Comprehensive Study Notes on Motion, Force, and Laws of Motion

Foundations of Kinematics and Motion Descriptors

• Historical Scientific Foundations:

  • The phenomenon of motion was placed on a rigorous scientific basis by two key scientists: Galileo Galilei and Isaac Newton.

• Relativity of Rest and Motion:

  • Rest and motion are relative terms rather than absolute conditions; an object's state depends entirely on the reference frame of the observer.

  • Example: Passengers inside a moving bus observe trees, buildings, and roadside bystanders moving backwards. Conversely, a bystander standing on the roadside sees the bus and all passengers inside moving forward. However, each passenger inside the bus observes fellow passengers sitting still without moving.

• Specifying Location and Reference Points:

  • To describe the position of an object, a reference point called the origin must be specified.

  • Example: Stating that a city library is located $2\,\text{km}$ north of the railway station establishes the railway station as the reference origin.

• Distance vs. Displacement:

  • Distance: The total length of the actual path covered by a body, regardless of its direction of motion. It is a scalar quantity, always positive or zero, and depends on the specific path chosen between two points.

  • Displacement: The shortest straight-line distance measured from the initial position to the final position of an object in a given direction. It is a vector quantity, can be positive, negative, or zero, and is independent of the path taken.

  • Key Relationships:

  • The magnitude of displacement between two points is always less than or equal to the distance covered ($\text{Displacement} \le \text{Distance}$).

  • Displacement can be zero even if distance is non-zero, provided the object returns to its initial starting point.

  • Example (Vertical Throw): If a ball is thrown vertically upwards from position $P$ to highest point $Q$ and returns to $P$, the net displacement is $0$, while the total distance travelled is $2 \times \text{distance}(P, Q)$.

  • Example (Circular Track): A jogger completing one full lap around a circular park returns to the starting point, resulting in a displacement of $0$

• Speed, Velocity, and Acceleration:

  • Speed: The rate of change of distance ($\text{Speed} = \frac{\text{Distance}}{\text{Time}}$). It is a scalar quantity with the SI unit $\text{m\,s}^{-1}$ and is always positive.

  • Speed Comparison: $30\,\text{m/s}$ is greater than $30\,\text{km/h}$, because $30\,\text{m/s} = 30 \times \frac{18}{5}\,\text{km/h} = 108\,\text{km/h}$.

  • Velocity: The rate of change of displacement in a specified direction ($\text{Velocity} = \frac{\text{Displacement}}{\text{Time}}$). It is a vector quantity with the SI unit $\text{m\,s}^{-1}$ and can be positive, negative, or zero.

  • Average Speed: Total distance covered divided by total time taken ($\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$). Unit is $\text{m\,s}^{-1}$.

  • Average Velocity: Total displacement divided by total time taken ($\text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Total Time}}$).

  • Condition for Equality: The magnitude of average velocity equals average speed if and only if the total distance travelled equals the magnitude of displacement (i.e., motion in a straight line without changing direction).

  • Instantaneous Velocity: The velocity of a body at any specific instant during its motion (e.g., a speedometer reading $40\,\text{km\,h}^{-1}$ at a specific second).

  • Uniform Velocity: An object covers equal distances in equal time intervals in a constant direction (e.g., moving at $40\,\text{km\,h}^{-1}$ due west).

  • Non-Uniform (Variable) Velocity: An object covers unequal distances in equal time intervals, or its direction of motion changes continuously (e.g., a revolving fan operating at constant speed).

• Angular and Circular Motion Definitions:

  • One Radian: The angle subtended at the centre of a circle by an arc whose length is equal to the radius of the circle.

  • Relation Between Linear and Angular Velocity:     $\text{Linear velocity} = \text{Angular velocity} \times \text{Radius of circular path}$

• Measurement Instruments:

  • Odometer (Odograph): A instrument in automobiles that measures total distance covered based on the perimeter of the rotating wheel.

Equations of Motion and Graphical Analysis

• Acceleration Dynamics:

  • Acceleration ($a$): The rate of change of velocity ($a = \frac{v - u}{t}$).

  • Meaning of $2\,\text{m/s}^2$: The velocity of the object increases by $2\,\text{m/s}$ every second.

  • Uniform Acceleration: Velocity increases or decreases by equal amounts in equal intervals of time along a straight-line path (e.g., a freely falling object).

  • Non-Uniform Acceleration: Velocity changes at variable rates over equal time intervals (e.g., a bus accelerating away from a bus stop).

  • Negative Acceleration (Retardation/Deceleration): Occurs when final velocity is less than initial velocity as speed decreases over time (e.g., applying brakes to a moving truck).

  • Uniform Linear Motion: Motion along a straight path with constant speed, meaning acceleration is zero ($a = 0$); uniform linear motion cannot be accelerated.

• Graphical Analysis of Motion:

  • Distance-Time ($x-t$) Graphs:

  • Straight Line with Positive Slope: Represents uniform speed/motion.

  • Curved Line: Represents non-uniform motion.

  • Horizontal Line Parallel to Time Axis: Represents an object at rest ($v = 0$).

  • Functions: Shows position at any instant, distance covered over time, and speed/velocity.

  • Velocity-Time ($v-t$) Graphs:

  • Horizontal Line Parallel to Time Axis: Represents uniform velocity (zero acceleration).

  • Straight Inclined Line: Represents uniform acceleration.

  • Area Under Curve: The area enclosed beneath a velocity-time graph equals the total displacement ($\text{Area} = \text{Velocity} \times \text{Time} = \text{Displacement}$).

• Derivation of First Equation of Motion ($v = u + at$):

  • Consider an object moving with uniform acceleration $a$, initial velocity $u \neq 0$ at point $A$ ($OA = u$), and final velocity $v$ at point $B$ ($BC = v$) over time $t$ ($OC = t$).

  • Line $AD$ is drawn parallel to time axis $OC$.

  • Change in velocity in time $t$ is given by $BD = BC - CD = BC - OA = v - u$.

  • From the definition of acceleration from the graph:     $a = \frac{\text{Change in velocity}}{\text{Time Taken}} = \frac{BD}{AD} = \frac{BD}{OC} = \frac{BD}{t}$

  • Rearranging yields $BD = at$.

  • Substituting $BD = v - u$ gives:     $v - u = at \implies v = u + at$

• Full Set of Kinematic Equations:

  1. $v = u + at$

  2. $s = ut + \frac{1}{2}at^2$

  3. $v^2 - u^2 = 2as$

Force, Acceleration, and Types of Forces

• Concept of Force:

  • Definition: A push or pull exerted on an object that produces or tends to produce acceleration in the body.

  • SI Unit: Newton ($\text{N}$).

  • Definition of One Newton: The amount of force that produces an acceleration of $1\,\text{m/s}^2$ when applied to a mass of $1\,\text{kg}$ ($1\,\text{N} = 1\,\text{kg\,m/s}^2$).

• Effects of Force on a Body:

  • Changes the speed of a body.

  • Changes the direction of motion of a body.

  • Changes the shape or size of a body.

• Balanced vs. Unbalanced Forces:

  • Balanced Forces: Opposing forces acting on a body that are equal in magnitude. They result in a net force of zero and do not alter the state of rest or uniform motion (e.g., equal teams in a tug-of-war).

  • Unbalanced Forces: Opposing forces acting on a body that are unequal in magnitude. They produce a net force greater than zero, altering the state of rest, speed, or direction of motion (e.g., pushing a resting crate to make it slide).

• Frictional Force:

  • The resistive force that always acts opposite to the direction of motion between two contacting surfaces.

Newton's Laws of Motion

• Newton's First Law of Motion (Law of Inertia):

  • Statement: An object remains in its state of rest or of uniform motion in a straight line unless acted upon by an external unbalanced force.

  • Conceptual Focus: Establishes the necessity of an external unbalanced force to alter a body's velocity.

• Newton's Second Law of Motion:

  • Statement: The rate of change of momentum of an object is directly proportional to the applied unbalanced force and takes place in the direction of the force.

  • Formula: $F = ma$.

  • Conceptual Focus: Provides the quantitative measure of force required to produce a specific acceleration.

• Newton's Third Law of Motion:

  • Statement: To every action, there is always an equal and opposite reaction, and they act on two different bodies ($F_{AB} = -F_{BA}$).

  • Conceptual Focus: Explains the interrelationship between internal and external forces during interaction.

• Interrelation Among the Three Laws:

  • The First Law defines when motion changes (need for unbalanced force).

  • The Second Law determines how much force is needed to produce an acceleration.

  • The Third Law explains how forces interact between two bodies.

Inertia, Momentum, and Conservation Laws

• Inertia:

  • Definition: The natural inherent tendency of an object to resist any change in its state of rest or uniform motion.

  • Measure of Inertia: The mass of an object is the measure of its inertia. SI unit is $\text{kg}$. Higher mass corresponds to greater inertia (e.g., a $20\,\text{kg}$ mass has greater inertia than a $10\,\text{kg}$ mass; a stone has more inertia than a same-sized rubber ball; a train has more inertia than a bicycle; a 5-rupee coin has more inertia than a 1-rupee coin).

  • Types and Examples of Inertia:

  • Inertia of Rest: Tendency to stay at rest.

    • Vigorous shaking of a tree branch causes weakly attached leaves to detach and fall due to inertia of rest.

    • Beating a carpet with a stick moves the carpet forward while dust particles remain at rest due to inertia, detaching them.

    • Passengers jerk backward when a resting bus suddenly accelerates forward.

  • Inertia of Motion: Tendency to maintain uniform linear velocity.

    • Passengers fall forward when a fast bus applies sudden brakes.

    • Jumping out of a moving bus causes feet to stop upon touching the ground while the upper body moves forward due to inertia, leading to fall (must run forward to prevent injury).

    • Luggage placed on a bus roof slips backward during sudden forward acceleration and forward during sudden braking; hence, tying with a rope is necessary.

  • Inertia of Direction: Tendency to maintain a straight line path.

    • When a car takes a sharp turn at high speed, passengers slip outward due to inertia resisting directional change.

    • Mountain roads are inclined inward at sharp turns to counteract the vehicle's tendency to fall outward due to directional inertia.

• Momentum ($p$):

  • Definition: The product of an object's mass ($m$) and its velocity ($v$) ($p = mv$).

  • Properties: It is a vector quantity pointing in the direction of velocity. SI unit is $\text{kg\,m/s}$.

  • Determining Factors: Momentum depends directly on both mass and velocity.

• Law of Conservation of Momentum:

  • Statement: The total sum of momenta of two colliding objects before collision equals the total sum of momenta after collision, provided no external unbalanced force acts on the system.

  • Mathematical Derivation:

  • Consider two bodies $A$ and $B$ with masses $m_A, m_B$, initial velocities $u_A, u_B$, and final velocities $v_A, v_B$ after colliding for time $t$.

  • Rate of change of momentum of $A$ ($F_{BA}$, force exerted by $B$ on $A$):       $F_{BA} = \frac{m_A(v_A - u_A)}{t}$

  • Rate of change of momentum of $B$ ($F_{AB}$, force exerted by $A$ on $B$):       $F_{AB} = \frac{m_B(v_B - u_B)}{t}$

  • By Newton's Third Law, $F_{AB} = -F_{BA}$:       $\frac{m_B(v_B - u_B)}{t} = -\frac{m_A(v_A - u_A)}{t}$

  • Multiplying by $t$ and expanding yields:       $m_B v_B - m_B u_B = -m_A v_A + m_A u_A$

  • Rearranging gives the conservation equation:       $m_A u_A + m_B u_B = m_A v_A + m_B v_B$

• Practical Applications of Motion Laws and Momentum:

  • Cricket Fielder: Fielders pull hands backward while catching a high-speed ball to increase impact time $t$, reducing deceleration $a$ and lowering impact force $F = ma$.

  • High Jump Event: Athletes land on cushioned or sand beds to extend collision time, reducing momentum change rate and landing force.

  • Karate Strike: A karate player strikes an ice slab with high velocity in an extremely short time interval, exerting huge force to break it.

  • Athletic Starting Blocks: Athletes push backward on solid foot supports; the support pushes them forward with equal and opposite force for a quick start.

  • Seat Belts: Restrain forward body movement during sudden deceleration, extending stopping time to prevent impact injury.

  • Kicking Stone vs. Football: Kicking a football (low mass/inertia) yields large acceleration; kicking a heavy stone (high mass/inertia) exerts equal and opposite reaction force, hurting the foot.

  • Jumping on Concrete: Stopping abruptly on a hard concrete surface results in a short stopping time, high retardation, and high opposing force causing injury.

  • Swimming: Swimmer pushes water backward (action); water pushes swimmer forward (reaction).

  • Gun Recoil: Gun pushes bullet forward (action); bullet exerts equal backward force on gun (recoil reaction).

  • Rocket Propulsion: Hot exhaust gases expelled downward (action); gases exert upward force on rocket (reaction).

  • Horse and Cart: Horse exerts backward force on ground; ground exerts forward reaction force on horse, driving cart forward.

  • Fireman Hose: High-velocity water ejected forward exerts equal backward force on hose, making stability difficult for fireman.

Solved Numerical Problems and Conceptual Applications

• Kinematics Problems:

  • Cyclist Average Speed: Cyclist covers $3\,\text{m/s}$ for $t\,\text{s}$ and $5\,\text{m/s}$ for $t\,\text{s}$.

  • $\text{Total Time} = 2t\,\text{s}$, $\text{Total Distance} = 3t + 5t = 8t\,\text{m}$.

  • $\text{Average Speed} = \frac{8t}{2t} = 4\,\text{m/s}$.

  • Bus Deceleration: Bus speed drops from $80\,\text{km/h}$ ($22.22\,\text{m/s}$) to $60\,\text{km/h}$ ($16.66\,\text{m/s}$) in $5\,\text{s}$.

  • $a = \frac{16.66 - 22.22}{5} = -1.112\,\text{m\,s}^{-2}$.

  • Train Acceleration: Train reaches $40\,\text{km/h}$ ($11.11\,\text{m/s}$) from rest in $10\,\text{min}$ ($600\,\text{s}$).

  • $a = \frac{11.11}{600} = 0.0185\,\text{m\,s}^{-2}$ (or $240\,\text{km\,h}^{-2}$).

  • Bus Uniform Acceleration: Starts from rest ($u = 0$), $a = 0.1\,\text{m\,s}^{-2}$ for $2\,\text{min}$ ($120\,\text{s}$).

  • $v = 0 + (0.1)(120) = 12\,\text{m/s}$.

  • $S = (0)(120) + \frac{1}{2}(0.1)(120)^2 = 720\,\text{m}$.

  • Train Stopping Distance: $u = 90\,\text{km/h} = 25\,\text{m/s}$, $v = 0$, $a = -0.5\,\text{m\,s}^{-2}$.

  • $v^2 - u^2 = 2aS \implies 0^2 - 25^2 = 2(-0.5)S \implies S = 625\,\text{m}$.

  • Trolley Acceleration: Down incline with $a = 2\,\text{cm\,s}^{-2}$ ($0.02\,\text{m\,s}^{-2}$), $t = 3\,\text{s}$.

  • $v = 0 + (2)(3) = 6\,\text{cm\,s}^{-1} = 0.06\,\text{m/s}$.

  • Racing Car Distance: $u = 0$, $a = 4\,\text{m\,s}^{-2}$, $t = 10\,\text{s}$.

  • $S = \frac{1}{2}(4)(10)^2 = 200\,\text{m}$.

  • Train Crossing Bridge: Speed $72\,\text{km/h} = 20\,\text{m/s}$, bridge length $500\,\text{m}$, time $1\,\text{min}$ ($60\,\text{s}$).

  • $\text{Total Distance} = 20 \times 60 = 1200\,\text{m}$.

  • $\text{Train Length} = 1200\,\text{m} - 500\,\text{m} = 700\,\text{m}$.

  • Farmer Square Boundary: Field side $10\,\text{m}$ (perimeter $40\,\text{m}$), covered in $40\,\text{s}$ ($\text{speed} = 1\,\text{m/s}$). In $2\,\text{min } 20\,\text{s}$ ($140\,\text{s}$):

  • $\text{Distance} = 140\,\text{m}$; $\text{Rounds} = \frac{140}{40} = 3.5$.

  • Final position is diagonally opposite corner: $s = \sqrt{10^2 + 10^2} = 10\sqrt{2} = 14.14\,\text{m}$.

  • Spaceship Signal: Time $5\,\text{min}$ ($300\,\text{s}$), speed of light $3 \times 10^8\,\text{m/s}$.

  • $\text{Distance} = (3 \times 10^8) \times 300 = 9 \times 10^{10}\,\text{m}$.

  • Upward Stone Throw: $u = 5\,\text{m/s}$, downward $a = -10\,\text{m\,s}^{-2}$, v = 0$.\n * v^2 - u^2 = 2as \implies 0 - 25 = 2(-10)s \implies s = 1.25\,\text{m}.\n * v = u + at \implies 0 = 5 - 10t \implies t = 0.5\,\text{s}.\n * **Athlete Circular Lap:** Diameter 200\,\text{m}$($\text{perimeter} = 200\pi\,\text{m}$) in$40\,\text{s}$. Total time$140\,\text{s}$($3.5 rounds).\n * \text{Distance} = 3.5 \times 200\pi = 2200\,\text{m}.\n * \text{Displacement} = \text{Diameter} = 200\,\text{m}.\n * **Joseph Jogging:** 300\,\text{m}$straight from A to B in$150\,\text{s}$, then$100\,\text{m}$back to C in$60\,\text{s}.\n * From A to B: \text{Avg Speed} = \frac{300}{150} = 2\,\text{m/s}$;$\text{Avg Velocity} = \frac{300}{150} = 2\,\text{m/s}.\n * From A to C: \text{Avg Speed} = \frac{400}{210} = 1.9\,\text{m/s}$;$\text{Avg Velocity} = \frac{200}{210} = 0.95\,\text{m/s}.\n * **Abdul Trip:** Outbound speed 20\,\text{km/h}$, return speed$30\,\text{km/h}$over distance$d$.

  • $t_1 = \frac{d}{20}$, $t_2 = \frac{d}{30}$.

  • $\text{Avg Speed} = \frac{2d}{\frac{d}{20} + \frac{d}{30}} = \frac{2d}{\frac{5d}{60}} = 24\,\text{km/h}$.

  • Motorboat Distance: $u = 0$, $a = 3.0\,\text{m\,s}^{-2}$, $t = 8.0\,\text{s}$.

  • $S = \frac{1}{2}(3)(8)^2 = 96\,\text{m}$.

  • Car Braking Comparison: Car 1 ($52\,\text{km/h} = 14.44\,\text{m/s}$) stops in $5\,\text{s}$; Car 2 ($3\,\text{km/h} = 0.83\,\text{m/s}$) stops in $10\,\text{s}$.

  • $\text{Displacement Car 1} = \frac{1}{2} \times 5 \times 14.44 = 36\,\text{m}$.

  • $\text{Displacement Car 2} = \frac{1}{2} \times 10 \times 0.83 = 4.15\,\text{m}$.

  • Car 1 travels farther after braking.

  • Artificial Satellite Speed: Radius $r = 42250\,\text{km}$, period $t = 24\,\text{h}$.

  • $\text{Circumference} = 2\pi r = 265571.42\,\text{km}$.

  • $\text{Speed} = \frac{265571.42}{24} = 11065.4\,\text{km/h}$.

• Dynamics and Conservation Problems:

  • Rifle Recoil: Rifle mass $4\,\text{kg}$ ($4000\,\text{g}$), bullet mass $50\,\text{g}$, bullet velocity $35\,\text{m/s}$.

  • $m_1 v_1 + m_2 v_2 = 0 \implies (50)(35) + 4000 v_2 = 0 \implies v_2 = -0.4375\,\text{m/s}$.

  • Two-Body Collision: $m_1 = 100\,\text{g}$, $m_2 = 200\,\text{g}$, $u_1 = 2\,\text{m/s}$, $u_2 = 1\,\text{m/s}$, $v_1 = 1.67\,\text{m/s}$.

  • $\text{Initial Momentum} = (100)(2) + (200)(1) = 400\,\text{g\,m\,s}^{-1}$.

  • $(100)(1.67) + 200 v_2 = 400 \implies 200 v_2 = 233 \implies v_2 = 1.165\,\text{m/s}$.

  • Football Velocity Changes: A football player kicks ball to teammate (1), teammate kicks to goalkeeper (2), goalkeeper stops ball (3), goalkeeper kicks to team player (4). Velocity changes 4 times.

  • Truck Acceleration and Force: Truck mass $7\,\text{tonnes}$ ($7000\,\text{kg}$), distance $400\,\text{m}$ in $20\,\text{s}$ from rest.

  • $400 = 0 + \frac{1}{2} a (20)^2 \implies a = 2\,\text{m\,s}^{-2}$.

  • $F = ma = 7000 \times 2 = 14000\,\text{N}$.

  • Stone on Ice: Mass $1\,\text{kg}$, $u = 20\,\text{m/s}$, stops ($v = 0$) in $50\,\text{m}$.

  • $0^2 - 20^2 = 2 a (50) \implies a = -4\,\text{m\,s}^{-2}$.

  • $F = ma = 1 \times (-4) = -4\,\text{N}$ (Friction force is $4\,\text{N}$ opposing motion).

  • Train Engine Acceleration: Engine $8000\,\text{kg}$, 5 wagons $2000\,\text{kg}$ each ($\text{total mass} = 18000\,\text{kg}$). Engine force $40000\,\text{N}$, track friction $5000\,\text{N}$.

  • $\text{Net Accelerating Force} = 40000\,\text{N} - 5000\,\text{N} = 35000\,\text{N}$.

  • $\text{Acceleration} = \frac{35000}{18000} = 1.94\,\text{m\,s}^{-2}$.

  • Automobile Braking Force: Mass $1500\,\text{kg}$, negative acceleration $-1.7\,\text{m\,s}^{-2}$.

  • $F = ma = 1500 \times (-1.7) = -2550\,\text{N}$.

  • Moving Cabinet: Moving a wooden cabinet at constant velocity using $200\,\text{N}$ horizontal force requires an opposing friction force of $-200\,\text{N}$ ($a = 0$).

  • Head-On Sticky Collision: Two objects of $1.5\,\text{kg}$ each moving in opposite directions at $2.5\,\text{m/s}$ stick together.

  • \text{Initial Momentum} = (1.5)(2.5) + (1.5)(-2.5) = 0$.\n * Combined mass 3.0\,\text{kg}$, final velocity$v = 0\,\text{m/s}.\n * **Parked Truck Push Logic:** A student claims equal and opposite reaction forces cancel each other when pushing a massive parked truck. *Correction:* The truck fails to move because the applied force is smaller than the maximum static friction force between tires and road, not because action-reaction forces cancel (action and reaction act on *different* bodies).\n * **Hockey Ball Momentum Change:** Mass 200\,\text{g}$($0.2\,\text{kg}$),$u = 10\,\text{m/s}$, struck to return at$v = -5\,\text{m/s}.\n * \text{Initial Momentum} = 0.2 \times 10 = 2\,\text{kg\,m\,s}^{-1}.\n * \text{Final Momentum} = 0.2 \times (-5) = -1\,\text{kg\,m\,s}^{-1}.\n * \text{Change in Momentum} = -1 - 2 = -3\,\text{kg\,m\,s}^{-1}$(reduces by$3\,\text{kg\,m\,s}^{-1}$or$3000\,\text{g\,m\,s}^{-1}).\n * **Bullet Penetration into Block:** Bullet mass 10\,\text{g}$($0.01\,\text{kg}$),$u = 150\,\text{m/s}$, stops in$0.03\,\text{s}.\n * 0 = 150 + a(0.03) \implies a = -5000\,\text{m\,s}^{-2}.\n * \text{Penetration Distance } s = \frac{0^2 - 150^2}{2(-5000)} = 2.25\,\text{m}.\n * \text{Exerted Force } F = (0.01)(-5000) = -50\,\text{N}.\n * **Inelastic Block Collision:** 1\,\text{kg}$object at$10\,\text{m/s}$collides and sticks to stationary$5\,\text{kg} block.\n * \text{Total Initial Momentum} = 1 \times 10 + 5 \times 0 = 10\,\text{kg\,m\,s}^{-1}.\n * \text{Total Final Momentum} = 10\,\text{kg\,m\,s}^{-1}.\n * Combined mass 6\,\text{kg}$, final velocity$v = \frac{10}{6} = 1.66\,\text{m/s}$($1.67\,\text{m/s}).\n * **Object Momentum and Force:** Mass 100\,\text{kg}$, accelerated from$5\,\text{m/s}$to$8\,\text{m/s}$in$6\,\text{s}.\n * \text{Initial Momentum} = 100 \times 5 = 500\,\text{kg\,m\,s}^{-1}.\n * \text{Final Momentum} = 100 \times 8 = 800\,\text{kg\,m\,s}^{-1}.\n * a = \frac{8 - 5}{6} = 0.5\,\text{m\,s}^{-2}.\n * \text{Applied Force } F = 100 \times 0.5 = 50\,\text{N}.\n * **Insect Windshield Collision Debates:**\n * *Kiran's Claim* (Insect suffered greater momentum change): **False**; momentum change is equal in magnitude for both insect and car.\n * *Akhtar's Claim* (Car velocity exerted greater force): **Partially Right**; large mass produces high impact force, but forces are equal and opposite by Third Law.\n * *Rahul's Claim* (Both felt same force and same momentum change): **Completely Right**; equal/opposite force by Third Law, equal momentum change magnitude by Conservation Law.\n * **Dumb-bell Floor Impact:** Mass 10\,\text{kg}$dropped from$80\,\text{cm}$($0.8\,\text{m}$),$a = 10\,\text{m\,s}^{-2}.\n * v^2 - 0 = 2(10)(0.8) = 16 \implies v = 4\,\text{m/s}.\n * \text{Transferred Momentum} = mv = 10 \times 4 = 40\,\text{kg\,m\,s}^{-1}.\n * **Cubic Distance Table Analysis:** Distance values given as d = t^3$($0, 1, 8, 27, 64, 125, 216, 343\,\text{m}$for$t = 0$to$7\,\text{s}).\n * Acceleration is **increasing** over time.\n * Net force acting on the object is **increasing** over time (F = ma).\n * **Three Persons Pushing Car:** Two persons push 1200\,\text{kg}$car at uniform velocity ($a = 0$). Adding a third person produces$a = 0.2\,\text{m\,s}^{-2}.\n * \text{Force of Third Person} = ma = 1200 \times 0.2 = 240\,\text{N}.\n * Since effort is equal, each person pushes with a force of **240\,\text{N}**.\n * **Hammer on Nail:** Mass 500\,\text{g}$($0.5\,\text{kg}$),$u = 50\,\text{m/s}$, stopped in$0.01\,\text{s}.\n * a = \frac{0 - 50}{0.01} = -5000\,\text{m\,s}^{-2}.\n * \text{Force of Hammer on Nail} = 0.5 \times (-5000) = -2500\,\text{N}.\n * \text{Force of Nail on Hammer} = +2500\,\text{N} (equal and opposite reaction).\n * **Car Slowing Down:** Mass 1200\,\text{kg}$,$u = 90\,\text{km/h}$($25\,\text{m/s}$),$v = 18\,\text{km/h}$($5\,\text{m/s}$),$t = 4\,\text{s}.\n * a = \frac{5 - 25}{4} = -5\,\text{m\,s}^{-2}.\n * \text{Initial Momentum} = 1200 \times 25 = 30000\,\text{kg\,m\,s}^{-1}.\n * \text{Final Momentum} = 1200 \times 5 = 6000\,\text{kg\,m\,s}^{-1}.\n * \text{Change in Momentum} = 6000 - 30000 = -24000\,\text{kg\,m\,s}^{-1}.\n * \text{Force Magnitude} = 1200 \times 5 = 6000\,\text{N}$$.