Chapter 1: Physical World
Key Terms and Concepts Defined and Explained
• Physics:
◦ Physics is a basic discipline in the category of Natural Sciences, which also includes Chemistry and Biology.
◦ The word "Physics" originates from a Greek word meaning nature, and its Sanskrit equivalent, "Bhautiki," refers to the study of the physical world.
◦ Broadly, physics can be described as the study of the basic laws of nature and their manifestation in different natural phenomena.
◦ Humans have always been curious about the world, observing the physical environment, seeking patterns, and building tools to interact with nature, which eventually led to modern science and technology.
◦ Physics is exciting due to the elegance and universality of its basic theories, showing how a few concepts and laws can explain a vast range of physical quantities. It also offers the thrill of imaginative experiments to uncover nature's secrets, verify or refute theories, and apply physical laws to create useful devices.
• Science and the Scientific Method:
◦ Science is a systematic attempt to understand natural phenomena in as much detail and depth as possible, and to use the gained knowledge to predict, modify, and control phenomena.
◦ It involves exploring, experimenting, and predicting based on observations of the world around us. The initial step is the curiosity to learn and unravel nature's secrets.
◦ The scientific method comprises several interconnected steps: systematic observations, controlled experiments, qualitative and quantitative reasoning, mathematical modelling, prediction, and verification or falsification of theories.
◦ While speculation and conjecture have a place in science, a scientific theory must ultimately be verified by relevant observations or experiments to be acceptable.
• Principal Thrusts in Physics:
◦ Unification: This involves attempting to explain diverse physical phenomena in terms of a few concepts and laws. The goal is to see the physical world as a manifestation of universal laws across different domains and conditions. For example, Newton's law of gravitation describes the fall of an apple, the Moon's orbit, and planetary motion around the Sun. Similarly, Maxwell's equations govern all electric and magnetic phenomena. Efforts to unify fundamental forces (Section 1.4) reflect this quest.
◦ Reductionism: This approach seeks to derive the properties of a bigger, more complex system from the properties and interactions of its constituent simpler parts. It is considered to be at the heart of physics. For instance, thermodynamics deals with bulk systems using macroscopic quantities (temperature, internal energy, entropy), but kinetic theory and statistical mechanics later interpreted these in terms of the properties of molecular constituents, relating temperature to the average kinetic energy of molecules.
• Domains of Physics (Scope):
◦ The scope of physics is truly vast, covering a tremendous range of magnitudes for physical quantities like length, mass, and time.
◦ Macroscopic Domain: This includes phenomena at laboratory, terrestrial, and astronomical scales. Classical Physics primarily deals with this domain.
▪ Mechanics: Founded on Newton's laws of motion and gravitation, it concerns the motion (or equilibrium) of particles, rigid and deformable bodies, and general systems of particles. Examples include rocket propulsion and wave propagation.
▪ Electrodynamics: Deals with electric and magnetic phenomena associated with charged and magnetic bodies. Its basic laws were given by Coulomb, Oersted, Ampere, and Faraday, and encapsulated by Maxwell's equations. Problems include current-carrying conductors in magnetic fields and radio wave propagation.
▪ Optics: Focuses on phenomena involving light, such as the working of telescopes and microscopes, and colours exhibited by thin films.
▪ Thermodynamics: Contrasting with mechanics, it addresses systems in macroscopic equilibrium, concerned with changes in internal energy, temperature, and entropy through external work and heat transfer. Examples include the efficiency of heat engines and refrigerators.
◦ Microscopic Domain: This includes atomic, molecular, and nuclear phenomena. Quantum Theory is the currently accepted framework for explaining phenomena at these minute scales (atoms, nuclei, elementary particles) and their interactions. Classical physics is inadequate for this domain.
◦ Mesoscopic Physics: An exciting emerging field of research that lies intermediate between the macroscopic and microscopic domains, dealing with a few tens or hundreds of atoms.
• Fundamental Forces in Nature:
◦ There are four fundamental forces that govern diverse phenomena in both the macroscopic and microscopic worlds.
◦ Gravitational Force:
▪ It is the force of mutual attraction between any two objects by virtue of their masses.
▪ It is a universal force, meaning every object experiences it due to every other object in the universe.
▪ It governs large-scale phenomena like the motion of the Moon and artificial satellites around Earth, planets around the Sun, and the formation and evolution of stars and galaxies.
▪ It is always attractive.
▪ It has an infinite range and a relative strength of 10<sup>-39</sup> (the weakest of the four).
◦ Electromagnetic Force:
▪ Operates among charged particles.
▪ It can be attractive or repulsive, unlike gravity, due to charges coming in two varieties (positive and negative).
▪ It has an infinite range and a relative strength of 10<sup>-2</sup>.
▪ It underlies macroscopic forces like tension, friction, normal force, and spring force, and governs chemical reactions and material properties.
◦ Strong Nuclear Force:
▪ The strongest force in nature, with a relative strength of 1.
▪ Has a short, nuclear size range (around 10<sup>-15</sup> m).
▪ Operates among nucleons (protons and neutrons) and heavier elementary particles.
▪ It binds protons and neutrons in a nucleus.
◦ Weak Nuclear Force:
▪ It is not as weak as gravitational force, but much weaker than strong nuclear and electromagnetic forces.
▪ Its range is exceedingly small, on the order of 10-16 m.
▪ Operates among sub-nuclear particles, particularly electrons and neutrinos.
▪ Responsible for certain nuclear processes like beta (β)-decay.
• Nature of Physical Laws and Conserved Quantities:
◦ Physicists seek to discover laws that summarise observed facts, often as mathematical equations.
◦ A remarkable fact is that some special physical quantities remain constant over time in physical phenomena; these are called conserved quantities.
◦ Understanding these conservation principles is crucial for describing phenomena quantitatively.
◦ Examples include: conservation of total mechanical energy (for motion under external conservative force), total linear momentum, total angular momentum, and charge.
◦ Some conservation laws are valid for one fundamental force but not others.
◦ Conservation laws have a deep connection with symmetries of nature.
▪ The constancy of natural laws with respect to translation (displacement) in time is equivalent to the law of conservation of energy.
▪ The homogeneity of space (laws of nature are the same everywhere) leads to the conservation of linear momentum.
▪ The isotropy of space (no preferred direction) underlies the law of conservation of angular momentum.
▪ Conservation laws for charge and other elementary particle attributes are related to abstract symmetries.
• Hypothesis, Axioms, and Models:
◦ All physics and mathematics are based on assumptions, which are variously called hypotheses, axioms, or postulates.
◦ A hypothesis is a supposition without assuming its truth; it can be verified but not proved. For example, Newton's law of gravitation was a hypothesis.
◦ An axiom is a self-evident truth.
◦ A model is a theory proposed to explain observed phenomena. Bohr's model of the hydrogen atom is an example.
Exam-Style Questions and Answers
Q1: What are the primary aims of science, and how does the scientific method achieve these aims? A1: Science aims for a systematic attempt to understand natural phenomena in as much detail and depth as possible, and to use this knowledge to predict, modify, and control phenomena. This is achieved through the scientific method, which involves interconnected steps: systematic observations, controlled experiments, qualitative and quantitative reasoning, mathematical modelling, prediction, and ultimately, verification or falsification of theories through observation or experiment.
Q2: Differentiate between the principles of 'unification' and 'reductionism' as principal thrusts in physics, providing an example for each. A2:
• Unification is the effort to explain diverse physical phenomena using a few basic concepts and laws, seeing the physical world as manifestations of universal laws in different contexts. For example, Newton's law of gravitation unifies the fall of an apple on Earth with the motion of the Moon around Earth and planets around the Sun.
• Reductionism is the approach of deriving the properties of complex systems from the properties and interactions of their simpler constituent parts. For example, thermodynamics deals with macroscopic quantities like temperature, but kinetic theory and statistical mechanics reduce temperature to the average kinetic energy of the molecular constituents of the system.
Q3: List the four fundamental forces in nature and briefly describe one characteristic for each. A3: The four fundamental forces are:
1. Gravitational Force: Always attractive and has an infinite range.
2. Electromagnetic Force: Can be attractive or repulsive, acting between charged particles, and has an infinite range.
3. Strong Nuclear Force: The strongest force, responsible for binding nucleons (protons and neutrons) in a nucleus, with a very short range of ~10<sup>-15</sup> m.
4. Weak Nuclear Force: Involved in processes like β-decay, much weaker than the strong and electromagnetic forces, and has an exceedingly small range of ~10<sup>-16</sup> m.
Q4: Explain the connection between conservation laws and symmetries in nature, providing an example. A4: Conservation laws are deeply connected with symmetries of nature. A symmetry implies that a physical law remains unchanged under a certain transformation. For instance:
• The law of conservation of energy is equivalent to the symmetry of nature with respect to translation (displacement) in time – meaning the laws of nature do not change over time.
• The law of conservation of linear momentum arises from the homogeneity of space – meaning the laws of nature are the same everywhere in the universe.
• The law of conservation of angular momentum is underpinned by the isotropy of space – meaning there is no intrinsically preferred direction in space.
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Chapter 2: Units and Measurement
Key Terms and Concepts Defined and Explained
• Physical Quantity: Something that can be measured.
• Unit: A certain basic, arbitrarily chosen, internationally accepted reference standard used for comparison when measuring a physical quantity.
• Fundamental (Base) Units: Units for fundamental or base quantities.
◦ There are seven SI base units: metre (m) for length, kilogram (kg) for mass, second (s) for time, ampere (A) for electric current, kelvin (K) for thermodynamic temperature, mole (mol) for amount of substance, and candela (cd) for luminous intensity.
◦ There are also two supplementary units: radian (rad) for plane angle and steradian (sr) for solid angle, both dimensionless quantities.
• Derived Units: Units of all other physical quantities, expressed as combinations of the base units.
• System of Units: A complete set of units, including both base and derived units.
◦ Historically, systems like CGS (centimetre, gram, second), FPS (foot, pound, second), and MKS (metre, kilogram, second) were used.
◦ The International System of Units (SI) is the currently internationally accepted system, adopted in 1971, and uses a decimal system for simple conversions.
• Measurement of Length:
◦ Direct Methods: Include using a metre scale (10<sup>-3</sup> m to 10<sup>2</sup> m), vernier callipers (accuracy 10<sup>-4</sup> m), and screw gauge/spherometer (accuracy 10<sup>-5</sup> m).
◦ Indirect Methods (for Large Distances):
▪ Parallax Method: Used for large distances (e.g., planets, stars). It involves observing an object from two different points (the basis) and measuring the angle (parallax) subtended by the basis at the object. The distance can then be calculated using trigonometry (D = b/θ, where b is the basis and θ is the parallax angle in radians).
◦ Estimation of Very Small Distances (e.g., Size of a Molecule):
▪ Optical microscopes are limited by the wavelength of visible light (~4000-7000 Ã…).
▪ Electron microscopes use electron beams (which also behave as waves with much smaller wavelengths) to achieve resolutions up to 0.6 Ã…, enabling resolution of atoms and molecules. Tunnelling microscopy offers even better resolution.
• Range of Lengths: Varies from the size of a tiny nucleus (10<sup>-14</sup> m or less) to the entire observable universe (10<sup>26</sup> m), a factor of 10<sup>40</sup> difference.
◦ Special units: fermi (f) = 10<sup>-15</sup> m, angstrom (Ã…) = 10<sup>-10</sup> m, astronomical unit (AU) = 1.496 × 10<sup>11</sup> m (average Sun-Earth distance), light year (ly) = 9.46 × 10<sup>15</sup> m, parsec = 3.08 × 10<sup>16</sup> m.
• Measurement of Mass: Mass is a basic property of matter, independent of temperature, pressure, or location. The SI unit is kilogram (kg). Masses range from electron (10<sup>-30</sup> kg) to the known observable universe (10<sup>55</sup> kg).
• Measurement of Time: Uses an atomic standard based on periodic vibrations in a cesium atom (cesium clock). A second is defined by 9,192,631,770 periods of radiation from cesium-133.
• Accuracy, Precision, and Errors in Measurement:
◦ Error: The uncertainty in any measurement.
◦ Accuracy: How close the measured value is to the true value of the quantity.
◦ Precision: To what resolution or limit the quantity is measured. A precise measurement may not be accurate.
◦ Systematic Errors: Errors that consistently tend in one direction (positive or negative). Sources include instrumental errors (imperfect design, calibration, zero error), imperfections in experimental technique, and personal errors (e.g., parallax error). These can be minimised by improving instruments, using correct techniques, and removing personal bias.
◦ Random Errors: Errors that occur irregularly and are random in sign and magnitude. Sources include unpredictable fluctuations in experimental conditions (e.g., temperature, voltage), personal errors (e.g., random judgment errors), and least count error (resolution of the instrument). These can be reduced by repeating observations many times and taking the arithmetic mean.
◦ Absolute Error (Δa): The magnitude of the difference between an individual measurement and the true value (or arithmetic mean). It is always positive.
◦ Mean Absolute Error (Δa_mean): The arithmetic mean of all the absolute errors.
◦ Relative Error: The ratio of the mean absolute error to the mean value (Δa_mean / a_mean).
◦ Percentage Error (δa): Relative error expressed in percent: (Δa_mean / a_mean) × 100%.
◦ Combination of Errors:
▪ Sum/Difference (Z = A ± B): The absolute error in the final result is the sum of the absolute errors in individual quantities: ΔZ = ΔA + ΔB.
▪ Product/Quotient (Z = AB or A/B): The relative error in the final result is the sum of the relative errors in individual quantities: ΔZ/Z = ΔA/A + ΔB/B.
▪ Quantity Raised to a Power (Z = A^k): The relative error is k times the relative error in the individual quantity: ΔZ/Z = k (ΔA/A). For Z = A^p B^q / C^r, ΔZ/Z = p(ΔA/A) + q(ΔB/B) + r(ΔC/C).
• Significant Figures (Digits):
◦ The reliable digits plus the first uncertain digit in a measurement. They indicate the precision of measurement.
◦ Rules for counting: All non-zero digits are significant. Zeros between two non-zero digits are significant. Leading zeros (to the left of the first non-zero digit) are not significant. Trailing zeros in a number without a decimal point are not significant; in a number with a decimal point, they are significant. Multiplying/dividing factors that are exact numbers have infinite significant digits.
◦ Rules for arithmetic operations: The final result should not have more significant figures than the original data with the least number of significant figures. For addition/subtraction, the result should be rounded to the same number of decimal places as the number with the fewest decimal places. For multiplication/division, the result should contain the same number of significant figures as the original number with the least number of significant figures.
• Dimensions of Physical Quantities:
◦ The nature of a physical quantity described by its fundamental base quantities.
◦ The seven base quantities are the seven dimensions of the physical world: Length [L], Mass [M], Time [T], Electric Current [A], Thermodynamic Temperature [K], Luminous Intensity [cd], and Amount of Substance [mol].
◦ The dimensions are the powers (exponents) to which the base quantities are raised to represent that quantity.
• Dimensional Formulae and Equations:
◦ Dimensional Formula: The expression showing how base quantities represent the dimensions of a physical quantity (e.g., Volume: [M⁰ L³ T⁰]).
◦ Dimensional Equation: An equation formed by equating a physical quantity with its dimensional formula (e.g., [V] = [M⁰ L³ T⁰]).
• Dimensional Analysis and its Applications:
◦ Principle of Homogeneity of Dimensions: Only physical quantities with the same dimensions can be added or subtracted. Both sides of a mathematical equation must have the same dimensions.
◦ Checking Dimensional Consistency: If an equation fails this test, it is wrong. If it passes, it is not necessarily correct, but it is dimensionally correct. It does not verify numerical constants.
◦ Deducing Relations among Physical Quantities: The method can be used if the dependence of a physical quantity on others (up to three) is known in a product type of dependence. However, it cannot determine dimensionless constants.
Exam-Style Questions and Answers
Q1: Define 'base unit' and 'derived unit', providing one example of each from the SI system. A1:
• A base unit is a unit for a fundamental or base quantity, chosen as a reference standard. Example: metre (m) for length.
• A derived unit is a unit obtained for derived quantities, expressed as combinations of base units. Example: m/s (metre per second) for speed, derived from length and time [Appendix A 6.1].
Q2: A student measures the period of oscillation of a simple pendulum five times, obtaining readings of 2.63 s, 2.56 s, 2.42 s, 2.71 s, and 2.80 s. Calculate the mean period, the mean absolute error, and the percentage error. A2:
1. Mean Period (T_mean): T_mean = (2.63 + 2.56 + 2.42 + 2.71 + 2.80) / 5 = 13.12 / 5 = 2.624 s. Rounded to two decimal places (resolution of measurement), T_mean = 2.62 s.
2. Absolute Errors (ΔT):
◦ ΔT₁ = |2.63 - 2.624| = 0.006 s
◦ ΔT₂ = |2.56 - 2.624| = 0.064 s
◦ ΔT₃ = |2.42 - 2.624| = 0.204 s
◦ ΔT₄ = |2.71 - 2.624| = 0.086 s
◦ ΔT₅ = |2.80 - 2.624| = 0.176 s
3. Mean Absolute Error (ΔT_mean): ΔT_mean = (0.006 + 0.064 + 0.204 + 0.086 + 0.176) / 5 = 0.536 / 5 = 0.1072 s. Rounded to two significant figures, ΔT_mean = 0.11 s.
4. Percentage Error (δT): δT = (ΔT_mean / T_mean) × 100% = (0.1072 / 2.624) × 100% = 4.085%. Rounded to two significant figures, δT = 4.1%.
Q3: The resistance R is given by R = V/I, where V = (100 ± 5)V and I = (10 ± 0.2)A. Calculate the percentage error in R. A3:
• Percentage error in V = (ΔV/V) × 100% = (5/100) × 100% = 5%.
• Percentage error in I = (ΔI/I) × 100% = (0.2/10) × 100% = 2%.
• For a quotient (Z = A/B), the relative error is the sum of individual relative errors. Therefore, the total percentage error in R is the sum of the percentage errors in V and I.
• Percentage error in R = 5% + 2% = 7%.
Q4: The period of oscillation of a simple pendulum is given by T = 2Ï€√(L/g). If the measured value of L is 20.0 cm with 1 mm accuracy and the time for 100 oscillations (t) is found to be 90 s using a wristwatch with 1 s resolution, what is the percentage error in the determination of g? A4: The formula for g is derived from T = 2Ï€√(L/g) as g = 4Ï€²L/T². The period of one oscillation T = t/n, where n = 100 oscillations. The relative error in T (ΔT/T) is equal to the relative error in t (Δt/t) because n is an exact number.
• Accuracy in L = 1 mm = 0.1 cm. So, ΔL = 0.1 cm.
• Relative error in L (ΔL/L) = 0.1 / 20.0 = 0.005.
• Resolution of wristwatch = 1 s. So, Δt = 1 s.
• Relative error in t (Δt/t) = 1 / 90 = 0.0111.
• Using the rule for errors in quantities raised to a power (Z = A^p B^q / C^r => ΔZ/Z = p(ΔA/A) + q(ΔB/B) + r(ΔC/C)): Δg/g = (ΔL/L) + 2(ΔT/T). Since ΔT/T = Δt/t, Δg/g = (ΔL/L) + 2(Δt/t) = 0.005 + 2(1/90) = 0.005 + 0.0222 = 0.0272.
• Percentage error in g = (Δg/g) × 100% = 0.0272 × 100% = 2.72%.
• Rounding to the level of precision of the inputs, this is approximately 3%.
Q5: The position of a particle is given by x = a + bt², where a = 8.5 m and b = 2.5 m s⁻². (a) What are its dimensions of 'a' and 'b'? (b) What is its velocity at t = 0 s and t = 2.0 s? A5: (a) According to the principle of homogeneity of dimensions, quantities added together must have the same dimensions. Since x is a position (length), its dimension is [L].
• Therefore, the dimension of 'a' must be [L].
• The dimension of 'bt²' must also be [L]. Since [t²] is [T²], the dimension of 'b' must be [L]/[T²] or [L T⁻²]. (b) Velocity is the time rate of change of position, v = dx/dt. Given x = a + bt², dx/dt = d(a)/dt + d(bt²)/dt [Appendix 3.1, Formulae for derivatives]. Since 'a' is a constant, d(a)/dt = 0. d(bt²)/dt = b * d(t²)/dt = b * 2t = 2bt [Appendix 3.1, Formulae for derivatives: d(u^n)/dx = n u^(n-1) du/dx]. So, v = 2bt. Given b = 2.5 m s⁻²:
• At t = 0 s, v = 2 * (2.5 m s⁻²) * (0 s) = 0 m s⁻¹.
• At t = 2.0 s, v = 2 * (2.5 m s⁻²) * (2.0 s) = 10 m s⁻¹.
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Chapter 3: Motion in a Straight Line
Key Terms and Concepts Defined and Explained
• Motion: Change in position of an object with time. If one or more coordinates of an object change with time, it is in motion; otherwise, it is at rest with respect to the chosen frame of reference.
• Rectilinear Motion: Motion of objects along a straight line. In one-dimensional motion, directions can be specified by + and – signs.
• Point Object: An approximation valid when the size of the object is much smaller than the distance it moves in a reasonable duration of time.
• Kinematics: The study of ways to describe motion without delving into the causes of motion.
• Frame of Reference: A coordinate system (e.g., X-, Y-, Z-axes with an origin) along with a clock, used to specify position and time. Description of an event depends on the chosen frame of reference.
• Position: Specified using a reference point (origin) and a set of axes. In straight-line motion, positions to the right of the origin are positive, and to the left are negative.
• Path Length: The total distance traversed by an object along its actual path. It is a scalar quantity (magnitude only, no direction). It is generally greater than or equal to the magnitude of displacement.
• Displacement (Δx): The change in position, defined as the difference between the final and initial positions (Δx = x₂ - x₁). It has both magnitude and direction (a vector quantity, though signs suffice in 1D). Its magnitude can be zero even if path length is not.
• Uniform Motion along a Straight Line: Occurs when an object covers equal distances in equal intervals of time.
• Position-Time (x-t) Graph: A powerful tool to represent and analyse motion.
◦ For a stationary object, it's a straight line parallel to the time axis.
◦ For uniform motion, it's a straight line inclined to the time axis.
• Average Velocity (v_avg): The change in position (displacement Δx) divided by the time interval (Δt) in which the displacement occurs. v_avg = Δx/Δt.
◦ It is a vector quantity.
◦ Can be positive, negative, or zero depending on the sign of displacement.
◦ On an x-t graph, it is the slope of the line connecting the initial and final positions.
• Average Speed: The total path length travelled divided by the total time interval.
◦ Always positive.
◦ Generally, average speed is greater than or equal to the magnitude of the average velocity. Equality holds only if motion is along a straight line in the same direction.
• Instantaneous Velocity (v) / Velocity: The limit of the average velocity as the time interval (Δt) becomes infinitesimally small (Δt → 0). v = dx/dt.
◦ It is the rate of change of position with respect to time at that instant.
◦ On an x-t graph, it is the slope of the tangent to the curve at that instant.
• Instantaneous Speed (Speed): The magnitude of instantaneous velocity. Instantaneous speed is always equal to the magnitude of instantaneous velocity.
• Acceleration (a): The rate of change of velocity with time.
◦ Average Acceleration (a_avg): The change in velocity (Δv) divided by the time interval (Δt): a_avg = Δv/Δt. On a v-t graph, it is the slope of the straight line connecting the points corresponding to (v₂, t₂) and (v₁, t₁).
◦ Instantaneous Acceleration (a): The limit of the average acceleration as the time interval (Δt) goes to zero: a = dv/dt. On a v-t graph, it is the slope of the tangent to the curve at that instant.
◦ Acceleration can be positive, negative, or zero.
◦ A change in velocity (and thus acceleration) can result from a change in speed, a change in direction, or both.
◦ The area under a velocity-time graph represents the displacement over a given time interval.
• Uniformly Accelerated Motion (Kinematic Equations):
◦ These equations relate displacement (x), time (t), initial velocity (v₀), final velocity (v), and constant acceleration (a).
◦ Assuming x₀ = 0 at t = 0:
▪ v = v₀ + at
▪ x = v₀t + ½at²
▪ v² = v₀² + 2ax
◦ If initial position is x₀:
▪ x = x₀ + v₀t + ½at²
▪ v² = v₀² + 2a(x - x₀)
◦ These equations are algebraic (quantities may be positive or negative) and applicable for one-dimensional motion with constant acceleration.
• Free-Fall: Motion of an object under the sole influence of gravity, neglecting air resistance. Acceleration due to gravity (g ≈ 9.8 m s⁻²) is taken as constant for heights small compared to Earth's radius.
• Stopping Distance: The distance a vehicle travels before stopping when brakes are applied. It is proportional to the square of the initial velocity (d_s = -v₀² / 2a).
• Reaction Time: The time a person takes to observe, think, and act in response to a situation.
• Relative Velocity:
◦ Velocity of object B relative to object A (v_BA) is v_B - v_A.
◦ Velocity of object A relative to object B (v_AB) is v_A - v_B.
◦ Therefore, v_AB = -v_BA.
Exam-Style Questions and Answers
Q1: A car starts from rest and accelerates uniformly at 2.0 m s⁻². At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11 s? (Neglect air resistance.) A1: First, find the velocity of the truck (and thus the initial horizontal velocity of the stone) at t = 10 s. Given: u = 0 m/s (starts from rest), a_truck = 2.0 m s⁻², t_truck = 10 s. Using v = u + at: v_truck = 0 + (2.0 m s⁻²)(10 s) = 20 m s⁻¹.
Now consider the motion of the stone. At t = 10 s (when dropped), the stone has:
• Initial horizontal velocity (v_ox) = v_truck = 20 m s⁻¹ (constant, neglecting air resistance).
• Initial vertical velocity (v_oy) = 0 m s⁻¹ (dropped).
• Acceleration in the vertical direction (a_y) = g = 10 m s⁻² (downwards, if upward is positive, then a_y = -10 m s⁻²). The question does not specify a positive direction, but gravity acts downwards. Let's assume downwards is positive for vertical motion for simplicity, so a_y = 10 m s⁻².
• Time elapsed after dropping (Δt_stone) = 11 s - 10 s = 1 s.
(a) Velocity of the stone at t = 11 s:
• Horizontal component of velocity (v_x) = v_ox = 20 m s⁻¹.
• Vertical component of velocity (v_y) = v_oy + a_y Δt_stone = 0 + (10 m s⁻²)(1 s) = 10 m s⁻¹. The velocity of the stone at t = 11 s is a vector with components (20 m s⁻¹ horizontally, 10 m s⁻¹ vertically downwards). Magnitude of velocity = √(v_x² + v_y²) = √(20² + 10²) = √(400 + 100) = √500 ≈ 22.4 m s⁻¹. Direction: tanθ = v_y / v_x = 10 / 20 = 0.5. So, θ = tan⁻¹(0.5) ≈ 26.6° below the horizontal.
(b) Acceleration of the stone at t = 11 s: Once the stone is dropped, the only force acting on it (neglecting air resistance) is gravity. Therefore, its acceleration is solely due to gravity. a = 10 m s⁻² vertically downwards. This acceleration is constant throughout its free fall.
Q2: Explain why the average speed of an object over a given time interval is always greater than or equal to the magnitude of its average velocity over the same interval. When are they equal? A2:
• Average speed is defined as the total path length travelled divided by the total time interval.
• The magnitude of average velocity is the magnitude of the displacement divided by the time interval.
• The path length traversed by an object is always greater than or equal to the magnitude of its displacement between the same two points.
• Since time intervals are identical for both, this means the average speed will always be greater than or equal to the magnitude of the average velocity.
• They are equal only when the object moves along a straight line and does not change its direction during the course of motion. In this specific case, the path length equals the magnitude of the displacement.
Q3: A train A moves north with a speed of 54 km h⁻¹, and train B moves south with a speed of 90 km h⁻¹. What is the velocity of train B with respect to train A? A3: First, convert speeds to m/s:
• Train A speed (v_A) = 54 km h⁻¹ = 54 * (1000 m / 3600 s) = 15 m s⁻¹.
• Train B speed (v_B) = 90 km h⁻¹ = 90 * (1000 m / 3600 s) = 25 m s⁻¹. Choose the positive x-axis direction from south to north.
• Velocity of train A (v_A) = +15 m s⁻¹ (since it moves north).
• Velocity of train B (v_B) = -25 m s⁻¹ (since it moves south). The velocity of train B with respect to train A (v_BA) is given by: v_BA = v_B - v_A. v_BA = (-25 m s⁻¹) - (+15 m s⁻¹) = -40 m s⁻¹. This means train B appears to train A to move with a speed of 40 m s⁻¹ from north to south (due to the negative sign).
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Chapter 4: Motion in a Plane
Key Terms and Concepts Defined and Explained
• Scalars: Quantities with magnitude only. They are specified completely by a single number along with the proper unit. Examples include distance, speed, mass, temperature, time, and density. Rules for combining scalars are those of ordinary algebra.
• Vectors: Quantities that have both magnitude and direction, and obey the triangle law or parallelogram law of addition. Examples include displacement, velocity, acceleration, and force.
◦ Represented by bold face type (e.g., v) or an arrow over a letter (e.g., →v).
◦ Magnitude of a vector is its absolute value (e.g., |v| = v).
• Position Vector (r): A vector that describes the position of an object relative to a chosen origin. In a plane, r = xî + yĵ.
• Displacement Vector (Δr): The straight line joining the initial and final positions of an object. It does not depend on the actual path taken. Δr = r' - r = (x' - x)î + (y' - y)ĵ = Δxî + Δyĵ. The magnitude of displacement is either less than or equal to the path length.
• Equality of Vectors: Two vectors are equal if, and only if, they have the same magnitude and the same direction. Displacing a vector parallel to itself leaves it unchanged (free vectors).
• Multiplication of Vectors by Real Numbers: Multiplying a vector A by a real number λ results in a new vector λA. Its magnitude is |λ| times the magnitude of A. Its direction is the same as A if λ is positive, and opposite if λ is negative. The dimension of λA is the product of the dimensions of λ and A.
• Vector Addition (Graphical Method):
◦ Head-to-Tail Method (Triangle Law): To find A + B, place the tail of B at the head of A. The resultant vector R is drawn from the tail of A to the head of B.
◦ Parallelogram Law: If two vectors A and B are represented as adjacent sides of a parallelogram, their sum R is represented by the diagonal passing through their common tail.
◦ Commutative Law: Vector addition is commutative: A + B = B + A.
◦ Associative Law: Vector addition is associative: (A + B) + C = A + (B + C).
• Null (Zero) Vector (0): A vector with zero magnitude. Its direction cannot be specified. Properties: A + 0 = A, λ0 = 0, 0A = 0. It represents zero displacement when initial and final positions coincide.
• Vector Subtraction: Defined as the sum of A and the negative of B: A - B = A + (-B).
• Resolution of Vectors: Expressing a vector as a sum of two or more component vectors along chosen directions.
◦ Unit Vector: A vector of unit magnitude that points in a particular direction. It has no dimension or unit and is used only to specify direction. Unit vectors along x, y, z axes are î, ĵ, k̂ respectively. |î| = |ĵ| = |k̂| = 1.
◦ A vector A in x-y plane can be resolved into components: A = A_xî + A_yĵ. A_x = A cos θ, A_y = A sin θ, where θ is the angle with the x-axis.
◦ Magnitude: A = √(A_x² + A_y²). Direction: tanθ = A_y / A_x.
◦ In 3D: A = A_xî + A_yĵ + A_zk̂. Magnitude: A = √(A_x² + A_y² + A_z²).
• Vector Addition (Analytical Method): Easier and more accurate than graphical methods. If R = A + B, then R_x = A_x + B_x, R_y = A_y + B_y, R_z = A_z + B_z. This method extends to any number of vectors.
• Instantaneous Velocity (v) in 2D: The time derivative of the position vector: v = dr/dt. Its direction is tangential to the path at that point, in the direction of motion. Components: v_x = dx/dt, v_y = dy/dt. Magnitude: v = √(v_x² + v_y²). Direction: tanθ = v_y / v_x.
• Instantaneous Acceleration (a) in 2D: The time derivative of the velocity vector: a = dv/dt. Components: a_x = dv_x/dt, a_y = dv_y/dt. In 2D/3D, velocity and acceleration vectors may have any angle between them (unlike 1D where they are always collinear).
• Motion in a Plane with Constant Acceleration:
◦ If acceleration a is constant, then v = v₀ + at.
◦ Position vector: r = r₀ + v₀t + ½at².
◦ This implies that motion in a plane can be treated as two separate simultaneous one-dimensional motions with constant acceleration along two perpendicular directions.
• Relative Velocity in Two Dimensions:
◦ Velocity of object A relative to object B: v_AB = v_A - v_B.
◦ Velocity of object B relative to object A: v_BA = v_B - v_A.
◦ v_AB = -v_BA.
• Projectile Motion: The motion of an object in flight after being thrown or projected, assuming air resistance is negligible.
◦ Considered as two simultaneous motions: horizontal (no acceleration) and vertical (constant acceleration due to gravity, -g).
◦ Equations (assuming initial position (0,0) and projection angle θ₀ with horizontal):
▪ Horizontal position: x = (v₀ cos θ₀)t
▪ Vertical position: y = (v₀ sin θ₀)t - ½gt²
▪ Horizontal velocity: v_x = v₀ cos θ₀ (constant)
▪ Vertical velocity: v_y = v₀ sin θ₀ - gt
◦ Path of a Projectile: A parabola (y = ax + bx² form).
◦ Time of Maximum Height (t_m): Time when v_y = 0. t_m = (v₀ sin θ₀)/g.
◦ Maximum Height (h_m): h_m = (v₀² sin²Î¸₀)/(2g).
◦ Total Time of Flight (T_f): Time until it returns to the same initial height. T_f = 2t_m = (2v₀ sin θ₀)/g.
◦ Horizontal Range (R): Horizontal distance travelled during time of flight. R = (v₀² sin 2θ₀)/g. Maximum range occurs at θ₀ = 45°.
• Uniform Circular Motion: Motion of an object along a circular path at a constant speed.
◦ Velocity direction continuously changes, implying acceleration.
◦ Centripetal Acceleration (a_c): Directed towards the centre of the circle. Magnitude: a_c = v²/R. It is not a constant vector as its direction changes.
◦ Angular Speed (ω): Time rate of change of angular displacement (Δθ/Δt).
◦ Relation between linear and angular speed: v = Rω.
◦ Centripetal acceleration in terms of angular speed: a_c = Rω².
◦ Time Period (T): Time for one revolution. v = 2Ï€R/T.
◦ Frequency (ν): Number of revolutions per second (1/T). ω = 2πν, v = 2Ï€Rν, a_c = 4Ï€²Î½²R.
Exam-Style Questions and Answers
Q1: Rain is falling vertically with a speed of 35 m s⁻¹. A woman rides a bicycle with a speed of 12 m s⁻¹ in the east to west direction. In which direction should she hold her umbrella? A1: Let v_r be the velocity of rain (vertically downwards) and v_b be the velocity of the bicycle (east to west). Both are with respect to the ground. The woman experiences the velocity of rain relative to her (which is moving with the bicycle). This is v_rb = v_r - v_b. We need to find the direction of v_rb. Let's consider a coordinate system where vertically downwards is the positive y-axis and east-to-west is the positive x-axis.
• v_r = (0 î - 35 ĵ) m s⁻¹ (assuming down is negative, or if down is positive, it's (0 î + 35 ĵ) m s⁻¹)
• v_b = (12 î + 0 ĵ) m s⁻¹ (east to west) So, v_rb = (0 - 12) î + (35 - 0) ĵ = -12 î + 35 ĵ m s⁻¹ (using vertical upwards as positive y, and east as positive x). Let's re-align to match example 4.6's interpretation: Let v_r be along the vertical (downwards) and v_b be horizontal (east to west). The resultant relative velocity v_rb will have components:
• Horizontal component (from -v_b) = 12 m s⁻¹ (westward).
• Vertical component (from v_r) = 35 m s⁻¹ (downwards). The angle θ with the vertical (towards the west) is given by: tan θ = |v_b| / |v_r| = 12 / 35 ≈ 0.343. θ = tan⁻¹(0.343) ≈ 19°. Therefore, the woman should hold her umbrella at an angle of approximately 19° with the vertical towards the west.
Q2: A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball? (Neglect air resistance). A2: The maximum horizontal range (R_max) of a projectile for a given initial speed v₀ occurs when the projection angle θ₀ = 45°. R_max = (v₀² sin(2 * 45°))/g = v₀²/g. Given R_max = 100 m, so v₀²/g = 100 m. The maximum height (h_max) a ball can be thrown occurs when it is thrown vertically upwards (θ₀ = 90°). h_max = (v₀² sin²(90°))/(2g) = v₀²/(2g). Substitute v₀²/g = 100 m into the equation for h_max: h_max = (1/2) * (v₀²/g) = (1/2) * 100 m = 50 m. So, the cricketer can throw the ball 50 m high.
Q3: A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of the acceleration of the stone? A3: This is uniform circular motion. Radius R = 80 cm = 0.80 m. Number of revolutions = 14. Time taken = 25 s. First, calculate the frequency (ν) and angular speed (ω): Frequency ν = revolutions / time = 14 / 25 s⁻¹ = 0.56 s⁻¹. Angular speed ω = 2πν = 2Ï€ * 0.56 rad s⁻¹ ≈ 3.5186 rad s⁻¹. Now, calculate the magnitude of centripetal acceleration (a_c): a_c = Rω². a_c = (0.80 m) * (3.5186 rad s⁻¹)² ≈ 0.80 * 12.38 ≈ 9.904 m s⁻². The direction of centripetal acceleration is always towards the centre of the circle.
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Chapter 5: Laws of Motion
Key Terms and Concepts Defined and Explained
• Aristotle's Fallacy: The incorrect idea that an external force is always required to keep a body in motion. Galileo disproved this, leading to modern science.
• Inertia: The property of a body to resist a change in its state of rest or uniform motion.
• Newton's First Law of Motion (Law of Inertia): If the net external force on a body is zero, a body at rest remains at rest, and a body in motion continues to move with uniform velocity. It implies that constant velocity (including zero velocity) is the natural state of motion in the absence of external forces. If a body is at rest or in uniform linear motion, it means the various external forces acting on it cancel out to a zero net external force.
• Momentum (p): The product of a body's mass (m) and velocity (v): p = mv. It is a vector quantity. It quantifies the effect of force on motion.
• Newton's Second Law of Motion: States that the rate of change of momentum of a body is proportional to the applied force and takes place in the direction in which the force acts.
◦ Mathematically: F = dp/dt.
◦ For a body of fixed mass m, this simplifies to F = ma, where a is the acceleration.
◦ F represents the net external force on the body.
◦ It is a vector law, implying components of force relate to components of acceleration (e.g., F_x = ma_x). This means force changes velocity only along its direction.
◦ It's a local relation: force at an instant determines acceleration at that instant, not by historical motion.
◦ Consistent with the First Law: if F=0, then a=0.
• Impulse: The product of force and time duration, which equals the change in momentum.
◦ Impulse = FΔt = Δp.
◦ Useful when a large force acts for a very short duration, producing a finite change in momentum that is measurable.
• Newton's Third Law of Motion: "To every action, there is always an equal and opposite reaction".
◦ It means forces always occur in pairs between two interacting bodies.
◦ The force on body A by body B is equal in magnitude and opposite in direction to the force on body B by body A.
◦ Action and reaction forces are simultaneous; there is no cause-effect relationship implied.
◦ Crucially, action and reaction forces act on different bodies, so they cannot cancel each other out. Internal action-reaction forces within a system do sum to zero.
• Conservation of Momentum: A direct consequence of Newton's Second and Third Laws.
◦ If the total external force acting on a system of particles is zero, then the total linear momentum of the system is constant (conserved).
◦ Mathematically: If F_ext = 0, then dP/dt = 0, so P = Constant.
◦ This also means the velocity of the centre of mass remains constant if F_ext = 0.
• Equilibrium of a Particle: A particle is in equilibrium if the net external force acting on it is zero (∑F = 0).
◦ This implies that the sum of the components of all forces in each direction (x, y, z) is zero (e.g., ∑F_x = 0, ∑F_y = 0, ∑F_z = 0).
• Common Forces in Mechanics:
◦ Gravitational Force: Force of attraction due to gravity, acts at a distance.
◦ Contact Forces: Forces arising from physical contact between objects (solids or fluids). Fundamentally, these arise from electrical forces at the microscopic level.
▪ Normal Reaction (N): Component of contact force normal (perpendicular) to the surfaces in contact.
▪ Friction (f): Component of contact force parallel to the surfaces in contact, which opposes impending or actual relative motion between the surfaces.
• Static Friction (f_s): Opposes impending relative motion (when bodies are at rest relative to each other but a force tries to move them). Its magnitude adjusts itself up to a maximum value: f_s ≤ µ_s N, where µ_s is the coefficient of static friction.
• Kinetic Friction (f_k): Opposes actual relative motion (when bodies are sliding past each other). It is typically constant: f_k = µ_k N, where µ_k is the coefficient of kinetic friction. Generally, µ_k < µ_s.
• Friction is largely independent of contact area.
◦ Tension (T): Force transmitted through a string, rope, or similar connector, always pulling along the length of the string.
◦ Viscous Force/Air Resistance: Contact forces between solids and fluids.
◦ Buoyant Force: Upward force on a solid immersed in a fluid, equal to the weight of the fluid displaced.
• Centripetal Force: The force required to provide the centripetal acceleration (v²/R) for a body moving in a circle of radius R with uniform speed v.
◦ F_c = mv²/R.
◦ It is always directed towards the centre.
◦ It is not a new kind of force but is provided by existing forces like tension, gravity, friction, etc..
• Free-Body Diagram: A diagram that shows a chosen system and all external forces acting on it by the remaining part of the assembly or other agencies. Internal forces are not included. Crucial for solving mechanics problems systematically.
Exam-Style Questions and Answers
Q1: State Newton's First Law of Motion. What is its significance in defining the concept of force? A1: Newton's First Law of Motion states that if the net external force on a body is zero, a body at rest continues to remain at rest and a body in motion continues to move with a uniform velocity. This law, also known as the Law of Inertia, is significant because it defines the natural state of motion as constant velocity (including zero velocity) in the absence of any external force. It implies that a force is required only to change the state of motion (i.e., to cause acceleration), not to maintain uniform motion.
Q2: A body of mass 0.40 kg moving initially with a constant speed of 10 m s⁻¹ to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. (a) What is the acceleration of the body? (b) What is the velocity of the body at t = 25 s (assuming the force starts at t=0)? A2: Given: Mass (m) = 0.40 kg, Initial speed (u) = 10 m s⁻¹ (north), Force (F) = 8.0 N (south), Time (Δt) = 30 s. Let's define north as the positive direction. So, u = +10 m s⁻¹ and F = -8.0 N.
(a) Acceleration (a): Using Newton's Second Law, F = ma. a = F/m = (-8.0 N) / (0.40 kg) = -20 m s⁻². The acceleration is 20 m s⁻² directed towards the south.
(b) Velocity at t = 25 s: Using the kinematic equation v = u + at, where 'a' is the acceleration calculated above and 't' is the time interval from when the force started (t=0) to 25 s. v = (10 m s⁻¹) + (-20 m s⁻²)(25 s) = 10 - 500 = -490 m s⁻¹. The velocity of the body at t = 25 s is 490 m s⁻¹ directed towards the south.
Q3: Explain with an example why action and reaction forces do not cancel each other, even though they are equal in magnitude and opposite in direction. A3: Action and reaction forces, though equal and opposite, do not cancel each other because they always act on different bodies. For forces to cancel, they must act on the same body. Example: Consider a book resting on a table.
• Action: The force of the book on the table (due to gravity and contact).
• Reaction: The normal force of the table on the book (pushing upwards). Here, the action force acts on the table, and the reaction force acts on the book. Since they act on different bodies, they cannot cancel each other. The book is at rest because the gravitational force on the book is balanced by the normal force from the table, both acting on the book.
Q4: A block of mass 2 kg rests on a soft horizontal floor. An iron cylinder of mass 25 kg is placed on top of the block. The floor yields steadily, and the block and cylinder together go down with an acceleration of 0.1 m s⁻². What is the action of the block on the floor? (Take g = 10 m s⁻²). A4: Consider the system as the block and the cylinder combined. Total mass (M) = mass of block + mass of cylinder = 2 kg + 25 kg = 27 kg. The system is accelerating downwards with a = 0.1 m s⁻². Forces acting on the system:
1. Weight of the system (W) acting downwards = Mg = (27 kg)(10 m s⁻²) = 270 N.
2. Normal force (R') from the floor on the block (system) acting upwards. Applying Newton's Second Law (F_net = Ma). Let's take downwards as positive. W - R' = Ma. 270 N - R' = (27 kg)(0.1 m s⁻²). 270 - R' = 2.7 N. R' = 270 N - 2.7 N = 267.3 N. The action of the block on the floor is the force exerted by the system (block + cylinder) on the floor. According to Newton's Third Law, this action force is equal in magnitude and opposite in direction to the normal force exerted by the floor on the block (R'). Therefore, the action of the block on the floor is 267.3 N vertically downwards.
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Chapter 6: Work, Energy, and Power
Key Terms and Concepts Defined and Explained
• Work (W):
◦ In physics, work is done by a force on a body over a certain displacement.
◦ For a constant force F acting on an object undergoing a displacement d, work is defined as the product of the component of the force in the direction of the displacement and the magnitude of the displacement.
◦ Mathematically, W = (F cos θ)d = F ⋅ d (scalar product).
◦ No work is done if:
▪ Displacement is zero (e.g., pushing a rigid wall, weightlifter holding a mass).
▪ Force is zero (e.g., block moving on a smooth horizontal table without friction).
▪ Force and displacement are mutually perpendicular (θ = 90°, cos 90° = 0) (e.g., gravitational force on a block moving horizontally, Earth's gravity on the Moon if orbit is perfectly circular).
◦ Work can be positive (θ between 0° and 90°) or negative (θ between 90° and 180°, e.g., work done by friction).
◦ Dimensions: [ML²T⁻²]. SI unit: joule (J).
◦ For a variable force F(x) in one dimension, the work done is the definite integral of force over displacement: W = ∫ F(x)dx from x_i to x_f. Geometrically, it's the area under the force-displacement curve.
• Kinetic Energy (K): The energy an object possesses due to its motion.
◦ For an object of mass m moving with velocity v, its kinetic energy is K = ½mv².
◦ It is a scalar quantity and is always positive.
◦ Dimensions: [ML²T⁻²]. SI unit: joule (J).
• Work-Energy Theorem (WE Theorem): States that the change in kinetic energy of a body is equal to the work done on it by the net force.
◦ K_f - K_i = W_net.
◦ This theorem is an integral form of Newton's second law and holds for both constant and variable forces. It is a scalar equation, so directional information from Newton's second law is integrated out.
• Potential Energy (PE or V): The 'stored' energy a body possesses by virtue of its position or configuration. When released, this energy can manifest as kinetic energy.
◦ Applicable only to conservative forces.
◦ Change in potential energy (ΔV) for a conservative force is equal to the negative of the work done by the force. ΔV = -F(x)Δx.
◦ Dimensions: [ML²T⁻²]. Unit: joule (J).
◦ The zero of potential energy is arbitrary and chosen for convenience, but must be consistently adhered to.
◦ Gravitational Potential Energy (V_g): For a particle of mass m at height h near Earth's surface (where g is constant), V_g(h) = mgh (if V=0 at h=0). More generally, for a particle of mass m at distance r from Earth's center (mass M_E), V(r) = -G M_E m / r (if V=0 as r → ∞).
◦ Elastic Potential Energy of a Spring (V_s): For an ideal spring with spring constant k and extension/compression x, V_s(x) = ½kx² (if V=0 at x=0, equilibrium position).
• Conservative Force: A force for which:
◦ The work done by it on an object is path-independent and depends only on the initial and final positions.
◦ The work done by the force in a closed path (loop) is zero.
◦ It can be derived from a scalar potential energy function V(x) by F(x) = -dV(x)/dx.
◦ Examples: Gravitational force, spring force.
• Non-Conservative Force: A force for which the work done depends on the path taken, and work done in a closed path is not zero. Example: Friction.
• Law of Conservation of Mechanical Energy: The total mechanical energy (sum of kinetic and potential energies, K+V) of a system remains constant if the only forces doing work on it are conservative. If non-conservative forces are present, mechanical energy is not conserved (it may be converted to other forms like heat).
• Various Forms of Energy: Besides mechanical (kinetic + potential), energy exists in forms such as heat, light, sound, chemical energy, electrical energy, nuclear energy.
• Equivalence of Mass and Energy: Einstein's relation E = mc², where c is the speed of light. This shows that mass can be converted into energy and vice versa.
• Law of Conservation of Energy: The total energy of an isolated system does not change. Energy may be transformed from one form to another, but the total energy can neither be created nor destroyed. The universe as a whole can be considered an isolated system, so its total energy is conserved.
• Power (P): The time rate at which work is done or energy is transferred.
◦ Average Power (P_av): P_av = W/t (Work / total time).
◦ Instantaneous Power (P): P = dW/dt (rate of work done).
◦ Can also be expressed as the dot product of force and instantaneous velocity: P = F ⋅ v.
◦ It is a scalar quantity. Dimensions: [ML²T⁻³].
◦ SI unit: watt (W) = 1 J s⁻¹.
◦ Other units: horsepower (hp) = 746 W. kilowatt-hour (kWh) is a unit of energy, not power (1 kWh = 3.6 × 10⁶ J).
• Collisions: Events where two bodies interact for a short duration, causing a change in their momenta.
◦ Conservation of Linear Momentum: In all collisions, the total linear momentum of the system is conserved. This is because the impulsive forces between colliding objects are internal forces, which cancel out.
◦ Elastic Collision: A collision in which total kinetic energy is conserved (before and after collision). In an elastic collision between two equal masses where one is initially at rest, the first mass comes to rest, and the second mass moves off with the initial speed of the first.
◦ Inelastic Collision: A collision in which total kinetic energy is not conserved (it is lost, often converted to heat, sound, or deformation energy). The final kinetic energy is always less than the initial kinetic energy. Linear momentum is still conserved.
Exam-Style Questions and Answers
Q1: A cyclist comes to a skidding stop in 10 m. During this process, the force on the cycle due to the road is 200 N and is directly opposed to the motion. (a) How much work does the road do on the cycle? (b) How much work does the cycle do on the road? A1: Given: Displacement (d) = 10 m, Force (F) = 200 N (frictional force from road on cycle). (a) Work done by the road on the cycle: The force (frictional force) is directly opposed to the motion, so the angle (θ) between the force and displacement is 180°. Work done (W_r) = F ⋅ d = Fd cos θ. W_r = (200 N)(10 m) cos(180°) = 2000 J * (-1) = -2000 J. This negative work done by the resistive force brings the cycle to a halt, as per the Work-Energy Theorem.
(b) Work done by the cycle on the road: According to Newton's Third Law, the cycle exerts an equal and opposite force of 200 N on the road. However, the road undergoes no displacement. Since work done is F ⋅ d, and d = 0 for the road, the work done by the cycle on the road is zero.
Q2: Differentiate between a conservative and a non-conservative force. Give one example of each. A2:
• Conservative Force:
◦ The work done by a conservative force on an object is path-independent; it depends only on the initial and final positions.
◦ The work done by a conservative force in a closed path (loop) is zero.
◦ A potential energy function can be associated with a conservative force.
◦ Example: Gravitational force.
• ** collides with a horizontally mounted spring of spring constant 6.25 × 10³ N m⁻¹. Calculate the maximum compression of the spring, assuming the road is smooth (frictionless).** A3: Given: Mass (m) = 1000 kg, Initial speed (v) = 18.0 km/h, Spring constant (k) = 6.25 × 10³ N m⁻¹. First, convert initial speed to m/s: v = 18.0 km/h = 18.0 * (1000 m / 3600 s) = 5.0 m s⁻¹. Since the road is smooth (frictionless), mechanical energy is conserved during the collision. At maximum compression, the car momentarily comes to rest, so its kinetic energy is entirely converted into the potential energy stored in the spring. Initial kinetic energy of the car (K_i) = ½mv². K_i = ½ * (1000 kg) * (5.0 m s⁻¹)² = ½ * 1000 * 25 = 12500 J. Final kinetic energy (K_f) = 0 J (at max compression). Initial potential energy of spring (V_i) = 0 J (unstretched). Final potential energy of spring (V_f) = ½kx_m² (where x_m is max compression). By conservation of mechanical energy (K_i + V_i = K_f + V_f): 125001800 kg (elevator + passengers) and is moving up with a constant speed of 2 m s⁻¹. The frictional force opposing the motion is 4000 N. Determine the minimum power delivered by the motor to the elevator in watts.** A4: Given: Total mass (m) = 1800 kg, Constant speed (v) = 2 m s⁻¹, Frictional force (F_f) = 4000 N. Since the elevator is moving at a constant speed, the net acceleration is zero, implying the net force is zero. Therefore, the upward force delivered by the motor must balance the total downward forces. The downward forces are:
1. Weight of the elevator and passengers (W) = mg = (1800 kg)(10 m s⁻²) = 18000 N. (Assuming g = 10 m/s², as often used in textbook examples for simplicity, or 9.8 m/s² if specified.)
2. Frictional force (F_f) = 4000 N. Total downward force (F_total_down) = W + F_f = 18000 N + 4000 N = 22000 N. To move at constant speed, the force supplied by the motor (F_motor) must be equal to this total downward force: F_motor = 22000 N. Power (P) delivered by the motor is given by P = F ⋅ v. Since the force is in the direction of motion, cos θ = 1. P = F_motor * v = (22000 N)(2 m s⁻¹) = 44000 W. The minimum power delivered by the motor is 44,000 watts.
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Chapter 7: Systems of Particles and Rotational Motion
Key Terms and Concepts Defined and Explained
• Rigid Body: Ideally, a body for which the distances between its constituent particles do not change, even when forces are applied.
• Translational Motion (Pure Translation): Motion where all particles of the body move with the same velocity at any instant of time. The body's orientation remains unchanged.
• Rotational Motion (Pure Rotation): Motion where a rigid body is fixed (pivoted) at one point or along a line, and every particle of the body moves in a circle.
◦ The circle lies in a plane perpendicular to the axis of rotation and has its centre on the axis.
◦ Particles on the axis remain stationary.
◦ In rotation about a fixed axis, every point has the same angular velocity at any instant.
• Combination of Translation and Rotation: When a body is not fixed and undergoes both translational and rotational motion (e.g., rolling motion).
• Centre of Mass (CM or C): A point that behaves as if the entire mass of the system is concentrated there, and all external forces act at that point. *
◦ The velocity of the CM (V) is given by: V = dR/dt = (∑ m_i v_i) / M = P/M, where P is the total linear momentum of the system.
◦ The acceleration of the CM (A) is given by: A = dV/dt = F_ext / M.
◦ This means the CM of a system moves as if all the mass is concentrated there and all external forces are applied there. Internal forces do not affect the motion of the CM.
• Linear Momentum of a System of Particles (P): The vector sum of the linear momenta of all individual particles in the system. P = ∑ p_i = ∑ m_i v_i.
◦ Newton's Second Law for a System of Particles: dP/dt = F_ext.
◦ Conservation of Total Linear Momentum: If the total external force (F_ext) on a system of particles is zero, then the total linear momentum of the system (P) is constant. Consequently, the velocity of the centre of mass remains constant.
• Vector Product (Cross Product) of Two Vectors (a × b):
◦ Results in a vector (c) perpendicular to the plane containing a and b.
◦ Magnitude of c = c = ab sin θ, where θ is the angle between a and b.
◦ Direction is given by the right-handed screw rule or right-hand rule.
◦ Properties: a × b = - (b × a) (anti-commutative), a × a = 0, distributive law (a × (b + c) = a ×v) of a particle at position r from the origin (on the axis): v = ω × r. The magnitude is v = ωr_⊥, where r_⊥ is the perpendicular distance from the axis.
• Angular Acceleration (α): The time rate of change of angular velocity. α = dω/dt. For a fixed axis, it's a scalar equation α = dω/dt.
• Moment of Force (Torque, Ï„): The rotational analogue of force in linear motion.
◦ For a force F acting on a particle at position r (from the origin O), the moment of force about O is defined as the vector product: Ï„ = r × F.
◦ Magnitude: Ï„ = rF sin θ, where θ is the angle between r and F.
◦ Dimensions: [ML²T⁻²]. SI unit: **newton metre (N515].
• Angular Momentum of a System of Particles (L): The vector sum of the angular momenta of all individual particles in the system. L = ∑ l_i = ∑ (r_i × p_i).
• Relation between Torque and Angular Momentum: The time rate of change of total angular momentum of a system about a point is equal to the the total external torque on the system about the same point. Ï„_ext = dL/dt.
• Conservation of Angular Momentum: If the total external torque (Ï„_ext) on a system of particles is zero, then the total angular momentum of the system (L) is conserved (remains constant). For a rigid body rotating about a fixed axis of symmetry, L_z = Iω = constant, if Ï„_ext = 0.
• Principle of Moments (for a lever): For rotational equilibrium, the sum of moments about the fulcrum must be zero. Load arm × load = effort arm × effort.
• Centre of Gravity (CG): That point of an extended body where the total gravitational torque on the body is zero. For a body where g is uniform, the CG coincides with the CM.
• Moment of Inertia (I): The rotational analogue of mass in linear motion. It resists a change in rotational motion.
◦ For a system of particles rotating about an axis, I = ∑ m_ inertia about an axis perpendicular to its plane (I_z) is equal to the sum of its moments of inertia about two perpendicular axes concurrent with the perpendicular axis and lying in the plane of the body (I_x, I_y). I_z = I_x + I_y.
◦ Theorem of Parallel Axes: Applicable to a body of any shape. States that the moment of inertia about any axis (I_z') is equal to the sum of the moment of inertia about a parallel axis passing through its centre of mass (I_z) and the product of its mass (M) and the square of the distance (a) between the two parallel axes. I_z' = I_z + Ma².
• Dynamics of Rotational Motion:
◦ Rotational analogue of F=ma is Ï„ = Iα (Torque = Moment of Inertia × Angular Acceleration).
◦ Instantaneous power: P = τω [54For a body rolling down an inclined plane, final velocity v = √[2gh / (1 + k²/R²)], where k is the radius of gyration. (Radius of gyration k is defined by I = Mk²).
Exam-Style Questions and Answers
Q1: Explain the concept of the centre of mass for a system of particles. How does the motion of the centre of mass relate to the external forces acting on the system? A1: The centre of mass (CM) of a system of particles is a conceptual point whose motion describes the overall translational motion of the entire system. Its position vector is defined as R = (∑ m_i r_i) / M, where m_i is the mass of the i-th particle, r_i is its position vector, and M is the total mass of the system. The motion of the centre of mass is governed by Newton's Second Law for a system of particles: MA = F_ext and passing through the centre), and rotating with angular speeds ω₁ and ω₂ are brought into contact face to face with their axes of rotation coincident. (a) What is the angular speed of the two-disc system? (b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? Take ω₁ ≠ ω₂. A2: (a) This scenario represents a collision or coupling of rotating bodies. Since there are no external torques acting on the combined system (the forces involved in bringing them into contact are internal), the total angular momentum of the system is conserved. Initial total angular momentum (L_initial) = I₁ω₁ + I₂ω₂. Let the final common angular speed of the combined system be ω_f. Final total angular momentum (L_final) = (I₁ + I₂)ω_f. By conservation of angular momentum: L_initial = L_final. I₁ω₁ + I₂ω₂ = (I0, 581]. Substitute ω_f from part (a): K_final = ½(I₁ + I₂) [(I₁ω₁ + I₂ω₂) / (I₁ + I₂)]² K_final = ½ (I₁ω₁ + I₂ω₂) ² / (I₁ + I₂). To show K_final < K_initial, consider the difference: K_initial - K_final = ½I₁ω₁² + ½I₂ω₂² - ½ (I₁ω₁ + I₂ω₂) ² / (I₁ + I₂) = ½ [ (I₁ω₁² + I₂ω₂²)(I₁ + I₂) - (I₁ω₁ + I₂ω₂) ² ] / (I₁ + I₂) = ½ [ I₁²Ï‰₁² + I₁I₂ω₁² + I₁I₂ω₂² + I₂²Ï‰₂² - (I₁²Ï‰₁² + I₂²Ï‰₂² + 2I₁I₂ω₁ω₂) ] / (I₁ + I₂) = ½ [ I₁I₂ω₁² + I₁I₂initial - K_final > 0, which means K_final < K_initial.
Account for the energy loss: The loss in kinetic energy is due to energy dissipation in frictional contact between the two discs as they are brought into contact and adjust to a common angular speed. This energy is typically converted into heat and sound, which are non-mechanical forms of energy.
Q3: State and explain the Theorem of Parallel Axes for moments of inertia. to a body of any shape. It is highly useful because it allows us to calculate the moment of inertia about any axis once the moment of inertia about a parallel axis through the centre of mass is known.
Q4: A solid cylinder of mass 20 kg and radius 0.25 m rotates about its axis with an angular speed of 100 rad s⁻¹. What is the kinetic energy associated with the rotation of the cylinder? What is the magnitude of the angular momentum of the cylinder about its axis? A4: Given: Mass (M) = 20 kg, Radius (R) = 0.25 m, Angular speed (ω) = 100 rad s⁻¹. For a solid cylinder rotating about its own axis, the moment of inertia (I) is given by I = ½MR² [532, Table 7.1 (7)]. I = ½ * (20 kg) * (0.25 m)² = ½ * 20 * 0.0625 = 10 * 0.0625 =, the magnitude of angular momentum is given by L = Iω. L = (0.625 kg m²)(100 rad s⁻¹) = 62.5 J s.
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Chapter 8: Gravitation
Key Terms and Concepts Defined and Explained
• Newton's Law of Universal Gravitation: States that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.
◦ Mathematically: F = G m₁m₂/r², where F is the gravitational force, m₁ and m₂ are the masses, r is the distance between their centres, and G is the Universal Gravitational Constant.
◦ G has a value of approximately 6.672 × 10⁻¹¹ N m² kg⁻².
◦ It is a **central the entire mass of the shell is concentrated at its centre.
◦ For a hollow spherical shell of uniform density, the force of attraction on a point mass situated inside the shell is zero.
• Acceleration due to Gravity (g): The acceleration experienced by an object due to Earth's gravitational force.
◦ On the Earth's surface: g = GM_E/R_E², where M_E is Earth's mass and R_E is Earth's radius.
◦ m₂ separated by a distance r, the gravitational potential energy is conventionally given by V = -G m₁m₂/r, with V=0 as r → ∞.
◦ The total potential energy for a system of particles is the sum of energies for all possible pairs of constituent particles (superposition principle).
◦ Near Earth's surface (constant g), the potential energy difference is ΔV ≈ mgΔh.
• Escape Speed (v_e): The minimum speed an object needs to be projected from a point (e.g., Earth's surface) so that it escapes Earth's gravitational influence and does not fall back.
◦ v_e = √(2GM_E/R_E).
◦ It is independent of the mass of the escaping body and its direction of projection. It depends on the gravitational potential at the launch point.
• Earth Satellites Energy of an Orbiting Satellite (E): E = K + V = ½mv² - GM_E m/r = -GM_E m/(2r) for a circular orbit. * The total energy is always negative for a bound (closed) orbit. * Kinetic energy (positive) is half the magnitude of potential energy (negative).
• Kepler's Laws of Planetary Motion: Describe the motion of planets around the Sun [ a central force.
◦ Law of Periods (Third Law): The square of the time period of revolution (T) of a planet is proportional to the cube of the semi-major axis (a) of its elliptical orbit. T² ∝ a³.
• Geostationary Satellites: Satellites in circular orbits in the equatorial plane with a time period of 24 hours, appearing fixed from any point on Earth. They orbit at approx 4.22 × 10⁴ km from Earth's centre.
• Polar Satellites: Satellites that orbit in a north-south direction, passing over the Earth's poles. They are useful for remote sensing, meteorology, and environmental studies.
• Weightlessness: The phenomenon experienced when an object is in free fall, meaning there is no normal force supporting it. This is not because gravitational force is absent or small, but because both the object₁m₂/r²**, where G is the universal gravitational constant. For an extended object like the Earth and a point mass, the law is not directly applicable because forces from different parts of the extended object would not all be in the same direction. However, for special cases involving spherically symmetric bodies, a simple law results:
1. The force of attraction between a hollow spherical shell of uniform density and a point mass situated outside is as if the entire mass of the shell is concentrated at its centre.
2. The escape speed (v_e) is the minimum speed with which an object must be projected from a point (e.g., Earth's surface) to just escape the gravitational influence of the Earth and never fall back. The escape speed from Earth's surface is given by the formula v_e = √(2GM_E/R_E).
• Does it depend on the mass of the body? No. As seen from the formula, the mass 'm' of the projectile does not appear in the equation [614, 632 laws describe the motion of planets around the Sun:
1. Law of Orbits (First Law): All planets move in elliptical orbits with the Sun situated at one of the foci of the ellipse.
2. Law of Areas (Second Law): The line that joins any planet to the Sun sweeps equal areas in equal intervals of time. This implies that a planet moves faster when it is closer to the Sun and slower when it is farther away.
A4: Given: Mass of satellite (m) = 400 kg, Initial orbital radius (r_i) = 2R_E, Final orbital radius (r_f) = 4R_E. The total energy of a satellite in a circular orbit of radius r is E = -G M_E m / (2r). Let G M_E = g R_E². So, E = -g R_E² m / (2r). Initial energy (E_i). ΔE = (-g R_E m / 8) - (-g R_E m / 4) = (-g R_E m / 8) + (2g R_E m / 8) = g R_E m / 8. Now, substitute values: Earth's radius R_E ≈ 6.4 × 10⁶ m, g ≈ 9.8 m s⁻². ΔE = (9.8 m s⁻² * 6.4 ×