Eg 1: Chapter 2.8 Weight 5kg Definition: The force of gravity acting on an object Formula : W = mg m = mass g = gravity (9.81 msβ»²) W = mg g= π π Characteristic Mass Weight Symbol?: m W Unit?: Kg N Type?: Physical Quantity?: Quantity of Gravitation matter al force Base quantity Influenced not by gravity?: influenced by gravity Earth = 600 N Moon = 1/6 x 600 N = 100 N Derived quantity influenced by gravity Calculate the weight of the brick Answer: W = mg W = (5)(9.81) W = 49.05 N Simplifying the statement in question: object’s mass = 10kg object’s weight is 150 on an unknown planet (a) w = mg w = (10)(9.81) w = 98.1 N (b) The planet is larger because the strength of the gravitational field is greater 150 g = π 10 g= = 15 N kgβ»¹ π 15 N kgβ»¹ > 9.81 N kgβ»¹ Astronaut’s suit mass = 81.65kg 4. An astronaut of mass 60kg is assigned to explore the Moon. What is the astronaut’s weight on the Moon’s surface? Calculate the weight of the astronaut’s suit in Moon Astronaut’s weight in Earth: W = 60kg x 9.81 N kgβ»¹ W = 588.6 N Astronaut’s weight on Moon using Astronaut’s weight in Earth: 1 6 W = x (588.6 N) = 98.1 N Charle’s law = Volume is directly proportional to absolute temperature for a fixed mass of gas at constant pressure Chapter 4.4 Gas Laws Since k = constant T = absolute temperature (K) V = volume of gas (m³) π Formula: π If gas experiences a change in volume and temperature from condition 1 to condition 2, π π = π, Condition 1 of gas = Condition 2 of gas = Then, =π πβ πβ = πβ πβ πβ πβ =π =π πβ πβ 1) A gas has a volume of 100 ml at a temperature of 36°C. What is the temperature if the volume is 50 ml? Solution: Fixed mass of gas heated at constant pressure Temperature point Temperature π/°C Temperature π»/K Absolute zero -273 0 Melting ice 0 273 Steam 100 373 πβ πβ = πβ πβ 100ππ 50ππ = 309πΎ πβ Change 36°C to K: 36 + 273 = 309 100 x Tβ = 309 x 50 100Tβ = 15450 15450 154.5 – 273 = -118.5°C Tβ = 100 (If asked for °C) Tβ = 154.5K 2) If the volume is 250 ml at 100°C, what is the volume at 0°C πβ πβ = πβ πβ 250ππ πβ = 373πΎ 273πΎ 373 x Vβ = 250 x 273 373Vβ = 68250 68250 Vβ = 373 Vβ = 183ml Change 100°C to K: 100 + 273 = 373 Chapter 5.2 Damping and Resonance An oscillation with its amplitude decreasing with time shows that the system experiences a gradual loss of energy. Finally the oscillation stops. This phenomenon is known as damping. Oscillating systems experience loss of energy due to: External Damping Oscillating system loses energy to overcome friction or air resistance Internal Damping Oscillating system loses energy because of the stretching and compression of the vibrating particles in the system Damping is the reduction in amplitude in an oscillating system due to loss of energy. During damping, the oscillating frequency remains constant while the oscillating amplitude decreases The effect of damping can be overcome by applying periodic external force on the oscillating system. The periodic external force transfers energy into the oscillating system to replace the energy lost. The system is said to be in a forced oscillation. Chapter 5.4 Refraction of Waves Refraction of waves is the change in direction of propagation of waves caused by the change from one medium to another. Characteristic From deep water region to shallow water region Smart info • • • Speed of water wave is influenced by depth of water Speed of sound wave is influenced by density of air Speed of light wave is influenced by optical density of medium From shallow water region to deep water region Angle of incidence Angle of incidence Angle of and angle of > angle of incidence < refraction refraction angle of refraction Wavelength Decreasing Increasing Frequency No change No change Wave Speed Decreasing Increasing Direction of propagation Refracted towards Refracted away the normal from the normal Chapter 5.4 Refraction of Waves Chapter 5.4 Refraction of Waves • • • • Figure 5.41 shows phenomenon of refraction of sea waves The cape is the shadow water region The bay is the deep water region Away from the shoreline, the wavefront of the water is almost straight and parallel ο Because the water waves move at a uniform speed • When the water front of the water propagates to the cape ο The speed of the water waves decreases causing the wavelength to be shorter • Wavefront of water approaching the bay moves at a higher speed and the wavelength is longer ο This causes the wavefront to curve and follow the shape of the shoreline • • • Refraction of water waves causes water wave energy to converge towards the cape Water wave energy diverges from the bay and spread out to a wider region Thus, the amplitude of waves at bay is smaller than at the cape From the formula of speed of wave v=fπ, In deep region vβ = fπβ …………(1) In shallow region vβ = fπβ …………(2) π£β π£β π£β πβ (1) ÷ (2) gives that is π£β πβ = = πβ πβ a) v = fπ a) 30=10π a) π=3cm π£β πβ a) vβ = fπβ OR a) 30=10πβ a) πβ=3cm b) π£β = πβ 30 3ππ = π£β 1.5ππ (c) (d) Frequency: Shallow region = Deep region Wavelength : Shallow region < deep region Wave speed : Shallow region < deep region Chapter 6 Light and Optics Everything relating to Concave mirror ο If the inner surface of the part that has been cut reflects light, the mirror is a concave mirror Optical Term Explaination Principal axis Straight line passing through the centre of curvature, C and pole of the spherical mirror, P Centre of curvature, C Centre of sphere which produces a concave or convex mirror Radius of curvature of mirror, r Distance between the pole of spherical mirror, P and the centre of curvature, C Focal point, F A point on the principal axis of the spherical mirror, • For concave mirror, light rays which are parallel to the principal axis will converge at this point • For convex mirror, light rays which are parallel to the principal axis appear to diverge from this point Object distance, u Distance between object and the pole of spherical mirror, P Image distance, v Distance between image and the pole of spherical mirror, P Focal length, π Distance between focal point, F and the pole of spherical mirror, P Chapter 5.6 Interference of waves Interference with Coherent Source of Waves Interference of waves = superposition of two or more waves from a coherent source of wave ο Two sources of waves are coherent when the frequency of both waves is the same and the phase difference is constant. ο Superposition of waves produces constructive interference and destructive interference. Constructive interference occurs when: Two crests are in superposition Two troughs are in superposition Destructive interference occurs when a crest and a trough are in superposition to produce zero combined displacement Interference of water waves ο Produced by two coherent sources, Sβ and Sβ in a ripple tank Chapter 5.6 Interference of waves Interference of light waves ο The pattern is formed on the screen with light from a laser pen ο Diffracted light waves that appear from the double-slit are coherent ο Superposition of waves from the double-slit produces a pattern made up of bright fringes and dark fridges • Constructive interference produces bright fringes • Destructive interference produces dark fringes Interference of sound waves ο Sound waves cannot be seen ο In constructive interference regions, Observer can only head loud sounds ο In destructive interference regions, Observer can only hear soft sounds Drawing patterns of Wave Interference ο Interference of water, light, sound waves can be analysed by drawing their patterns of interference ο Points P & Q = antinodes, the points where constructive interference occurs ο Point R = a node, the point where destructive interference occurs Formula to determine wavelength of Interference of water, light, sound waves : π = ππ₯ π· Applications of Interference of Waves in Daily Life Interference of water waves Bulbous bow generates water waves which interfere destructively with the water waves around the hull. This causes the water around the ship to become calmer and thus, reduces water drag. Interference of light waves Anti-reflective coating on the lens surface causes reflected light to interfere destructively. This coating helps make vision clearer and prevents image formation on the glasses' lenses Interference of sound waves Microphone and transmitter system in head phones used on aeroplanes produces sound waves which interfere destructively with the surrounding noise. (a) Diffraction (b) Interference (c) Bright fringes result from constructive interference Dark fringes result from destructive interference ππ₯ (d) π = π· (0.30÷1000)(4.6 ÷1000) π= 2.5 = 5.52 x 10β»β· Chapter 3 Gravitation, Subtopic 3.3 Man-Made Satelites, Escape velocity Escape Velocity Escape Velocity is achieved when the minimum kinetic energy of an object is able to overcome its gravitational potential energy. As such: ο Escape velocity, v is the minimum velocity needed by an object on the surface of the Earth to overcome the gravitational force and escape to outer space ο If the distance of an object from the centre of the Earth is r, the mass of the object is m, and the mass of the Earth is M, then the object possesses πΊππ gravitational potential energy, U = − . π Figure 3.38 shows an object launched at escape velocity, v. ο This object can overcome gravitational force and move an infinite distance from the Earth ο Escape velocity, v of an object depends on the mass of the Earth, M and distance, r of the object from the centre of the Earth. ο Escape velocity does not depend on mass of the object, m Benefits and Implications of Escape velocity Solving problems involving escape velocity Definition of the minimum velocity required to overcome the force of gravity and escape into space. Solution: Past velocity, Depends on: 1. Earth's mass, M/Planet’s mass and 2. Distance, r object from the centre of the earth Calculate the velocity at the surface of the earth if the mass of the earth is 5.97 π₯ 1024 and its radius is 6.37π₯10βΆ. v= 2(6.67 π₯ 10−11 )(5.97π₯1024 ) 6.37π₯10βΆ = 11180 m sβ»¹ = 11.2 km sβ»¹ Calculate the escape velocity if it was on Jupiter. v= Mass of Jupiter = 1898.13 x 10²β΄ Radius of Jupiter = 7.15 x 10β· 2(6.67 π₯ 10−11 )(1898.13 x 10²β΄) 7.15 x 10β· = 59509.7 m sβ»¹ Benefit of high launch velocity (11200m/s) 1. Earth maintains a layer of atmosphere 2. Gas molecules do not escape into space 3. Airplanes can fly higher. ο reason: The linear speed of the object is less/lower than the escape velocity from the Earth Implications/adverse effects 1. Greenhouse effect due to CO2 (trapping heat) 2. Rocket launch needs a lot of fuel to produce a large engine thrust ο Hydrogen gas can escape into space because the mass is very small and can reach a high escape velocity Solution: 4 . No, because the escape velocity depends on the mass of the earth and the distance of the object from the center of the earth. 5. v = v= (a) v = 2πΊπ π 10−11 )(5.97π₯1024 ) 2(6.67 π₯ 6.37π₯106 + 700000 = 7504.8 m sβ»¹ Solution: 2πΊπ π 2(6.67 π₯ 10−11 )(5.68π₯1026 ) v= 6.03π₯10β· v= 35448.14 m sβ»¹ (b) The particles escaping into the outer space is impossible (hard to escape) because the escape velocity is very high Chapter 4.2 Specific heat capacity Application of heat capacity in daily life Heat capacity -Sand heats up quickly because of the small heat capacity -water heats up slowly because of the high heat capacity Example: 2kg 1kg Eg: Water 5000J = 1°C 5000J Sand 100J = 50°C Heat capacity, C is the amount of heat required to raise the temperature of an object by 1°C C= π Δπ quantity of heat (J) temperature change (°C) Unit: J °Cβ»¹ 4200J Heat raised 1°C 8400J Heat raised 1°C? What is the quantity of heat at 10°C when the water is 0.5kg? Solution: 4200J x 0.5kg = 2100 2100 x 10°C = 21000J Specific heat capacity, c ο The quantity of heat required to raise the temperature of 1kg of an object by 1°C Example: Specific heat capacity of water, c = 4200 J/kg/°C ο 4200J of heat is required to raise 1kg by 1°C π c = πΔπ Heat(J) Perubahan suhu (°C) Δ = change Mass(kg) -The more mass, the more heat capacity Q = πcΔπ heat absorbed/released Unit : J kgβ»¹ °Cβ»¹ 1. Calculate the heat required to raise the temperature of 0.2 kg of cooking oil by 50°C (Specific heat capacity of cooking oil = 800) ο The value specific heat capacity is high, the temperature rises slightly (slow heating) Q = πcΔπ = 0.2 π₯ 800 π₯ 50 = 8000π½ m c Calculate the temperature rise of 2kg of iron and 2kg of concrete if supplied with 5000J of heat. (Specific heat capacity of iron = 450, concrete = 850) Δπ 2. Calculate the heat absorbed by 1.5 kg of water when heated from 26°C to boiling. (Specific heat capacity of water, c = 4200) 1.5x(100-26)x4200 = 466200π½ Application of Specific heat capacity Iron Q = πcΔπ 2ππ π₯ 450 π₯ Δπ = 5000π½ Δπ = 5.6 °C Concrete Q = πcΔπ 2ππ π₯ 850 π₯ Δπ = 5000π½ Δπ = 2.94 °C Conclusion: If the specific heat capacity is high, the temperature change/temperature rise is small Chapter 2.5 Momentum v= d v t= t π π‘ π π£ d=v*t Momentum for Paragliding (essay) Paraglider wings redirect a mass of air (m) from the apparent wind against the undisturbed wind, which decelerates (a) the re-directed airflow. This action creates turbulence and a backwards force (Force = ma). The equal and opposite forward force provides thrust to push the paraglider ahead. The paraglider steals momentum from the wind by slowing it down. This process is similar to how a kitesurfer, sailboat, and glider generate a force Boats sailing into the wind on a close haul tack at a positive sail AOA, passively re-directs a mass of air each second (m/dt) from the apparent wind (relative airflow), helped by the Coanda effect on the leeward side of the sail. The re-directed airflow pushes against undisturbed apparent wind at the trailing edge of the sail, creating turbulence. The turbulence provides something to push against and causes the re-directed airflow to decelerate to a reduced velocity (dv). This action generates a backward force, as described by the equation: See Fig. III-d. Force BACK = m/dt * dv The inertia arising from the wind slowing down allows for the reactive equal and opposite force (Thrust). The thrust generated pushes the sail ahead. This dynamic is summarized by the equations: See Fig. I-a. Force BACK = Force FORWARDS (Thrust) = m/dt * dv Simplified to: Thrust = m/dt * dv Chapter 6 Light and optic, Subtopic 6.2 Total Internal Reflection Prism periscope: • • • • • Used to see objects behind a barrier. Made up of two right angle prisms fitted at both ends of a long tube. Light rays from an object travel along the normal to the side AB of the upper prism (pass through the opening of the periscope). The light rays reach side AC without refraction. The angle of incidence is 45° and is larger than the critical angle of the prism, which is 42 °. Therefore, total internal reflection happens at side AC and the light rays are reflected downwards. The reflected light rays travel downwards along the normal to side DE of the lower prism. Once again, the reflected light rays experience total internal reflection at side DF. Finally, the reflected light rays emerge from side EF without refraction and enter the eyes of observer. The image formed is upright and of the same size as the object. 2.6 Force, Newton’s Second Law of Motion (essay) According to newton's second law of motion, Acceleration is produced when a force acts on a mass. The greater the mass is, the grater the acceleration is needed to move forward. This law basically states that a force applied to the objects changes its velocity overtime in the direction of the force that is applied, the acceleration is directly proportional to the force, as an example, if pushing on an object, causing it to accelerate, and then you push, the same object three times harder, the acceleration will be three times greater and the acceleration is inversely proportional to the mass of an object, if you push equally on two objects, and one of the objects has five times more mass than the other, it will accelerate at one fifth of the object. Expression of Newton’s Second law of motion Change of momentum = ππ£ − ππ’ Rate of change of momentum = (ππ£ −ππ’) π‘ SUV's have the greater mass than the car. So SUV's need more force than light car to move forward in the direction the force is applied, we can say that SUV's need more force to act than a car, for example driving an SUV and light car at the same velocity, the force needed for car would be less than the SUV and the car would run faster than a SUV as it need less force than an SUV. SUV requires more fuel than a normal car. A Car could run faster than an SUV and it even requires less fuel. Therefore, the more mass the object has, it requires more force to make it move forward and to act on it. Newton's second law of motion states that the heavier objects require more force to get them moving but smaller objects or lighter objects require less force to actually get them moving. The formula that is used is force equals mass times acceleration. For example if you throw a 5lb weight stone and a 2lb weight with the same amount of force the 2lbs weight will travel faster than the 5lbs weight, That is because there is less mass to be moved, A truck hits a car and allows the car to move forward. The truck provides the force and the car’s mass, and the acceleration is how quickly the car (mass) moves forward. The larger and heavier the car mass is, the more force it takes to move forward. If the car has very less mass it move's forward quicker than the car with more mass. Thus, we can say that more people would prefer a car than a SUV's because of it's mass and fuel efficiency that is required which is very less in consumption in a car than compared to SUV's.
