AP Physics Chapter 3 Motion in Two Dimensions

1. AP Physics Chapter 3Motion in Two Dimensions

2. Chapter 3: Motion in Two Dimensions 3.1 Components of Motion 3.2 Vector Addition and Subtraction 3.3 Relative Velocity 3.4 Projectile Motion

3. Learning Objectives for Chapter 3 • Students should be able to add, subtract, and resolve displacement and velocity vectors so they can: 1. Determine components of a vector along two specified, mutually perpendicular axes. (x and y) 2. Determine the net displacement of a particle or the location of a particle relative to another. 3. Determine the change in velocity of a particle or the velocity of one particle relative to another. • Students should understand the motion of projectiles in a uniform gravitational field so they can: (freefall) 1. Write down expressions for the horizontal and vertical components of velocity and position as functions of time, and sketch or identify graphs of these components. X: zero acceleration, constant velocity; y: constant acceleration downward

4. Homework for Chapter 3 • Read Chapter 3 • Vector Component WS 3.A • Relative Velocity WS 3.B • Projectile Motion HW 3: AP Classroom

5. 3.1 Components of Motion Vectors are drawn as arrows on the coordinate plane. - the length of the arrow corresponds to the magnitude - the vector points to its direction N Vector A has a magnitude of 3 meters and a direction of 60°above the positive x-axis. W E y 90° A S 3.0 m 3.0 m B Vector B has a magnitude of 3 meters and a direction of β° above the negative x-axis. 60° β x 0° 180° 1.5 m C 270° Vector C has a magnitude of 1.5 meters and a direction of 270°.

6. 3.1 Components of Motion • To analyze motion, a vector can be broken down into x- and y- components. y Given a vector v with a magnitude of a directed an angle Ө above the horizontal vx = a cos Ө vy = a sin Ө 5 units y- component Ө = 53° x example: Find the components of v if its magnitude is 5 units and Ө = 53°. vx = a cos Ө = 5 cos 53° = 3 units vy = a sin Ө = 5 sin 53° = 4 units x- component Note: The magnitude of the vector v is v = √ vx2 + vy2Why?

7. 3.1 Components of Motion • Kinematics problems in two-dimensions can be solved by: • resolving the displacement, velocity, and acceleration vectors into their respective components. • using the three constant acceleration equations separately for the x and y directions. • using the Pythagorean theorem to find the magnitude of your resultant vector. • using an inverse trig function to find the angle of your resultant vector

8. 3.1 Components of Motion Example 1: A boat travels with a speed of 5.0 m/s in a straight path on a still lake. Suddenly, a steady wind pushes the boat perpendicularly to its straight line path with a speed of 3.0 m/s for 5.0 s. Relative to its position just when the wind started to blow, where is the boat at the end of this time? Solution: List givens for x and y directions separately. x-direction y-direction Unknowns: x and y vo = 5.0 m/s vo = 3.0 m/s a = 0 a = 0 Common to both: t = 5.0 s Analysis: Both motions are motion with constant velocity. Choose the straight path of the boat as the x axis and the direction of the wind as the y axis. Sketch: wind y d x Ө boat

9. x-direction y-direction Unknowns: x and y vo = 5.0 m/s vo = 3.0 m/s a = 0 a = 0 xo = 0 yo = 0 Common to both: t = 5.0 s x-component: (use equation 2) x = vot + ½ at2 = (5.0 m/s)(5.0 s) + 0 = 25 m y-component: (use equation 2) y = vot + ½ at2 = (3.0 m/s)(5.0 s) + 0 = 15 m Now, d = √ x2 + y2 = √ (25 m)2 + (15 m)2 = 29 m And Ө = tan-1 y = 15m = 31° x 25 m wind y d x Ө boat

10. 3.2 Vector Addition and Subtraction vector addition – combining vector quantities by using one of several techniques • geometric methods (triangle, parallelogram, polygon) • vector components and the analytical component method vector subtraction – a special case of vector addition • A - B = A + (-B) • subtracting a vector is the same as adding a negative vector resultant – the overall effect of combining vectors; the vector sum Example: Add vectors A + B + C = R Solution: Use tip-to-tail method

11. SCALE: 1 cm = 5 m 3.2 Vector Addition and Subtraction • Analytical (Trigonometry) Method • When vectors make a right triangle with each other, • use the Pythagorean theorem to find the magnitude of your resultant vector. • use the inverse trig function to find the angle of your resultant vector. A = side adjacent to angle Ө O = side opposite to angle ӨSOH – CAH - TOA H = hypotenuse of triangle sin Ө = O cos Ө = A tan Ө = O H H A Ө = sin-1OӨ = cos-1AӨ = tan-1O H H A H O Ө A

12. 3.2 Vector Addition and Subtraction Example: Eric leaves the base camp and hikes 11 km, north and then hikes 11 km east. Determine Eric's resulting displacement and express in the magnitude-angle form. Solution:  Magnitude of displacement = 16 km

13. 3.2 Vector Addition and Subtraction

14. 3.2 Vector Addition and Subtraction

15. Please Do Now In two minutes, write as many examples of vectors that you can..

16. 3.2 Vector Addition and Subtraction • Analytical Component Method • Resolve the vectors to be added into their x and y components. Include directional signs (plus or minus) in the components. • 2. Add, algebraically, all the x- components together and all the y- components together to get the x and y components of the resultant vector. • 3. Express the resulting vector using • The component form using unit vectors , e.g. A = Axx + Ayyor • In the magnitude-angle form, e.g., A = √Ax2 + Ay2, Ө = tan-1Ay • Ax unit vector – has a magnitude of one and no units. It indicates a vector’s direction only. ex: R = (11 km) x + (11 km) y 11 km, E R 11 km, N

17. Please Do Now In 1 minute, write as many examples of vetors that you can.

18. WS 3.A Vector Components

19. 3.3 Relative Velocity • “Everything is relative…” • Physical phenomena can be observed from different frames of reference. ex: The velocity of a ball tossed by a passenger in a moving car will be measured differently by a passenger on the car than by an observer on Earth. • Your reference is the origin, or zero point, of your coordinate system. • Any velocity we measure is relative. ex: The velocity of a moving car is measured relative to the ground. ex: The revolving motion of the Earth around the Sun is relative to the Sun. ex: moving sidewalk in an airport • Relative velocity can be determined with vector addition or subtraction. • Standard symbols are used to represent relative velocity. ex: vcg (where c stands for car and g stand for ground) means the velocity of a car relative to the ground.

20. 3.3 Relative Velocity • When velocities are along a straight line in the same or opposite direction and all have the same reference, we can find relative velocities by vector subtraction. vAB = vA – vB vAB = - vBA vAC = vAB + vBC Mythbusters Example: b = ball, t = truck, g = ground velocity of truck relative to ground is vtg velocity of ball relative to truck is vbt What is the velocity of the ball with respect to the ground? vbg = ? vbg = vbt + vtg = 30 – 30 = 0 m/s

21. 3.3 Relative Velocity • Example: A student walks on a treadmill moving at 4.0 m/s and remains at the same place in the gym. • What is the student’s velocity relative to the gym floor? • zero • b) What is the student’s speed relative to the treadmill? • s = student, t = treadmill, and g = ground • Unknown: vstKnowns: vsg = 0, vtg = -4.0 m/s. • vsg= vst + vtg→ vst = vsg – vtg = 0 – (- 4.0 m/s) = 4.0 m/s • c) What is the treadmill’s speed relative to the student? • vts = -vst= - 4.0 m/s

22. 3.3 Relative Velocity Practice Problem 1: A train travels at 60 m/s to the east with respect to the ground. A businessman on the train runs at 5 m/s to the west with respect to the train. Find the velocity of the man with respect to the ground. t = train m = man g = ground vtg = vmt = vmg = vmt + vtg

23. 3.3 Relative Velocity Practice Problem 2: An airplane flies at 250 m/s to the east with respect to the air. The air is moving at 15 m/s to the east with respect to the ground. Find the velocity of the airplane with respect to the ground. p = plane a = air g = ground vpa = vag = vpg = vpa + vag

24. 3.3 Relative Velocity Practice Problem 3: An airplane flies at 250 m/s to the east with respect to the air. The air is moving at 35 m/s to the north with respect to the ground. Find the velocity of the air with respect to the ground. p = plane a = air g = ground vpa = vag = vpg = vpa + vag

25. WS 3.BRelative Velocity

26. 3.4 Projectile Motion projectile – an object upon which the only force is gravity. The key to projectile motion problems is recognizing the vertical and horizontal components of motion are independent. • The vertical motion, or y- component is free fall, so a = -9.80 m/s2 • The horizontal motion, or x- component, has zero acceleration. Why? • Time is the common factor, and can be use to link the two equations. Use kinematics equations 1 & 2 for the y- direction: vy = vyo + a t y = yo + vyo t + ½ ay t2 Use the same equations for the x- direction: (remember, a is zero) vx = vxo x = vxo t

27. 3.4 Projectile Motion x- component of initial velocity vxo = vo cos Ө y- component of initial velocity vyo = vo sin Ө

28. Example: A package is dropped from an airplane traveling with a constant horizontal speed of 120 m/s at an altitude of 500 m. What is the horizontal distance the package travels before hitting the ground (range)? Solution: Given: horizontal motion vertical motion vxo = 120 m/s vyo = 0 y = -500 m Find: range Since the range is given by x = vxo t , we have to find the time of flight t first. From the vertical motion, we use y = vyot + ½ ayt2 So, -500 m = 0 + ½ (-9.80 m/s2) t2, solving, t = 10.1 s Therefore, x = (120 m/s)(10.1 s) = 1.21 x 103 m = 1.21 km Hint: The quantities such as initial velocities and displacements have to be treated independently. For example, the initial horizontal velocity is 120 m/s and the initial vertical velocity is zero. The 120 m/s can only be used in the horizontal motion and the 0 m/s can only be used in the vertical motion. A common mistake is to mix up these quantities or not treat them as independent. x vo = vxo y 500 m range

29. Example: A golfer hits a golf ball with a velocity of 35 m/s at an angle of 25° above the horizontal. If the point where the ball is hit and the point where the ball lands are at the same level, a) how long is the ball in the air? b) what is the range of the ball? • Solution: • Given: horizontal motion vertical motion • vox = vo cos Өvoy = vosin Ө • = (35 m/s) cos 25° = (35 m/s) sin 25° • = 31.7 m/s = 14.8 m/s • Find: a) t b) x • On landing, y = 0; and from y = voyt + ½ at2, we have • 0 = (14.8 m/s)t + 1/2(-9.80 m/s2) t2. Solve using the quadratic equation: t = 0 or 3.02 s. • The landing position is the second root, so the flight time is t = 3.0 s. y vo vyo Ө x vxo range

30. Example: A golfer hits a golf ball with a velocity of 35 m/s at an angle of 25° above the horizontal. If the point where the ball is hit and the point where the ball lands are at the same level, a) how long is the ball in the air? b) what is the range of the ball? • Solution: • Given: horizontal motion vertical motion • vxo = vocosӨvyo = vosin Ө • = (35 m/s) cos 25° = (35 m/s) sin 25° • = 31.7 m/s = 14.8 m/s • Solution which avoidsusing the quadratic equation: • Find the time to the peak at the half way point (when vy = 0) and then double that time. Use kinematics equation #1: • vy = voy + at, so we have • 0 = 14.8 m/s + (-9.80 m/s2) t • t= 3.0 s. • b) x = vxot = (31.7 m/s)(3.02 s) = 96 m y vo vyo Ө x vxo range

31. Check for Understanding:

32. vy = vyo + at ; vy = 0 at peak. x= vxo t

33. Practice Problems from Textbook p.95: 71, 70. p. 96:77. • HW 3: AP Classroom