Physics Reviewer

Review Material for PHY10 : DO NOT BRING DURING ACTUAL EXAMINATION

PHY10 : Review

A. Vectors Scalar Quantity - a quantity which has 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 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 Resultant, R- is the vector sum of 2 or more vectors whose effect is the same as the given set of vectors. Ex: A small boat travels 80.0 km north and then travels 60.0 km east in 1.0 hr. What is the boats displacement for one-hour trip? What is the boat’s average speed during the one-hour trip? Unit Vectors A. Scalar Product or Dot Product of vectors A and B = AB cos θ B. Vector Product or Cross Product of vectors A and B =AB sin θ

B. Kinematics It is the study of how object moves without regard to the cause of motion. EQUATIONS for motion along a straight path (x-axis) VF = VO + at s = VOt + ½ at2 VF2 = VO2 + 2as

Where: VF – final velocity VO – initial velocity s – horizontal displacement/distance traveled a – constant acceleration h – vertical displacement/ height

Note : if a = 0 ; V = constant EQUATIONS for motion along vertical axis (y-axis) VF = VO + gt h = VOt + ½ gt2

g – gravitational acceleration = 9.8 m/s 2 Sign Convention (from origin) : s : (+) → & h : (+) ↑ V : (+) → OR (+) ↑ a : (+) if speeding up OR (+) ↑, hence g = – 9.8 m/s2

VF2 = VO2 + 2gh Note : if object is free-fall or dropped VO = 0 EQUATIONS for Projectile Motion (xaxis) VOX = VO cos θ : VOX – Horizontal

EQUATIONS for Projectile Motion (y-axis)

Component of VO

VO

VOX = VFX = VnX

Note : Constant Horizontal

VOY = VO sin θ VFY = VOY + gt

Velocities

s = Vx t Page 1 of 4

h = VOYt + ½ gt2 VFY2 = VOY2 + 2gh

: VOY – Vertical Component of

Review Material for PHY10 : DO NOT BRING DURING ACTUAL EXAMINATION Sign Convention (from origin) : s : (+) → & h : (+) ↑ V : (+) → OR (+) ↑ g = – 9.8 m/s2 Other Important Equations R = [VO 2 sin (2θ)] /g tR = R/VOX

Note g = + 9.8 m/s2 H = VOY2/(2g) = (VO sin θ 2 )/(2g) tH = VOY /g = (VO sin θ)/g = tR/2

Ex: A projectile is fired from a gun and has initial horizontal and vertical components of velocity equal to 30 m/s and 40 m/s, respectively. a. What is the initial speed of the projectile? b. Approximately how long does it take the projectile to reach the highest point? c. What is the speed of the projectile at the highest point in its trajectory? d. Approximately what is the maximum horizontal distance reached by the projectile

C. Forces & Laws of Motion 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 the 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 [ΣFX=0 & ΣFY = 0] Ex: A 2-kg picture frame is supported by two wires where each wire makes 30° with the horizontal. What is the tension in each wire? 

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 = Fnet / m] or [Fnet = ma] ΣF =ma where : ΣFX= maX & ΣFY = 0 (if motion is relatively or purely horizontal) ΣFX= 0 & ΣFY = maY (if motion is relatively or purely vertical) Ex: An object moves due to a net force of 37.5 N. It starts at rest and accelerates at 2.5 m/s 2. What is its mass? A. 10 kg

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B. 25 kg

C. 15 kg

D. 20 kg

Review Material for PHY10 : DO NOT BRING DURING ACTUAL EXAMINATION  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 object. It indicates that forces come in pairs – an action force and reaction force Friction Force (f) – force acting between the body and its surface of contact which is acting parallel but opposite to the direction of motion [ fαN or f = μN ] Types:  Static Friction, fs – exists when the body is at rest but has the tendency to slide.  Kinetic Friction, fk – opposing force when the body is in motion Ex: Three identical blocks are pulled or pushed across a horizontal surface by a force F, as shown in the drawings. The force F in each case has the same magnitude. Which figure will give the least kinetic frictional force?

Ex: A 10-kg block is set moving with an initial speed of 6 m/s on a rough horizontal surface. If the friction force is 20 N, approximately how far does the block travel before it stops?

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

aC = v2/R v= ω/R ω= 2π/T T = 1/f

Where: 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)

FC = 4mRπ2f2 Newton’s Second Law of Motion Applications Page 3 of 4

Review Material for PHY10 : DO NOT BRING DURING ACTUAL EXAMINATION 1. Horizontal Circle Where: 1.1 Flat Curve v – linear or tangential velocity (m/s) : vMAX – vMAX = √(μSRg) maximum velocity

μS – static coefficient of the road surface 1.2 Banked Curve R – radius of curvature (m) tan β = v2 / (Rg) β – banking angle (°) β = tan –1 [v2 / (Rg)] 2. Vertical Circle (Non Uniform) @ Lowest @ Highest Point Point 2 aCmax = vMAX /R aCmin = vMIN2/R FC is maximum FC is minimum

Ex: A 10 kg garbage bag is whirled into a uniform circular motion. The diameter of this circle is 50m. If the centripetal force was measured to be 84,100 dynes, what is the linear speed of the garbage bag? A. 1.45 m/s B. 12.5 m/s C. 20.56 m/s D. 17.677 m/s

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. F = Gm1m2 / (R2) Where: m1 & m2 – masses of the two particles F – force of attraction (N) (kg) R – distance between the two G – gravitational constant = 6.67 x10-11 N m2 particles (m) /kg2 Determining the Acceleration Due to Gravity, g 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 which is the weight of the object on that planet. F=W G m MPlanet/ R2 = mg g = G Mplanet/ R2 Ex: If the earth were three times farther from the sun than it is now, the gravitational force exerted on it by the sun would be a. b. c. d.

three times as large as it is now nine times as large as it is now one-third as large as it is now one-ninth as large as it is now

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