Olive Defence

Physics · CDS

PC02 — Kinematics & Laws of Motion

📖 PC02 · CDS General Science — Physics 🎯 CDS Level : High Priority

This is the highest-yield chapter in CDS Physics. Kinematics describes how objects move; Newton's laws explain why. Expect 2–3 questions from here in every CDS paper — covering equations of motion, projectile motion, Newton's three laws, impulse-momentum, and friction.

📌 CDS regularly tests: Equations of motion (especially v² = u² + 2as); projectile range and height; complementary angles giving same range; gun-bullet recoil (conservation of momentum); Newton's 2nd law numericals; friction types; circular motion centripetal force; apparent weight in a lift.

Topics at a Glance

① Scalars & Vectors

Distance vs displacement; speed vs velocity

② Equations of Motion

v=u+at; s=ut+½at²; v²=u²+2as

③ Motion Graphs

s-t and v-t graphs; slope; area

④ Projectile Motion

Range, height, time of flight; 45° max range

⑤ Newton's Laws & Friction

Inertia; F=ma; action-reaction; μ types

⑥ Momentum & Circular Motion

Conservation of momentum; centripetal force

1. Scalars, Vectors & Equations of Motion

1.1

Types of Motion & Equations Under Uniform Acceleration

📍 Scalars vs Vectors

Scalar: magnitude only — distance, speed, mass, time, energy, temperature
Vector: magnitude + direction — displacement, velocity, force, acceleration, momentum
Vector addition: triangle law or parallelogram law
Resultant of two equal vectors at angle θ: R = 2A cos(θ/2)

📍 Distance vs Displacement

Distance = total path length (scalar; always positive)
Displacement = shortest path from start to end (vector; can be negative)
Speed = distance / time (scalar)
Velocity = displacement / time (vector)
Average speed ≥ average velocity (always)

⚡ Three Equations of Motion (Uniform Acceleration)

v = u + at ← find velocity after time t s = ut + ½at² ← find displacement in time t v² = u² + 2as ← find velocity given displacement (no t needed) sₙ = u + ½a(2n − 1) ← displacement in the nth second Free fall (downward): use a = +g = 9.8 m/s² ≈ 10 m/s² Throw upward: use a = −g (decelerating); at max height v = 0 Time to go up = time to come down (symmetric motion under gravity)

Use v² = u² + 2as whenever time is not given and not required — saves significant calculation time in CDS.

2. Motion Graphs

2.1

Reading s-t and v-t Graphs

CDS often gives a graph and asks for velocity, acceleration, or displacement

Fig. 1 — Motion graph rules: slope of s-t gives velocity; slope of v-t gives acceleration; area under v-t gives displacement. These are direct CDS graph-reading questions.

3. Projectile Motion

3.1

Two-Dimensional Motion Under Gravity

⚡ Projectile Formulae

Horizontal component: uₓ = u cosθ (constant — no horizontal force) Vertical component: uᵧ = u sinθ (reduces under gravity) Time of flight: T = 2u sinθ / g Maximum height: H = u² sin²θ / 2g Horizontal range: R = u² sin 2θ / g Maximum range at θ = 45°: Rₘₐₓ = u²/g At 45°: H = R/4 (height is exactly one-quarter of range) Complementary angles give same range: 30° & 60° → same R; 20° & 70° → same R; 15° & 75° → same R At any time t: x = u cosθ·t; y = u sinθ·t − ½gt²

Key insight: Horizontal and vertical motions are completely independent. A ball thrown horizontally and a ball dropped vertically from the same height both hit the ground at exactly the same time — horizontal speed has zero effect on the time of fall.

4. Newton's Laws of Motion & Friction

4.1

Three Laws, Impulse, Momentum & Friction Types

① First Law — Inertia

Body at rest stays at rest; body moving stays moving — until net external force acts
Inertia ∝ mass
Examples: passenger jerks forward when bus brakes; coin on card; seat belts

② Second Law — F = ma

F = ma (net force = mass × acceleration)
F = dp/dt (rate of change of momentum)
Impulse = F × t = Δp = m(v−u)
Apparent weight in lift: N = m(g±a)

③ Third Law — Action-Reaction

Every action has equal and opposite reaction — on different bodies
Gun recoil; rocket propulsion; swimming; boat-oar
They never cancel (act on different objects)

⚡ Impulse, Momentum & Friction

Momentum: p = mv [MLT⁻¹] Impulse: J = F·t = Δp [MLT⁻¹] (area under F-t graph) Conservation of momentum (when net external force = 0): m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (for all collisions) Gun-bullet example: Initial momentum = 0; m_b·v_b + m_g·v_g = 0 → gun recoils Friction: f_max = μₛN (limiting static friction; μₛ = coefficient of static friction) Kinetic friction = μₖN (while sliding) Order of magnitude: μₛ > μₖ > μ_rolling Friction is independent of surface area of contact

⚠ CDS Traps in Kinematics & Laws: (1) Complementary angles give same range — 30° and 60°, NOT 30° and 45°. (2) At max height of projectile: v_vertical = 0, but v_horizontal = u cosθ (not zero). (3) Friction is independent of area — only depends on normal force and surfaces. (4) Inertia is measured by mass, not weight. (5) In free fall, apparent weight = 0 (weightlessness), actual weight unchanged.

5. Circular Motion

5.1

Uniform Circular Motion — Speed Constant, Direction Always Changing

⚡ Circular Motion Formulae

Angular velocity: ω = 2π/T = 2πf (rad/s) Linear velocity: v = ωr Centripetal acc: aₒ = v²/r = ω²r (directed toward centre) Centripetal force: Fₒ = mv²/r (provided by: gravity/tension/friction) Work done by centripetal force = 0 (F always ⊥ v)

Centrifugal force is a pseudo-force felt in a rotating (non-inertial) frame — it does not exist in an inertial frame. A passenger in a turning car feels pushed outward — that sensation is inertia, not a real outward force.

📝 CDS PYQ

Kinematics & Laws of Motion

Q1. A gun of mass 3 kg fires a bullet of mass 30 g with velocity 300 m/s. The recoil velocity of the gun is:

(a) 1 m/s
(b) 3 m/s
(c) 0.3 m/s
(d) 30 m/s

Answer: (b) 3 m/s
By conservation of momentum (initial momentum = 0):
0 = m_b·v_b + m_g·v_g → 0 = 0.030 × 300 + 3 × v_g → v_g = −9/3 = −3 m/s (negative = opposite to bullet). Gun recoils at 3 m/s.

Q2. A ball is thrown at 45° and travels horizontal range R. The maximum height reached is:

(a) R
(b) R/2
(c) R/4
(d) 2R

Answer: (c) R/4
At θ = 45°: H = u²sin²45°/2g = u²/4g and R = u²sin90°/g = u²/g. Therefore H = R/4. This specific ratio at 45° is directly and repeatedly tested in CDS — commit it to memory.

Q3. A body starts from rest and acquires velocity v in time t. The distance covered is:

(a) vt
(b) vt/2
(c) 2vt
(d) vt²/2

Answer: (b) vt/2
From rest: u = 0; final velocity = v; time = t. From v = u + at → a = v/t. Distance s = ut + ½at² = 0 + ½(v/t)t² = vt/2. Alternatively: s = (u+v)/2 × t = (0+v)/2 × t = vt/2. The average velocity method is fastest here.

Q4. The horizontal range of a projectile is the same for angles of projection:

(a) 20° and 60°
(b) 30° and 60°
(c) 30° and 45°
(d) 45° and 60°

Answer: (b) 30° and 60°
Range R = u²sin2θ/g is the same for complementary angles (θ and 90°−θ). 30° + 60° = 90° → complementary → same range. Similarly 20° & 70°, 15° & 75°. Neither 20°&60°, nor 30°&45°, nor 45°&60° are complementary pairs.

📚 Formula Sheet — PC02

📊 Equations of Motion

v = u + at
s = ut + ½at²
v² = u² + 2as
sₙ = u + ½a(2n−1)
Avg velocity = (u+v)/2

🌠 Projectile

T = 2u sinθ/g
H = u²sin²θ/2g
R = u²sin2θ/g; max at 45°
Complementary angles → same R
At 45°: H = R/4

👪 Newton's Laws

F = ma; F = dp/dt
p = mv; J = FΔt = Δp
μₛ > μₖ > μ_rolling
f = μN (independent of area)
Lift: N = m(g±a)

🔄 Circular Motion

v = ωr; ω = 2πf = 2π/T
aₒ = v²/r (centripetal, inward)
Fₒ = mv²/r
Work done by centripetal force = 0
Centrifugal = pseudo-force only

⚡ Quick Revision — PC02

🚨 Key Traps

Max height at 45°: v_horizontal ≠ 0
Dropped + thrown horizontal: same fall time
Inertia ∝ mass (not weight)
Friction: area doesn't matter
H = R/4 always at 45°

🌠 Projectile Facts

Max range at 45°
Complementary angles: same range
30° & 60°; 20° & 70°; 15° & 75°
Horizontal: constant velocity
Vertical: uniform acceleration g

📊 Graph Rules

s-t slope = velocity
v-t slope = acceleration
v-t area = displacement
Horizontal v-t = constant velocity
Curved s-t = non-uniform motion

📝 Practice Exercise

E-01

A body is thrown vertically upward with 20 m/s. Maximum height (g=10 m/s²):

(a) 10 m
(b) 20 m
(c) 40 m
(d) 5 m

E-02

A 5 kg body accelerates at 4 m/s². The net force acting on it is:

(a) 1.25 N
(b) 20 N
(c) 9 N
(d) 0.8 N

E-03

A ball is projected at 60° with velocity 20 m/s. Time of flight (g=10 m/s²):

(a) 2 s
(b) 2√3 s
(c) 4 s
(d) 1 s

Answers: E-01: (b) 20 m [v²=u²-2gH; 0=400-20H; H=20m] | E-02: (b) 20 N [F=ma=5×4=20 N] | E-03: (b) 2√3 s [T=2u sinθ/g=2×20×sin60°/10=4×(√3/2)=2√3 s]

📝 Mock Tests 🎯 Subject Quizzes ✈️ Telegram