Chapter 3

Kinematics in Two Dimensions

Vectors

© 2006, B.J. LiebSome figures electronically reproduced by permission of Pearson Education, Inc., Upper Saddle River, New Jersey Giancoli, PHYSICS,6/E © 2004.

AP Physics Motion in two dimensions

Module 5

Vectors

Giancoli, Sec 3-1, 2, 3, 4

Module 5 - 1

Vectors•A vector has magnitude as well as direction.

•Examples: displacement, velocity, acceleration, force, momentum

•A scalar has only magnitude

•Examples: time, mass, temperature

Module 5 - 2

Vector Addition – One Dimension

A person walks 8 km East and then 6 km East.

Displacement = 14 km East

A person walks 8 km East and then 6 km West.

Displacement = 2 km

Module 5 - 3

Vector Addition

21 DDDR

22

21 DDDR

Example 1: A person walks 10 km East and 5.0 km North

kmkmkmDR 2.11)0.5()0.10( 22

RD

D2sin

0121 5.26)2.11

0.5(sin)(sin

km

km

D

D

R

Order doesn’t matter

Module 5 - 4

Graphical Method of Vector AdditionTail to Tip Method

1V

2V

3V

RV

Module 5 - 5

Graphical Method of Vector AdditionTail to Tip Method

1V

2V

3V

RV

1V

2V

3V

Module 5 - 6

Parallelogram Method

Module 5 - 7

Subtraction of Vectors

Negative of vector has same magnitude but points in the opposite direction.

For subtraction, we add the negative vector.

Module 5 - 8

Multiplication by a Scalar

A vector V can be multiplied by a scalar c; the result is a vector cV that has the same direction but a magnitude cV. If c is negative, the resultant vector points in the opposite direction.

Module 5 - 9

Adding Vectors by Components

Any vector can be expressed as the sum of two other vectors, which are called its components. Usually the other vectors are chosen so that they are perpendicular to each other.

Module 5 - 10

Trigonometry Review

Opposite

Adjacent

Hypotenuse

Hypotenuse

Oppositesin

Hypotenuse

Adjacentcos

cos

sin

Adjacent

Oppositetan

Module 5 - 11

Adding Vectors by Components

If the components are perpendicular, they can be found using trigonometric functions.

sinVVy

cosVVx

V

Vy

V

Vx

cos

sin

Adj

Opptan

Hypotenuse

Oppositesin

Hypotenuse

Adjacentcos

Module 5 - 12

Adding Vectors by Components

The components are effectively one-dimensional, so they can be added arithmetically:

Module 5 - 13

Signs of Componentsy

x

y

x

R

R

y

x

R

R

y

x

R

R

y

x

R

R

Module 5 - 14

3-4 Adding Vectors by Components

Adding vectors:

1. Draw a diagram; add the vectors graphically.

2. Choose x and y axes.

3. Resolve each vector into x and y components.

4. Calculate each component using sines and cosines.

5. Add the components in each direction.

6. To find the length and direction of the vector, use:

V

Vysin

Module 6

Vector Problems and Relative Velocity

Giancoli, Sec 3- 4, 8

Giancoli, PHYSICS,6/E © 2004. Electronically reproduced by permission of Pearson Education, Inc., Upper Saddle River, New Jersey

The following is an excellent lecture on this material.

Module 6 - 1

Example 2

BAR

ARB

XXXARB

A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 150 cm and makes a angle of 1200 with the positive x-axis. The resultant displacement has a magnitude of 140 cm and is directed at an angle of 35.00 to the positive x axis. Find the magnitude and direction of the second displacement.

120cos)150(35cos)140( cmcmBx cmBx 190

yyyARB

120sin)150(35sin)140( cmcmBy

cmBy 6.49

cmA 150

cmR 140A

0120 R

035

B

Module 6 - 2

Example 2

BAR

ARB

XXXARB

Alternative Solution. In the solution below, the angles for vector A are measured from the negative x axis. In this case, we have to assign the signs for the components. The answer is the same.

60cos)150(35cos)140( cmcmBx

cmBx 190

yyyARB

60sin)150(35sin)140( cmcmBy

cmBy 6.49

cmA 150

cmR 140R

035

BA

060

Module 6 - 3

Example 2 Continued

22

yxBBB

22 )6.49()190( cmcmB cm196

A man pushing a mop across a floor causes it to undergo two displacements. The first has a magnitude of 150 cm and makes a angle of 1200 with the positive x-axis. The resultant displacement has a magnitude of 140 cm and is directed at an angle of 35.00 to the positive x axis. Find the magnitude and direction of the second displacement.

cm

cm

196

6.49sin

196

6.49sin 1

6.14

Module 6 - 4

Relative Velocity

•Will consider how observations made in different reference frames are related to each other.

A person walks toward the front of a train at 5 km / h (VPT). The train is moving 80 km / h with respect to the ground (VTG). What is the person’s velocity with respect to the ground (VPG)?

TGPTPG VVV

hkmhkmhkmVPG /85/80/5

Module 6 - 5

Relative Velocity

•Boat is aimed upstream so that it will move directly across.

•Boat is aimed directly across, so it will land at a point downstream.

•Can expect similar problems with airplanes.

Module 6 - 6

Example 6 An airplane is capable of flying at 400 mi/h in still air. At what angle should the pilot point the plane in order for it to travel due east, if there is a wind of speed 50.0 mi/h directed due south? What is the speed relative to the ground?

PAV

AGV

PGV

01 18.7400

0.50sin

hmi

hmi

hmi

hmiVPG 397)18.7(cos)400( 0

PA

AG

V

Vsin

PA

AG

V

V1sinNorth of East

Module 6 - 7

Vectors

Vector Addition Applet

Module 7

Projectile Motion

Giancoli, Sec 3-5, 6, 7

Module 7 - 1

3-5 Projectile Motion

A projectile is an object moving in two dimensions under the influence of Earth's gravity; its path is a parabola.

Module 7 - 2

Projectile Motion

•Neglect air resistance

•Consider motion only after release and before it hits

•Analyze the vertical and horizontal components separately (Galileo)

•No acceleration in the horizontal, so velocity is constant

•Acceleration in the vertical is – 9.8 m/s2 due to gravity and thus velocity is not constant.

•Object projected horizontally will reach the ground at the same time as one dropped vertically

Module 7 - 3

Equations for Projectile Motion

00 xvv

tvxx x00

tgvv yy 0

)(2 02

02 yygvv yy

200 2

1tgtvyy y

Horizontal Vertical

ax=0 ay = - g

vx= constant

Module 7 - 4

Initial Velocity

cos00 vvx

•If the ball returns to the y = 0 point, then the velocity at that point will equal the initial velocity.

•At the highest point, v0 y = 0 and v = vx0

sin00 vvy

Module 7 - 5

Example 3A

)0.50)(cos0.18(0

smv

x s

m6.11

)0.50)(sin0.18(0

smv

y

A football is kicked at an angle of 50.00 above the horizontal with a velocity of 18.0 m / s. Calculate the maximum height. Assume that the ball was kicked at ground level and lands at ground level.

sm8.13

tgvv yy 0 0

g

vt y 0

280.9

8.13

sm

sm

s41.1

2

0max 2

1gttvyy

yo

22max )41.1)(8.9(

2

1)41.1)(8.13(0 s

smss

my

my 7.9max

at top:

Module 7 - 6

Level Horizontal Range

g

vt y02

•Range is determined by time it takes for ball to return to ground level or perhaps some other vertical value.

•If ball hits something a fixed distance away, then time is determined by x motion

•If the motion is on a level field, when it hits: y = 0

2200 2

100

2

1tgtvtgtvyy yoy

Solving we find

We can substitute this in the x equation to find the range R

g

v

g

vv

g

vvtvxR yoxy

xox00

2000

0

cossin22)

2(

Module 7 - 7

Level Horizontal Range

g

vR

2sin20

We can use a trig identity 2sincossin2

•Greatest range: = 450

• = 300 and 600 have same range.

Caution– the range formula has limited usefulness. It is only valid when the projectile returns to the same vertical position.

)1545( 00

Module 7 - 8

Example 3B

)41.1)(2( st s82.2

A football is kicked at an angle of 50.00 above the horizontal with a velocity of 18.0 m / s. Calculate the range. Assume that the ball was kicked at ground level and lands at ground level.

Assume time down = time up

Could also use range formula

m

sm

sm

g

vR 33

/8.9

)50()2(sin)/18(2sin2

0220

tvxxRx 00

)82.2)(6.11(0 ssm

m33

For Range:

Module 7 - 9

Example 4A

)0.50)(cos0.18(0

smv

x s

m6.11

)0.50)(sin0.18(0

smv

y

A football is kicked at an angle of 50.00 above the horizontal with a velocity of 18.0 m / s. The football hits a window in a house that is 25.0 m from where it was kicked. How high was the window above the ground.

sm8.13

tvxx 0

0xv

xt

smm

6.11

0.25 s16.2

2

00 2

1gttvyy

y

22 )16.2)(8.9(

2

1)16.2)(8.13(0 s

smss

my

my 9.6

Time to hit the window:

Module 7 - 10

Example 4 B

st 16.2gtvv

yy

0

)16.2()8.9()8.13( 2 ss

ms

mvy

What is the final velocity and angle of the football that hit the window in Example 4 A.

sm37.7

smv

x6.11

22 )37.7()6.11( sm

smv

sm7.13

x

y

v

vtan

6.11

37.7tan 1 4.32 below x axis

Module 7 - 12

Example 5. (35) A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below. If the plane is traveling horizontally with a speed of 250 km /h (69.4 m / s) how far in advance of the recipients (horizontal distance) must the goods be dropped (Fig. 3–37a)? .

smv

v

x

y

/4.69

0

0

0

Coordinate system is 235 m below plane

02

10235 2 tgmy mx

ssmtvxx xo

481

)93.6()/4.69(00

ssm

mt 93.6

/8.9

)235()2(2

Module 7 - 13

Sec. 3-7 Projectile Motion Is ParabolicIn order to demonstrate that projectile motion is parabolic, the book derives y as a function of x. When we do, we find that it has the form:

This is the equation for a parabola.