Vectors and Two Dimensional Motion

Unit 2

Lesson 1 : Some Properties of Vectors

Adding Vectors

A

BR = A + B

Resultant (R) is drawn from the tail of the first vector to the tip of the last

vector

Commutative Law of Addition

When two vectors are added, the sum is independent of the order of the addition.

A + B = B + A

A

B

RB

A

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.

20.0 km

35.0 km

R

Negative of a Vector

The vector that when added to A gives zero for the vector sum.

A + (-A) = 0

A

-A

A and –A have the same magnitude but point in opposite directions

Subtracting Vectors

We define the operation A – B as vector –B added to vector A.

A – B = A + (-B)

AB

-B

C = A - B

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.

Lesson 2 : Components of a Vector and Unit Vectors

A

Ax

Ay

A = Ax + Ay

Ax = A cos

Ay = A sin

Signs of the Components Ax and Ay

Ax positive

Ay positive

Ax positive

Ay negative

Ax negative

Ay positive

Ax negative

Ay negative

Unit Vectors

A unit vector is a dimensionless vector having a magnitude of exactly 1.

Units vectors specify a given direction in space.

i

i (x direction)j

j (y direction)

kk (z direction)

Ax i = Axi x^ ^

Ay j = Ay j x^ ^

A = Ax i + Ay j^ ^

y

x

(x,y)

r

Position Vector (r)

r = x i + y j^ ^

Vector Addition Using Unit Vectors

R = A + B = (Ax i + Ay j ) + ( Bx i + By j )^ ^ ^ ^

R = (Ax + Bx ) i + ( Ay + By ) j^ ^

Rx = Ax + Bx

Ry = Ay + By

Given : A = Ax i + Ay j^ ^

B = Bx i + By j^ ^

AB

Since R = Rx2 + Ry

2

R = (Ax + Bx)2 + (Ay + By)2

(magnitude)

tan = Ry

Rx

tan = Ay + By

Ax + Bx

(direction)

Example 1

Find the magnitude and direction of the position vector below.

r = 10 i – 6 j^ ^x

y

Example 2

b) find the magnitude and direction of the resultant.

Given the vectors

A = -7 i + 4 j

B = 5 i + 9 j

^ ^

^ ^

a) find an expression for the resultant A + B in terms of unit vectors.

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.

Lesson 3 : Vector Multiplication

Vector x Vector

Dot Product(scalar product)

Cross Product(vector product)

X

Dot Product

A

B

To what extent are these two vectors in the same direction ?

Answer : Dot Product

A

B

A cos

When vectors are parallel, dot product is a maximum.

When vectors are perpendicular, dot product is a minimum.

A . B = AB cos

A . B = (AxBx + AyBy)

A . A = (Ax2 + Ay

2) = A2

Example 1

Find the angle between the two vectors

A = -7 i + 4 j

B = -2 i + 9 j

^ ^

^ ^

Example 2

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 3

Find the component of A = 5 i + 6 j

that lies along the vector B = 4 i – 8 j.^ ^

^ ^

Cross Product

The vector product a x b produces a third vector c whose magnitude is

C = AB sin

The cross product is maximum when vectors are perpendicular.

The cross product is minimum (0) when vectors are parallel.

Direction of the Cross Product

The direction of c is perpendicular to the plane that contains a and b.

Right-Hand Rule

1. Place vectors a and b tail-to-tail.

3. Pretend to place your right hand around that line so that your fingers sweep a

into b through the smaller angle between them.

2. Imagine a perpendicular line to their plane where they meet.

4. Your outstretched thumb points in the direction of c.

Order of Cross Product is Important

Commutative law does not apply to a vector product.

A x B = -B x A

In unit-vector notation :

A x B = (Axi + Ayj + Azk) x (Bxi + Byj + Bzk)^ ^ ^ ^ ^ ^

Example 4

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 5

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

^ ^ ^ ^

Lesson 4 : Projectile Motion

To describe motion in two dimensions precisely, we use the position vector, r.

r(t1)r(t2)

r

r = r(t2) – r(t1)

vav =r

t

v =r

tlim =

dr

dtt 0

v =dr

dt=

dx

dt+i

^ dy

dtj^

a =dv

dt=

d2r

dt2

Example 1

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 2

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.

Analyzing Projectile Motion

In projectile motion, the horizontal motion and the vertical motion are independent of

each other. Neither motion affects the other.

vvy

vx

X-Direction Constant Velocity

Y-Direction Constant

Acceleration

Initial x and y Components

v i

vix

viy

vix = vi cos

viy = vi sin

Horizontal Motion Equations

Vertical Motion Equations

x = vix t

vx = vix

vy = viy - gt

y = ½ (vy + viy) t

y = viy t – ½ gt2

vy2 = viy

2 – 2 gy

Upward and toward right is +

ay = -g

Proof that Trajectory is a Parabola

x = vix t

t = xvix

y = viy t – ½ gt2

y = viy ( ) – ½ g ( )2xvix

xvix

y = viy

vix( )x + (-

g

2vix

)x2

(equation of a parabola)

Maximum Height of a Projectile

vy = viy - gt

0 = vi sin - gt (at peak)

t =vi sin

g(at peak)

y = viy t – ½ gt2

( )h = (vi sin)vi sin

g

vi sing

- ½ g

2

h =vi

2 sin2

2g

Horizontal Range of a Projectile

x = R = vix t

R = vi cos 2t (twice peak time)

t =vi sin

g(at peak)

R = vi cos 2vi sing

sin 2 = 2sincos(trig identity)

R =vi

2 sin 2

g

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 ?

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 ?

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 ?

Example 6 : 1985 #1A 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.

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.

a) Calculate the total time from launch until the projectile hits the ground at point C.

Case I : The projectile follows the path shown by the curved line in the following diagram.

Example 7 : The Monkey GunProve 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.

xdart = (vi cos)(t)

ydart = (vi sin)(t) – ½ gt2

xmonkey = x

ymonkey = y – ½ gt2

When the two objects collide :

xmonkey = xdart and ymonkey = ydart

(vi cos)(t) = x

t = x

(vi cos)

ydart = (vi sin)(t) – ½ gt2 = ymonkey = y – ½ gt2

y = (vi sin)(t)

y = (vi sin) x

(vi cos)= x (tan)

Lesson 5 : Uniform Circular Motion

Object moves in a circular path with constant speed

Object is accelerating because velocity vector changes

v

v

Centripetal Acceleration

The direction of v is toward the center of the circle

vf

-vi

v = vf - vi

v

vivf

v

Since the magnitude of the velocity is constant, the acceleration vector can only

have a component perpendicular to the path.

a

v

vr

r=

(similar triangles)

v =v r

r

a = v

t

a =

vr

t

r

ac = v2

r

centripetal acceleration

(center-seeking)

Speed in Uniform Circular Motion

vPeriod (T)

time required for one complete revolution

T =2r

v

v =2r

T

Lesson 6 : Tangential and Radial Acceleration

Velocity changing in direction and magnitude

at = tangential acceleration (changes speed)

ar = radial acceleration (changes direction)

at =dv

dt

ar =v2

r

perpendicular components

of a ar

ata

a = (ar)2 + (at )2 (magnitude of a)

a =dv

dt-

^ v2

rr

(total acceleration)

a = at + ar

Example 1The diagram below represents the total acceleration of a particle moving clockwise in a circle of radius 2.50 m at a

certain instant of time. At this instant, find

a) the radial acceleration

b) the speed of the particle

c) the tangential acceleration

Example 2

A ball swings in a vertical circle at the end of a rope 1.5 m long. When the ball is 36.9o

past the lowest point on its way up, its total acceleration is (-22.5 i + 20.2 j) m/s2.

At that instant,

^ ^

a) sketch a vector diagram showing the components of its acceleration

b) determine the magnitude of its radial acceleration

c) determine the speed and velocity of the ball

Example 3

A boy whirls a stone on a horizontal circle of radius 1.5 m and at height 2.0 m above level ground. The string breaks, and the

stone flies off horizontally and strikes the ground after traveling a horizontal

distance of 10. m. What is the magnitude of the centripetal acceleration of the stone

during the circular motion ?