Key Relationships

🟢 Lite — Quick Review (1h–1d)

Newton’s Three Laws of Motion form the bedrock of classical mechanics. Here is what you must recall:

First Law (Law of Inertia): A body at rest stays at rest; a body in motion continues in uniform motion unless an external force acts on it. Inertia is the property that resists changes in motion — mass directly measures inertia.

Second Law (F = ma): Force equals mass times acceleration. Units: 1 Newton = 1 kg·m/s². Also expressed as F = dp/dt (rate of change of momentum). Momentum p = mv is a vector quantity.

Third Law: Every action has an equal and opposite reaction. Action and reaction forces act on different bodies — they never cancel each other.

Must-know formulas: F = ma, p = mv, W = mg, J = FΔt.

TNPSC Quick Picks: Weight = mg (g = 9.8 m/s² on Earth). In a lift accelerating upward, apparent weight = m(g + a). Questions on conservation of momentum in collisions appear every year. Watch for confusion between mass (constant) and weight (varies with g).

🟡 Standard — Regular Study (2d–2mo)

Definitions and Physical Meaning

Newton’s First Law states that an isolated body maintains its state of rest or uniform linear motion indefinitely. The term inertia refers specifically to this resistance to any change in velocity. Heavier bodies have greater inertia — this is precisely why mass is defined as the quantitative measure of inertia.

Newton’s Second Law gives us two equivalent formulations. The common form F = ma applies when mass is constant. When mass changes (e.g., a rocket burning fuel), the more general form F = dp/dt must be used, where p = mv. Both expressions are dimensionally consistent with force measured in newtons (kg·m/s²).

Newton’s Third Law emphasises that forces always occur in pairs. If body A exerts force F on body B, then body B simultaneously exerts force –F on body A. Because these forces act on different bodies, they do not cancel.

Key Relationships

Quantity	Formula	Units
Force	F = ma	Newton (N) = kg·m/s²
Momentum	p = mv	kg·m/s
Weight	W = mg	Newton (N)
Impulse	J = FΔt = Δp	N·s = kg·m/s

Motion Equations (constant acceleration)

v = u + at
s = ut + ½at²
v² = u² + 2as

Apparent Weight in Elevators

When a lift accelerates upward with acceleration a, the normal reaction N = m(g + a), so apparent weight increases. When accelerating downward, N = m(g – a). If the cable snaps (free fall), N = 0 — the person experiences weightlessness.

Common TNPSC Trap

Students often confuse apparent weight with actual weight. Actual weight = mg always acts downward. Apparent weight is what a weighing scale reads — it equals the normal reaction force. Always draw a free-body diagram before solving lift problems.

🔴 Extended — Deep Study (3mo+)

Conservation of Momentum

For an isolated system (no external forces), total momentum remains constant. In a collision between two bodies: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. This holds regardless of whether the collision is elastic or inelastic — the difference is that kinetic energy is conserved only in perfectly elastic collisions.

Impulse-Momentum Theorem

Impulse J = FΔt equals the change in momentum: FΔt = Δp = mv – mu. This theorem explains why catching a ball with a bent elbow reduces the force on your hands — increasing the time interval Δt decreases the average force F.

Friction Forces

Friction opposes relative motion or impending motion. Static friction adjusts from zero up to a maximum value fₛ(max) = μₛN. Kinetic friction during sliding equals fₖ = μₖN. Note that the friction coefficient is applied to the normal force N, not directly to mg — on an inclined plane, N = mg cos θ.

Common Mistakes to Avoid

Mass units: Always convert mass to kilograms before using F = ma. Using grams gives answers 1000 times too large.
Directional signs: Momentum and velocity are vectors. Establish a positive direction and apply it consistently.
Third-law pairs: Action and reaction never act on the same body. When a block rests on a table, the weight of the block and the normal reaction are a third-law pair only if you trace them correctly — both act on the block.
Apparent weight in lifts: Remember the lift’s acceleration adds to g when going up, subtracts when coming down.

Worked Example

A 5 kg block rests on a horizontal surface (μₖ = 0.3). A horizontal force of 20 N is applied. Find the acceleration.

Solution: Normal force N = mg = 5 × 9.8 = 49 N. Kinetic friction fₖ = μₖN = 0.3 × 49 = 14.7 N. Net force = F – fₖ = 20 – 14.7 = 5.3 N. Acceleration a = F_net/m = 5.3/5 = 1.06 m/s².

TNPSC Group 1 Exam Strategy

Laws of Motion typically yields 2–4 MCQs in Prelims and 1 descriptive question in Mains. In Prelims, prioritize numerical problems on F = ma and momentum conservation. In Mains, prepare to write clearly on the distinction between mass and weight, and explain lift acceleration scenarios with free-body diagrams. Allocate approximately 3–4 minutes per numerical question.

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Key Relationships