We know that: tanθ= opposite/adjacent So if we know the opposite and adjacent side of the triangle, to solve for the angle, we will use: tan - 1 (opp/adj) = θ If we do this for problem on the last page for Mary s displacement, we would do: tan - 1 (opp/adj) = θ tan - 1 (7.0m/12.0m) = θ = θ 30. = θ Examples to try on your own: 1) While following the directions on a treasure map, a pirate walks 45.0m north and then turns and walks 7.5m east. What single straight- line displacement could the pirate have taken to reach the treasure? What angle would he have had to walk? (Answer: 46m, angle of 9.5 ) 2) Find the components of the velocity of a helicopter travelling 95km/h at an angle of 35 to the ground. (Answer: x- component: 78km/h and y- component: 54km/h) Vectors that are not perpendicular Vectors that are not perpendicular cannot be done as simply as those that are. Since the vectors are not perpendicular, a 90 angle does not exist for us to use. We must follow five steps to solve these problems. Also, we will notice that the angle is now given with directions behind it such as North of East or South of West. There is a diagram on the next page to help you figure out which angle you are dealing with.

3 N W of N E of N N of W N of E W S of W S of E E W of S E of S S Step 1: Draw a vector diagram with the two resultants you have been given. Step 2: Use the angle and the magnitude of the resultant vector, along with sine and cosine to find the x- and y- components of the resultant vector (you should now have two x- values and two y- values) Step 3: Add the x- components together and add the y- components together. Remember, if an x- component is to the west, it will be negative, as will a y- component to the south. Step 4: Draw a new vector diagram with your x- and y- component. Connect the two with a resultant vector. Step 5: Use Pythagorean Theorem to calculate the magnitude of the resultant vector. You must then use the tangent function to calculate the angle. Don t forget to record the direction of the angle (north of east, west of south, etc.). **Refer to page 91 for an example of this or your notes for one of the examples that we have done in class.

4 Section 3 - Projectile Motion Projectile Motion: - The motion of objects that are thrown or launched into the air are subject to gravity and called projectiles. - Unless thrown straight up, projectiles follow parabolic trajectories (a path in the shape of a parabola). Think about an Olympic running long- jumper. When they arrive at the jumping point and push off of the ground (or jump), they will have both an upward (vy) velocity and a forward velocity (vx). At some point, when the maximum height is reached, the jumpers vy will become a negative (downward), while vx will continue to be in the forward direction. Essentially, we are combining horizontal motion (which we did at the beginning of the year) with the motion of a falling object (which we just finished). Although it is tough to do sometimes, you have to look at the motion of projectiles in terms of the x- and y- components of the motion. Figure 3.4 from your textbook on page 95 shows us the path of two table- tennis balls. One that is dropped straight down and one that is given an initial horizontal velocity. What you will notice is that both balls will hit the ground at the exact same time (ignoring air resistance and if done in a vacuum). This is the key to being able to calculate projectile motion questions in a simplistic nature. The formulas you will need are as follows: a. For the vertical component of the motion (provided the object is not thrown downwards or upwards): b. For the horizontal component of the motion

5 Example: The Royal Gorge Bridge in Colorado rises 321m above the Arkansas River. Suppose you kick a rock horizontally off the bridge. The magnitude of the rock s horizontal displacement in 45.0m. Find the speed at which the rock was kicked. Step 1: Figure out what information you can obtain from the question and draw a diagram Step 2: Select the appropriate formula In our case, we are looking for the horizontal velocity of the rock. The horizontal velocity is going to remain constant through the entire motion of the rock as there is no air resistance. We can use the formula: Δx=vxΔt Step 3: Since we do not have our change in time for the horizontal motion of the rock, we can calculate the change in time from the vertical motion of the rock. (The horizontal and vertical times have to be identical since when the rock hits the ground in terms of the y- component, it also hits the ground in terms of the x- component) We can use the formula: t =!!!, where a = m/s2 Step 4: Take the formula for t and insert it into the equation from step 2. Insert the numbers from the question and solve. Final Answer: 5.56m/s 2. There are practice questions on page 97. The answers to each question are as follows: 1) 0.66 m/s 2) 4.9 m/s 3) 7.6 m/s 4) 5.6 m