Year 12 Physics Complete Study Guide

MEASUREMENTS AND UNCERTAINTIES

• Precision measurement devices are essential tools across fields such as physics, engineering, woodworking, metalworking, and medicine.

• Vernier Caliper:

  • Consists of a main (fixed) scale and a Vernier (movable) scale with two sets of jaws to measure external and internal dimensions of circular objects.

  • Main scale smallest division: $0.1\,\text{cm}$ ($1\,\text{mm}$).

  • Vernier scale resolution: $0.05\,\text{mm}$ ($0.005\,\text{cm}$).

  • Procedure for reading external size:

  1. Identify the main scale value immediately preceding the zero mark of the Vernier scale (e.g., between $1.1\,\text{cm}$ and $1.2\,\text{cm}$ gives a first reading of $1.1\,\text{cm}$).

  2. Identify the mark on the Vernier scale that aligns best with any mark on the main scale (e.g., mark $6.5$ corresponds to $0.065\,\text{cm}$).

  3. Sum the values: $1.1\,\text{cm} + 0.065\,\text{cm} = 1.165\,\text{cm}$.

• Micrometer Screw Gauge:

  • Precision tool used by engineers. Each full revolution of the ratchet moves the spindle face $0.5\,\text{mm}$ relative to the anvil face.

  • The ratchet is turned clockwise until the object is trapped between the anvil and spindle and begins clicking.

  • Procedure for reading:

  1. Main sleeve reading for full millimeters (e.g., $12.00\,\text{mm}$).

  2. Half-millimeter sleeve reading on bottom half of scale (e.g., $0.50\,\text{mm}$).

  3. Thimble scale reading for hundredths of a millimeter (e.g., $16$ divisions = $0.16\,\text{mm}$).

  4. Total measurement: $12.00\,\text{mm} + 0.50\,\text{mm} + 0.16\,\text{mm} = 12.66\,\text{mm}$.

• Types of Measurement Uncertainty:

  • Random Uncertainty: Measurement is equally likely to be larger or smaller than the true value.

  • Examples: Stopwatch reaction time variations during pendulum timing.

  • Sources: Imperfect observer, equipment readability limits, external environmental fluctuations.

  • Mitigation: Taking the average of multiple independent trials.

  • Systematic Uncertainty: Consistent bias due to a repeated defect in the device or technique.

  • Examples: Metre rule with worn ends, unzeroed dial instrument needle, consistently late human reaction times.

  • Sources: Constant observer error, zero offset error, improper calibration.

  • Parallax Error: Apparent shift in an object's position when viewed from different angles.

  • Mitigation: Viewing dials and scales directly at a perpendicular angle ($90^\circ$).

• Significant Figures Rules:

  • All non-zero digits are significant (e.g., $3.67$ has $3$ s.f.).

  • Zeros between non-zero digits are significant (e.g., $3005$ has $4$ s.f.; $2.3006$ has $5$ s.f.).

  • Zeros to the right of a non-zero digit in the decimal part are significant (e.g., $3.50$ has $3$ s.f.).

  • Zeros used solely to space the decimal point are non-significant (e.g., $0.036$ has $2$ s.f.).

  • Powers of $10$ in scientific notation carry no significance (e.g., $5.01 \times 10^6$ has $3$ s.f.).

• Quantifying Uncertainty:

  • Absolute Uncertainty (Absolute Error): Range of values within which the true value lies. For $25.4 \pm 0.1\,\text{cm}$, absolute uncertainty is $0.1\,\text{cm}$.

  • Percentage Uncertainty (Relative Error):     $\text{Percentage Uncertainty} = \frac{\text{Absolute Uncertainty}}{\text{Best Estimate}} \times 100\%$

  • Example: For $35.4 \pm 0.2\,\text{cm}$, relative uncertainty is:       $\frac{0.2\,\text{cm}}{35.4\,\text{cm}} \times 100\% = 0.6\%$

• Calculations with Uncertainties:

  • Addition and Subtraction: Add absolute uncertainties directly.

  • Example: $(20.4 \pm 0.5\,\text{mm}) + (32.3 \pm 0.5\,\text{mm}) = (20.4 + 32.3) \pm (0.5 + 0.5) = 52.7 \pm 1.0\,\text{mm}$.

  • Multiplication and Division:

  1. Convert absolute uncertainties to percentage uncertainties.

  2. Add percentage uncertainties together.

  3. Convert total percentage uncertainty back to absolute uncertainty.

  4. Round absolute uncertainty to one significant figure, and match the final measurement's decimal places to the uncertainty.

  • Example: Area of paper with width $5.63 \pm 0.15\,\text{mm}$ and length $64.2 \pm 0.7\,\text{mm}$:       $\%\text{ width error} = \frac{0.15}{5.63} \times 100\% = 2.66\%$       $\%\text{ length error} = \frac{0.7}{64.2} \times 100\% = 1.09\%$       $\text{Total }\%\text{ error} = 2.66\% + 1.09\% = 3.75\%$       $\text{Calculated Area} = 5.63 \times 64.2 = 361.446\,\text{mm}^2$       $\text{Absolute Error} = \frac{3.75}{100} \times 361.446 = 13.571\,\text{mm}^2$       $\text{Final Area (rounded to 3 s.f.)} = 361 \pm 10\,\text{mm}^2$

PHYSICAL RELATIONSHIPS AND GRAPHICAL ANALYSIS

• Linear experimental data follows the standard line equation:   $y = mx + c$   where $m$ is the gradient and $c$ is the y-intercept.

• Non-linear relationships produce curved graphs:

  • Power-of relationship: $y = k x^2$

  • Root-of relationship: $y = k \sqrt{x}$

• To establish a precise mathematical formula from curved experimental data, variables are transformed (e.g., squared) until a straight line is achieved.

• Linearizing Distance vs. Time Data:

  • Data set: $(t, d) = (0.0, 0.0), (1.0, 2.0), (2.0, 8.0), (3.0, 18.0), (4.0, 32.0), (5.0, 50.0)$

  • Squaring time values produces $t^2 = 0.0, 1.0, 4.0, 9.0, 16.0, 25.0\,\text{s}^2$.

  • Plotting $d$ against $t^2$ yields a straight line passing through $(0, 0)$ and $(25, 50)$.

  • Slope calculation:     $m = \frac{\Delta y}{\Delta x} = \frac{50\,\text{m} - 0\,\text{m}}{25\,\text{s}^2 - 0\,\text{s}^2} = 2.0\,\text{m/s}^2$

  • Mathematical expression: $d = 2.0 t^2$

• Direct and Inverse Square Relationships:

  • Direct Square Relationship ($A = k B^2$): If $B$ doubles, $A$ quadruples. If $B$ triples, $A$ increases by a factor of $9$.

  • Example: Kinetic Energy $E_k = \frac{1}{2} m v^2$. Doubling velocity quadruples $E_k$.

  • Inverse Square Relationship ($A = \frac{k}{B^2}$): If $B$ doubles, $A$ decreases to $\frac{1}{4}$ of its original value. If $B$ triples, $A$ decreases to $\frac{1}{9}$.

  • Example: Newton's Law of Gravitation:       $F = \frac{G m_1 m_2}{r^2}$

    • If distance $r$ doubles, F_{new} = \frac{1}{4} F$.\n - If masses m_1$and$m_2$double,$F_{new} = 4 F$.

    • If mass $m_1$ and distance $r$ both double, F_{new} = \frac{(2)(1)}{2^2} F = \frac{1}{2} F$.\n\n# VECTOR ANALYSIS AND RELATIVE MOTION\n\n- Vector quantities require both magnitude and directional orientation.\n- **Vector Subtraction**:\n - Used to find changes in vector quantities:\n    \Delta \mathbf{v} = \mathbf{v}f - \mathbf{v}_i = \mathbf{v}_f + (-\mathbf{v}_i)\n - Example 1: A car moving East at 20\,\text{m/s}$turns North at$20\,\text{m/s}.\n - Vector addition: \mathbf{v}_f\,\text{[North]} + (-\mathbf{v}_i)\,\text{[West]}\n - Magnitude via Pythagoras: R = \sqrt{20^2 + 20^2} = 28.3\,\text{m/s}\n - Direction via trigonometry: \theta = \tan^{-1}\left(\frac{20}{20}\right) = 45^\circ\n - Resultant change in velocity: 28.3\,\text{m/s}\text{ at N } 45^\circ \text{ W}.\n - Example 2: A ball thrown at 6\,\text{m/s}$against a wall rebounds at$6\,\text{m/s}:\n - Magnitude via Pythagoras: X = \sqrt{6^2 + 6^2} = \sqrt{72} = 8.48\,\text{m/s}.\n- **Vector Components**:\n - Resolving a vector \mathbf{F}$at angle$\theta above horizontal:\n - Horizontal Component: F_h = F \cos(\theta)\n - Vertical Component: F_v = F \sin(\theta)\n - Example: A 60\,\text{N}$force pulled at$30^\circ above horizontal:\n - F_h = 60 \cos(30^\circ) = 52\,\text{N}\n - F_v = 60 \sin(30^\circ) = 30\,\text{N}\n- **Non-Perpendicular Vector Addition**:\n - **Sine Rule**: \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}\n - **Cosine Rule**: c^2 = a^2 + b^2 - 2ab \cos(C)\n - Example 1: Two 70\,\text{N}$forces acting at$60^\circ to each other:\n - Internal triangle angle: 180^\circ - 60^\circ = 120^\circ\n - F_R^2 = 70^2 + 70^2 - 2(70)(70) \cos(120^\circ) = 4900 + 4900 - (-4900) = 14700\n - F_R = \sqrt{14700} = 121.24\,\text{N}\n - Example 2: 6.0\,\text{N}$South and$9.0\,\text{N}$at S$60^\circ W:\n - Total Force magnitude: F_T^2 = 9^2 + 6^2 - 2(9)(6) \cos(120^\circ) = 81 + 36 + 54 = 171 \implies F_T = 13.1\,\text{N}\n - Angle direction via Sine Rule: \frac{13.1}{\sin(120^\circ)} = \frac{6}{\sin(\theta)} \implies \sin(\theta) = 0.397 \implies \theta = 23.4^\circ\n - Reference direction: 60^\circ - 23.4^\circ = 36.6^\circ \implies 13.1\,\text{N}\text{ S } 36.6^\circ \text{ W}.\n- **Relative Velocity in One Dimension**:\n - Vector subtraction formula: v{A \text{ rel } B} = v_A - v_B\n - Scenario: You walk West at 8\,\text{km/h}$, train moves West at$40\,\text{km/h}$, car moves East at$30\,\text{km/h}.\n - Train relative to You: v_{T \text{ rel } Y} = 40\,\text{km/h West} - 8\,\text{km/h West} = 32\,\text{km/h West}.\n - Car relative to You: v_{C \text{ rel } Y} = 30\,\text{km/h East} - (-8\,\text{km/h East}) = 38\,\text{km/h East}.\n - Train relative to Car: v_{T \text{ rel } C} = 40\,\text{km/h West} - (-30\,\text{km/h West}) = 70\,\text{km/h West}.\n- **Relative Velocity in Two Dimensions**:\n - **Boat/River Problems**:\n - Key vectors: v_{BW}$(boat rel water),$v_{WS}$(water rel shore),$v_{BS} (boat rel shore).\n - Canoeist example (v_{BW} = 5\,\text{m/s}$, river flowing$v_{WS} = 3\,\text{m/s [E]}$, width =$220\,\text{m}):\n - Aimed Northward: v_{BS} = \sqrt{5^2 + 3^2} = 5.83\,\text{m/s}.\n - Direction angle: \theta = \tan^{-1}\left(\frac{3}{5}\right) = 31^\circ \implies \text{N } 31^\circ \text{ E}.\n - Crossing time: t = \frac{D}{v_{BW}} = \frac{220\,\text{m}}{5\,\text{m/s}} = 44\,\text{s}.\n - Downstream landing distance: \Delta d_x = v_{WS} \times t = 3 \times 44 = 132\,\text{m [E]}.\n - Heading to land directly across: \sin(\theta) = \frac{v_{WS}}{v_{BW}} = \frac{3}{5} \implies \theta = 36.87^\circ \implies \text{N } 36.87^\circ \text{ W}.\n - **Airplane/Wind Problems**:\n - True Airspeed (TAS): Speed relative to air.\n - Groundspeed: Speed relative to ground (vector sum of TAS and wind speed).\n - Pilot flies TAS 400\,\text{km/h}$North with crosswind$300\,\text{km/h} from East:\n - Resultant groundspeed: R = \sqrt{400^2 + 300^2} = 500\,\text{km/h}.\n - Direction angle: \theta = \tan^{-1}\left(\frac{300}{400}\right) = 36.9^\circ \implies \text{N } 36.9^\circ \text{ W}.\n - Heading to fly due North at 400\,\text{km/h}$groundspeed: Pilot must head N$36.9^\circ$E at$500\,\text{km/h}.\n\n# FORCES AND NEWTON'S LAWS OF MOTION\n\n- Force (\mathbf{F}$) is a vector quantity with SI unit Newton ($\text{N}).\n- **Resultant Force** (\mathbf{F}r): A single force replacing multiple concurrent forces with equivalent effect.\n- **Newton's Second Law of Motion**:\n  \mathbf{F} = m \mathbf{a}\n - Example 1: m = 5000\,\text{kg}$,$F = 600\,\text{N} \implies a = \frac{600}{5000} = 0.12\,\text{m/s}^2\n - Example 2: a = 13\,\text{m/s}^2$,$F = 50\,\text{N} \implies m = \frac{50}{13} = 3.85\,\text{kg}\n- **Equilibrium in Two Dimensions**:\n - Occurs when net resultant force and net moment equal zero:\n    \sum F_x = 0, \quad \sum F_y = 0, \quad \sum M_A = 0\n - Pendulum Example: Pendulum weight 20.0\,\text{N}$pulled horizontally by force$F$to$40^\circ from vertical:\n - \sum F_y = 0 \implies T \cos(40^\circ) = 20.0\,\text{N} \implies T = \frac{20.0}{\cos(40^\circ)} = 26.1\,\text{N}\n - \sum F_x = 0 \implies F = T \sin(40^\circ) = 26.1 \sin(40^\circ) = 16.8\,\text{N}\n - Engine Crane Example: Engine weight 2000\,\text{N}$suspended by chain at$30^\circ to vertical:\n - Initial state: T{\text{chain}} = F_g = 2000\,\text{N}.\n - Final pulled state: \frac{F_{\text{applied}}}{F_g} = \tan(30^\circ) \implies F_{\text{applied}} = 2000 \tan(30^\circ) = 1155\,\text{N}.\n - Chain tension: T_{\text{chain}} = \sqrt{2000^2 + 1155^2} = 2310\,\text{N}.\n- **One-Dimensional Force Systems**:\n - **Masses Pulled by Strings**:\n - Acceleration: a = \frac{F}{m_1 + m_2}\n - Tension in connecting string B: T_B = m_2 a = \frac{m_2 F}{m_1 + m_2}\n - Numerical Example: m_1 = 5\,\text{kg}$,$m_2 = 3\,\text{kg}$,$F = 16\,\text{N}:\n      a = \frac{16}{5 + 3} = 2\,\text{m/s}^2, \quad T_B = (3\,\text{kg})(2\,\text{m/s}^2) = 6\,\text{N}\n - Friction Example: m_1 = 20\,\text{kg}$,$m_2 = 10\,\text{kg}$,$F_{\text{applied}} = 50\,\text{N}$,$F_{\text{friction}} = 14\,\text{N}:\n      F_{\text{net}} = 50 - 14 = 36\,\text{N} \implies a = \frac{36}{30} = 1.2\,\text{m/s}^2\n - **Bodies in Contact**:\n - Acceleration: a = \frac{F}{m_1 + m_2}\n - Contact force between bodies: F_{\text{contact}} = \frac{m_2 F}{m_1 + m_2}\n - Numerical Example: F = 7.5\,\text{N}$,$m_1 = 5\,\text{kg}$,$m_2 = 10\,\text{kg}:\n      a = \frac{7.5}{15} = 0.5\,\text{m/s}^2, \quad F_{\text{on } m_2} = (5\,\text{kg})(0.5\,\text{m/s}^2) = 2.5\,\text{N}\n - **Horizontal Acceleration via Gravity (Pulley Table System)**:\n - Mass m_1$on frictionless horizontal surface connected to hanging mass$m_2:\n      T = m_1 a, \quad m_2 a = m_2 g - T \implies a = \frac{m_2 g}{m_1 + m_2}\n - Numerical Example: m_1 = 4\,\text{kg}$,$m_2 = 6\,\text{kg}$,$g = 10\,\text{m/s}^2$,$F_w = 60\,\text{N}:\n      a = \frac{60}{4 + 6} = 6\,\text{m/s}^2, \quad T = 4 \times 6 = 24\,\text{N}\n - **Masses Over Pulley (Atwood Machine)**:\n - For m_1 < m_2:\n      a = \frac{(m_2 - m_1)g}{m_1 + m_2}, \quad T = m_1 g + m_1 a\n - Numerical Example: m_1 = 20\,\text{kg}$,$m_2 = 25\,\text{kg}$,$g = 10\,\text{m/s}^2.\n      a = \frac{(25 - 20)(10)}{20 + 25} = 1.11\,\text{m/s}^2\n      T = (20)(10) + (20)(1.11) = 222.22\,\text{N}\n- **Two-Dimensional Force Systems**:\n - **Acceleration Down Slopes**:\n - Gravity component parallel to slope: F_{||} = mg \sin(\theta) \implies a = g \sin(\theta)\n - Gravity component perpendicular to slope: F_{\perp} = mg \cos(\theta) = R\n - Example: M = 4\,\text{kg}$, ramp angle$\theta = 30^\circ$,$g = 10\,\text{m/s}^2:\n      a = 10 \sin(30^\circ) = 5\,\text{m/s}^2\n - **Slopes and Pulleys**:\n - Case 1: If m_1 g \sin(\theta) > m_2 g$,$m_1 slides down ramp:\n      a = \frac{m_1 g \sin(\theta) - m_2 g}{m_1 + m_2}\n - Case 2: If m_2 g > m_1 g \sin(\theta)$,$m_2 falls vertically:\n      a = \frac{m_2 g - m_1 g \sin(\theta)}{m_1 + m_2}\n - Numerical Example: m_1 = 5.0\,\text{kg}$on$30^\circ$ramp,$m_2 = 20.0\,\text{kg}$hanging,$g = 9.8\,\text{m/s}^2.\n      a = \frac{20(9.8) - 5(9.8)\sin(30^\circ)}{5 + 20} = \frac{196 - 24.5}{25} = 6.86\,\text{m/s}^2\n      T = m_2 g - m_2 a = (20)(9.8) - (20)(6.86) = 58.8\,\text{N}\n\n# STATICAL EQUILIBRIUM AND MOMENTS\n\n- **Torque (Moment)**: Turning effect generated by a force about a pivot point.\n  \text{Torque} = F \times d_{\perp}\n  where d_{\perp} is the perpendicular distance from pivot to force line of action.\n- **Couple**: Two equal and oppositely directed parallel forces acting at a separation distance, producing pure rotation.\n  \text{Moment of Couple} = F \times d\n- **Conditions for Complete Equilibrium**:\n 1. \sum \mathbf{F} = 0 (Translational equilibrium; zero acceleration).\n 2. \sum \text{Torque} = 0 (Rotational equilibrium; clockwise moments = anticlockwise moments about any arbitrary pivot).\n- **Scenarios**:\n - **Painter on Scaffolds**:\n - Painter weight 875\,\text{N}$($865\,\text{N}$in force balance) standing$1\,\text{m}$from end A of a$3\,\text{m}$plank ($4\,\text{m} support distance):\n - Taking moments about pivot B:\n      \text{Anticlockwise moments} = F_A \times 4\n      \text{Clockwise moments} = 865 \times 3 = 2595\,\text{N\,m}\n      4 F_A = 2595 \implies F_A = 649\,\text{N}\n      F_A + F_B = 865\,\text{N} \implies F_B = 865 - 649 = 216\,\text{N}\n - **Firefighter on Ladder**:\n - Ladder length 8\,\text{m}$, weight$W_L = 355\,\text{N}$acting at midpoint ($4\,\text{m}$), leaning against smooth wall at angle$50^\circ$to horizontal. Firefighter weight$W_F = 875\,\text{N}$standing$6.30\,\text{m} up ladder.\n - Rotational equilibrium about base:\n      \sum \text{Torque} = (N_{\text{wall}} \times 8 \sin(50^\circ)) - (875 \times 6.30 \cos(50^\circ)) - (355 \times 4 \cos(50^\circ)) = 0\n      6.13 N_{\text{wall}} - 3543 - 913 = 0 \implies 6.13 N_{\text{wall}} = 4456 \implies N_{\text{wall}} = 727\,\text{N}\n - **Diver on Diving Board**:\n - Man weight 530\,\text{N}$at right end of$3.90\,\text{m}$weightless board bolted at left end, supported by fulcrum$1.40\,\text{m} from bolt.\n - Taking moments about bolt:\n      (F_{\text{fulcrum}} \times 1.40) - (530 \times 3.90) = 0 \implies 1.40 F_{\text{fulcrum}} = 2067 \implies F_{\text{fulcrum}} = 1476\,\text{N}\text{ (up)}\n - Vertical equilibrium:\n      -F_{\text{bolt}} + F_{\text{fulcrum}} - 530 = 0 \implies -F_{\text{bolt}} + 1476 - 530 = 0 \implies F_{\text{bolt}} = 946\,\text{N}\text{ (down)}\n\n# KINEMATICS AND PROJECTILE MOTION\n\n- **Kinematic Equations of Motion** (for uniform acceleration in a straight line):\n 1. v = u + at\n 2. d = ut + \frac{1}{2}at^2\n 3. v^2 = u^2 + 2ad\n  where u$is initial velocity,$v$is final velocity,$a$is constant acceleration,$t$is time, and$d is displacement.\n- **Calculations**:\n - Example 1: u = 6\,\text{m/s}$,$a = 2\,\text{m/s}^2\n - Distance after t = 3\,\text{s}$:$d = (6)(3) + \frac{1}{2}(2)(3)^2 = 18 + 9 = 27\,\text{m}\n - Velocity after d = 20\,\text{m}$:$v^2 = 6^2 + 2(2)(20) = 36 + 80 = 116 \implies v = 10.8\,\text{m/s}\n - Example 2: Ball rolling up slope stops (v = 0\,\text{m/s}$) in$d = 16.0\,\text{m}$with$u = 4.0\,\text{m/s}:\n    0^2 = 4.0^2 + 2 a (16.0) \implies -16 = 32 a \implies a = -0.50\,\text{m/s}^2\n- **Vertical Motion Under Gravity** (g = 10\,\text{m/s}^2):\n - Upward movement: a = -10\,\text{m/s}^2$; peak velocity$v = 0\,\text{m/s}.\n - Example 1: Object thrown upwards at u = 30\,\text{m/s}:\n - Max height: 0^2 = 30^2 + 2(-10)d \implies 20d = 900 \implies d = 45\,\text{m}\n - Time to peak: 0 = 30 - 10t \implies t = 3\,\text{s}\n - Example 2: Rock dropped from rest (u = 0\,\text{m/s}$) from a$50\,\text{m} cliff:\n - Distance fallen in t = 2.0\,\text{s}$:$d = \frac{1}{2}(10)(2)^2 = 20\,\text{m}\n - Time to fall total 50\,\text{m}$:$50 = \frac{1}{2}(10)t^2 \implies t^2 = 10 \implies t = 3.16\,\text{s}\n- **Full Projectile Motion**:\n - Parabolic trajectory governed purely by gravity acting vertically downwards.\n - Velocity components for launch angle \theta$and speed$v:\n    v_x = v \cos(\theta), \quad v_y = v \sin(\theta)\n - Horizontal velocity remains constant throughout flight.\n - At maximum height H_{\text{max}}$, vertical velocity$v_y = 0\,\text{m/s}.\n - Total flight time t_{\text{total}} = 2 t_{\text{peak}}.\n - Horizontal Range R = v_x \times t_{\text{total}}.\n - Worked Example: Rocket fired at v = 1000\,\text{m/s}$at$\theta = 30^\circ:\n - Components: v_x = 1000 \cos(30^\circ) = 866\,\text{m/s}$,$v_y = 1000 \sin(30^\circ) = 500\,\text{m/s}\n - Peak Height: 0^2 = 500^2 + 2(-10)d \implies 10d = 250000 \implies d = 25000\,\text{m}\n - Time to Peak: 0 = 500 - 10t \implies t = 50\,\text{s} \implies t_{\text{total}} = 100\,\text{s}\n - Range: R = 866\,\text{m/s} \times 102\,\text{s} = 88332\,\text{m}\n- **Half Projectile Motion**:\n - Horizontal launch (u_y = 0\,\text{m/s}$) from height$h:\n - Worked Example: m = 6\,\text{kg}$launched horizontally at$v_x = 20\,\text{m/s}$from height$h = 25\,\text{m}:\n - Time of flight: 25 = \frac{1}{2}(10)t^2 \implies t = \sqrt{5} = 2.24\,\text{s}\n - Impact vertical velocity: v_y^2 = 0^2 + 2(10)(25) = 500 \implies v_y = 22.36\,\text{m/s}\n - Impact total speed: v = \sqrt{v_x^2 + v_y^2} = \sqrt{20^2 + 22.36^2} = \sqrt{400 + 500} = 29.8\,\text{m/s}\n\n# MOMENTUM, IMPULSE, AND COLLISIONS\n\n- **Momentum** (\mathbf{p}): Vector quantity equal to product of mass and velocity.\n  \mathbf{p} = m \mathbf{v} \quad (\text{units: kg\,m/s})\n - Golf ball example (35\,\text{g}$at$10\,\text{m/s}$):$p = (0.035\,\text{kg})(10\,\text{m/s}) = 0.35\,\text{kg\,m/s}.\n - Trolley example (20\,\text{kg}$at$0.85\,\text{m/s}$South):$p = 17\,\text{kg\,m/s South}.\n - Ship example (40000\,\text{tonne} = 4 \times 10^7\,\text{kg}$at$0.2\,\text{m/s}$):$p = 8 \times 10^6\,\text{kg\,m/s}.\n- **Change in Momentum** (\Delta \mathbf{p}):\n  \Delta \mathbf{p} = \mathbf{p}f - \mathbf{p}_i = m \mathbf{v}_f - m \mathbf{v}_i\n - Rebound Example: 2\,\text{kg}$ball strikes wall at$3\,\text{m/s}$left, rebounds at$3\,\text{m/s} right:\n    \Delta p = (2)(3) - (2)(-3) = 6 + 6 = 12\,\text{kg\,m/s Right}\n- **Impulse**:\n  \text{Impulse} = \mathbf{F} \Delta t = \Delta \mathbf{p}\n - Satellite Thruster Example: m = 300\,\text{kg}$,$v_i = 5000\,\text{m/s}$,$v_f = 6000\,\text{m/s}$, Thruster force$F = 1500\,\text{N}:\n    \Delta p = 300(6000 - 5000) = 300000\,\text{kg\,m/s}\n    1500 \times \Delta t = 300000 \implies \Delta t = 200\,\text{s}\n- **Collisions in Two Dimensions**:\n - **Elastic Collision**: Total Kinetic Energy is conserved before and after collision.\n - **Inelastic Collision**: Total Kinetic Energy is not conserved (energy converted to heat/sound/deformation). Completely inelastic if bodies stick together.\n - Example 1 (Identical Masses 1.5\,\text{kg} Elastic Collision):\n - Ball A moving at 3.0\,\text{m/s}$hits stationary Ball B. Ball A deflects at$2.0\,\text{m/s}$at$90^\circ to Ball B.\n - Initial momentum: p{\text{init}} = 1.5 \times 3.0 = 4.5\,\text{kg\,m/s}.\n - Momentum A after: p_A = 1.5 \times 2.0 = 3.0\,\text{kg\,m/s}.\n - Momentum B after via Pythagoras: p_B = \sqrt{4.5^2 - 3.0^2} = 3.35\,\text{kg\,m/s} \implies v_B = \frac{3.35}{1.5} = 2.2\,\text{m/s}.\n - Angles: \theta_A = \cos^{-1}\left(\frac{3.0}{4.5}\right) = 48.2^\circ, \quad \theta_B = \sin^{-1}\left(\frac{3.0}{4.5}\right) = 42^\circ.\n - Example 2 (Inelastic Two-Car Junction Collision):\n - Police car 800\,\text{kg}$East collides with car$500\,\text{kg}$North, sticking together at common velocity$12\,\text{m/s}$at$\theta = 30^\circ North of East.\n - Combined final momentum: P = (800 + 500)(12) = 15600\,\text{kg\,m/s}.\n - Initial velocity car A (East): 800 v_A = 15600 \cos(30^\circ) = 13510 \implies v_A = 16.9\,\text{m/s}.\n - Initial velocity car B (North): 500 v_B = 15600 \sin(30^\circ) = 7800 \implies v_B = 15.6\,\text{m/s}.\n - Example 3 (Mass Explosion / Splitting):\n - 30\,\text{kg}$mass splits into$10\,\text{kg}$(piece A at$8\,\text{m/s}$,$30^\circ$above horizontal) and$20\,\text{kg}$(piece B at$5\,\text{m/s}$,$60^\circ below horizontal).\n - Momentum A = 10 \times 8 = 80\,\text{kg\,m/s}$; Momentum B =$20 \times 5 = 100\,\text{kg\,m/s}.\n - Perpendicular vector sum: p_{\text{total}}^2 = 80^2 + 100^2 = 16400 \implies p_{\text{total}} = 128.06\,\text{kg\,m/s}.\n - Initial velocity: u = \frac{128.06}{30} = 4.27\,\text{m/s}.\n\n# CIRCULAR MOTION AND GRAVITATION\n\n- **Fundamental Terms**:\n - Period (T): Time taken for one complete rotation/revolution (SI unit: seconds).\n - Frequency (f$): Number of rotations per second (SI unit: Hertz,$\text{Hz}).\n    f = \frac{1}{T}, \quad T = \frac{1}{T}\n - Converts from revolutions per minute (rpm): f\,\text{(Hz)} = \frac{\text{rpm}}{60}.\n- **Tangential Speed**:\n  v = \frac{2 \pi r}{T}\n - Propeller Example: f = 50\,\text{Hz}$($T = 0.02\,\text{s}$), radius$r = 85\,\text{cm} = 0.85\,\text{m}:\n    v = \frac{2 \pi (0.85)}{0.02} = 267\,\text{m/s}\n- **Centripetal Force and Acceleration**:\n - Centripetal force acts continuously at right angles to velocity, directed inward toward the center of curvature.\n    F_c = \frac{m v^2}{r}\n    a_c = \frac{v^2}{r}\n - Hammer throw Example: m = 8.0\,\text{kg}$,$v = 12\,\text{m/s}$,$r = 1.5\,\text{m}:\n    F_c = \frac{(8.0)(12)^2}{1.5} = 768\,\text{N}\text{ (inward)}\n    a_c = \frac{12^2}{1.5} = 96\,\text{m/s}^2\n- **Newton's Universal Law of Gravitation**:\n  F = \frac{G m_1 m_2}{r^2}\n  where Universal Gravitational Constant G = 6.673 \times 10^{-11}\,\text{N\,m}^2\text{kg}^{-2}.\n - Example 1: Two 1\,\text{kg}$masses separated by$10\,\text{cm}$($0.1\,\text{m}):\n    F = \frac{(6.673 \times 10^{-11})(1)(1)}{(0.1)^2} = 6.673 \times 10^{-9}\,\text{N}\n - Example 2: If distance between two masses exerting 18\,\text{N} is halved:\n    F_{\text{new}} = 18 \times 4 = 72\,\text{N}\n\n# ENERGY, WORK, POWER, AND THERMAL PHYSICS\n\n- **Work** (W): Scalar quantity measuring energy transfer.\n  W = F d \cos(\theta)\n - Example: Force 12\,\text{N}$at$60^\circ$over$3\,\text{m}:\n    W = 12 \cos(60^\circ) \times 3 = 18\,\text{J}\n- **Mechanical Energy Forms**:\n - Gravitational Potential Energy: P.E. = mgh\n - Kinetic Energy: K.E. = \frac{1}{2} m v^2\n - Power (P$): Rate of performing work ($1\,\text{Watt} = 1\,\text{J/s}):\n    P = \frac{W}{t}\n- **Elastic Potential Energy and Hooke's Law**:\n - Restoring/Extending Force: F = kx\n - Stored Energy: E.P.E. = \frac{1}{2} k x^2\n - Mass 0.5\,\text{kg}$extending spring by$25\,\text{cm}$($0.25\,\text{m}):\n    k = \frac{mg}{x} = \frac{(0.5)(10)}{0.25} = 20\,\text{N/m}\n    E.P.E. = \frac{1}{2}(20)(0.25)^2 = 0.63\,\text{J}\n- **Thermal Energy Quantities**:\n - Heat Energy (Q): Total energy contained by all particles of a body.\n - Specific Heat Capacity (c$): Heat energy required to raise$1\,\text{kg}$of a substance by$1^\circ\text{C}$(or$1\,\text{K}).\n    Q = m c \Delta T\n - Water specific heat capacity: c = 4200\,\text{J\,kg}^{-1}\text{K}^{-1}.\n - Heating 100\,\text{g}$($0.1\,\text{kg}$) water from$10^\circ\text{C}$to$15^\circ\text{C}:\n      Q = (0.1)(4200)(5) = 2100\,\text{J}\n - Latent Heat (L$): Energy needed to change the state of$1\,\text{kg}$of substance without temperature change ($Q = m L).\n - Latent heat of fusion of ice = 336000\,\text{J/kg}.\n - Energy released when 1\,\text{g}$steam at$100^\circ\text{C}$condenses and cools to$20^\circ\text{C}$($L_v = 2250000\,\text{J/kg}):\n      Q = m L_v + m c \Delta T = (0.001 \times 2250000) + (0.001 \times 4200 \times 80) = 2250 + 336 = 2586\,\text{J}\n- **Conservation of Energy Scenarios**:\n - **Sandbag and Recoil Bullet**:\n - 30\,\text{g}$bullet fired at$400\,\text{m/s}$embeds in$10\,\text{kg} suspended sandbag.\n - Combined velocity after impact:\n      (0.03)(400) + (10)(0) = (0.03 + 10) v \implies 12 = 10.03 v \implies v = 1.2\,\text{m/s}\n - Vertical recoil height h:\n      \frac{1}{2} m v^2 = m g h \implies h = \frac{v^2}{2g} = \frac{1.2^2}{2(10)} = 0.072\,\text{m} = 7.2\,\text{cm}\n - **Bungee Jumper**:\n - 75\,\text{kg}$jumper drops from bridge; head touches river at$30\,\text{m}$drop. Rope natural length =$10\,\text{m}.\n - Total potential energy loss: P.E. = mgh = (75)(10)(30) = 22500\,\text{J}.\n - Extension x = 30 - 10 = 20\,\text{m}.\n - Spring constant calculation:\n      22500 = \frac{1}{2} k (20)^2 \implies 200 k = 22500 \implies k = 112.5\,\text{N/m} \approx 110\,\text{N/m}\n\n# ENVIRONMENTAL PHYSICS AND GREENHOUSE EFFECT\n\n- **Albedo**:\n - The ratio of reflected light from a surface relative to the total incident solar shortwave radiation.\n - High albedo surfaces (bright/reflective): fresh snow, ice.\n - Low albedo surfaces (dark/absorbent): open water, dense forests.\n - Lower albedo leads to increased solar heat absorption in oceans.\n- **Greenhouse Effect**:\n - Natural atmospheric process where gases trap long-wave heat radiation emitted by the Earth's surface.\n - Concept originated in 1824 with Joseph Fourier.\n - Mechanics:\n 1. Shortwave solar radiation passes through atmospheric gases unimpeded.\n 2. Earth absorbs shortwave light and re-radiates it as long-wave infrared (IR) radiation.\n 3. Greenhouse gases absorb long-wave IR radiation and re-emit it in all directions, including back toward Earth's surface.\n - Primary Greenhouse Gases: Water vapour (\text{H}2\text{O}$), Carbon dioxide ($\text{CO}_2$), Methane ($\text{CH}_4$), Nitrous oxide ($\text{N}_2\text{O}$), Ozone ($\text{O}_3).\n - Atmospheric Absorption Spectral Profile: Atmosphere absorbs high-energy UV and long-wave IR, while letting visible light pass to the surface.\n - Responses to Climate Change:\n - Adaptation: Building seawalls, relocating communities.\n - Mitigation: Reducing greenhouse emission rates via regulatory policies.\n - Geo-engineering: Active technological manipulation of the global climate system to offset warming.\n\n# FLUID MECHANICS\n\n- Fluid Mechanics branches into Fluid Dynamics (fluids in motion) and Fluid Statics (fluids at rest).\n- **Bernoulli's Principle**:\n - "The pressure in a fast-moving stream of fluid is lower than in a slower stream of fluid."\n - **Applications**:\n - Aeroplane Wing (Aerofoil): Shape forces air to travel faster over the curved top surface than the flat bottom, generating low pressure on top and high pressure below to produce vertical lift.\n - Venturi Meter: Instrument measuring pressure drop across a constricted pipe section to calculate fluid flow rate.\n - Perfume Atomizer: Squeezing bulb accelerates air past top of tube, dropping local pressure. Higher atmospheric pressure in reservoir forces liquid up the tube into the stream.\n- **Viscosity**:\n - Measure of a fluid's internal resistance to flow or deformation under shear/tensional stress.\n - Kinematic Viscosity: Measure of flow volume over time (units: stokes or \text{m}^2/\text{s}).\n - Dynamic (Absolute) Viscosity: Internal friction measurement under mechanical stress (SI unit: Pascal-seconds \text{Pa\,s}, or poise).\n- **Fluid Statics and Pressure**:\n - Pressure Conversions: 1\,\text{atm} = 1.01325 \times 10^5\,\text{Pa} = 760\,\text{mmHg} = 1.01325\,\text{bar}.\n - Gauge Pressure (P{\text{gauge}}): Pressure relative to ambient atmospheric pressure.\n - Absolute Pressure: P_{\text{total}} = P_{\text{gauge}} + P_{\text{atmosphere}}.\n - Hydrostatic Pressure in Uniform Fluid:\n    P_{\text{gauge}} = \rho g h\n    P_{\text{total}} = P_{\text{atmosphere}} + \rho g h\n - Diver Example: Gauge pressure = 515\,\text{kPa}$:$P_{\text{total}} = 101\,\text{kPa} + 515\,\text{kPa} = 616\,\text{kPa}.\n - Swimming Pool Depth Example (h = 2\,\text{m}$,$\rho = 1000\,\text{kg/m}^3):\n    P_{\text{total}} = 101300 + (1000)(9.8)(2) = 120900\,\text{Pa} = 1.209 \times 10^5\,\text{Pa}\n\n# IDEAL GAS LAWS AND KINETIC THEORY\n\n- **Kinetic Theory of Gases Postulates**:\n 1. Gases consist of small particles in continuous, random linear motion colliding with each other and container walls.\n 2. Particle size is negligible relative to total gas volume (large inter-particle spacing makes gases highly compressible).\n 3. No intermolecular forces or interactions exist between particles.\n 4. All collisions are perfectly elastic (no net kinetic energy lost during impacts).\n 5. Average kinetic energy of particles is directly proportional to absolute temperature (K.E. \propto T).\n- **Gas Laws**:\n - **Boyle's Law** (T$constant):$P \propto \frac{1}{V} \implies P_1 V_1 = P_2 V_2\n - Example: Air at 10^5\,\text{Pa}$in$12\,\text{cm}^3$pump volume extended to$24\,\text{cm}^3:\n      10^5 \times 12 = P_2 \times 24 \implies P_2 = 5.0 \times 10^4\,\text{Pa}\n - **Charles's Law** (P$constant):$V \propto T \implies \frac{V_1}{T_1} = \frac{V_2}{T_2}\n - **Pressure Law** (V$constant):$P \propto T \implies \frac{P_1}{T_1} = \frac{P_2}{T_2}\n- **Derivation of Pressure from Kinetic Theory**:\n - For N$molecules of mass$m$moving with average squared speed$\bar{v}^2$in volume$V:\n    P = \frac{1}{3} \frac{N m \bar{v}^2}{V}\n - Root-Mean-Square Speed: v_{rms} = \sqrt{\bar{v}^2}\n - Combined Ideal Gas Law equation:\n    P V = n R T\n    where R = 8.31\,\text{J\,mol}^{-1}\text{K}^{-1}$and$n is number of moles.\n - Combined ratio formula: \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\n\n# GEOMETRICAL OPTICS AND LIGHT MODELS\n\n- Light is electromagnetic radiation moving at speed c = 3 \times 10^8\,\text{m/s}.\n- **Laws of Reflection**:\n 1. Angle of incidence equals angle of reflection (\theta_i = \theta_r).\n 2. Incident ray, reflected ray, and normal lie on the same plane.\n- **Refraction and Snell's Law**:\n - Refraction occurs due to change in light velocity across media boundaries.\n - Ray bends toward normal entering denser medium; bends away entering less dense medium.\n - Snell's Law:\n    n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\n - Example: Ray traveling from air (n_1 = 1.0$) to glass ($n_2 = 1.5$) at$\theta_1 = 35^\circ:\n    1.0 \sin(35^\circ) = 1.5 \sin(\theta_2) \implies \sin(\theta_2) = 0.3824 \implies \theta_2 = 22.5^\circ\n- **Critical Angle and Total Internal Reflection**:\n - Critical Angle (\theta_c$): Angle of incidence producing an angle of refraction of$90^\circ when moving from denser to less dense medium.\n    \sin(\theta_c) = \frac{n_2}{n_1}\n - Total Internal Reflection occurs when incident angle exceeds critical angle (\theta_i > \theta_c).\n- **Selected Refractive Indices**:\n - Diamond: 2.24$(or$2.4)\n - Ruby: 1.76\n - Flint glass: 1.65\n - Crown glass: 1.52\n - Perspex: 1.49\n - Paraffin oil: 1.44\n - Ethanol: 1.36\n - Water: 1.33\n - Ice: 1.31\n - Air: 1.00\n- **Particle Model of Light**:\n - Treats light as a stream of particles (photons) moving at high speed.\n - Explains sharp shadows, ray crossing without interaction, and inverse square law of illumination.\n - **Failures of Particle Model**:\n 1. Cannot explain partial reflection and partial refraction occurring simultaneously at a boundary.\n 2. Cannot explain diffraction (bending around obstacles).\n 3. Incorrectly predicts light speed is higher in denser media (Foucault measured light speed in water to be slower than in air).\n 4. Cannot explain wave interference patterns (bright and dark bands).\n\n# WAVE MOTION AND INTERFERENCE\n\n- **Boundary Behaviors of String Waves**:\n - Heavy to Light String: Fast transmitted pulse, small reflected pulse in-phase (same way up).\n - Light to Heavy String: Slow transmitted pulse, reflected pulse inverted (180^\circ phase shift).\n - Wave velocity depends on string tension, independent of amplitude.\n- **Principle of Superposition**:\n - Displacements of overlapping waves add algebraically.\n - Constructive Superposition: In-phase waves meet to produce doubled amplitude.\n - Destructive Interference: Crest meets trough resulting in zero or reduced amplitude.\n- **Ripple Tank Phenomena**:\n - Nodal Lines: Undisturbed lines formed by continuous destructive interference.\n - Antinodes: Regions formed by continuous constructive interference (crest meets crest, or trough meets trough).\n - Refraction across shallow water: Wave speed and wavelength decrease, while frequency remains constant.\n    n_{12} = \frac{\sin(\theta_1)}{\sin(\theta_2)} = \frac{v_1}{v_2} = \frac{\lambda_1}{\lambda_2}\n- **Young's Double Slit Experiment (Wave Model of Light)**:\n - Performed by Thomas Young in 1810 to prove wave nature of light via diffraction and interference.\n - Path Difference Formula:\n    \text{Path Difference} = d \sin(\theta)\n - Constructive Interference (Bright Bands / Antinodes):\n    d \sin(\theta) = n \lambda \quad (n = 0, 1, 2, 3, \dots)\n - Destructive Interference (Dark Bands / Nodes):\n    d \sin(\theta) = \left(n - \frac{1}{2}\right) \lambda \quad (n = 1, 2, 3, \dots)\n - Small angle approximation (\sin(\theta) \approx \tan(\theta) = \frac{x}{L}):\n    x = \frac{n \lambda L}{d}\n - Example: First bright band position for n = 1$, slit spacing$d = 1\,\text{mm} = 10^{-3}\,\text{m}$, screen distance$L = 2\,\text{m}$, wavelength$\lambda = 10^{-6}\,\text{m}:\n    x = \frac{(1)(10^{-6})(2)}{10^{-3}} = 2 \times 10^{-3}\,\text{m} = 2\,\text{mm}\n- **Electromagnetic Spectrum Relationship**:\n  c = f \lambda\n  where speed of light in vacuum c = 3 \times 10^8\,\text{m/s}. Higher frequency waves carry higher energy.\n\n# ELECTROSTATICS\n\n- Symbol for charge is q$; SI unit is Coulomb ($\text{C}).\n  1\,\text{C} = 6.25 \times 10^{18}\text{ electrons}\n  \text{Charge of 1 electron } (e) = 1.60 \times 10^{-19}\,\text{C}\n- **Coulomb's Law**:\n  F = \frac{k q_1 q_2}{r^2}\n  where electrostatic constant k = 9 \times 10^9\,\text{N\,m}^2\text{C}^{-2}$.

  • Example 1: $q_1 = +3 \times 10^{-9}\,\text{C}$, $q_2 = -5 \times 10^{-9}\,\text{C}$, $r = 2\,\text{m}$:     $F = \frac{(9 \times 10^9)(3 \times 10^{-9})(5 \times 10^{-9})}{2^2} = 3.37 \times 10^{-8}\,\text{N}\text{ (attractive)}$

  • Example 2 (Comparison between Electrostatic and Gravitational forces for 2 electrons $10^{-10}\,\text{m}$ apart):

  • Electrostatic Force:       $F_e = \frac{(9 \times 10^9)(1.6 \times 10^{-19})^2}{(10^{-10})^2} = 2.3 \times 10^{-8}\,\text{N}\text{ (repulsive)}$

  • Gravitational Force:       $F_g = \frac{(6.67 \times 10^{-11})(9.1 \times 10^{-31})^2}{(10^{-10})^2} = 5.54 \times 10^{-51}\,\text{N}$

• Electric Field ($\mathbf{E}$):   $E = \frac{F}{q}$

  • Field near a point charge:     $E = \frac{k Q}{r^2}$

  • Lint example: Test charge $1.60 \times 10^{-19}\,\text{C}$ experiences force $3.2 \times 10^{-9}\,\text{N}$:     $E = \frac{3.2 \times 10^{-9}}{1.60 \times 10^{-19}} = 2 \times 10^{10}\,\text{N/C}$

  • Fly example: Charge $3.0 \times 10^{-10}\,\text{C}$ at distance $0.02\,\text{m}$:     $E = \frac{(9.0 \times 10^9)(3.0 \times 10^{-10})}{(0.02)^2} = 6800\,\text{N/C}\text{ (radially outwards)}$

  • Two point charges ($Q_1 = +3\,\text{nC}$ at origin, $Q_2 = -4\,\text{nC}$ at $40\,\text{cm}$) evaluated at $10\,\text{cm}$ from $Q_1$:     $E_1 = \frac{(9 \times 10^9)(3 \times 10^{-9})}{(0.1)^2} = 2700\,\text{N/C}$     $E_2 = \frac{(9 \times 10^9)(4 \times 10^{-9})}{(0.3)^2} = 400\,\text{N/C}$     $E_{\text{total}} = 2700 + 400 = 3100\,\text{N/C}\text{ (directed towards } Q_2)$

• Electric Potential Energy ($E_p$):   $E_p = \frac{k Q_1 Q_2}{r}$

  • Potential energy between $7\,\text{nC}$ and $20\,\text{nC}$ charges at $2\,\text{cm}$ ($0.02\,\text{m}$):     $E_p = \frac{(9 \times 10^9)(7 \times 10^{-9})(20 \times 10^{-9})}{0.02} = 6.3 \times 10^{-5}\,\text{J}$

• Electric Potential Difference ($V$):   $V = \frac{W}{q}$

  • Measured in Volts ($1\,\text{V} = 1\,\text{J/C}$).

  • Example: Energy $W = 600\,\text{J}$ to move $2\,\text{C}$ charge gives $V = \frac{600}{2} = 300\,\text{V}$.

• Field Between Parallel Charged Plates:

  • Uniform electric field strength: $E = \frac{V}{d}$

  • Work done moving charge: $W = F d = E q d = V q$

  • Example: Plates separated by $3\,\text{cm}$ ($0.03\,\text{m}$) with $V = 12\,\text{V}$:     $E = \frac{12}{0.03} = 400\,\text{V/m} \quad (\text{or } 400\,\text{N/C})$     $F_{\text{electron}} = E q = (400)(1.6 \times 10^{-19}) = 6.4 \times 10^{-17}\,\text{N}$

  • Electron acceleration across potential difference $100\,\text{V}$:     $V q = \frac{1}{2} m v^2 \implies v = \sqrt{\frac{2 V q}{m}} = \sqrt{\frac{2(100)(1.6 \times 10^{-19})}{9.11 \times 10^{-31}}} = 5.96 \times 10^6\,\text{m/s}$

• Millikan's Oil Drop Experiment:

  • Used a variable voltage across parallel plates to balance gravitational force on charged oil drops.

  • Force balance: $E q = m g \implies \frac{V q}{d} = m g \implies q = \frac{m g d}{V}$

  • Discovered charge quantization: $Q = n e$ where $n = 1, 2, 3, \dots$

  • Numerical Example: Oil drop mass $2.05 \times 10^{-12}\,\text{kg}$, plate separation $5\,\text{cm}$, potential $500\,\text{V}$:

  • Top plate polarity must be positive ($+$).

  • $E = \frac{500}{0.05} = 10000\,\text{N/C}$

  • $q = \frac{m g}{E} = \frac{(2.05 \times 10^{-12})(10)}{10000} = 2.05 \times 10^{-15}\,\text{C}$

• Electron Volt ($\text{eV}$):   $1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}$

CURRENT ELECTRICITY AND DC CIRCUITS

• Fundamental Equations:

  • Current: $I = \frac{\Delta q}{\Delta t}$ ($1\,\text{A} = 1\,\text{C/s}$)

  • Ohm's Law: $V = I R$

  • Power Dissipation: $P = I V = I^2 R = \frac{V^2}{R}$

• Series Circuits:

  • Total Resistance: $R_T = R_1 + R_2 + R_3 + \dots$

  • Current identical through all components: $I_T = I_1 = I_2 = I_3$

  • Voltage divides across components: $V_T = V_1 + V_2 + V_3$

  • Numerical Example ($15\,\Omega, 30\,\Omega, 10\,\Omega$ in series across $25\,\text{V}$):

  • $R_T = 15 + 30 + 10 = 55\,\Omega$

  • $I_T = \frac{25}{55} = 0.455\,\text{A}$

  • Voltage drops: $V_{15} = (0.455)(15) = 6.825\,\text{V}$, $V_{30} = (0.455)(30) = 13.65\,\text{V}$, $V_{10} = (0.455)(10) = 4.55\,\text{V}$

  • Power in $30\,\Omega$ resistor: $P = I^2 R = (0.455)^2(30) = 6.21\,\text{W}$

• Parallel Circuits:

  • Total Resistance: $\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots$

  • Voltage identical across all branches: $V_T = V_1 = V_2 = V_3$

  • Current divides across branches: $I_T = I_1 + I_2 + I_3$

  • Numerical Example ($10\,\Omega, 30\,\Omega, 15\,\Omega$ in parallel across $25\,\text{V}$):

  • $\frac{1}{R_T} = \frac{1}{10} + \frac{1}{30} + \frac{1}{15} = \frac{6}{30} = \frac{1}{5} \implies R_T = 5\,\Omega$

  • $I_T = \frac{25}{5} = 5\,\text{A}$

  • Branch currents: $I_{10} = \frac{25}{10} = 2.5\,\text{A}$, $I_{30} = \frac{25}{30} = 0.833\,\text{A}$, $I_{15} = \frac{25}{15} = 1.67\,\text{A}$

  • Power in $30\,\Omega$ resistor: $P = \frac{V^2}{R} = \frac{25^2}{30} = 20.83\,\text{W}$

• Combination Circuits:

  • Parallel branch ($15\,\Omega$ and $30\,\Omega$) connected in series with a $10\,\Omega$ resistor across $25\,\text{V}$:

  • $\frac{1}{R_p} = \frac{1}{15} + \frac{1}{30} = \frac{3}{30} \implies R_p = 10\,\Omega$

  • Total Resistance $R_T = 10 + 10 = 20\,\Omega$

  • Total Current $I_T = \frac{25}{20} = 1.25\,\text{A}$

  • Voltage drop across series $10\,\Omega$ resistor: $V = (1.25)(10) = 12.5\,\text{V}$

  • Voltage across parallel network: $V_p = 25 - 12.5 = 12.5\,\text{V}$

  • Branch currents: $I_{15} = \frac{12.5}{15} = 0.833\,\text{A}$, $I_{30} = \frac{12.5}{30} = 0.417\,\text{A}$

  • Power in $30\,\Omega$ resistor: $P = (12.5)(0.417) = 5.213\,\text{W}$

• Electrical Safety Devices:

  • Fuses: Contain a thin metal conductor designed to melt and interrupt current flow if rated threshold is exceeded (e.g., $0.25\,\text{A}$ rating).

  • Circuit Breakers: Mechanical reset-able switches triggering on current overload.

  • Ground-Fault Interrupter (GFI): Electronic safety device detecting micro-leakage current differences (as low as $5\,\text{mA}$) to prevent fatal electrocution.

  • Earthing System: Provides a zero-resistance safety pathway to ground.

ELECTROMAGNETISM AND MOTOR EFFECT

• Motor Effect: Current-carrying conductor in a magnetic field experiences a force.   $F = B I L \sin(\theta)$   where $B$ is magnetic field strength in Tesla ($\text{T}$), $I$ is current in Amperes ($\text{A}$), $L$ is wire length inside field in meters ($\text{m}$), and $\theta$ is orientation angle to field lines ($F_{\text{max}}$ at $\theta = 90^\circ$).

  • Numerical Example: Conductor length $0.4\,\text{m}$, current $10.6\,\text{A}$, field $0.003\,\text{T}$ at $90^\circ$:     $F = (0.003)(10.6)(0.4) = 0.13\,\text{N}$

• Fleming's Left Hand Rule: Used to determine force direction on current-carrying wire (Thumb = Force/Motion, First Finger = Magnetic Field, Second Finger = Current).

• DC Electric Motor:

  • Converts electrical energy to mechanical energy via interaction between permanent magnet field and coil field.

  • Components: Single/multi-turn armature coil, split-ring commutator (reverses current direction every half-turn to sustain continuous rotation), carbon brushes.

  • Speed increase methods: Increase magnetic field strength, increase number of coil turns, increase coil current.

• Force on a Moving Charged Particle in a Magnetic Field:   $F = q v B \sin(\theta)$

  • Direction determined via Right Hand Rule (Fingers = velocity, Curl = B-field, Thumb = Force on positive charge; opposite for negative charge).

  • Numerical Example: Electron ($q = 1.6 \times 10^{-19}\,\text{C}$) moving at $150\,\text{m/s}$ perpendicular to $80000\,\text{T}$ field:     $F = (1.6 \times 10^{-19})(150)(80000) = 1.92 \times 10^{-12}\,\text{N}$

• Circular Orbit in Magnetic Field:

  • Magnetic force acts as centripetal force doing zero work ($v$ remains constant):     $\frac{m v^2}{r} = q v B \implies r = \frac{m v}{q B}$

ELECTROMAGNETIC INDUCTION AND GENERATORS

• Induced Voltage in Moving Conductor:   $V = B L v \sin(\theta)$

  • Numerical Example: Wire length $0.3\,\text{m}$ moved at $50\,\text{m/s}$ through $0.015\,\text{T}$ field:     $V = (0.015)(0.3)(50) = 0.23\,\text{V}$

• Rotating Coil EMF:

  • Tangential speed of rotating loop: $v = \frac{2 \pi r N}{60}$

  • Total induced EMF for $N_{turns}$:     $\text{EMF} = N_{turns} B v l$

  • Numerical Example: Coil $N = 200\text{ turns}$, radius $r = 0.12\,\text{m}$, length $l = 0.23\,\text{m}$, $B = 0.06\,\text{T}$, rotated at $3000\,\text{rpm}$:     $v = \frac{2 \pi (0.12)(3000)}{60} = 37.70\,\text{m/s}$     $\text{Single turn } V = (0.06)(0.23)(37.70) = 0.52\,\text{V}$

• Generators:

  • Convert mechanical energy into electrical energy.

  • AC Generator (Alternator): Uses slip rings to output alternating current.

  • DC Generator: Uses a split-ring commutator to output unidirectional pulsed DC current.

• Lenz's Law:

  • "The polarity of induced EMF is such that it produces a current whose magnetic field opposes the change in magnetic flux that produced it."

  • Approaching North Pole induces North polarity on facing coil face (anticlockwise current) to repel approach.

  • Receding North Pole induces South polarity on facing coil face (clockwise current) to attract receding pole.

• Transformers:

  • Step-up or step-down voltage using mutual induction across a soft iron core.

  • Voltage-turn ratio equation:     $\frac{V_p}{V_s} = \frac{N_p}{N_s}$

  • Ideal Transformer Power Conservation ($100\%$ efficient):     $P_{\text{in}} = P_{\text{out}} \implies V_p I_p = V_s I_s \implies \frac{I_p}{I_s} = \frac{V_s}{V_p} = \frac{N_s}{N_p}$

  • Step-down Example: Mains voltage $230\,\text{V}$ reduced to $11.5\,\text{V}$ with $1200$ primary turns:     $N_s = 1200 \times \left(\frac{11.5}{230}\right) = 60\text{ turns}$

• Long Distance Electrical Transmission:

  • High voltage transmission minimizes current $I$, dramatically reducing line heat loss ($P_{\text{loss}} = I^2 R$).

ATOMIC PHYSICS AND RADIOACTIVITY

• Forms of Radiation:

  • Alpha ($\alpha$): Helium nucleus ($_2^4\text{He}$, charge $+2e$); stopped by a single sheet of paper; deflected as positive in B-field.

  • Beta ($\beta^-$): High-speed electron ($_{-1}^0 e$); stopped by$3\,\text{mm} of aluminum; deflected as negative in B-field.\n - **Gamma** (\gamma): High-energy electromagnetic photon; stopped by several centimeters of lead; undeflected in magnetic fields.\n- **Rutherford's Gold Foil Experiment (1911)**:\n - Performed by Ernest Rutherford, Hans Geiger, and Ernest Marsden targeting alpha particles at thin gold foil.\n - **Observations**:\n 1. Most alpha particles passed straight through undeflected.\n 2. Some alpha particles were deflected through large angles.\n 3. A tiny fraction reversed direction completely (bounced backward).\n - **Conclusions**:\n 1. Most of the atom's volume is empty space.\n 2. Positive charge and almost all atomic mass are concentrated in a tiny central region called the nucleus.\n - **Drawbacks of Planetary Model**: Accelerating orbiting electrons should continuously radiate energy, spiraling into the nucleus; model fails to explain electronic shell configurations.\n- **Nuclear Decay Equations**:\n - Alpha decay: {87}^{168}\text{Ir} \rightarrow {85}^{164}\text{Re} + 2^4\text{He}\n - Beta-minus decay: Neutron converts to proton emitting an electron (_{-1}^0 e).\n - Gamma decay: Nucleus drops from excited state without changing mass or atomic numbers.\n- **Half-Life** (T{1/2}):\n - Time required for half of the radioactive parent nuclei in a sample to decay.\n - Graphical Determination Example: Time required for count rate to drop from 400$to$200\text{ counts/min}$on a decay curve yields$T_{1/2} = 8\,\text{minutes}.\n\n# PHOTOELECTRIC EFFECT\n\n- Photoelectric effect occurs when light above a threshold frequency (f_0) strikes a metal surface, emitting photoelectrons.\n- Key experimental findings:\n 1. Emission occurs instantly if incident frequency f \ge f_0, regardless of light intensity.\n 2. Maximum kinetic energy of photoelectrons depends strictly on light frequency, not intensity.\n 3. Increasing light intensity increases the rate (number) of photoelectrons emitted per second.\n- **Photon Concept and Einstein's Photoelectric Equation**:\n - Light energy arrives in quantized packets called photons:\n    E = h f = \frac{h c}{\lambda}\n    where Planck's constant h = 6.63 \times 10^{-34}\,\text{J\,s}.\n - Einstein's Conservation Equation:\n    \text{Photon Energy } (E) = \text{Work Function } (\phi) + \text{Max Kinetic Energy } (E_k)\n    E_k = h f - \phi\n- **Important Terms and Definitions**:\n - Work Function (\phi$): Minimum energy needed to eject an electron from the metal surface ($\phi = h f_0).\n - Threshold Frequency (f_0): Minimum light frequency required for photoelectric emission.\n - Threshold Wavelength (\lambda_0$): Maximum wavelength capable of causing emission ($\lambda_0 = \frac{c}{f_0}).\n - Stopping Potential / Cut-off Voltage (V_{co}$): Retarding potential needed to reduce photoelectric current to zero ($E_k = e V_{co}).\n- **Linear Graph Properties** (E_k$vs$f):\n - Slope = Planck's constant (h).\n - Horizontal x-intercept = Threshold frequency (f_0).\n - Vertical y-intercept = Negative work function (-\phi).\n- **Calculations**:\n - Example 1: Radiation photon energy E = 1.6 \times 10^{-13}\,\text{J}:\n    f = \frac{E}{h} = \frac{1.6 \times 10^{-13}}{6.63 \times 10^{-34}} = 2.41 \times 10^{20}\,\text{Hz}\n    \lambda = \frac{c}{f} = \frac{3 \times 10^8}{2.41 \times 10^{20}} = 1.24 \times 10^{-12}\,\text{m}\n - Example 2: Metal work function \phi = 3.0\,\text{eV} = 3.0 \times (1.6 \times 10^{-19}\,\text{J}) = 4.8 \times 10^{-19}\,\text{J}:\n    f_0 = \frac{\phi}{h} = \frac{4.8 \times 10^{-19}}{6.63 \times 10^{-34}} = 7.24 \times 10^{14}\,\text{Hz}$$