herriman high honors physics chapter 3 kinematics in two dimensions vector addition and projectile...

Herriman High Honors Physics

Chapter 3Kinematics in Two

Dimensions

Vector Addition and Projectile Motion


Vectors and Scalars

Vector vs. Scalars Vectors have both magnitude and

direction Displacement, Velocity, Acceleration,

Force, and Momentum Scalars have only magnitude

Mass, Time, and Temperature


Vector Addition Two Methods

Graphical Method Requires a ruler and a protractor Process

Convert each vector into a line that fits a scale of your choosing

Draw Vectors head to tail, measuring the angle exactly

Draw a resultant Measure the resultant and the resulting angle Convert the measurement back into a vector

answer


Graphical Addition

A man drives 125 km west and then turns 45º North of West and travels an additional 100 km. What is his displacement?

Step one: Set Scale

1 cm = 25 Km

125 km = 5 cm

100 km = 4 cm

45°


Graphical Addition

Step Two: Move Vectors Head to Tail

Step Three: Draw Resultant from Tail of First Vector to the head of the last.

125 km = 5 cm

100 km = 4 cm

45°


Graphical Addition

Step Four: Measure the Measure the Resultant and the resulting angle.

Step Five: Using the Scale, convert the measurement into an answer.

125 km = 5 cm

100 km = 4 cm

45°

8.3 cm 8.3 cm * 25 Km/cm = 208

Km

19.9°

Final Answer: His displacement is 208 Km @ 19.9° North of West


Mathematical Addition Mathematical Addition of Vectors

Requires a basic knowledge of Geometry – You must know: Pythagorean Theorem a2 + b2 = c2

Sin θ = Opposite/Hypotenuse Cos θ = Adjacent/Hypotenuse Tan θ = Sin θ/ Cos θ

=Opposite/Adjacent ArcSin, ArcCos, ArcTan


Mathematical AdditionsVectors at Right Angles

(Easiest)

A plane flies due north at 200 m/s while the wind is blowing due west at 25 m/s. What is the resulting velocity of the plane with respect to the ground?

To Solve:1. Draw a rough sketch of the

original vectors2. Draw a parallelogram 3. Draw the Resultant4. Use pythagorean theorem to

find the magnitude of the velocity

5. Use Arctan to find the direction

1

2 & 3

200 m/s

25 m/s

200 m/s

25 m/s

4 & 5 55.20125200 22 cTanθ=(200/25)

θ=82.9° N of W

θ

Try: P. 89Practice A

Problems 2 & 4


Vectors at Right AnglesSample Problems

An archaeologist climbs the Great Pyramid in Giza, Egypt. The pyramid’s height is 136 meters and its with is 230 meters. What is the magnitude and direction of the displacement of the archaeologist after she has climbed from the bottom of the pyramid to the top?

Answer: Δx = 178 meters @ 49.8° with respect to the ground.

Try: P. 89Practice A

Problems 1 & 3


Resolving Vectors into Components

The opposite of adding to vectors at right angles is resolving a single vector into two components at right angles.

This process uses the trig functions Sin and Cosine

For example, you have a displacement vector of 100 Km @ 30° N of W and you want to know how far north and how far west you are of your starting point.

Solution1. Draw original Vector2. Draw Parallelogram3. Label the sides North and

West4. Use the definition of Sine and

Cosine to solve

Cos θ = W/100 km

W = 100 Cos 45° =70.7 Km

Sin θ = N/100 km

N = 100 Sin 45° =70.7 Km

45°N

W

Try: Practice B, P. 92

Problems 2 & 4


Putting it all TogetherMathematical Addition when Vectors

are Not at Right Angles

Draw a rough sketch of the original vectors

Resolve each vector into its Components using Sine and Cosine.

Reduce the component vectors into a system of two vectors by adding horizontal and vertical vectors

Draw a parallelogram Draw the Resultant Use Pythagorean Theorem to find the

magnitude, and tangent to find the direction of the resultant


Mathematical AdditionStep One: Draw Original Vectors

125 Km West 100 Km 45°

North of West

45º


Mathematical AdditionStep Two: Resolve Vectors into components

125 Km West 100 Km 45°

North of West

Cos θ = W/100 km

W = 100 Cos 45° =70.7 Km

Sin θ = N/100 km

N = 100 Sin 45° =70.7 Km


Mathematical AdditionStep Three: Reduce to a system of two vectors

125 Km West

70.7 Km North

70.7 Km West+ =195.7 Km West

70.7 Km North


Mathematical Addition Step 4: Draw Parallelogram Step 5: Draw Resultant Step 6:Use Pythagorean

Theorem to find magnitude and tangent to find the direction of the resultant.

Pythagorean Theorem: C2 = A2 + B2

C = SQRT(A2 + B2) = SQRT{(195.7)2+(70.7)2} =

208 km

Tan θ = 70.7/197.5 = 0.3579

Θ= ArcTan 0.3579 = 19.9° North or West

A

B

TRY: P. 94

Practice C

Problems 1 & 3


Graphical vs. Mathematical Methods

Graphical Method Less Complicated Math Good Approximation Not as exact as you would like

Mathematical Method No need for ruler or protractor Less time dedicated to sketch More Accurate result Mathematics is more difficult


Projectile Motion A projectile is any object thrown or

launched Uses the same kinematic equations

you have already learned Requires that you use two sets of

equations; one in the horizontal and one in the vertical

These two sets are related only by time


Projectile Motion(Launched or Thrown

Horizontally)

A projectile which is dropped or thrown horizontally has an initial velocity in the Y direction (V0y) = 0

Acceleration in the horizontal direction (ax) = 0; hence Velocity in the horizontal direction (Vx) is constant

This means that only one kinematic equation applies in the horizontal direction:

Xx = Vxt


Projectile Motion(Launched or Thrown Horizontally)

A ball is thrown horizontally off a building which is 200 m high with a velocity of 10 m/s

(a) How long does it take to reach the ground?

(b) How far from the building will it land?

(c) What is its velocity just before it hits the ground (remember magnitude and direction)


Solution X Direction

Vx= 10 m/s The only Equation is:

Xx=VxT

b) Solve for displacement in the X direction

Xx=(10 m/s)(6.4 s) = 64 m

Y Direction Viy= 0 m/s Ay = 9.8 m/s2

Xy= 200 m

A) Find TimeXy= ViyT + ½ayT2

200 m = 0(T)+½(9.8 m/s2)T2 4.64.9

200T s

C) 1. Find final velocity in the Y direction

Vfy = aT = (9.8 m/s2)(6.4 s) = 62.7 m/s

C) 2. Combine X & Y Components to get final velocity magnitude

and direction

49.637.6210 22 CTan θ=62.7/10 =6.27

θ= 81° above horizontalTry: p. 99, Practice D

Problems 1& 3


Projectile Motion(Launched at an Angle)

Similar to Launching Horizontally however:

Initial Velocity in the Y direction is Not zero.

You must break the initial velocity into components using Sine and Cosine

Final velocity, if the projectile falls to the same level, will have the same magnitude and direction as at the launch (assuming no air resistance).


Projectile Motion(Launched at an Angle)

An arrow is fired into the air with a velocity of 25 m/s at 30º above the horizontal.

(a) How high will it rise?(b) How long does it take to reach

the ground?(c) How far from the building will it

land?


Solution

X Direction

Xx= VxT

Xx= (21.7 m/s)(1.28 sec) = 27.68 m

Y Direction Viy= (25 m/s)(sin 30°)

=12.5 m/s Ay = -9.8 m/s2

Vfy = 0 m/s

Vfy2 = Viy

2 + 2AyXy

X= -(12.5 m/s)2/(2)(-9.8 m/s2) = 9.97 m

Vfy= Viy + AT T = -(12.5 m/s)/-9.8 m/s2

= 1.28 seconds

Vix= (25 m/s)(cos 30°) = 21.7 m/s

Try: P. 101

Practice E

Problems 2 & 4


Frame of ReferenceRelative Velocity

Velocity Measurements differ in different frames of reference.

For example if a dummy is dropped from a low flying airplane a person in the plane would have a different frame of reference than an observer watching from the ground.

The observer in the plane, because they are moving at the same horizontal speed as the dummy, would see the dummy fall straight down.

The observer on the ground, because they are standing still, would see the dummy following parabolic trajectory, moving forward while falling.


Frame of Reference(Moving along the same path)

Two cars are traveling in the same direction one at 60 km/h and the other at 90 km/h. How fast is the faster car moving with respect to the slower one? Vse= 60 km/h (se = slower w/respect to earth) Vfe = 90 km/h (fe = faster w/respect to earth) Vfs = (fs = faster w/respect to slower) Solve

Vfs = Vfe- Vse = 90 km/h – 60 km/hr= 30 km/h


Frame of Reference(Moving along different paths)

A boat heading north crosses a wide river with a velocity of 10 km/h relative to the water. The river has a uniform velocity of 5 km/h due east. Determine the boat’s resultant velocity relative to an observer on the shore.

Solve using vectors (demo) Pythagorean Theorem (magnitude) Tangent (direction)

Try: p. 105

Practice F

Problems 1 & 3

herriman high honors physics chapter 3 kinematics in two dimensions vector addition and projectile...

Documents

vector answer slide

displacement vector

north of west slide

scalars vectors

vectors head

arctan slide

original vectors

dimensions vector addition