chapter 1: representing motion - cabrillo collegejmccullough/physics11/files/midterm celebration...

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Chapter 1: Representing Motion Motion Diagrams: Displacement: f i x x x = total distance traveled = time interval Average speed Average velocity: avg x v t = 1 in = 2.540 cm 1 m = 100 cm = 3.281 ft 1 mi = 5280 ft = 1609 m 1 hr = 3600 s 1 day = 86,400 s 1 year = 365.25 days when converting between units, you multiply by conversion factors that equal 1 ex: 1 3600 1.0 1.0 2.24 1609 1 m m mi s mi s s m hr hr = = Trig Review: Pythagorean theorem: 2 2 2 a b c + = SOH CAH TOA sin opposite hypotenuse = cos adjacent hypotenuse = tan opposite adjacent = giga G 10 9 mega M 10 6 kilo k 10 3 centi c 10 -2 milli m 10 -3 micro µ 10 -6 nano n 10 -9 a b c You must know how to express numbers in scientific notation and with the correct number of significant figures.

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Page 1: Chapter 1: Representing Motion - Cabrillo Collegejmccullough/physics11/files/Midterm Celebration Review Equations.pdff i xi vv x xv t = = =+ ... ( ) cos ( ) sin xi i yi i vv vv θ

Chapter 1: Representing Motion

Motion Diagrams: Displacement: f ix x x∆ = −

total distance traveled = time interval

Average speed

Average velocity: avgxvt

∆=

1 in = 2.540 cm 1 m = 100 cm = 3.281 ft 1 mi = 5280 ft = 1609 m 1 hr = 3600 s 1 day = 86,400 s 1 year = 365.25 days

⇒ when converting between units, you multiply by conversion factors that equal 1

ex: 1 36001.0 1.0 2.241609 1

m m mi s mis s m hr hr

= =

Trig Review: Pythagorean theorem: 2 2 2a b c+ = SOH CAH TOA

sin oppositehypotenuse

= cos adjacenthypotenuse

= tan oppositeadjacent

=

giga G 109 mega M 106 kilo k 103 centi c 10-2 milli m 10-3 micro µ 10-6 nano n 10-9

a

b c

You must know how to express numbers in scientific notation and with the correct number of significant figures.

Page 2: Chapter 1: Representing Motion - Cabrillo Collegejmccullough/physics11/files/Midterm Celebration Review Equations.pdff i xi vv x xv t = = =+ ... ( ) cos ( ) sin xi i yi i vv vv θ

Adding Vectors Graphically: 1) Draw the first vector to the correct length and in the correct direction. 2) Draw the second vector (correct length and direction) starting at the tip of the first 3) The resultant vector starts at the tail of the first vector and ends at the tip of the second

Page 3: Chapter 1: Representing Motion - Cabrillo Collegejmccullough/physics11/files/Midterm Celebration Review Equations.pdff i xi vv x xv t = = =+ ... ( ) cos ( ) sin xi i yi i vv vv θ

Chapter 2: Motion in One Dimension ⇒ The slope of an object’s position-versus-time graph is the object’s velocity at that point in the motion. ⇒ The area under an object’s velocity-versus-time graph gives the object’s displacement. ⇒ An object speeds up if the velocity and acceleration point in the same direction; it slows down if the velocity and acceleration point in opposite directions

Velocity and Acceleration: yxx y x y

vx y vv v a at t t t

∆∆ ∆ ∆= = = =

∆ ∆ ∆ ∆

Equations of Constant Acceleration:

2

2 2

( ) ( )1( ) ( )2

( ) ( ) 2 ( )

x f x i x

f i x i x

x f x i x f i

v v a t

x x v t a t

v v a x x

= + ∆

= + ∆ + ∆

= + −

freefall: gravity is the only force acting on an object (no air resistance) Equations of Freefall:

2

2 2

( ) ( )1( ) ( )2

( ) ( ) 2 ( )

y f y i y

f i y i y

y f y i y f i

v v a t

y y v t a t

v v a y y

= + ∆

= + ∆ + ∆

= + −

Notes for Freefall:

⇒ ay = 9.80 m/s2 downward or a = –9.80 m/s2 if up is defined as +

⇒ vy = 0 m/s at the highest point

⇒ freefall motion is symmetric

Page 4: Chapter 1: Representing Motion - Cabrillo Collegejmccullough/physics11/files/Midterm Celebration Review Equations.pdff i xi vv x xv t = = =+ ... ( ) cos ( ) sin xi i yi i vv vv θ
Page 5: Chapter 1: Representing Motion - Cabrillo Collegejmccullough/physics11/files/Midterm Celebration Review Equations.pdff i xi vv x xv t = = =+ ... ( ) cos ( ) sin xi i yi i vv vv θ

Chapter 3: Vectors and Motion in Two Dimensions Adding Vectors Analytically: 1) Break each vector up into x- and y-components. 2) Add all x-components together to find Rx. Add all y-components together to find Ry. 3) Find the magnitude and direction of the resultant vector:

2 2

1tan

x y

y

x

R R R

RR

θ −

= +

=

4) Check that the angle is in the correct quadrant (if not add 180°) and that your magnitude seems reasonable. ⇒ The acceleration of an object down a frictionless plane is given by: sinxa g θ= ± Note: In two-dimensional motion, each component (x and y) can be treated separately. The x-component is independent of the y-component and vice versa. The two components are connected by time Δt. Projectile Motion: ⇒ In projectile motion, we assume no air resistance so:

ax = 0 m/s2 ay = -9.80 m/s2 (assuming up is +)

x-component of motion

( ) ( ) constant( )

x f x i

f i x i

v vx x v t

= =

= + ∆

2

2 2

y-component of motion

( ) ( )1( ) ( )2

( ) ( ) 2 ( )

y f y i y

f i y i y

y f y i y f i

v v a t

y y v t a t

v v a y y

= + ∆

= + ∆ + ∆

= + −

( ) cos( ) sin

x i i

y i i

v vv v

θθ

==

Page 6: Chapter 1: Representing Motion - Cabrillo Collegejmccullough/physics11/files/Midterm Celebration Review Equations.pdff i xi vv x xv t = = =+ ... ( ) cos ( ) sin xi i yi i vv vv θ
Page 7: Chapter 1: Representing Motion - Cabrillo Collegejmccullough/physics11/files/Midterm Celebration Review Equations.pdff i xi vv x xv t = = =+ ... ( ) cos ( ) sin xi i yi i vv vv θ

Chapter 4: Forces and Newton’s Laws of Motion

Newton’s Laws: Newton’s 1st law: An object at rest will remain at rest, an object in motion will remain in motion in a straight line at a constant speed unless acted upon by an unbalanced force. Newton’s 2nd law: F ma=∑

Newton’s 3rd law: For every action there is an equal but opposite reaction. ⇒ The unit of force is the Newton (N). 1 N = 1 kg m/s2 Some Particular Forces: Weight: w Tension: T

Normal Force: η Static Friction: sf

Kinetic Friction: kf

Rolling Friction: rf

Drag: D

Thrust: thrustF

You must draw free-body diagrams for all Newton’s laws problems.

Page 8: Chapter 1: Representing Motion - Cabrillo Collegejmccullough/physics11/files/Midterm Celebration Review Equations.pdff i xi vv x xv t = = =+ ... ( ) cos ( ) sin xi i yi i vv vv θ
Page 9: Chapter 1: Representing Motion - Cabrillo Collegejmccullough/physics11/files/Midterm Celebration Review Equations.pdff i xi vv x xv t = = =+ ... ( ) cos ( ) sin xi i yi i vv vv θ

Chapter 5: Applying Newton’s Laws

⇒ In equilibrium, the sum of the forces on an object is zero:

0 and 0x yx yF ma F ma= = = =∑ ∑

⇒ If the object is accelerating:

andx yx yF ma F ma= =∑ ∑

Mass and weight: w = mg

⇒ A person’s apparent weight (wapp) is the magnitude of the contact force that supports him. Friction: Static Friction: 0 < fs < fs,max where fs,max = µsN

(fs = fs,max only when the object is on the verge of slipping)

Kinetic Friction: fk = µkN

Rolling Friction: fr = µrN

Inclined Planes: ⇒ For inclined planes, we usually define the x-axis parallel to the incline and the y-axis perpendicular to the incline. ⇒ The weight then needs to be broken up into a component parallel to the incline (wx) and a component perpendicular to the incline (wy)

wx = mg sinθ

wy = mg cosθ

y

x θ

Page 10: Chapter 1: Representing Motion - Cabrillo Collegejmccullough/physics11/files/Midterm Celebration Review Equations.pdff i xi vv x xv t = = =+ ... ( ) cos ( ) sin xi i yi i vv vv θ
Page 11: Chapter 1: Representing Motion - Cabrillo Collegejmccullough/physics11/files/Midterm Celebration Review Equations.pdff i xi vv x xv t = = =+ ... ( ) cos ( ) sin xi i yi i vv vv θ

Chapter 9: Momentum Linear Momentum: p mv=

⇒ Note this is really two equations: andx x y yp mv p mv= = Impulse: aveJ F t= ∆

= area under curve of force versus time

Impulse-Momentum Theorem: ave

J p m v

F t p m v

= ∆ = ∆

∆ = ∆ = ∆

Conservation of Momentum: Conservation of Momentum: if the sum of the external forces on a system is zero ( )0extF =∑

, the momentum of a system remains constant.

1 2 3

if 0

... = sum of of all objects in systemext f iF P P

P p p p p

= → =

= + + +

⇒ in a collision between two objects:

1 1 2 2 1 1 2 2( ) ( ) ( ) ( )f i

x f x f x i x i

P Pm v m v m v m v

=

+ = +

⇒ if the collision is completely inelastic (objects stick together after the collision):

1 2 1 1 2 2( )( ) ( ) ( )x f x i x im m v m v m v+ = +

Page 12: Chapter 1: Representing Motion - Cabrillo Collegejmccullough/physics11/files/Midterm Celebration Review Equations.pdff i xi vv x xv t = = =+ ... ( ) cos ( ) sin xi i yi i vv vv θ