General Physics Reviewer

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Units Of Measurements Basic Conversion Factors A. LENGTH Metric 1 m = 1000 mm 1 m = 100 cm 1 km = 1000 m

Non-Metric 1 m = 1.0936 yards 1 m = 39.37 inches 1 cm = 0.3937 inch

B. MASS Metric 1 kg = 1000 g 1 g = 1000 mg

1 tonne (metric ton) = 1000 kg

Non-Metric 1 slug = 14.5939 kg 1 slug = 32.17405 pound-mass 1 pound-mass = 0.4536 kg 1 ton (US) or short ton = 907.185 kg 1 ton (UK) or long ton = 1,016.047 kg

C. TIME Metric 9 1 sec = 10 nanoseconds (ns) 6 1 sec = 10 microseconds (μs) 3 1 sec = 10 milliseconds (ms) 2 1 sec = 10 centiseconds (cs)

Non-Metric 1 minute = 60 sec 1 hour = 3,600 sec 1 day = 86,400 sec 1 Julian Year = 31,557,600 sec = 365.25 days

D. FORCE Metric 5 1 Newton (N) = 10 dynes

Non-Metric 1 pound-force = 4.448 Newtons

FOR REVIEW PURPOSES ONLY : KEEP AWAY DURING YOUR Y OUR FINAL EXAMINATION

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A. Vectors Scalar Quantity - a quantity which only magnitude but no direction Ex. Speed – 40 kph Vector Quantity - a quantity which has both magnitude and direction Ex. Velocity- 40 kph due north

A.1 Analytical Methods of Finding the Resultant A. Triangle Method – construct a triangle using the 2 given vectors and use concepts of trigonometry such as Cosine Law, Sine Law or Pythagorean Theorem to solve for the unknown vector.

B. Vector Resolution Method or Component Method - it is the process of resolving the vector to its vertical and horizontal components

FOR REVIEW PURPOSES ONLY : KEEP AWAY DURING YOUR FINAL EXAMINATION

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A.2 Vector Multiplication 1. Dot Product or Scalar Product

2. Cross Product or Vector Product

B. Kinematics It is the study of how object moves without regard to the cause of motion.

B.1 Calculus Based (Time dependent) Let : Position (x) be a time dependent quantity : thus x(t) Velocity (v) is the first derivative : v(t) = dx/dt 2 Acceleration (a) is the next derivative : a(t) = dv/dt = dx/d t It is therefore possible to reverse this order : Acceleration (a) be a time dependent quantity : thus a(t) Velocity (v) is the first integral : v(t) = ʃ a(t)dt 2 Position (x) is the next integral : x(t) = ʃ v(t)dt = ʃʃ a(t)d t Instantaneous Values are obtained by substituting the exact value of time (t) in the equations above Average velocity and acceleration values: v ave = [x F – x O ] /(t F – t O ) aave = [v F – v O ] /(t F – t O ) where :

x O = x(t O ) and x F = x(t F )

where :

v O = v(t O ) and v F = v(t F )


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B.2 Motions with constant acceleration EQUATIONS for motion along a straight path (x-axis) VF = VO + at s = VOt + ½ at 2 2 2 VF = VO + 2as Note : if a = 0 ; V = constant EQUATIONS for motion along vertical axis (y-axis) VF = VO + gt h = VOt + ½ gt 2 2

2

VF = VO + 2gh Note : if object is free-fall or dropped V O = 0

EQUATIONS for Projectile Motion (x-axis) VOX = VO cos θ : VOX – Horizontal Component of VO VOX = VFX = VnX Note : Constant Horizontal Velocities s = Vx t

Other Important Equations R = [VO 2 sin (2θ)] /g tR = R/VOX = 2tH

Where: V F – final velocity V O – initial velocity s – horizontal displacement/distance traveled a – constant acceleration h – vertical displacement/ height 2 g – gravitational acceleration = 9.8 m/s Sign Convention (from origin) : s : (+) → & h : (+) ↑ V : (+) → OR (+) ↑ a : (+) if speeding up OR (+) ↑, hence g = – 9.8 2 m/s

EQUATIONS for Projectile Motion (y-axis) VOY = VO sin θ : VOY – Vertical Component of VO Note : See sign conventions below VFY = VOY + gt 2 h = VOYt + ½ gt 2 2 VFY = VOY + 2gh 2 g = – 9.8 m/s 2

Note g = + 9.8 m/s H = VOY2/(2g) = (VO sin θ)2/(2g) tH = VOY /g = (VO sin θ)/g =( tR/2)

Sign Convention : Within Area A : (+) h and (+) VY’s

Note :

Within Area B : (+) h and ( –) VY’s

1.

Within Area C : ( –) h and ( –) VY’s

V Y’s are the vertical velocity component values

2.

Common level vertical velocity components have the same magnitude but opposite in direction


PHY10 : REVIEWER

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C. Forces & Laws of Motion C.1 Newton’s Laws of Motion Newton’s First Law of Motion  An object sitting at rest will remain at rest if the sum of all forces acting on the object is zero. Similarly, if an object is moving and the sum of the forces acting on the object is zero, then the object will continue to move in t he same direction with same speed. [ΣF = 0] First Condition of Equilibrium The body or system at this condition remains at rest or moves in a straight line with constant velocity. (Translational Equilibrium) Static Equilibrium refers to all objects at rest.  Dynamic Equilibrium refers to all objects / systems moving at constant velocity 

–

–

ΣF = R = 0 [ΣF =0 & ΣF = 0] X

Y

Newton’s Second Law of Motion (NSLM) If an unbalanced force acts on a body, it is accelerated by an amount proportional to the unbalanced force and in the same direction but inversely to its mass. [a = F net / m] or [F net = ma] 

ΣF =ma where :

ΣF X = ma X & ΣF Y = 0 (if motion is relatively or purely horizontal) ΣF X = 0 & ΣF Y = maY (if motion is relatively or purely vertical)

Newton’s Third Law of Motion Whenever one object exerts a force on a second object, the second object exerts a reaction force of equal magnitude but opposite direction on the first obje ct. It indicates that forces come in pairs – an action force and reaction force 

C.2 Contact Forces Friction Force (f) – force acting between the body and its surface of contact which is acting parallel but opposite to the direction of motion Normal Force (N) – surface reaction force due to the weight of the body. It is always perpendicular to the surface of contact .


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C.3. NSLM Equations for acceleration 3.a. Sliding Objects

a = g (sin θ – μ cosθ) 3.b. Atwood’s Machine 1: (No contact surface) if : m1 > m2

a = g (m1 – m2)/(m1 + m2)

; going to the side of m 1

T = m1(g – a) or T = m2(g + a) if : m2 > m1

a = g (m2 – m1)/(m1 + m2) ; going to the side of m 2 T = m2(g – a) or T = m1(g + a) 3.c Atwood’s Machine 2: (With contact surface) Motion of the system is dependent mainly on these three things:

1. Mass of each object 2. Angle of the inclined surface 3. Surface friction The equations here assumes that either the angle of inclined or the surface friction does NOT PREVENT motion(it means that the value of acceleration you get here are always positive)

Going to the side of m 1

a = g (m1sinθ – μm1cosθ – m2)/(m1 + m2)

;

T = m2(g + a)

Going to the side of m 2

a = g (m2 – μm1cosθ – m1sinθ)/(m1 + m2) ; T = m2(g – a) if θ = 0° Logically it goes to the side of m 2

a = g (m2 – μm1)/(m1 + m2)

;

T = m2(g – a) FOR REVIEW PURPOSES ONLY : KEEP AWAY DURING YOUR FINAL EXAMINATION

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D. Circular Motion Centripetal Force (FC) The net force that causes object to move in a circular path rather than a straight one. Centripetal means “towards the center”, so this goes towards the axis of rotation.

Centripetal Acceleration (aC) The acceleration associated with the centripetal force. It also goes towards the center of rotation. Also known as the “radial acceleration”

EQUATIONS FC = maC FC = mv2/R FC = mRω2 FC = 4mRπ2/T2 FC = 4mRπ2f 2

Where: 2

aC = v /R v = ωR ω = 2π/T T = 1/f

v – linear or tangential velocity (m/s) ω – angular velocity (rad/s) T – period of rotation, time for 1 complete revolution (seconds) f – frequency, number of revolutions at a given time (per second or hertz)

Newton’s Second Law of Motion Applications 1. Horizontal Circle : ΣFX= maC & ΣFY = 0 Where: 1.1 Flat Curve vMAX =

√ μ

S

Rg

v – linear or tangential velocity (m/s) v MAX – maximum velocity

μS – static coefficient of the road surface 1.2 Banked Curve 2 R – radius of curvature (m) tan β = v /(Rg) –1 2 β – banking angle (°) β = tan [v /(Rg)] 2. Vertical Circle (Non Uniform) : ΣFX= 0 & ΣFY = maC @ Lowest Point @ Highest Point 2 aCmax = vMAX /R aCmin = vMIN2/R FC is maximum F C is minimum


PHY10 : REVIEWER

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E. Newton’s Law of Gravitation -Every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them. 2

F = Gm1m2 / (R ) Where: m1 & m2 – masses of the two particles (kg) R – distance between the two particles (m)

F – force of attraction (N) -11 2 2 G – gravitational constant = 6.67 x10 N m /kg

E.1 Determining the Acceleration Due to Gravity, g 1. Consider an object of mass m falling near a planet’s surface, the force of attraction is the gravitational pull of the planet toward the object is the weight of the object on that planet.

F=W 2 G m MPlanet/ R = mg

2

g = G MPlanet / R

2. Using the size comparison between the planet’s mass and radius to that of earth’s 2 gX = gE(MX/ME)(RE/RX)

E.2 Satellite Motion (Constant Radius Orbit) In order to determine orbital period and orbital (radial) distances of satellites (both natural and artificial), the Gravitational force is equated to the Centripetal force.

FG = FC 2

(M2 is the orbiting object) GM1M2/R = M2aC 2 GM1/R = aC 2 2 2 GM1/R = 4π R/T Cross – multiply to simplify: 2

2 3

GM1T = 4π R

(From here you can derive most of the needed equations)

Where : mass of center planet (M1), radial distance (R) & orbital period (T) -11

Since : G = 6.67x10 2

3

2

2

2

2

Nm /kg and 4π = 39.4784, we can have this as a constant 11

2

M1T / R = 4π /G = 5.9188x10 kg /Nm 2

3

11

2

2

3

(M1T )/R = 5.9188x10 kg-s /m


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