Laws of Motion

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Newton’s Three Laws — The Foundation:

Newton’s First Law (Law of Inertia): A body continues in its state of rest or uniform motion in a straight line unless compelled by an external unbalanced force to change that state. Inertia is the tendency of a body to resist changes in its state of motion. Mass is the quantitative measure of inertia — more mass means more inertia.

Newton’s Second Law: The rate of change of momentum of a body equals the net force applied. In the form F = ma, this is valid when mass is constant. More fundamentally: F = dp/dt = d(mv)/dt. For variable mass (rocket propulsion, conveyor belt problems), this expands to F_ext = mdv/dt + v_rel dm/dt.

Newton’s Third Law: For every action, there is an equal and opposite reaction. Action and reaction act on different bodies — never on the same body. This is why internal forces cancel in ΣF = ma for a system: every internal force has an equal and opposite pair.

Free Body Diagrams — The #1 Skill:

A Free Body Diagram (FBD) shows ALL external forces acting on a body, drawn from the body’s centre of mass, with arrows pointing in the direction of force application. Rules:

1. Isolate ONE body at a time
2. Show only forces ON that body (never forces the body exerts on others)
3. Include gravity (weight = mg downward), normal reaction (perpendicular to surface), tension (along string/rod away from body), friction (opposing relative motion or impending motion), and applied forces
4. Choose convenient coordinate axes — usually along and perpendicular to the surface for inclined plane problems

⚡ JEE Tip: Draw an FBD for EVERY problem. JEE Advanced 2021 Qn 28 required three separate FBDs (block, wedge, pulley) to solve a wedge-block constraint problem. Missing even one force dooms the solution.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Types of Forces in Mechanics:

Gravitational force (weight): W = mg, acts vertically downward on every mass. On inclined planes, component along plane = mg sinθ.

Normal reaction (N): Contact force perpendicular to the surface. For a body on horizontal ground: N = mg. On an inclined plane: N = mg cosθ. For a body against a vertical wall: N is horizontal (opposes the force pushing the body into the wall).

Tension (T): Force transmitted through a string, rope, or rod. For an inextensible string, tension is the same at both ends. For a massless string, tension is uniform throughout. For a string with mass, tension varies along its length (only needed in advanced problems).

Friction (f): f ≤ μN. Static friction (f_s) adjusts from 0 to μ_sN to prevent relative motion — it is self-adjusting. Kinetic friction (f_k = μ_kN) applies when surfaces slide. Limiting friction is the maximum static friction: f_s(max) = μ_sN.

Spring force: F = -kx, where k is the spring constant (N/m) and x is the displacement from natural length. This is a restoring force — it opposes deformation.

Pulley Systems:

Fixed pulley: Changes direction of force but not magnitude. Mechanical advantage = 1 (ignoring friction/mass).

Movable pulley: Support comes from both ends of string. Mechanical advantage = 2 for an ideal movable pulley — the force needed to lift a weight W is W/2.

Combination systems: For multiple pulleys (block and tackle), T = W/n for ideal systems with n supporting strands.

⚡ JEE Tip: For Atwood’s machine (two masses connected by string over pulley): acceleration a = (m₁ - m₂)g/(m₁ + m₂). When the heavier mass descends, it pulls the lighter mass up. Always write constraint: if m₁ moves down by x, m₂ moves up by x (same string length). Watch for sign conventions!

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Pseudo Forces — Non-Inertial Reference Frames:

When analysing motion from a non-inertial (accelerating) frame, Newton’s second law requires adding a fictitious (pseudo) force: F_pseudo = -m × a_frame, where a_frame is the acceleration of the frame relative to an inertial observer. This “fake” force has no physical origin — it arises purely from the observer’s choice of reference frame.

In an accelerating lift: If the lift accelerates upward with acceleration a, apparent weight = m(g + a). If it accelerates downward (but not in free fall), apparent weight = m(g - a). If the cable snaps and the lift is in free fall, apparent weight = 0 (weightlessness).

On a horizontal turntable: A person of mass m standing at distance r from the centre experiences a centrifugal pseudo force F = mω²r outward in the rotating frame. In an inertial frame, this is simply the centripetal force mv²/r directed inward (the normal reaction from the floor provides this).

Banked curves without friction: For a vehicle on a frictionless banked curve, the horizontal component of the normal reaction provides the centripetal force: N sinθ = mv²/r. The banking angle satisfies tanθ = v²/(rg). This is how railway tracks and roads are banked to reduce reliance on friction.

Mass-Spring Systems — Detailed Analysis:

For a block of mass m attached to a spring of constant k, oscillating on a horizontal surface: acceleration a = -(k/m)x. The system undergoes Simple Harmonic Motion (SHM) with angular frequency ω = √(k/m).

Series springs: When two springs of constants k₁ and k₂ are connected in series, effective k_eq = (k₁k₂)/(k₁ + k₂). For n identical springs in series: k_eq = k/n.

Parallel springs: When two springs are connected in parallel, k_eq = k₁ + k₂. For n identical springs in parallel: k_eq = nk.

⚡ JEE Advanced Pattern: Pseudo force problems appear frequently in JEE Advanced Paper 2, often combined with circular motion or relative velocity. The 2022 paper included a problem where a bead slides on a rotating rod inside an accelerating car — students had to correctly identify the pseudo force direction and magnitude in the car’s frame and apply constraints for the bead’s motion along and perpendicular to the rod. The key insight: from the car’s frame, the bead experiences a fictitious force -m×a_car horizontally AND an outward centrifugal force mω²r as the rod rotates.

Common mistakes include adding pseudo forces when the frame is inertial (don’t!) and getting the sign wrong for the pseudo force (it always acts opposite to the frame’s acceleration).