Why this chapter matters for AAI ATC? Laws of Motion are the bedrock of understanding how aircraft behave in flight — thrust, drag, lift, weight, inertia in straight and turning flight, momentum during landing, friction on runways, and centripetal forces in holding patterns. Every controller must intuitively grasp these forces to predict aircraft behaviour.
Aristotle's Fallacy: Aristotle wrongly believed an external force is needed to keep a body in motion. The real culprit that stops moving bodies is friction, not the absence of force.
Galileo's Insight → Law of Inertia: On a frictionless surface, a body in motion continues forever with constant velocity. Rest and uniform motion are equivalent states — both require zero net force.
Newton's First Law: Every body continues to be in its state of rest or uniform motion in a straight line unless compelled by some external force to act otherwise.
First Law
If ΣF = 0 → a = 0 (body at rest OR constant velocity)
Inertia is the resistance of a body to change its state of motion. Greater mass = greater inertia.
⚡ Key Points
First Law defines a force — it's the agent that causes acceleration.
If net external force = 0, acceleration = 0 (not velocity).
A body can be in motion with zero acceleration (constant velocity).
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An aircraft in level cruise at constant speed has zero net force — thrust exactly equals drag.
A parked aircraft on the ramp stays stationary (inertia) until engines provide thrust to overcome static friction.
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[Add image here: Galileo's double inclined plane experiment; Book on table with W and N forces; Car at constant velocity with balanced forces]
🎯 Practice MCQs — Newton's First Law & Inertia
Q1 An astronaut in deep space far from all stars and with engines off experiences:
A Acceleration equal to 9.8 m/s²
B Zero acceleration
C Acceleration in the direction of previous motion
D Deceleration due to inertia
✔ Correct Answer: B — Zero acceleration
With no nearby stars and engines off, the net external force on the astronaut = 0.
By Newton's First Law: F = 0 → a = 0.
If the astronaut was moving, they continue at constant velocity; if at rest, they remain at rest. This is the classic application of N1L.
Q2 A book rests on a table. The weight W of the book is 20 N. The normal reaction N from the table is:
A 0 N (they cancel each other)
B 10 N
C 20 N (because net force = 0)
D 40 N
✔ Correct Answer: C — 20 N
The book is observed to be at rest → net force = 0 (by First Law).
Therefore N must equal W = 20 N.
Note: We don't say "W = N, so they cancel." We say: "Book is at rest → net force = 0 → N = W = 20 N."
Q3 A 5 kg block is sliding on ice with velocity 10 m/s. If the surface is perfectly frictionless, after 5 seconds the block will have a speed of:
A 0 m/s (inertia slows it)
B 5 m/s
C 10 m/s
D 50 m/s
✔ Correct Answer: C — 10 m/s
Perfectly frictionless → no net horizontal force → no acceleration.
By First Law: constant velocity. Speed after 5s = 10 m/s (unchanged).
This is Galileo's insight: on an ideal frictionless surface, motion never ceases.
4.5
Newton's Second Law & Momentum
Momentum (p): The product of mass and velocity. It is a vector quantity in the direction of velocity.
Momentum
p = mv | Units: kg·m/s = N·s
Newton's Second Law: The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction of force.
Second Law
F = dp/dt = ma | 1 N = 1 kg·m/s²
Component Form: Fₓ = maₓ, Fᵧ = maᵧ, F_z = ma_z. Force in one direction only changes velocity in that direction; perpendicular component remains unchanged.
⚡ Key Points
F = 0 implies a = 0 (consistent with First Law).
For same force, lighter body gets greater speed but same final momentum.
The Second Law is a local law — acceleration at an instant depends only on force at that instant, not on history.
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[Add image here: F = ma vector diagram; Projectile showing horizontal velocity unchanged while vertical velocity changes due to gravity]
🎯 Practice MCQs — Newton's Second Law & Momentum
Q1 A bullet of mass 0.04 kg is fired at 300 m/s and stops in a wooden block after traveling 0.15 m. The average resistive force on the bullet is:
A 6000 N
B 12000 N
C 4500 N
D 9000 N
✔ Correct Answer: B — 12000 N
Using v² = u² + 2as: 0 = 300² + 2a(0.15)
a = –90000/(0.30) = –300000 m/s²
F = ma = 0.04 × 300000 = 12000 N
Q2 A rocket of lift-off mass 20,000 kg is blasted upward with initial acceleration 5 m/s². The thrust force of the blast is: (g = 10 m/s²)
A 100,000 N
B 200,000 N
C 300,000 N
D 400,000 N
✔ Correct Answer: C — 300,000 N
Net upward force = F_thrust – mg
F_thrust – mg = ma
F_thrust = m(g + a) = 20000 × (10 + 5) = 20000 × 15 = 3,00,000 N
Q3 A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. The magnitude of acceleration of the body is:
A 1.4 m/s²
B 2.8 m/s²
C 2.0 m/s²
D 3.5 m/s²
✔ Correct Answer: C — 2.0 m/s²
Resultant force = √(8² + 6²) = √(64 + 36) = √100 = 10 N
a = F/m = 10/5 = 2.0 m/s²
Direction: tan⁻¹(6/8) = tan⁻¹(0.75) ≈ 36.87° with the 8N force
4.5b
Impulse
Impulse is useful when a large force acts for a very short time, producing a finite change in momentum. We cannot measure force and time separately, but their product is measurable.
An impulsive force is a large force acting for a very short time. Examples: bat hitting a ball, hammer driving a nail, explosion.
⚡ Key Points
Impulse = area under F-t graph.
A cricketer draws hands backward while catching → increases contact time → reduces force needed.
Automobile crumple zones work on impulse: Δt ↑ → F ↓ for same Δp.
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[Add image here: Cricketer drawing hands back while catching; F-t graph showing impulse as area under curve]
🎯 Practice MCQs — Impulse
Q1 A batsman hits back a ball straight to the bowler without changing its speed of 12 m/s. Mass of ball = 0.15 kg. The impulse imparted to the ball is:
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[Add image here: Gun-bullet recoil; Action-reaction pair diagram for two bodies A and B; Horse-cart action-reaction]
🎯 Practice MCQs — Newton's Third Law
Q1 A gun of mass 100 kg fires a shell of mass 0.02 kg at 80 m/s. The recoil speed of the gun is:
A 0.016 m/s
B 0.16 m/s
C 1.6 m/s
D 16 m/s
✔ Correct Answer: A — 0.016 m/s
By conservation of momentum (initial p = 0):
m_shell × v_shell + m_gun × v_gun = 0
0.02 × 80 + 100 × v_gun = 0
v_gun = –1.6/100 = –0.016 m/s (recoil direction)
Q2 Which statement about Newton's Third Law is CORRECT?
A Action and reaction act on the same body
B Action occurs before reaction
C Action and reaction are simultaneous and act on different bodies
D Action and reaction cancel each other
✔ Correct Answer: C
Action and reaction always occur simultaneously — no time delay.
They act on different bodies, so they cannot cancel each other.
If you consider both bodies together as a system, the action-reaction pair becomes internal forces that sum to zero.
Q3 A block of mass 2 kg rests on a table. The reaction force to the weight of the block (as per Newton's Third Law) is:
A Normal force of table on block (upward)
B Force of block on Earth (downward on Earth)
C Normal force of block on table
D Weight of table
✔ Correct Answer: B — Force of block on Earth
Weight of block = gravitational pull of Earth on block (action).
Reaction = gravitational pull of block on Earth (upward on Earth) = 2 × 10 = 20 N upward.
Note: Normal force (table on block) is NOT the reaction to weight — it is a different force pair (contact).
4.7
Conservation of Momentum
Law of Conservation of Momentum: The total momentum of an isolated system (no external force) remains constant. This follows directly from Newton's Second and Third Laws.
For two bodies A and B colliding: p'_A + p'_B = p_A + p_B
This holds for both elastic (KE conserved) and inelastic (KE not conserved) collisions.
⚡ Key Points
Isolated system = no net external force acting on the system.
Gun recoil, rocket propulsion, explosions — all examples of momentum conservation.
When a nucleus at rest disintegrates, two products must move in opposite directions to keep total p = 0.
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[Add image here: Gun-bullet recoil showing equal and opposite momenta; Two-body collision before and after with momentum vectors]
🎯 Practice MCQs — Conservation of Momentum
Q1 Two billiard balls each of mass 0.05 kg moving in opposite directions at 6 m/s collide and rebound with the same speed. The impulse on each ball is:
A 0.3 N·s
B 0.6 N·s
C 1.2 N·s
D 0 N·s
✔ Correct Answer: B — 0.6 N·s
For one ball: initial momentum = 0.05 × 6 = +0.3 N·s
After rebound: –0.05 × 6 = –0.3 N·s
Impulse = |Δp| = |–0.3 – 0.3| = 0.6 N·s
Q2 A 10 kg body moving at 5 m/s collides and sticks to a 5 kg body at rest. Their combined velocity is:
A 5 m/s
B 3.33 m/s
C 2.5 m/s
D 10 m/s
✔ Correct Answer: B — 3.33 m/s
Initial momentum = 10 × 5 + 5 × 0 = 50 N·s
After collision (perfectly inelastic): (10 + 5) × v = 50
v = 50/15 = 3.33 m/s
Q3 A nucleus at rest disintegrates into two fragments of masses 4u and 12u. If the heavier fragment moves at 100 m/s, the speed of the lighter fragment is:
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[Add image here: Static vs kinetic friction graph (frictional force vs applied force); Block on inclined plane with fₛ, N, mg components]
🎯 Practice MCQs — Friction
Q1 A box of mass 20 kg lies on a horizontal surface. μₛ = 0.4, μₖ = 0.3, g = 10 m/s². The maximum static friction force is:
A 60 N
B 80 N
C 120 N
D 40 N
✔ Correct Answer: B — 80 N
N = mg = 20 × 10 = 200 N
(fₛ)max = μₛ × N = 0.4 × 200 = 80 N
Once applied force exceeds 80 N, the box starts sliding. Then kinetic friction = 0.3 × 200 = 60 N.
Q2 A block on an inclined plane begins to slide when angle = 30°. The coefficient of static friction between block and plane is:
Q3 A train can have maximum acceleration of 1.5 m/s² due to static friction. The coefficient of static friction between train and track is: (g = 10 m/s²)
Centripetal Force: For a body in circular motion, a net inward force is always needed directed toward the centre. This is not a new type of force — it is provided by existing forces (tension, friction, gravity, normal force).
Centripetal Force
Fc = mv²/R = mω²R | (directed toward centre)
Aircraft in a banked turn — the horizontal component of lift provides centripetal force.
Standard rate turn for IFR: 3°/second, bank angle ≈ (v/10) + 7 degrees (approx rule).
Holding pattern radius depends on aircraft speed and bank angle — ATC assigns holding stack based on this.
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[Add image here: Car on level road (circular turn) with friction as centripetal force; Car on banked road with N and f components; Aircraft banked turn showing lift components]
🎯 Practice MCQs — Circular Motion & Banking
Q1 A car takes a circular turn of radius 50 m on a flat road. μₛ = 0.5, g = 10 m/s². The maximum speed without slipping is:
A 10 m/s
B 15 m/s
C 20 m/s
D 25 m/s
✔ Correct Answer: A — 10 m/s
v_max = √(μₛRg) = √(0.5 × 50 × 10) = √250 = 15.8 m/s ≈ 15 m/s
Wait — exact: √(0.5 × 50 × 10) = √250 ≈ 15.8 m/s → closest: B. Correct: B — 15 m/s (closest to 15.8)
Q2 A circular racetrack of radius 300 m is banked at 15°. The optimum speed (friction not needed) is: (g = 9.8 m/s²)