Final exam

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Jan 9, 2024

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Final Examination Semester 2 Year 2022-2023 Subject-Physics Course Code-SPH4U Date: March.25, 2023 Duration: 2 hr. Period:1 Page 1 of 7 Teacher: Shaista Maqbool Student Name: Muhammed Syed Instructions:  Label the diagrams properly if applicable.  Underline/ highlight Yellow the final answer  Do not spend too much time on one question.  Attached all numerical picture under the respective questions.  No hand writing for theory section, only you need to type answer under the question. Examination Marks Breakdown Part A-Knowledge and Understanding /13 Part B- Communication /13 Part C- Application /13 Part D- Thinking Inquiry /13 Total Marks /52 Formulae: p = h / λ p = mv

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Marks PART A - KNOWLEDGE and UNDERSTANDING (Total Marks 13) [Overall Expectation: B3] (2) 1. Describe the advantages and disadvantages of static and kinetic friction. Benefits: Because there is little friction, walking on a slippery surface is particularly difficult. Because ice has no friction, walking on it might be challenging. On paper, we can communicate. It is feasible for the atmosphere to move with the earth. Disadvantages: The primary drawback of friction is that it causes heat to build up in various machine components. As a result, some useful energy is lost as heat energy. In machines, we have to use extra force because of friction. It rejects the proposal. [Overall Expectation: D2] (5) 2. Microwaves with a wavelength of 1.5cm carry television signals using a sequence of relay towers. [1 mark for given/required, 1 mark for formula and answer, 1 mark for solution] a. Determine the frequency of the microwaves. (2) Given: λ= 1.5 cm = 0.015 m Required: f F = c/ λ = 3.00 x 10^8/0.015 F= 2.00 x 10^10 Hz

b. How much time does it take for a microwave signal to travel 5.0 x 10 3 km across Canada from St. John’s, Newfoundland, to Victoria, British Colombia? (2) Given: d= 5.0 x 10^3 km = 5.0 x 10^6 m; Required: t Time = distance / speed =5.0 x 10^6 / 3.00 x 10^8 t = 1.7 x 10^14 [Overall Expectation: B2] (6) Q3. A marble rolls off a table with a horizontal velocity of 1.93 m/s and onto the floor. The tabletop is 76.5 cm above the floor. Air resistance is negligible. a) Determine how long the marble is in the air. (2) h = vit + 1/1/2 at where a is the acceleration caused by gravity (-9.81 m/s2), vi is the starting vertical velocity (in this example, 0), and t is the amount of time the marble has been in the air. We can figure out t: h = 1/2 at2 and t = sqrt(2h/a). replacing the specified values: t = square root(2(0.765 m)/(9.81 m/s2)) = 0.3951 s As a result, the stone flies for 0.40 seconds b) Calculate the range of the marble. (2) We can use the kinematic equation to determine the marble's range:

x = vit + 1/1/2 at where x is the range (or horizontal distance covered), vi is the starting horizontal velocity (1.93 m/s in this case), an is the horizontal acceleration (0 because air resistance is minimal), and t is the duration of the marble's flight (which we just calculated to be 0.387 s). We can figure out x: x = vit = (0.38 s)(0.387 s)(1.93 m/s) = 0.746 m The marble's range is 0.75 m as a result. c) Calculate the velocity of the marble when it hits the floor.(2) Vy = (-g)(t) Vfy= Viy –gt = 0 – (9.8) (0.3951) Vfy= -3.872 m/s ▌Vf▐ = √(V fx ) 2 +(V fy ) 2 = √(1.93) 2 + (-3.872) 2 ▌V f ▐ = 4.3 m/s Now calculate angle below horizontal line which is: ∅ = tan − 1 (▌Vfy▐÷▐Vfx▐) = Tan -1 (3.872÷1.93) Teta = 64*

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PART B – COMMUNICATION (Total Marks 13) Marks [Overall Expectation: C1] (4) 1. What is the advantage of using hydroelectric power as an energy source over fossil fuel? Also explain some disadvantages of using hydroelectric power as an energy source . Hydroelectric Power Advantages: Renewable: Hydroelectric power relies on the natural water cycle, which continuously replenishes the water in rivers and reservoirs, making it a renewable energy source. Clean: Because it doesn't emit greenhouse gases or other air pollutants that worsen air pollution or contribute to climate change, hydroelectric power is a clean energy source. Reduced running expenses: After construction, a hydroelectric power plant's operational costs are comparatively low, making it a long-term, cost-efficient source of electricity. Fossil Fuel Advantages: Fossil fuels are a dependable source of energy since they can be easily stored, transported, and used to generate electricity whenever it's needed, regardless of the weather. Fossil fuels have a high energy density, which translates to a lot of energy packed into a small amount of space, making them an economical source of energy. Disadvantages: Restricted availability: Disadvantages: Fossil fuels are non-renewable resources, which means that they are limited and will eventually

Hydroelectric power plants must be built in areas with sufficient water flows that are strong and constant, which can restrict their availability in some regions. On river ecosystems and fish populations, hydroelectric power plants can have a detrimental effect on the environment, especially when building dams and reservoirs. High capital costs: The cost of building hydroelectric power plants can prevent them from being widely used. exhaust themselves. Environmental impact: The extraction, transportation, and combustion of fossil fuels can cause greenhouse gas emissions, air pollution, and oil spills, among other harmful environmental effects. Health risks: Using fossil fuels can increase your risk of developing cancer and respiratory diseases if you live in a location with high levels of air pollution. [Overall Expectation: F2] (4) 2. How fast would a rocket be travelling if a stationary observer noticed its length was contracted to 1/4 of its original proper length? Ls = Lm * sqrt(1 - v^2/c^2) where Ls is the contracted length as measured by the stationary observer, L_m is the proper length of the rocket, and c is the speed of light. In this problem, we are given that the contracted length is 1/4 of the proper length, so we can write: Ls = 1/4 Lm Substituting this into the length contraction formula, we get: 1/4 Lm = Lm * sqrt(1 - v^2/c^2) Solving for the ratio of the contracted length to the proper length, we get: 1/4 = sqrt(1 - v^2/c^2) Squaring both sides of the equation, we get: 1/16 = 1 - v^2/c^2 v^2/c^2 = 1 - 1/16 v^2/c^2 = 15/16 Taking the square root of both sides of the equation, we get: v/c = sqrt(15)/4

v = (sqrt(15)/4) * c Substituting the value of c, we get: v = (sqrt(15)/4) * (2.998 × 10^8 m/s) v = 2.595 × 10^8 m/s Rounding to two significant figures, we get: v ≈ 2.9 × 10^8 m/s Therefore, the velocity of the rocket would be approximately 2.9 × 10^8 m/s. (3) 3. Define the following terms: a) Time dilation: Time appears to pass slower for objects that are moving at high speeds relative to an observer. b) inertial frame of reference: A frame of reference in which a body not subject to any forces appears to be at rest or moving with a constant velocity. c) Simultaneity: The concept of two events occurring at the same time, according to a particular observer, but this is relative and depends on the observer's reference frame. [Overall Expectation: E2] (2) 4. Your teacher produces a light interference pattern on the wall. How will you identify whether the pattern is being created by a single, double, or multiple slit apparatus? We may carefully examine the pattern to determine whether a single, double, or multiple slit apparatus is producing the light interference pattern. In a single slit interference pattern, the bright fringe in the centre is surrounded by alternating dark and bright fringes on either side. When they depart from the centre, the dazzling fringes will lose some of their power. A double slit interference pattern will display a collection of parallel, evenly spaced light and dark fringes. The distance between the two slits and the light's wavelength will determine how many fringes are there.

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A multiple slit interference pattern will display a collection of regularly spaced, parallel, brilliant and dark fringes, along with extra interference patterns brought on by the multiple slits. The quantity and arrangement of the slits will determine the size and spacing of the fringes. We can identify if the pattern is being produced by a single, double, or multiple slit apparatus by carefully examining the pattern and noting the number and spacing of the fringes. PART C – APPLICATION (Total Marks 13) [Overall Expectation: ] (4) Q1. You are shopping for sunglasses and want to test whether a pair you like has polarizing lenses or simply darkened plastic lenses. Describe two ways you could quickly test them. To test whether a pair of sunglasses has polarizing lenses or simply darkened plastic lenses, I would hold two of the same kind of lenses together and rotate them relative to each other to look for changes in transmission intensity. I could also test the lenses by rotating one of the lenses in front of my eyes to see how reflections from different objects change. With either method, if light intensity changes depending on the angles of incidence and refraction, the lenses are polarized. [Overall Expectation: D1] (2) Q2. Wireless power charging for electric cars is in development. a) What are some advantages of this technology? Convenience: Wireless power charging eliminates the need for physical plug-ins, which is beneficial for those who park their electric vehicles on the street or in public lots. Safety: With wireless charging, there are no exposed electrical contacts, thus there is less chance of electric shock or other electrical hazards. Efficiency: Due to the potential reduction in energy loss caused by resistance in cables and connectors, wireless charging may be more effective than conventional plug-in charging. Increasing adoption of electric vehicles: Consumers may be more drawn to electric vehicles as a result of the convenience and security of wireless charging, which could lead to a rise in their use. b) What are the challenges in implementing this technology? Standardization: Since there are currently no widely acknowledged guidelines for wireless charging in electric vehicles, it may be challenging for various manufacturers to create systems that work together. Cost: The adoption of wireless charging systems may be hampered by the fact that these systems can be more expensive than conventional plug-in charging systems.

Efficiency: Although wireless charging has the potential to be more effective than conventional charging, existing systems can still experience energy loss owing to heat and other factors. Range: Since wireless charging is often slower than plug-in charging, it may be possible for electric vehicles' range to be limited. Infrastructure: Widespread deployment of wireless charging infrastructure can be expensive and time-consuming, which can delay adoption. [Overall Expectation: E3] (4) Q3.A very bright standard light bulb will not damage a person’s skin, even at close distances, assuming the hot bulb does not touch the skin. However, ultraviolet light from the distant Sun can cause sunburn, even when the day is not particularly bright. Interpret these effects in terms of the energy of photons. Our skin reacts differently to different radiation spectra, such as those produced by the Sun and a bright light bulb. According to Wien's rule, the Sun emits more high- energy, high-frequency radiation than a light bulb because it is considerably hotter. Even while the higher-energy radiation has considerably lower intensities, it can nevertheless cause skin injury whereas lower-energy radiation cannot. The photoelectric effect has a role in the cause. Higher-energy ultraviolet (UV) radiation may cause tissue damage by displacing an electron from some tissue components, just as it does for a metal in the photoelectric effect. Low-energy radiation will not cause this kind of damage. [Overall Expectation: D2] (3)Q4. An electron, initially stationary, is repelled from a negatively charged plate towards a positively charged plate. The electric potential difference between the plates is 2.40 x 10 2 V. Determine the kinetic energy of the electron halfway to the positively charged plate. (e = 1.60 x10 -19 C) V = 2.40 x 102 V Δ q = 1.60 x10-19 C

vi = 0 m/s Starting point: Ui = q Vi = (1.60 x 10-19 C)(0 V) = 0 J Δ Halfway point: Uf = q Vf = (1.60 x 10-19 C)(2.40 x 102 V/2) = 1.92 x 10-17 J Δ Ui = Uf + K K = Ui - Uf = 0 J - 1.92 x 10-17 J = -1.92 x 10-17 J |K| = 1.92 x 10^-17 J Therefore, the kinetic energy of the electron halfway to the positively charged plate is 1.92 x 10^-17 J. 1.92 × 10-17 J. PART D – THINKING /INQUIRY (Total Marks 13) [Overall Expectation: D2] (4) 1. An electron is moving at 2.5 × 10 6 m/s. What is the wavelength of a photon that has the same momentum as this electron? Given: V electron = 2.5 × 10 6 m/s Required: λ of photon with same momentum Momentum of electron: p = mv p = (9.11 × 10 -31 ) (2.5 × 10 6 ) p = 2.28 × 10 -24 kg - m/s Wavelength of photon: p = h / λ λ = h / p λ = (6.63 × 10 -34 ) / (2.28 × 10 -24 ) λ = 0.29 nm Momentum in one-dimension: [Overall Expectation: C2] (5) Q2.Two balls collide in a perfectly elastic collision. Ball 1 has mass 3.5 kg and is initially travelling at a velocity of 5.4 m/s [right]. It collides head-on with stationary ball 2 with mass 4.8 kg. Determine the final velocity of ball 2.

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Using the conservation of momentum principle, we have: m1v1 + m2v2 = m1v1' + m2v2' where m1 = 3.5 kg is the mass of ball 1, v1 = 5.4 m/s is the initial velocity of ball 1, m2 = 4.8 kg is the mass of ball 2, v2 = 0 m/s is the initial velocity of ball 2, v1' is the final velocity of ball 1, and v2' is the final velocity of ball 2. Since the collision is perfectly elastic, we also have: (1/2)m1v1^2 + (1/2)m2v2^2 = (1/2)m1v1'^2 + (1/2)m2v2'^2 Substituting the given values and solving for v1' and v2', we get: v1' = (m1v1 + m2v2 - m2v2')/m1 = (3.5 kg)(5.4 m/s) + (4.8 kg)(0 m/s) - (4.8 kg)(4.6 m/s)/3.5 kg = -2.286 m/s v2' = (m1v1 + m2v2 - m1v1')/m2 = (3.5 kg)(5.4 m/s) + (4.8 kg)(0 m/s) - (3.5 kg)(-2.286 m/s)/4.8 kg = 4.6 m/s Therefore, the final velocity of ball 2 is 4.6 m/s [right]. [Overall Expectation: E2] (4) Q3. For a single slit experiment apparatus, determine how far from the central fringe the first order violet ((λ = 350nm) and red (λ = 700nm) colors will appear if the screen is 10 m away and the slit is 0.050 cm wide. sin = m / b θ λ where is the angle between the line from the center of the slit to the bright θ fringe and the line from the center of the slit to the screen, is the wavelength λ of light, and b is the width of the slit. For the first order fringe, m = 1. For the violet light with = 350 nm: λ

sin = (1)(350×10^(-9)) / (0.050×10^(-3)) θ = sin^(-1)(7) ≈ 0.11 rad θ The distance from the central fringe to the first order fringe for the violet light can be found using: x = L tan θ where L is the distance from the slit to the screen. Thus, x = (10 m) tan(0.11 rad) ≈ 0.0070 m For the red light with = 700 nm, we have: λ sin = (1)(700×10^(-9)) / (0.050×10^(-3)) θ = sin^(-1)(14) ≈ 0.24 rad θ x = (10 m) tan(0.24 rad) ≈ 0.014 m Therefore, the first order violet color will appear about 0.0070 m from the central fringe and the first order red color will appear about 0.014 m from the central fringe.

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