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Vectors & 2D MotionVectors & 2D Motion

Mr. Finn

Honors Physics

Slide 2

OverviewOverview

1. Vectors– What are they– Operations

• Addition• Subtraction

2. Relative Velocity– boats and planes

3. Projectile Motion– Free fall + relative motion

Slide 3

1. Vectors1. Vectors

• What are they?– How to represent them?

• Math Operations– Graphical Vector Addition– Analytical Vector Addition

• cartesian coordinates

• trig: laws of sine/cosine

Slide 4

What are What are vectorsvectors

• Two types of physical quantities– ScalarsScalars: quantities with no “sense” of direction example: volume, temperature

– VectorsVectors: quantities with a direction example: displacement, force

• How to represent vectors– 1D: direction = +/- sign as forward/backward– 2D: direction = angle from reference line OR

two components

Slide 5

Representing VectorsRepresenting Vectors

• Graphically

• Analytically– Cartesian: (x, y)– Polar: (r, )

angle tail

head(“tip”)

magnit

ude r

€

rA = A(Textbook = bold)

x = r cos y = r sin

r = (x2 + y2)1/2

= tan-1(y/x)

x

y

Slide 6

PracticePractice

1. Break the following into componentsa) 12.0 m at 35º

b) 42 m/s at 160º

2. Find magnitude/directiona) (-2.4 m, +4.8 m)

b) (+500 m, -350 m)

Slide 7

Compass HeadingsCompass Headings

North

South

EastWest

60° North of East

22.5 West of North

Slide 8

Mathematical OperationsMathematical Operations

• Scalar arithmetic operations: +///• Vector arithmetic operations

– must be different two numbers vs. one

– vector addition = adding two arrows• cannot add magnitudes so “1 + 1 2”

• unit vectors at right angles = sum of 2

• Redefine arithmetic operation = addition– by drawing pictures

– by doing calculations

Slide 9

Graphical AdditionGraphical Addition“Tip-to-Tail” Method“Tip-to-Tail” Method

€

rA

€

rB

Reposition vectors so they are aligned “tip-to-tail”

€

rB

Sum of A and B is arrow whose tail starts at tail of A and whose head ends at tip of B.

€

rA +

r B

Slide 10

Graphical AdditionGraphical Addition“Parallelogram” Method“Parallelogram” Method

€

rA

€

rB

Reposition vectors so they are aligned “tail-to-tail”

Sum of A and B is diagonal of parallelogram from tail to head. Difference is other diagonal

€

rB

Complete parallelogram with two vectors as two adjacent sides

€

rA +

r B

Slide 11

PracticePractice

70º

2.0 m

4.0 m

AA

BB

Find: A + B (graphically)

Slide 12

SolutionSolution

length = 6.2 m at 22º

Slide 13

Analytical AdditionAnalytical Addition

• Break each vector into its components– A = (Ax, Ay)

– B = (Bx, By)

• Add corresponding components to get components of sum– A+B = (Ax+Bx, Ay+By)

• Convert to polar coordinates, if needed

Slide 14

Vector SubtractionVector Subtraction

• Inverse of vector A = -A– switch direction

• Subtract by adding inverse– A - B = A + (-B)– Subtraction is treated same as Addition

€

rA

€

− r

A

Slide 15

PracticePractice

• Add the following and find magnitude, direction:– A: (4.5 m, -8.2 m)– B: (5.3 m, -3.1 m)

• Subtract the preceding (A - B) and find magnitude, direction

Slide 16

2. Relative Velocity2. Relative VelocityApplication of Vector AdditionApplication of Vector Addition

• Examples of adding vectors– used displacement vectors “Race across

Desert”– now use velocity vectors file flight plan

• All motion is relative to FoR - Galileo– boat sailing across water– plane flying through air

media defines one FoR to

describe motionimplied second FoR is ground

relate two motions because 2 FoR are

moving relative to each other

Slide 17

Relative Velocity EquationRelative Velocity Equation

• Let: – A = air (water); G = ground; P = plane (boat)– / = “relative to”

• Relation among relative velocities:

€

rv P /G =

r v P / A +

r v A /G

plane relative to gnd(AKA ground speed)

plane relative to air(AKA air speed)air relative to gnd

(AKA wind speed)

Slide 18

Relative Velocity “Triangle”Relative Velocity “Triangle”

Destination

€

rv P / A

Represent equation graphically - as a triangle

€

rv A /G

€

rv P /G

Air = FoR?

Gnd = FoR

Slide 19

3. Projectile Motion3. Projectile Motion

• Acceleration = gravity, down– ax = 0

– ay = -9.8 m/s2

• Initial velocity at angle to vertical– vox = vo cos = constant

– voy = vo sin

vo

Free Fall in 2D or as seen in moving FoR

Free Fall in 2D or as seen in moving FoR

Slide 20

Kinematics VariablesKinematics Variables

• Position = Cartesian coordinates – x = xo + vox t– y = yo + voy t - 1/2 g t2

• Velocity– vx = vox = constantconstant

– vy = voy - gt

• Acceleration– ay = -g = constantconstant

Horizontal-vertical motions are independent

Horizontal-vertical motions are independent

Slide 21

Direction of AccelerationDirection of Acceleration

€

rv

€

ra

€

ra

€

rv

€

ra

€

rv

Slide 22

Relative Motion & ProjectilesRelative Motion & Projectiles

• Projectile motion– free fall seen from a moving frame of reference

• vertical motion = constant acceleration down

• horizontal = constant velocity

Slide 23

Motion – Ground FoR

Wagon moves relative to ground

Slide 24

Motion – Wagon FoR

Wagon moves relative to ground – so tree “moves” toward wagon.

Slide 25

A ball rolls off a level /horizontal table with an initial velocity of 3.0 m/s. The table is 0.50 m high. Where does the ball strike the ground - i.e., how far from the end of the table?

Example:Example:

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Numerical SolutionNumerical Solution

• time to fall from table to floor– horizontal-vertical motion independent

– h = 1/2 gt2 where h = 0.50 m, g = 9.8 m/s2

– t = 0.3194 s (avoid rounding intermediate results)

• horizontal distance moved during fall– x = vox t where vox = 3.0 m/s (table horizontal)

– x = 0.9583 m 0.96 m (proper SigFig in answer)

Slide 27

Baseball ProblemBaseball Problem

• Baseball is hit so that it leaves the bat at– initial speed of 45 m/s– angle of 30° above horizontal

• Find:– time in air– distance traveled (hit ground)

• Assume– no air resistance (?!)– ignore initial height (start on ground)

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Solution - Range EquationSolution - Range Equation

• Time in air - vertical velocity– voy = vo sin

– vtop = 0 0 = vo sin - gt

• t = vo sin / g

– T = total time in air = 2t = 2 vo sin / g• T = 4.59 s

Slide 29

Solution - continuedSolution - continued

• Distance traveled = “range”– R = x = xi + vix T = 0 + (vo cos )T

• But T = 2 vo sin / g

– R = (vo2/g)(2 cos sin )

• trig identity for sin 2 = 2 cos sin

– R = vo2/g sin 2

• R = 179 m

Slide 30

SummarySummary

• For projectiles– start and end on ground

• Time in air– T = 2 vo sin / g

• Range or distance traveled– R = vo

2/g sin 2

– ONLYONLY if xo = x = 0

R

Rocket ChallengeRocket Challenge

Slide 31

Harder ProblemHarder Problem

• Shoot cannon from castle wall– height of 20 m– velocity of 50 m/s– angle of 20° above horizontal

• Find– time in air– where hit ground below

• assume ground to be level

Do not use “Range” equation!!Do not use “Range” equation!!

4.41 s4.41 s

207 m207 m

Slide 32

SummarySummary

KinematicsKinematics ProjectileProjectile

LinearLinear

CircularCircular

a = constanta = constant(free fall a = -g)(free fall a = -g)

aaxx = 0 = 0aayy = -g = -g

direction = +/direction = +/

direction = vector, angledirection = vector, angle

Kinematics of Uniform Kinematics of Uniform AccelerationAcceleration

dot or scalar productdot or scalar product

Projectile Motion = free fall plus relative motionProjectile Motion = free fall plus relative motion

Next Key Question: What causes acceleration or the motion of an object to change?

€

rx =

r x i +

r v it + 1

2

r a t 2

r v =

r v i +

r a t

v 2 = v i2 + 2

r a ⋅Δ

r x