Lesson 4: Laws of Motion | State Board · Class 11 · Physics

Newton's laws and inertia

From daily life

When a car stops suddenly we lurch forward; the body wants to keep its motion. Link to earlier classes : Newton's laws, momentum and work-energy of classes 9-10 now go deeper with vectors and calculus.

Simple definition

A force changes a body's velocity or shape; Newton's laws relate force and motion.

In the figure : Left : F = ma; right : action-reaction.

Rule 1: First law and inertia — With no external force a body stays at rest or in uniform motion in a straight line. Inertia is this resistance to change; it is measured by mass m.

Easy : A bus starts suddenly ⇒ passengers lean back.
Medium : More mass, more inertia : a truck is harder to stop.
Exam (CET) : Measure of inertia? (1) mass (2) velocity (3) force (4) weight → answer 1)

Rule 2: Second law — F = dp/dt; for constant m F = m a. 1 N = 1 kg m s⁻²; dimension [M L T⁻²].

Easy : F = 20 N, m = 4 kg ⇒ a = 5 m/s².
Medium : m = 2 kg, a = 3î + 4ĵ ⇒ F = 6î + 8ĵ; |F| = 10 N.
Exam (CET) : F doubles and m doubles ⇒ a? (1) same (2) double (3) half (4) four times → answer 1)

Rule 3: Third law — Every action has an equal and opposite reaction : F₁₂ = − F₂₁. They act on two different bodies, so they do not cancel each other.

Easy : Gun recoil : force on the bullet forward, on the gun backward.
Medium : While walking the foot pushes the ground back; the ground pushes us forward.
Exam (CET) : Action and reaction act? (1) on different bodies (2) on the same body (3) cancel (4) differ in magnitude → answer 1)

Derivation : From F = dp/dt to F = ma

Step 1 : Momentum p = m v.
Step 2 : F = d(mv)/dt.
Step 3 : m constant ⇒ F = m dv/dt.
Step 4 : dv/dt = a ⇒ F = m a.

Remember

F = dp/dt = ma; F₁₂ = −F₂₁; 1 N = 1 kg m s⁻².

Common mistake / CET trap

Action and reaction act on different bodies, so they do not cancel; inertia is not a force.

Shortcut

Memory clue : "First = inertia; second = ma; third = pair".

Spot the difference

Law	Formula / idea	Example
First	F = 0 ⇒ v constant	bus stopping suddenly
Second	F = ma	truck heavy, car light
Third	F₁₂ = −F₂₁	rocket

Practice questions

Easy

1What is inertia?
2When is F = ma valid?
3State the third law.

Medium

4m = 5 kg, F = 30 N : a?
5m = 3 kg, a = 4 m/s² : F?
6F = 12 N, a = 3 m/s² : m?

Challenging

7m = 2 kg, F = 6î + 8ĵ : |a|?
8The horse pulls the cart; why does it move despite action-reaction?

Show answers (try them yourself first)

1Resistance to change
2When m is constant
3Equal and opposite reaction
46 m/s²
512 N
64 kg
7a = 5 m/s²
8The forces act on different bodies; friction from the ground decides

Types of forces and work done by a force

From daily life

Pushing a box, the hand, the floor and gravity — many forces act together.

Simple definition

Work = force × displacement along the force; energy is the capacity to do work.

In the figure : N, mg, F, f : a free-body diagram.

Rule 1: Types of forces — Gravitational (mg), normal N, tension T, friction f, spring force (F = −kx), electric/magnetic. Show every force in a free-body diagram (FBD).

Easy : Block at rest : N = mg (m = 5 kg ⇒ 50 N).
Medium : Spring k = 200 N/m, x = 0.1 m ⇒ F = 20 N.
Exam (CET) : The force in a rope is? (1) tension (2) normal (3) friction (4) gravity → answer 1)

Rule 2: Work done by a force — W = F · s = F s cosθ (J = N m). F ⟂ s ⇒ W = 0. Against gravity : W = mgh.

Easy : F = 50 N, θ = 60°, s = 10 m ⇒ W = 250 J.
Medium : m = 4 kg, h = 3 m ⇒ W = 4 × 10 × 3 = 120 J.
Exam (CET) : Work is zero when? (1) F ⟂ s (2) F ∥ s (3) θ = 0 (4) large s → answer 1)

Rule 3: Work-energy theorem and power — W = ½mv² − ½mu²; KE = ½mv²; PE = mgh; P = W ÷ t = F v (W).

Easy : m = 2 kg, F = 10 N, s = 5 m, u = 0 ⇒ v = √(2W/m) = 7.07 m/s.
Medium : F = 20 N, v = 3 m/s ⇒ P = 60 W.
Exam (CET) : 1 hp = ? (1) 746 W (2) 1000 W (3) 100 W (4) 10 W → answer 1)

Derivation : Work-energy theorem

Step 1 : W = F s = m a s.
Step 2 : v² = u² + 2as ⇒ a s = (v² − u²) ÷ 2.
Step 3 : W = m (v² − u²) ÷ 2.
Step 4 : W = ½mv² − ½mu² = ΔKE.

Remember

W = F s cosθ; W = ΔKE; P = Fv; PE = mgh.

Common mistake / CET trap

Centripetal force does no work in circular motion; weight (N) and mass (kg) differ.

Shortcut

Memory clue : "Work = force × distance along the force".

Spot the difference

Quantity	Formula	Unit / dimension
W	F s cosθ	J; M L² T⁻²
KE	½mv²	J
PE	mgh	J
P	W/t = Fv	W; M L² T⁻³

Practice questions

Easy

1Formula for work?
2Work-energy theorem?
3Unit of power?

Medium

4F = 30 N, s = 4 m, θ = 0° : W?
5m = 1 kg, v = 10 : KE?
6W = 600 J, t = 20 s : P?

Challenging

7m = 2 kg, u = 3, v = 5 : W?
8m = 60 kg climbs 10 m in 20 s : P?

Show answers (try them yourself first)

1F s cosθ
2W = ΔKE
3watt
4120 J
550 J
630 W
716 J
8300 W

Applications of Newton's laws

From daily life

In a lift going up your feet feel heavy; going down, light — the force changes, not the mass.

Simple definition

Applications : draw an FBD and write F = ma for each body.

In the figure : m₁ > m₂ ⇒ m₁ goes down, m₂ goes up.

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Applications of Newton's laws

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Rule 1: Object in a lift — R = m (g + a) accelerating up; R = m (g − a) accelerating down; free fall ⇒ R = 0 (weightless).

Easy : m = 60 kg, a = 2 : up ⇒ R = 720 N; down ⇒ 480 N.
Medium : Constant speed ⇒ R = mg = 600 N.
Exam (CET) : R in a freely falling lift? (1) 0 (2) mg (3) 2mg (4) infinite → answer 1)

Rule 2: Atwood machine — m₁ > m₂ : a = (m₁ − m₂) g ÷ (m₁ + m₂); T = 2 m₁ m₂ g ÷ (m₁ + m₂).

Easy : m₁ = 3, m₂ = 2 ⇒ a = 2 m/s²; T = 24 N.
Medium : m₁ = m₂ ⇒ a = 0; T = mg.
Exam (CET) : If m₁ = 2m₂, a = ? (1) g/3 (2) g/2 (3) g (4) 0 → answer 1)

Rule 3: Inclined plane — Smooth : a = g sinθ; N = mg cosθ. Rough : a = g (sinθ − μ cosθ).

Easy : θ = 30°, smooth ⇒ a = 5 m/s²; N = mg cos30° = 0.866 mg.
Medium : μ = 0.2, θ = 30° ⇒ a = 3.27 m/s².
Exam (CET) : On a slope N = ? (1) mg cosθ (2) mg sinθ (3) mg (4) 0 → answer 1)

Derivation : Atwood : a and T

Step 1 : m₁ : m₁g − T = m₁a.
Step 2 : m₂ : T − m₂g = m₂a.
Step 3 : Add : (m₁ − m₂) g = (m₁ + m₂) a ⇒ a = (m₁ − m₂) g ÷ (m₁ + m₂).
Step 4 : T = m₂ (g + a) = 2 m₁ m₂ g ÷ (m₁ + m₂).

Remember

R = m(g ± a); a = (m₁ − m₂)g/(m₁ + m₂); a = g sinθ.

Common mistake / CET trap

In a lift the true weight stays, the apparent weight (R) changes; T is equal on both sides (light string).

Shortcut

Memory clue : "Up = heavy; down = light".

Spot the difference

Case	Formula
Lift up (a)	m(g + a)
Lift down (a)	m(g − a)
Smooth slope	g sinθ
Rough slope	g(sinθ − μcosθ)

Practice questions

Easy

1Lift at constant speed : R?
2Formula for a in an Atwood machine?
3a on a smooth incline?

Medium

4m = 50, lift up with a = 1 : R?
5m₁ = 5, m₂ = 3 : a?
6θ = 37° smooth : a? (sin37° = 0.6)

Challenging

7m₁ = 5, m₂ = 3 : T?
8θ = 30°, μ = 0.3 : a?

Show answers (try them yourself first)

1mg
2(m₁ − m₂)g/(m₁ + m₂)
3g sinθ
4550 N
52.5 m/s²
66 m/s²
737.5 N
82.40 m/s²

Momentum, impulse and conservation

From daily life

A fast ball is hard to stop, a slow one easy — mass × velocity is momentum.

Simple definition

Momentum = mass × velocity; impulse = force × short time.

In the figure : MV + mv = 0.

Rule 1: Momentum and impulse — p = m v (kg m/s); impulse J = F Δt = Δp. Area under the F–t graph = impulse.

Easy : m = 0.15 kg, v = 20 m/s ⇒ p = 3 kg m/s.
Medium : Ball 0.15 kg, 20 m/s reversed to 20 m/s : Δp = 6 kg m/s; Δt = 0.02 s ⇒ F = 300 N.
Exam (CET) : Unit of impulse? (1) N s (2) N (3) J (4) W → answer 1)

Rule 2: Conservation of momentum — With zero external force Σp is constant : m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.

Easy : Gun 5 kg, bullet 0.05 kg, 400 m/s ⇒ recoil V = -4 m/s.
Medium : 3 kg at 4 m/s sticks to 1 kg at rest : v = 12/4 = 3 m/s.
Exam (CET) : Total momentum of gun + bullet? (1) 0 (2) mv (3) MV (4) infinite → answer 1)

Rule 3: Rocket and tricks — A rocket throws gas backward ⇒ moves forward. Lengthening Δt lowers F (cushion, cricket catch).

Easy : Pulling the hands back while catching raises Δt and lowers F.
Medium : Δp = 6, Δt = 0.1 ⇒ F = 60 N (0.02 s ⇒ 300 N).
Exam (CET) : If Δt increases, F? (1) decreases (2) increases (3) same (4) 0 → answer 1)

Derivation : Impulse-momentum relation

Step 1 : F = dp/dt.
Step 2 : F dt = dp; integrate ∫F dt = ∫dp.
Step 3 : ∫F dt = p₂ − p₁ = Δp.
Step 4 : Constant F : J = F Δt = Δp.

Remember

p = mv; J = FΔt = Δp; external force 0 ⇒ Σp constant.

Common mistake / CET trap

Momentum is a vector, fix signs for direction; momentum may be conserved while energy is not.

Shortcut

Memory clue : "Impulse = F × Δt = Δp".

Spot the difference

Quantity	Formula	Unit
Momentum	m v	kg m/s
Impulse	F Δt = Δp	N s
Conservation	Σp before = Σp after	—

Practice questions

Easy

1What is momentum?
2What is impulse?
3When is momentum conserved?

Medium

4m = 2, v = 8 : p?
5F = 50 N, Δt = 0.2 s : J?
6Δp = 12, Δt = 0.03 : F?

Challenging

7Gun 4 kg, bullet 20 g at 500 m/s : recoil?
82 kg at 6 m/s and 4 kg at 1 m/s (same direction) stick : v?

Show answers (try them yourself first)

1m v
2F Δt = change of momentum
3When the external force is zero
416 kg m/s
510 N s
6400 N
72.5 m/s
8v = 16/6 = 2.67 m/s

Collisions

From daily life

Carrom coins swap energy when they hit; clay balls stick.

Simple definition

A collision is a brief interaction between two bodies.

In the figure : Before : m₁ with u₁; after : both move.

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Collisions

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Rule 1: Types of collision — Elastic : momentum and KE conserved, e = 1. Inelastic : momentum conserved, KE lost, 0 < e < 1. Perfectly inelastic : bodies stick, e = 0.

Easy : Car-car crash : inelastic; billiard balls : nearly elastic.
Medium : e = (v₂ − v₁) ÷ (u₁ − u₂) : relative speed after ÷ before.
Exam (CET) : Conserved in all collisions? (1) momentum (2) KE (3) velocity (4) e → answer 1)

Rule 2: Elastic collision in 1-D — v₁ = [(m₁ − m₂) u₁ + 2 m₂ u₂] ÷ (m₁ + m₂); v₂ = [(m₂ − m₁) u₂ + 2 m₁ u₁] ÷ (m₁ + m₂). Equal masses ⇒ velocities swap.

Easy : m₁ = m₂ = 2, u₁ = 6, u₂ = 0 ⇒ v₁ = 0, v₂ = 6.
Medium : m₁ = 1, m₂ = 3, u₁ = 8, u₂ = 0 ⇒ v₁ = -4, v₂ = 4 m/s.
Exam (CET) : Elastic, equal masses, second at rest : the first afterwards? (1) stops (2) reverses (3) doubles (4) same → answer 1)

Rule 3: Perfectly inelastic collision — v = (m₁u₁ + m₂u₂) ÷ (m₁ + m₂); KE loss = ½ [m₁m₂ ÷ (m₁ + m₂)] (u₁ − u₂)².

Easy : m₁ = m₂ = 2, u₁ = 6, u₂ = 0 ⇒ v = 3; KE 36 J → 18 J; loss 18 J.
Medium : A bullet 0.02 kg at 300 m/s embedded in a 1.98 kg block ⇒ v = 3 m/s.
Exam (CET) : e in a perfectly inelastic collision? (1) 0 (2) 1 (3) > 1 (4) ½ → answer 1)

Derivation : Elastic collision v₁, v₂ (with u₂ = 0)

Step 1 : Momentum : m₁u₁ = m₁v₁ + m₂v₂.
Step 2 : KE : ½m₁u₁² = ½m₁v₁² + ½m₂v₂² ⇒ m₁(u₁² − v₁²) = m₂v₂².
Step 3 : Divide : u₁ + v₁ = v₂ (relative velocity reverses).
Step 4 : Hence v₁ = (m₁ − m₂) u₁ ÷ (m₁ + m₂), v₂ = 2 m₁ u₁ ÷ (m₁ + m₂).

Remember

Σp always conserved; elastic : KE conserved, e = 1; stuck : e = 0.

Common mistake / CET trap

Lost KE does not mean lost momentum; keep the direction signs.

Shortcut

Memory clue : "Equal m = velocities swap".

Spot the difference

Type	Momentum	KE	e
Elastic	conserved	conserved	1
Inelastic	conserved	lost	0 to 1
Perfectly inelastic	conserved	maximum loss	0

Practice questions

Easy

1What is an elastic collision?
2Formula for e?
3What is a perfectly inelastic collision?

Medium

4m₁ = m₂ = 1, u₁ = 4, u₂ = 0 (elastic) : v₁, v₂?
53 kg at 5 m/s + 2 kg at rest (stick) : v?
6u₁ = 10, u₂ = 0, v₁ = 2, v₂ = 6 : e?

Challenging

73 kg at 5 m/s sticks to 2 kg at rest : KE loss?
8m₁ = 2, m₂ = 1, u₁ = 3, u₂ = 0 elastic : v₁, v₂?

Show answers (try them yourself first)

1Both momentum and KE conserved
2(v₂ − v₁)/(u₁ − u₂)
3Bodies stick and move together
4v₁ = 0, v₂ = 4
53 m/s
6e = 0.4
715.0 J
8v₁ = 1, v₂ = 4 m/s

Torque, couple and equilibrium

From daily life

A door is hard to push near the hinge and easy at the edge — torque.

Simple definition

Torque is the turning effect of a force; equilibrium means no change in translation or rotation.

In the figure : F₁ d₁ = F₂ d₂; support N = F₁ + F₂.

Rule 1: Torque — τ = r × F; magnitude τ = r F sinθ = F × perpendicular distance. Unit N m; dimension [M L² T⁻²].

Easy : Spanner 0.3 m, F = 40 N perpendicular ⇒ τ = 12 N m.
Medium : r = 0.5 m, F = 20 N, θ = 30° ⇒ τ = 5.0 N m.
Exam (CET) : Torque on a door is maximum when? (1) force far from the hinge and perpendicular (2) near (3) parallel (4) zero → answer 1)

Rule 2: Couple — Two equal, opposite, non-collinear forces : τ = F × d; zero net force, only rotation (steering wheel, tap).

Easy : F = 10 N, d = 0.2 m ⇒ τ = 2 N m.
Medium : Steering diameter 0.4 m, 5 N per hand ⇒ τ = 5 × 0.4 = 2 N m.
Exam (CET) : Net force of a couple? (1) 0 (2) 2F (3) F (4) F/2 → answer 1)

Rule 3: Equilibrium — ΣF = 0 (translational) and Στ = 0 (rotational). Lever : F₁ d₁ = F₂ d₂.

Easy : See-saw : 40 kg at 1.5 m ⇒ on the other side at 1 m, m = 60 kg.
Medium : Support reaction N = (40 + 60) × 10 = 1000 N.
Exam (CET) : In equilibrium? (1) ΣF = 0 and Στ = 0 (2) only ΣF = 0 (3) only Στ = 0 (4) neither → answer 1)

Derivation : Principle of the lever

Step 1 : Take Στ = 0 about the fulcrum.
Step 2 : F₁ is anticlockwise, F₂ is clockwise.
Step 3 : F₁ d₁ − F₂ d₂ = 0.
Step 4 : F₁ d₁ = F₂ d₂; N = F₁ + F₂.

Remember

τ = r F sinθ; couple τ = F d; ΣF = 0, Στ = 0.

Common mistake / CET trap

d is the perpendicular distance of the line of force from the pivot; both forces of a couple give a single τ.

Shortcut

Memory clue : "Long arm = easy turning".

Spot the difference

Property	Torque	Couple
Forces	one or many	two equal-opposite
Net F	can be non-zero	zero
Formula	r F sinθ	F d

Practice questions

Easy

1Formula for torque?
2What is a couple?
3Two conditions of equilibrium?

Medium

4r = 0.4, F = 25 N perpendicular : τ?
5F = 8 N, d = 0.5 : couple torque?
6F₁ = 50 N, d₁ = 0.6 : F₂ when d₂ = 0.3?

Challenging

7A 50 kg child at 2 m; where must a 40 kg child sit to balance?
81 m rod of 10 kg with 20 kg at one end : where is the pivot?

Show answers (try them yourself first)

1r F sinθ
2Two equal, opposite, non-collinear forces
3ΣF = 0, Στ = 0
410 N m
54 N m
6100 N
72.5 m
8At x from the 20 kg end : 20 x = 10 (1 − x) ⇒ x = 0.33 m

Centre of mass and centre of gravity

From daily life

A plank balances on one finger at one point — the centre of mass.

Simple definition

The centre of mass is the point where the mass can be treated as concentrated.

In the figure : The CM lies nearer the heavier m₂.

Rule 1: Centre of mass — x_cm = Σ mᵢ xᵢ ÷ Σ mᵢ (same for y, z). Shifts toward the heavier body. For a symmetric uniform body it is the geometric centre.

Easy : 2 kg at x = 0, 3 kg at x = 5 ⇒ x_cm = (0 + 15)/5 = 3 m.
Medium : Uniform rod : at the middle.
Exam (CET) : CM of two equal masses? (1) midway (2) on one (3) outside (4) infinite → answer 1)

Rule 2: Motion of the CM — v_cm = Σ mv ÷ M; zero external force ⇒ CM moves uniformly; F_ext = M a_cm.

Easy : An exploding shell : the fragments scatter but the CM follows the old path.
Medium : 2 kg at 3 m/s and 3 kg at −2 m/s ⇒ v_cm = (6 − 6)/5 = 0.
Exam (CET) : With zero external force the CM velocity is? (1) constant (2) rising (3) always zero (4) changing → answer 1)

Rule 3: Centre of gravity — The point where the whole weight can be taken to act. In uniform g, CG = CM. A lower CG gives more stability.

Easy : Heavy luggage at the bottom of a bus ⇒ less chance of toppling.
Medium : Empty bottle : middle; full : lower.
Exam (CET) : To increase stability? (1) lower the CG (2) raise it (3) smaller base (4) less mass → answer 1)

Derivation : CM of two particles

Step 1 : Positions x₁, x₂; masses m₁, m₂.
Step 2 : CM is the weighted-average point : m₁(x_cm − x₁) = m₂(x₂ − x_cm).
Step 3 : x_cm (m₁ + m₂) = m₁x₁ + m₂x₂.
Step 4 : x_cm = (m₁x₁ + m₂x₂) ÷ (m₁ + m₂).

Remember

x_cm = Σmx/Σm; v_cm = Σmv/M; F_ext = M a_cm.

Common mistake / CET trap

The CM lies between the particles (nearer the heavier); measure all distances from one reference point.

Shortcut

Memory clue : "CM leans toward the heavy one".

Spot the difference

Property	CM	CG
Based on	mass	weight
Formula	Σmx/Σm	Σ(mg)x/Σ(mg)
If g varies	same	may shift

Practice questions

Easy

1What is the centre of mass?
2Formula for CM?
3What is CG?

Medium

41 kg at x = 0, 4 kg at x = 5 : x_cm?
52 kg at 4 m/s and 2 kg at 0 : v_cm?
6m = 3 at (0, 0) and 3 at (4, 0) : CM?

Challenging

71 m rod of 2 kg with 2 kg at one end : CM?
8What happens to the CM when a shell explodes?

Show answers (try them yourself first)

1The point where total mass may be taken as concentrated
2Σmx/Σm
3The point where weight acts
4x_cm = 4 m
52 m/s
6(2, 0)
70.25 m from the loaded end
8The CM keeps its original path

Quick revision of the whole lesson

Concept	Key formula / fact	Unit / dimensions
Second law	F = dp/dt = ma	N; M L T⁻²
Work	W = F s cosθ = ΔKE	J; M L² T⁻²
Power	P = F v	W; M L² T⁻³
Atwood	a = (m₁ − m₂)g/(m₁ + m₂)	m/s²
Impulse	J = F Δt = Δp	N s
Collision	e = (v₂ − v₁)/(u₁ − u₂)	—
Torque	τ = r F sinθ	N m
Centre of mass	x_cm = Σmx/Σm	m

🎯 Most important points for the exam

Inertia = mass; F = ma; action-reaction on different bodies.
Lift : R = m(g ± a).
Impulse = Δp; larger Δt ⇒ smaller F.
Momentum is conserved in all collisions; elastic : KE conserved.
Equilibrium : ΣF = 0 and Στ = 0.
A couple has zero net F.
CM leans to the heavier particle; no external force ⇒ v_cm constant.

Study mantra

Draw the FBD → F = ma → conserve momentum/energy → ΣF, Στ = 0.