Laws of Motion revision notes

Aristotle's Fallacy and Galileo's Law of Inertia

• Aristotle’s Fallacy: The ancient Greek thinker Aristotle theorized that an external force is necessary to keep a body in uniform motion. This is false. A moving toy car comes to rest not because it inherently requires a driving force, but because the external force of friction from the floor opposes its motion.
• Galileo’s Insight: Galileo studied motion on inclined planes and deduced that uniform motion (constant velocity) is equivalent to a state of rest, as both have zero net external force.JEE TIPIf an object is moving uniformly, the applied external force exactly cancels opposing forces like friction or viscous drag.
• Galileo's Double Inclined Plane Experiment: A ball released on a smooth inclined plane will climb a facing plane to exactly its initial height. If the second plane is made perfectly horizontal, the ball will travel an infinite distance forever.
• Ideas in Ancient Indian Science: Thinkers like Bhaskara introduced concepts analogous to instantaneous velocity, and the Vaisesika theory of motion proposed vega (the tendency to move in a straight line), which closely parallels the concept of inertia.
• The Law of Inertia: Inertia is a body's "resistance to change". A body remains in its state of rest or uniform motion unless compelled by an external force.

Newton's First Law of Motion

• Statement: Every body continues to be in its state of rest or of uniform motion in a straight line unless compelled by some external force to act otherwise.
• Mathematical Implication: If the net external force on a body is zero ($\sum F = 0$), its acceleration is zero ($a = 0$).
• Application in Terrestrial Phenomena: Because gravity and friction act everywhere on Earth, an object observed at rest or moving with uniform velocity must have a net force of zero—meaning all individual external forces acting on it perfectly cancel out.

Newton's Second Law of Motion and Momentum

• Momentum ($\mathbf{p}$): Defined as the product of a body’s mass and velocity: $\mathbf{p} = m\mathbf{v}$. It is a vector quantity.
• Statement: The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts.
• Mathematical Form: $\mathbf{F} = \frac{d\mathbf{p}}{dt} = m\mathbf{a}$.
• Important Properties:
  • Vector Nature: The law is equivalent to three independent Cartesian equations: $F_x = m a_x$, $F_y = m a_y$, and $F_z = m a_z$.JEE TIPA force applied perpendicular to velocity alters only the direction, not the magnitude, of the corresponding velocity component.
  • Applicability to Systems: The law applies to a single particle, a rigid body, or a system of particles. For systems, $\mathbf{F}$ is the total external force, and $\mathbf{a}$ is the acceleration of the center of mass (internal forces cancel out).
  • Local Relation: Acceleration at a specific point and instant is determined purely by the force at that point and instant.JEE TIPA body carries no "memory" of past acceleration. For example, a stone dropped from an accelerating train instantly loses horizontal acceleration and experiences only downward acceleration due to gravity.
  • Kinematic Equations Context: If a particle's motion is defined by $y = ut + \frac{1}{2}gt^2$, double differentiation yields $a = g$, meaning the constant net force acting on it is $F = mg$.

Impulse and Impulsive Forces

• Impulse: When a large force acts for a very short duration (e.g., bat hitting a ball, ball bouncing off a wall), calculating separate force and time is difficult, but their product is measurable.
• Formula: Impulse = Force $\times$ time duration = Change in momentum ($\Delta \mathbf{p}$).
• Critical Assumption: During the short action time of an impulsive force, the position of the body is assumed to not change appreciably.
• Real-world Application: A cricketer drawing his hands backward while catching a fast ball increases the time interval ($\Delta t$) of the catch, thereby reducing the rate of change of momentum and minimizing the force exerted on the hands.

Newton's Third Law of Motion

• Statement: To every action, there is always an equal and opposite reaction.
• Key Characteristics:
  • Forces always occur in pairs; a single isolated force is impossible.
  • $\mathbf{F}_{AB} = -\mathbf{F}_{BA}$ (Force on A by B is equal and opposite to Force on B by A).
  • Action and reaction forces act on different bodies. They DO NOT cancel each other out when considering the free-body diagram of a single body.
  • There is no cause-and-effect relationship; the forces arise simultaneously.

Conservation of Linear Momentum

• Principle: The total momentum of an isolated system of interacting particles is conserved.
• Derivation: Follows from the 2nd and 3rd laws. During a collision between bodies A and B, the mutual forces are equal and opposite ($\mathbf{F}_{AB} = -\mathbf{F}_{BA}$). Their impulse is $\mathbf{F}_{AB} \Delta t = \Delta \mathbf{p}_A$ and $\mathbf{F}_{BA} \Delta t = \Delta \mathbf{p}_B$. Therefore, $\Delta \mathbf{p}_A + \Delta \mathbf{p}_B = 0$, meaning the total initial momentum equals total final momentum.
• Scope: Applies to both elastic and inelastic collisions.

Equilibrium of a Particle

• Definition: A particle is in equilibrium when the net external force acting on it is zero (it is either at rest or moving with uniform velocity).
• Concurrent Forces:
  • For two forces: $\mathbf{F}_1 = -\mathbf{F}_2$.
  • For three forces: $\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = 0$. Geometrically, they can be represented by the sides of a closed triangle taken in order.
  • For $n$ forces: Represented by a closed $n$-sided polygon with arrows in the same sense.

Common Forces in Mechanics

• Fundamental Origin: In macroscopic mechanics, non-gravitational contact forces (friction, normal, tension, spring) fundamentally arise from microscopic electrical forces between the charged constituents (atoms and molecules) of different bodies.
• Normal Reaction ($N$): The contact force component perpendicular to the surfaces in contact.
• Tension ($T$): Restoring force generated in a stretched string.JEE TIPTension is assumed uniform throughout a string only if the string is considered massless and inextensible.
• Spring Force: An ideal spring generates a restoring force $\mathbf{F} = -k\mathbf{x}$, where $k$ is the spring constant and $\mathbf{x}$ is displacement.JEE TIPThis linear relationship is an approximation strictly valid only for small displacements.

Friction (Static, Kinetic, and Rolling)

• Definition: The component of contact force parallel to the surfaces, opposing impending or actual relative motion.
• Static Friction ($f_s$): Opposes impending relative motion (motion that would occur without friction).
  • It is a self-adjusting force. If no external force is applied, $f_s = 0$.
  • Maximum limit: $(f_s)_{max} = \mu_s N$, where $\mu_s$ is the coefficient of static friction.
  • Independent of the area of contact.
• Kinetic/Sliding Friction ($f_k$): Opposes actual relative motion.
  • Formula: $f_k = \mu_k N$, where $\mu_k$ is the coefficient of kinetic friction.
  • $\mu_k$ is generally less than $\mu_s$, and is approximated as independent of relative velocity.
• Rolling Friction: Resistance encountered when a body (sphere/ring) rolls without slipping. Originates from the microscopic deformation of surfaces, creating a finite contact area rather than a point. Rolling friction is much smaller (by 2 or 3 orders of magnitude) than static/kinetic friction.
• Reduction Techniques: Lubricants, ball bearings, or compressed air cushions.
• Angle of Repose ($\theta_{max}$): The maximum incline angle before a block slides: $\tan \theta_{max} = \mu_s$. Independent of mass.

Circular Motion

• Centripetal Force ($f_c$): The required radial force toward the center to maintain uniform circular motion: $f_c = \frac{mv^2}{R}$.
• Car on a Level Road: Centripetal force is provided solely by static friction.
  • Condition to avoid slipping: $v_{max} = \sqrt{\mu_s R g}$. Independent of car mass.
• Car on a Banked Road: Banking reduces reliance on friction by utilizing the horizontal component of the normal reaction ($N \sin \theta$).
  • Optimum Speed ($v_0$): Speed at which NO friction is required (minimizes tyre wear): $v_0 = \sqrt{R g \tan \theta}$.
  • Maximum Permissible Speed ($v_{max}$): $v_{max} = \sqrt{R g \left(\frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta}\right)}$.

Standard Derivations & Step-by-Step Problem Solving

Mechanics FBD Strategy:

1. Draw a schematic of the entire assembly.
2. Choose a specific part as the "system".
3. Draw a Free-Body Diagram (FBD) showing the system isolated. Include ALL forces acting ON the system FROM the environment.
4. Include known forces (magnitudes/directions) and define unknowns.
5. Apply Newton's Third Law when creating subsequent FBDs for connected parts: if FBD A shows force $\mathbf{F}$ from B, FBD B MUST show force $-\mathbf{F}$ from A.

Derivation: Maximum Speed of a Car on a Banked Road:

1. Assume a car on a road banked at angle $\theta$.
2. Forces acting: Weight ($mg$), Normal Reaction ($N$), Frictional force down the bank ($f$).
3. Vertical Equilibrium (no vertical acceleration): $N \cos \theta = mg + f \sin \theta$
4. Horizontal Centripetal Force: $N \sin \theta + f \cos \theta = \frac{mv^2}{R}$
5. To find maximum speed ($v_{max}$), substitute the limiting value of static friction: $f = \mu_s N$.
6. Substitute into vertical equation: $N \cos \theta = mg + \mu_s N \sin \theta \implies N = \frac{mg}{\cos \theta - \mu_s \sin \theta}$
7. Substitute $N$ and $f$ into horizontal equation: $\left(\frac{mg}{\cos \theta - \mu_s \sin \theta}\right) (\sin \theta + \mu_s \cos \theta) = \frac{mv_{max}^2}{R}$
8. Divide numerator and denominator by $\cos \theta$: $v_{max} = \sqrt{R g \left[ \frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta} \right]}$

Derivation: Impulse of Oblique Collision (Billiard Ball bouncing off a wall):

1. A ball of mass $m$ and speed $u$ strikes a wall at angle $\theta$ to the normal and rebounds elastically.
2. Resolve momentum into $x$ (normal) and $y$ (parallel) components.
3. Initial: $p_{xi} = m u \cos\theta$, $p_{yi} = -m u \sin\theta$.
4. Final: $p_{xf} = -m u \cos\theta$, $p_{yf} = -m u \sin\theta$.
5. Change in momentum (Impulse): $\Delta p_x = -2mu \cos\theta$, $\Delta p_y = 0$.
6. Result: The impulse (and thus force) exerted by the wall is entirely normal to the surface. Parallel velocity is unaffected.

Key Concepts & Definitions

Inertia:

The inherent property of a material body to resist change in its state of rest or uniform motion.

Momentum (p\mathbf{p}p):

Product of mass and velocity.

Impulse:

Large force acting over a short time; equal to change in momentum.

Concurrent Forces:

Forces whose lines of action intersect at a single point; evaluated in equilibrium conditions.

Impending Motion:

Motion that would take place under an applied force if friction were perfectly absent.

Important Graphs & Diagrams

• Position-Time Graph for Impulsive Force: If a particle experiences an impulsive force, its $x-t$ graph displays a sharp kink (a sudden change in slope). The slope before the kink is initial velocity ($u$), and after is final velocity ($v$). The impulse is simply $m(v - u)$.
• Free-Body Diagrams (FBDs): Essential tool mapping single bodies isolated in space. Represents only forces acting ON the body.
• Banked Road Resolution: Both Normal reaction $N$ and friction $f$ must be resolved into horizontal ($N \sin\theta + f \cos\theta$) and vertical ($N \cos\theta - f \sin\theta$) components to solve for $v$.

Previous Year JEE Topics

• Two-Block Systems: Problems involving a block on a cart, calculating maximum acceleration before sliding via $a \le \mu_s g$.
• Banked Circular Tracks: Complex evaluations of car stability involving friction and variable radii.
• Impulse-Momentum Theorem: Graph interpretations (Area of F-t graph) and oblique collisions with walls determining the impulse imparted.
• Conservation of Momentum: Explosions of shells/nuclei where fragments fly in opposite directions to conserve initial zero momentum.
• Variable Mass & Systems: FBD mapping identifying subtle action-reaction pairs in coupled, multi-body accelerating frames.

Top 10 JEE MCQ Traps

• [JEE TIP] Trap 1 - The Equilibrium Action-Reaction Fallacy:

  • Misconception: The normal reaction force ($N$) acting upward on a resting block and its downward weight ($mg$) constitute a Newton's third law action-reaction pair.
  • Correct Understanding: Action-reaction pairs must act on two completely different bodies and have the exact same physical origin. The normal force and weight act on the same single body and arise from entirely different mechanisms (electromagnetic vs. gravitational). They only happen to balance each other numerically ($N = mg$) due to static vertical equilibrium, which breaks completely in an accelerating frame.

• [JEE TIP] Trap 2 - Absolute Motion vs. Relative Friction:

  • Misconception: Frictional forces always act as a resistive barrier that opposes the absolute physical motion of a body.
  • Correct Understanding: Friction strictly opposes relative motion or the tendency of relative motion between two surfaces in contact. Static friction routinely acts as the driving mechanism that causes absolute motion. For instance, when a truck accelerates, it is static friction pointing forward on the wheels that drives the truck, and static friction on a crate in the truck bed that accelerates the crate alongside it.

• [JEE TIP] Trap 3 - The Continuous Force Propulsion Myth:

  • Misconception: Sustaining a continuous, steady physical movement at a constant velocity demands the continuous application of a non-zero net external force.
  • Correct Understanding: According to Newton's First Law, uniform motion along a straight line strictly requires the net external force to be exactly zero ($F_{\text{net}} = 0$). Ongoing forces applied in daily life are merely required to cancel out opposing external resistive forces like friction or air drag to keep the net vector sum at zero.

• [JEE TIP] Trap 4 - Action-Reaction Temporal Latency:

  • Misconception: The action force must be applied first, which then triggers a microscopic time delay before the reaction force responds.
  • Correct Understanding: There is absolutely no cause-and-effect chronological delay. Both forces arise simultaneously and vanish simultaneously as a single mutual interaction. Labelling one force as "action" and the other as "reaction" is purely arbitrary.

• [JEE TIP] Trap 5 - Centripetal Force FBD Ghost Injection:

  • Misconception: When a body travels along a curved path, centripetal force must be drawn as an independent, newly applied extra arrow on its Free Body Diagram (FBD).
  • Correct Understanding: Centripetal force is not a distinct physical interaction; it is merely a geometric label given to the net real force pointing toward the center of curvature. It is provided by real forces like tension, gravity, normal reaction, or friction. Drawing "centripetal force" as an extra force alongside these real interactions is a fatal error that leads to severe double-counting in your system equations.

• [JEE TIP] Trap 6 - The Static Friction Equation Lockdown:

  • Misconception: The magnitude of the static friction force acting on a surface is universally calculated by plugging numbers into the expression $f_s = \mu_s N$.
  • Correct Understanding: Static friction is a passive, self-adjusting force governed by an inequality condition ($f_s \le \mu_s N$). It changes its magnitude and direction dynamically to perfectly match the opposing net external force. The expression $\mu_s N$ computes exclusively the limiting (maximum) static friction. If the applied force is less than this threshold, the actual static friction is exactly equal to the applied force, not $\mu_s N$.

• [JEE TIP] Trap 7 - Constant Speed Momentum Confusion:

  • Misconception: If an object maintains a perfectly constant scalar speed throughout its path, its momentum remains constant, and the net force acting on it must be zero.
  • Correct Understanding: Momentum is a vector quantity ($\mathbf{p} = m\mathbf{v}$). Even if the speed is perfectly uniform, any alteration in the direction of travel—such as in uniform circular motion—constitutes a continuous change in velocity and momentum. Deflecting this direction requires a non-zero net external force operating perpendicular to the motion.

• [JEE TIP] Trap 8 - The Normal Force Gravity Identity:

  • Misconception: The normal reaction force ($N$) exerted by a surface on an object is structurally fixed and always defaults to $N = mg$.
  • Correct Understanding: The normal force is a constraining force that must be extracted manually by setting up a perpendicular force balance equation. It drops to $mg\cos\theta$ on a smooth incline angled at $\theta$, drops when an external upward lifting force is applied, spikes when an object is pressed downward, and changes drastically inside accelerating frames like elevators ($m(g \pm a)$).

• [JEE TIP] Trap 9 - Impulsive Impact Spatial Displacement:

  • Misconception: A powerful impulsive force—such as a bat striking a ball—causes a massive physical displacement of the object during the timeline of the impact.
  • Correct Understanding: Impulsive forces operate over an infinitesimally brief temporal window ($\Delta t \to 0$). Because this duration is so small, the physical spatial position coordinate of the object remains virtually frozen and unchanged during the collision itself. The impulse acts purely to generate an instantaneous, discrete step-change in the momentum vector ($\int \mathbf{F} dt = \Delta \mathbf{p}$).

• [JEE TIP] Trap 10 - The Instantaneous Rest Zero Force Illusion:

  • Misconception: At the exact microsecond a moving body comes to a momentary stop and is at rest, the net force operating on it must drop to zero.
  • Correct Understanding: An instantaneous velocity of zero does not mathematically imply a zero acceleration or a zero net force. For example, at the absolute peak trajectory of a vertically projected ball, its velocity is momentarily $v = 0$, but the net force is rigidly locked at $F_{\text{net}} = mg$ downward, generating a continuous non-zero acceleration of $a = g$.