Vectors and Two Dimensional Motion

1. Vectors and Two Dimensional Motion Unit 2

2. R = A + B B A Lesson 1 : Some Properties of Vectors Adding Vectors Resultant (R) is drawn from the tail of the first vector to the tip of the last vector

3. A B R B A Commutative Law of Addition When two vectors are added, the sum is independent of the order of the addition. A + B = B + A

4. 35.0 km R 20.0 km Example 1 A car travels 20.0 km due north and then 35.0 km in a direction 60.0o west of north. Find the magnitude and direction of the car’s resultant displacement.

5. A -A Negative of a Vector The vector that when added to A gives zero for the vector sum. A + (-A) = 0 A and –A have the same magnitude but point in opposite directions

6. B A C = A - B -B Subtracting Vectors We define the operation A – B as vector –B added to vector A. A – B = A + (-B)

7. Multiplying a Vector by a Scalar When vector A is multiplied by a positive scalar quantity m, then the product mA is a vector with the same direction of A and magnitude mA. When vector A is multiplied by a negative scalar quantity -m, then the product -mA is a vector directed opposite A and magnitude mA.

8. A Ay q Ax Ax = A cosq Ay = A sinq Lesson 2 : Components of a Vector and Unit Vectors A = Ax + Ay

9. Ax negative Ax positive Ay positive Ay positive Ax negative Ax positive Ay negative Ay negative Signs of the Components Ax and Ay

10. Units vectors specify a given direction in space. (x direction) ^ ^ ^ ^ ^ ^ k j i j k i (y direction) (z direction) Unit Vectors A unit vector is a dimensionless vector having a magnitude of exactly 1.

11. ^ ^ Ax i = Axi x ^ ^ Ay j = Ay j x ^ ^ A = Axi + Ayj y (x,y) ^ ^ r r = x i + y j x Position Vector (r)

12. Given : A B ^ ^ ^ ^ B = Bxi + Byj A = Axi + Ayj ^ ^ ^ ^ R = A + B = (Axi + Ayj ) + ( Bxi + Byj ) ^ ^ R = (Ax + Bx ) i + ( Ay + By ) j Rx = Ax + Bx Ry = Ay + By Vector Addition Using Unit Vectors

13. R = (Ax + Bx)2 + (Ay + By)2 (magnitude) Ry tan q = Rx Ay + By tan q = (direction) Ax + Bx Since R = Rx2 + Ry2

14. y ^ ^ x r = 10 i – 6 j Example 1 Find the magnitude and direction of the position vector below.

15. Given the vectors ^ ^ A = -7 i + 4 j ^ ^ B = 5 i + 9 j Example 2 a) find an expression for the resultant A + B in terms of unit vectors. b) find the magnitude and direction of the resultant.

16. Example 3 A hiker begins a trip by first walking 25.0 km southeast from her car. She stops and sets up her tent for the night. On the second day, she walks 40.0 km in a direction 60.0o north of east. a) Determine the components of the hiker’s displacement for each day. b) Determine the components of the hiker’s resultant displacement (R) for the trip. c) Find an expression for R in terms of unit vectors.

17. Vector x Vector Dot Product Cross Product (scalar product) (vector product) Lesson 3 : Vector Multiplication X

18. To what extent are these two vectors in the same direction ? A q B A . B = AB cosq A q B A cosq Dot Product Answer : Dot Product When vectors are parallel, dot product is a maximum. When vectors are perpendicular, dot product is a minimum.

19. A . A = (Ax2 + Ay2) = A2 A . B = (AxBx + AyBy)

20. Find the angle between the two vectors ^ ^ A = -7 i + 4 j ^ ^ B = -2 i + 9 j Example 1

21. Two vectors r and s lie in the x-y plane. Their magnitudes are 4.50 and 7.30, respectively, and their directions are 320o and 85.0o, respectively, as measured counterclockwise from the +x axis. What is the value of r . s ? Example 2

22. ^ ^ Find the component of A = 5 i + 6 j that lies along the vector B = 4 i – 8 j. ^ ^ Example 3

23. The vector product a x b produces a third vector c whose magnitude is C = AB sinq Cross Product The cross product is maximum when vectors are perpendicular. The cross product is minimum (0) when vectors are parallel.

24. The direction of c is perpendicular to the plane that contains a and b. Direction of the Cross Product

25. 1. Place vectors a and b tail-to-tail. 4. Your outstretched thumb points in the direction of c. Right-Hand Rule 2. Imagine a perpendicular line to their plane where they meet. 3. Pretend to place your right hand around that line so that your fingers sweep a into b through the smaller angle between them.

26. A x B = -B x A In unit-vector notation : ^ ^ ^ ^ ^ ^ A x B = (Axi + Ayj + Azk) x (Bxi + Byj + Bzk) Order of Cross Product is Important Commutative law does not apply to a vector product.

27. Vector A lies in the x-y plane, has a magnitude of 18 units, and points in a direction 250o from the + x axis. Vector B has a magnitude of 12 units and points along the +z axis. What is the vector product c = a x b ? Example 4

28. ^ ^ ^ ^ If A = 3 i – 4 j and B = -2 i + 3 k, what is c = a x b ? Example 5

29. To describe motion in two dimensions precisely, we use the position vector, r. r(t1) r(t2) Dr = r(t2) – r(t1) Dr Lesson 4 : Projectile Motion

30. Dr dr v = lim = Dt dt Dt  0 ^ ^ dr dx dy v = = i + j dt dt dt dv d2r a = = dt dt2 Dr vav = Dt

31. An object is described by the position vector ^ ^ r(t) = (3t3 - 4t) i + (1 – ½ t2) j Find its velocity and acceleration for arbitrary times. Example 1

32. A rabbit runs across a parking lot. The coordinates of the rabbit’s position as functions of time t are given by x = -0.31t2 + 7.2t + 28 y = 0.22t2 – 9.1t + 30 a) Find its velocity v at time t = 15s in unit- vector notation and magnitude-angle notation. b) Find its acceleration a at time t = 15s in unit- vector notation and magnitude- angle notation. Example 2

33. vy v vx X-Direction Constant Velocity Y-Direction Constant Acceleration Analyzing Projectile Motion In projectile motion, the horizontal motion and the vertical motion are independent of each other. Neither motion affects the other.

34. vi viy q vix vix = vi cosq viy = vi sinq Initial x and y Components

35. Upward and toward right is + ay = -g Horizontal Motion Equations Vertical Motion Equations vy = viy - gt vx = vix Dy = ½ (vy + viy) t Dx = vix t Dy = viy t – ½ gt2 vy2 = viy2 – 2 gDy

36. Dx t = vix Dx Dx Dy = viy ( ) – ½ g ( )2 vix vix viy g ( ) (- ) y = x + x2 vix 2vix (equation of a parabola) Proof that Trajectory is a Parabola Dx = vix t Dy = viy t – ½ gt2

37. Maximum Height of a Projectile vi sinq vi sinq vi sinq (at peak) t = g g g 2 ( ) h = (vi sinq) vi2 sin2q - ½ g h = 2g vy = viy - gt 0 = vi sinq - gt (at peak) Dy = viy t – ½ gt2

38. Horizontal Range of a Projectile vi sinq (at peak) t = g 2vi sinq R = vi cosq g vi2 sin 2q R = g Dx = R = vix t R = vi cosq 2t (twice peak time) sin 2q = 2sinqcosq (trig identity)

39. Example 3 A ball rolls off a table 1.0 m high with a speed of 4 m/s. How far from the base of the table does it land ?

40. Example 4 An arrow is shot from a castle wall 10. m high. It leaves the bow with a speed of 40. m/s directed 37o above the horizontal. a) Find the initial velocity components. b) Find the maximum height of the arrow. c) Where does the arrow land ? d) How fast is the arrow moving just before impact ?

41. Example 5 A stone is thrown from the top of a building upward at an angle of 30o to the horizontal with an initial speed of 20.0 m/s. a) If the building is 45.0 m high, how long does it take the stone to reach the ground ? b) What is the speed of the stone just before it strikes the ground ?

42. Case I : The projectile follows the path shown by the curved line in the following diagram. Example 6 : 1985 #1 A projectile is launched from the top of a cliff above level ground. At launch the projectile is 35 m above the base of the cliff and has a velocity of 50 m/s at an angle of 37o with the horizontal. Air resistance is negligible. Consider the following two cases and use g = 10 m/s2, sin 37o = 0.60, and cos 37o = 0.80. a) Calculate the total time from launch until the projectile hits the ground at point C. b) Calculate the horizontal distance R that the projectile travels before it hits the ground. c) Calculate the speed of the projectile at points A, B, and C.

43. Prove that the monkey will hit the dart if the monkey lets go of the branch (free-fall starting from rest) at the instant the dart leaves the gun. Example 7 : The Monkey Gun

44. When the two objects collide : Dx t = xmonkey = xdart and ymonkey = ydart (vi cosq) Dx Dy = (vi sinq) (vi cosq) xdart = (vi cosq)(t) xmonkey = Dx ydart = (vi sinq)(t) – ½ gt2 ymonkey = Dy – ½ gt2 (vi cosq)(t) = Dx ydart = (vi sinq)(t) – ½ gt2 = ymonkey = Dy – ½ gt2 Dy = (vi sinq)(t) = Dx (tanq)

45. v v Lesson 5 : Uniform Circular Motion Object moves in a circular path with constant speed Object is accelerating because velocity vector changes

46. Centripetal Acceleration vi -vi vf Dv vf The direction of Dv is toward the center of the circle Dv = vf - vi

47. Dr Dv = r v (similar triangles) v a Since the magnitude of the velocity is constant, the acceleration vector can only have a component perpendicular to the path.

48. Dv v2 a = ac = Dt r vDr r a = Dt centripetal acceleration (center-seeking) v Dr Dv = r

49. Speed in Uniform Circular Motion v Period (T) time required for one complete revolution 2pr T = v 2pr v = T

50. Velocity changing in direction and magnitude Lesson 6 : Tangential and Radial Acceleration at = tangential acceleration (changes speed) ar = radial acceleration (changes direction)