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L-3: Laws of Motion

Physics — Class 12 · NIOS Code 312 · Module 1 · Source: 312_Physics_Eng_Lesson3.pdf

Why Do Things Move? — Starting Point

In the previous lesson you learnt to describe motion using displacement, velocity, and acceleration. But a deeper question remains: what makes an object move? Why does a ball rolling on the ground eventually stop? From everyday experience we know that pushing or pulling is needed to change an object's position. A football must be kicked, a cricket ball must be hit hard — muscular activity is involved and its effect is visible.

Yet many causes of motion are not visible. What makes raindrops fall? What makes Earth orbit the Sun? In this lesson you discover that force causes motion. Newton showed that force and motion are intimately connected. The laws of motion are fundamental — they help us understand walking, driving, collisions, rocket propulsion, and countless everyday phenomena.

NIOS learning objectives: explain inertia; state and illustrate Newton's three laws; explain conservation of momentum; understand equilibrium of concurrent forces; define coefficients of friction; distinguish static, kinetic, and rolling friction; suggest methods of reducing friction; and apply Newton's laws using free body diagrams.

3.1 Concepts of Force and Inertia

Stationary objects remain where placed unless forced to move. Similarly, an object moving with constant velocity must be forced to change its state of motion. The property by which an object resists change in its state of rest or uniform motion in a straight line is called inertia. Think of it as laziness built into matter — objects prefer to keep doing what they are already doing.

Mass is the measure of inertia. A heavy truck at low speed is harder to stop than a light cricket ball at high speed because the truck has far more mass. Inertia is a remarkable property. Without it, your books could mingle with your sibling's notes and your wardrobe could drift to a friend's house. However, rest and uniform motion are not absolute: an object at rest for one observer may appear in motion to another (Lesson 1). Observations confirm that change in velocity occurs only when a net force acts on the body.

3.1.1 Force and Its Effects

Force is familiar from daily life — pulling, pushing, kicking, hitting. Though invisible, its effects are seen. Forces can: (a) change the shape and size of an object (a balloon deforms); (b) influence motion — start motion, stop motion, or change direction or speed; (c) rotate a body about an axis (covered in Lesson 7).

Force is a vector quantity. When several forces act simultaneously, a net equivalent force is found by vector addition (Lesson 1). Motion is characterised by displacement and velocity. Velocity may increase (free fall) or decrease (ball on horizontal surface). A net non-zero force is required to change state of motion.

  • If net force is parallel to velocity → speed increases.
  • If net force is opposite to velocity → speed decreases.
  • If net force is perpendicular to velocity → speed stays constant but direction changes (uniform circular motion).

Student tip: A body does NOT always move in the direction of net force. If it already has velocity, force may only bend the path. Acceleration always aligns with net force; velocity may not.

3.1.2 Galileo's Experiments and the First Law

When you roll a marble on a smooth floor it stops — friction reduces velocity to zero. To keep it moving at constant velocity you must continuously apply force. Similarly, a trolley at constant velocity must be continuously pushed or pulled to balance friction.

Galileo showed that without external force, a body continues in rest or uniform straight-line motion. On an inclined plane moving downward, a body accelerates; moving upward, it retards. On a horizontal plane with no friction, it would move with uniform speed forever.

Galileo's Inclined Planes (Fig 3.x) PQ ↓ RS ↑ horizontal → ∞ ball path
Fig 3.x — Ball rolls down PQ, rises on RS; flatter RS → travels farther (First Law)

In his thought experiment, a ball rolls down plane PQ and rises on plane RS to nearly the same height. As RS becomes less inclined, the ball travels farther. When RS is horizontal, the ball keeps moving indefinitely if friction is absent.

Newton's First Law: A body continues in a state of rest or of uniform motion in a straight line unless acted upon by a net external force.

Because motion is relative, measurements must be made with respect to a frame of reference. An inertial frame is one in which a body in translatory motion has constant velocity when no net external force acts. A frame fixed to Earth is inertial for practical purposes.

3.2 Concept of Momentum

A fielder struggles to stop a fast cricket ball (small mass, large velocity) but also struggles with a slow truck (large mass, small velocity). Both mass and velocity matter when studying force and motion.

The product of mass m and velocity v is linear momentum p:

p = mv
p = Linear momentum (kg·m·s⁻¹)
m = Mass (kilograms, kg)
v = Velocity (m·s⁻¹)
Momentum is a vector — direction same as velocity.

Momentum changes when magnitude, direction, or both change. Example: a freely falling 2 kg object has zero momentum at t = 0; at t = 1 s, v = 9.8 m/s downward so p = 19.6 kg·m/s downward; at t = 2 s, p = 39.2 kg·m/s downward. Gravitational force causes this continuous increase.

When a 0.2 kg ball strikes a wall at 10 m/s and rebounds at 10 m/s, magnitude of momentum is unchanged but direction reverses. If initial momentum is +2 kg·m/s, final is −2 kg·m/s, so change Δp = −4 kg·m/s. The wall exerts force on the ball.

3.3 Newton's Second Law of Motion

A body at constant velocity has constant momentum; the first law implies no net external force. For a freely falling body, momentum increases with time — connecting force, time, and momentum change.

Second Law (momentum form): The rate of change of momentum is directly proportional to the net force, and change occurs in the direction of the net force.

F = Δp / Δt
F = Net external force (Newtons, N)
Δp = Change in momentum (kg·m·s⁻¹)
Δt = Time interval (seconds, s)
SI unit of force: kg·m·s⁻² = newton (N)

Writing p = mv and assuming constant mass:

F = ma
F = Net Force (Newtons, N)
m = Mass (kilograms, kg)
a = Acceleration (m·s⁻²)
Hover tip: Mass resists acceleration. The more mass, the more force needed for the same acceleration.

Example: A 0.4 kg ball rolling at 20 m/s stops in 10 s. |F| = m|Δv/Δt| = 0.4 × 20/10 = 0.8 N opposite to motion.

Example: A 10 kg body at 10 m/s is stopped by 50 N opposing force. t = m(v₀ − v)/F = 10 × 10/50 = 2 s.

Exam note: For rockets where mass changes with time, use F = d(mv)/dt — the more general form. F = ma alone is for constant mass only.

3.4 Forces in Pairs — Third Law and Impulse

When Earth pulls an object, does the object pull Earth? When you push an almirah, does it push back? Actions between two bodies are always mutual. Forces exist in pairs — both are real interaction forces.

Newton's Third Law: When two objects interact, the force exerted by one on the other is equal in magnitude and opposite in direction to the force exerted by the second on the first.

F₁₂ = −F₂₁
F₁₂ = Force on object 1 due to object 2
F₂₁ = Force on object 2 due to object 1
Equal magnitude, opposite direction. Act on DIFFERENT bodies — they do NOT cancel on one body.
Third Law — Book on Table Book F₁=mg ↓ F₂ ↑ reaction F₁₂ = −F₂₁ Different bodies!
Fig 3.x — Action & reaction on different bodies (do not cancel)

A book on a table exerts downward force F₁ = mg on the table; the table exerts upward F₂ on the book. F₁ and F₂ do not cancel because they act on different bodies. Action and reaction exist simultaneously, not one after the other.

Impulse

Impulse = F·Δt = Δp
F = Force (N)
Δt = Time duration (s)
Δp = Change in momentum (kg·m·s⁻¹)
SI unit: N·s. Impulse is a vector. Explains why fielders draw hands back when catching.

Action-Reaction Concept Map

                    NEWTON'S THIRD LAW
                           |
            +--------------+--------------+
            |              |              |
     Forces exist      Equal          Opposite
       in PAIRS       magnitude       direction
            |              |              |
            v              v              v
    +---------------+  F₁₂ = -F₂₁   Act on DIFFERENT
    |  Examples:    |              bodies (no cancel)
    +---------------+
    | Man kicks     |  Action: foot on ball
    |  football     |  Reaction: ball on foot
    +---------------+
    | Earth-Moon    |  Action: Earth pulls moon
    |               |  Reaction: moon pulls Earth
    +---------------+
    | Ball hits     |  Action: ball on wall
    |  wall         |  Reaction: wall on ball
    +---------------+
    | High jumper   |  Action: jumper pushes ground
    |               |  Reaction: ground pushes jumper UP
    +---------------+

3.5 Conservation of Momentum

When two bodies interact and mutual interaction is the only force, the vector sum of their momenta remains unchanged. For a closed (isolated) system — no external forces — total momentum is constant. Individual momenta may change due to internal mutual forces.

p_total = constant
Isolated system: Δp_A + Δp_B = 0
Total momentum before = total momentum after
Applies to collisions, explosions, gun recoil, rockets, nuclear reactions.

Derivation: From second law Δp = FΔt. For bodies A and B: F_AB = −F_BA, so Δp_A/Δt = −Δp_B/Δt, hence Δp_A + Δp_B = 0.

Gun Recoil — Momentum Conservation Gun (M) bullet v₁ → ← recoil v₂ mv₁ + Mv₂ = 0
Fig 3.x — Bullet forward momentum balanced by gun recoil backward

(a) Gun recoil: mv₁ + Mv₂ = 0, so v₂ = −(m/M)v₁. Since m ≪ M, recoil is much slower than the bullet.

(b) Collision: Two trolleys (mass m each) at v hit three stationary trolleys. 2mv = 5mv′ → v′ = (2/5)v.

(c) Explosion: Bomb at rest has zero momentum; fragments fly opposite ways so vector sum stays zero.

(d) Rocket propulsion: Escaping gases provide thrust via momentum conservation.

3.5.3 Equilibrium of Concurrent Forces

Forces acting simultaneously at one point are concurrent forces. They are in equilibrium when their resultant is zero: F₁ + F₂ + F₃ = 0. Graphically, the vector sum of any two must equal and oppose the third. For a point object in static equilibrium, the vector sum of all forces must be zero.

3.6 Friction

When a batsman hits a ball along the ground, it eventually stops — friction opposes motion and changes momentum. Friction is a contact force, parallel to surfaces and opposite to attempted or actual motion. Friction is a necessary evil: it wastes energy yet lets us walk, drive, and brake.

3.6.1 Static and Kinetic Friction

Static vs Kinetic Friction (Fig 3.x) F_ext → f_s ← At rest: f_s = F_ext (up to f_s max) f_k ← Sliding: f_k = μ_k F_N
Fig 3.x — Static friction adjusts until max; then kinetic friction opposes motion

A block on a horizontal surface does not move until applied force F_ext exceeds a limit. While at rest, static friction f_s matches F_ext up to maximum f_s(max). Beyond this, the block slides and kinetic friction f_k acts. Starting motion needs more force than maintaining it: f_s(max) > f_k.

f_s(max) = μ_s F_N
f_s(max) = Maximum static friction (N)
μ_s = Coefficient of static friction (dimensionless)
F_N = Normal force (N)
While f_s < f_s(max): f_s = F_ext. Independent of contact area.
f_k = μ_k F_N
f_k = Kinetic friction (N)
μ_k = Coefficient of kinetic friction
F_N = Normal force (N)
Generally μ_s > μ_k. On horizontal surface F_N = mg.

Example: 2 kg block, μ_s = 0.25 → f_s(max) = 0.25 × 2 × 9.8 = 4.9 N.

Example: 5 kg block, μ_k = 0.1, pulled by 10 N. f_k = 4.9 N; F_net = 5.1 N; a = 1.02 m/s².

3.6.2 Rolling Friction and Reducing Friction

Rolling friction is much smaller than sliding friction (~1/100 for steel on steel). Methods to reduce friction: wheels and ball bearings; lubricants (grease, oil); compressed air between surfaces; streamlined shapes for fluid friction. Cars are most fuel-efficient around 40–45 km/h due to air resistance.

3.7 Free Body Diagram Technique

Block on Inclined Plane — FBD mg mg sinθ N f ΣF = ma along incline
Fig 3.x — Free body diagram: weight components, normal, friction

Newton's laws become easier with free body diagrams (FBD) — diagrams showing all forces on an isolated object.

         HOW TO SOLVE FBD PROBLEMS
         =========================
                    |
                    v
         [1] Draw neat diagram of system
                    |
                    v
         [2] ISOLATE the object of interest
             (this is the "free body")
                    |
                    v
         [3] Mark ALL external forces with arrows
             touching the body; show line of action
                    |
                    v
         [4] Apply Newton's 2nd Law:
             ΣF = ma   (or ΣFx = max, ΣFy = may)
                    |
                    v
         [5] Count unknowns vs independent equations
             Need equal number for complete solution
                    |
                    v
         [6] Solve + CHECK extreme cases
             (e.g. m1 = m2, m1 >> m2 for pulleys)

Example — Two blocks on smooth surface: Blocks m₁ and m₂ connected by string; m₂ pulled by F. a = F/(m₁ + m₂). Tension T = m₁F/(m₁ + m₂).

Atwood Machine m₁ m₂ m₁ goes down ↓ m₂ goes up ↑ a = (m₁−m₂)g/(m₁+m₂)
Fig 3.x — Atwood machine: unequal masses accelerate under gravity

Example — Atwood machine: m₁ > m₂ over frictionless pulley. a = (m₁ − m₂)g/(m₁ + m₂); T = 2m₁m₂g/(m₁ + m₂).

Example — Trolley with friction: M = 10 kg, m = 2 kg, μ_k = 0.02 gives a = 1.47 m/s², T = 16.66 N.

3.8 Inertial and Non-Inertial Frames

An inertial frame is stationary or moves at constant velocity; Newton's laws hold directly. A non-inertial frame accelerates — add a pseudo force (−ma). Centrifugal force appears in rotating frames. Water tilts in a starting train because effective gravity in the train frame is g − a.

Summary

  • Inertia resists change in rest or uniform motion; mass measures inertia.
  • First Law: Uniform motion or rest continues unless net external force acts.
  • Second Law: F = Δp/Δt; for constant mass F = ma.
  • Third Law: F₁₂ = −F₂₁ on different bodies.
  • Conservation: Isolated system → total momentum constant.
  • Friction: f_s(max) = μ_s F_N; f_k = μ_k F_N; rolling ≪ kinetic.
  • FBD before every numerical — the most reliable exam strategy.
20 cards · click any card to flip
Inertia
Property of a body to resist change in state of rest or uniform motion in a straight line. Mass is the measure of inertia.
Newton's First Law
A body remains in rest or uniform straight-line motion unless acted upon by a net external force.
Inertial frame
Reference frame where a body in translatory motion has constant velocity if no net external force acts. Earth-fixed frame is approximately inertial.
Linear momentum
p = mv. Vector quantity; SI unit kg·m·s⁻¹. Direction same as velocity.
Newton's Second Law
Rate of change of momentum is proportional to net force. For constant mass: F = ma. 1 N = 1 kg·m·s⁻².
Impulse
F·Δt = Δp. Effect of force over short time. SI unit N·s. Drawing hands back while catching reduces force for same momentum change.
Newton's Third Law
F₁₂ = −F₂₁. Forces in pairs, equal and opposite, acting on different bodies simultaneously — they do NOT cancel.
Conservation of momentum
In an isolated system (no net external force), total momentum remains constant. Individual momenta may change.
Gun recoil
mv₁ + Mv₂ = 0 → v₂ = −(m/M)v₁. Recoil velocity much smaller than bullet velocity since m ≪ M.
Concurrent forces
Forces acting simultaneously at one point. In equilibrium when vector sum is zero: F₁ + F₂ + F₃ = 0.
Static friction fₛ
Opposes impending motion. Self-adjusts up to fₛ(max) = μₛFₙ. Independent of contact area.
Kinetic friction fₖ
fₖ = μₖFₙ when body slides. Generally μₛ > μₖ. Acts opposite to direction of motion.
Rolling friction
Much smaller than sliding friction (~1/100 for steel on steel). Wheels and ball bearings exploit this.
Normal force Fₙ
Perpendicular contact force. On horizontal surface Fₙ = mg; on incline Fₙ = mg cos θ.
Free Body Diagram
Diagram showing all external forces on an isolated object. Apply ΣF = ma after drawing.
Atwood machine
Two masses over pulley: a = (m₁−m₂)g/(m₁+m₂); T = 2m₁m₂g/(m₁+m₂) when m₁ > m₂.
Friction — necessary evil
Opposes motion and wastes energy, yet enables walking, driving, and braking.
Non-inertial frame
Accelerating reference frame. Use pseudo force (−ma); centrifugal force in rotation.
Rocket propulsion
Conservation of momentum: escaping gas gives thrust; a = −(ṁ/M)v as fuel burns.
Galileo's conclusion
Without external force, body continues in rest or uniform motion. On horizontal plane with no friction → uniform speed forever.

Q1. Which physical quantity is a measure of the inertia of a body?

Q2. The SI unit of force, one newton, is equivalent to:

Q3. Linear momentum p of a body of mass m moving with velocity v is defined as:

Q4. Newton's first law of motion is also known as the law of:

Q5. For a block on a horizontal surface, the maximum force of static friction is given by:

Q6. In an isolated system of colliding bodies with no external force, which quantity remains constant?

Q7. According to Newton's third law, when a book rests on a table:

Q8. A gun of mass M fires a bullet of mass m with velocity v₁. The recoil velocity v₂ is:

Q9. In the free body diagram technique, Newton's second law is applied as:

Q10. Rolling friction between steel wheels and steel rails is approximately:

p = mv
F = Δp / Δt
F = ma
F₁₂ = −F₂₁
Impulse = F·Δt = Δp
p_total = constant
mv₁ + Mv₂ = 0
F₁ + F₂ + F₃ = 0
f_s(max) = μ_s F_N
f_k = μ_k F_N
ΣF = ma
a = F / (m₁ + m₂)
T = m₁F / (m₁ + m₂)
a = (m₁ − m₂)g / (m₁ + m₂)
T = 2m₁m₂g / (m₁ + m₂)

1. Formulas & Definitions

Unlock the full Ch 3 study guide — definitions, derivations, why each formula works, history, and deep understanding for every equation.

p = mv

Definition: Linear momentum is mass times velocity — a vector in the direction of motion.

Derivation

Defined as p = mv from NIOS Sec 3.2. Momentum changes when mass, speed, or direction changes.

Variables

p = momentum (kg·m·s⁻¹) · m = mass (kg) · v = velocity (m·s⁻¹)

Why it works

Both mass AND speed matter when stopping motion. A light fast ball and a heavy slow truck can have similar momentum — that is why both are hard to stop.

Historical context

Descartes introduced momentum ideas; Newton unified force and momentum change in the Principia (1687).

Deep understanding

Momentum is conserved in isolated systems because internal forces cancel in pairs (Third Law). It is often easier than tracking forces in collisions.

2. Diagrams & Visuals

Momentum — mass × velocity m v → p = mv

Color-coded visual · step-by-step breakdown below

  1. Identify mass m in kg
  2. Identify velocity v (magnitude + direction)
  3. Multiply: p = mv
  4. State unit: kg·m·s⁻¹

3. Solved Examples

Basic

Q: 2 kg stone at rest. Find momentum.

Solution: v = 0 → p = mv = 2 × 0 = 0

Answer: p = 0

Intermediate

Q: 60 kg person walks at 1.0 m/s east. Momentum?

Solution: p = 60 × 1.0

Answer: p = 60 kg·m·s⁻¹ east

Advanced

Q: 0.2 kg ball hits wall at 10 m/s, rebounds at 10 m/s. Find Δp.

Solution: Δp = m(v_f − v_i) = 0.2(−10 − 10)

Answer: Δp = −4 kg·m·s⁻¹

Exam

Q: NIOS Ex 3.1: 60 kg at 1 m/s and 40 kg at 1.5 m/s toward each other. Find p₁, p₂.

Solution: p₁ = 60, p₂ = 40×(−1.5) = −60

Answer: Equal magnitude, opposite signs

F = Δp / Δt

Definition: Net force equals the rate of change of momentum.

Derivation

Newton's Second Law (general form). Proportionality constant k = 1 in SI units gives F = Δp/Δt.

Variables

F = net force (N) · Δp = momentum change · Δt = time (s)

Why it works

Force is not just a push — it is how quickly momentum changes. Short Δt with large Δp means huge force (hard catch).

Historical context

Newton published this in Philosophiæ Naturalis Principia Mathematica (1687), replacing Aristotle's idea that force maintains velocity.

Deep understanding

This is the form to use when mass changes (rockets) or for impulses. F = ma is a special case when m is constant.

2. Diagrams & Visuals

Force = rate of momentum change Δp ÷ Δt = F Bigger Δt → smaller F for same Δp

Color-coded visual · step-by-step breakdown below

  1. Find initial and final momentum
  2. Calculate Δp = p_f − p_i
  3. Measure time interval Δt
  4. F = Δp/Δt (include direction)

3. Solved Examples

Basic

Q: Momentum changes by 10 kg·m·s⁻¹ in 2 s. Find F.

Solution: F = 10/2

Answer: F = 5 N

Intermediate

Q: 0.4 kg ball: 20 m/s → 0 in 10 s. Find |F|.

Solution: Δp = 0.4×(−20) = −8; F = −8/10

Answer: |F| = 0.8 N

Advanced

Q: 0.16 kg ball caught from 20 m/s in 0.04 s.

Solution: Δp = −3.2; F = −3.2/0.04

Answer: |F| = 80 N

Exam

Q: Same ball caught in 0.20 s instead. Compare force.

Solution: F = −3.2/0.20 = −16 N

Answer: 5× gentler — Terminal Q9 idea

F = ma

Definition: Net force on a body equals mass times acceleration (constant mass).

Derivation

From F = Δp/Δt with p = mv and constant m: F = m(Δv/Δt) = ma.

Variables

F = net force (N) · m = mass (kg) · a = acceleration (m·s⁻²)

Why it works

Heavier objects need more force for the same acceleration — mass resists change in motion (inertia).

Historical context

Galileo studied acceleration; Newton quantified F ∝ a with mass as proportionality constant.

Deep understanding

Acceleration aligns with net force, not velocity. Perpendicular F bends path; opposite F slows down.

2. Diagrams & Visuals

F = ma — force causes acceleration F → m, a a = F/m more m → less a

Color-coded visual · step-by-step breakdown below

  1. Draw FBD — sum all forces
  2. Find net force ΣF
  3. Use ΣF = ma along motion direction
  4. Solve for unknown

3. Solved Examples

Basic

Q: 5 N net force on 2 kg block. Find a.

Solution: a = F/m = 5/2

Answer: a = 2.5 m/s²

Intermediate

Q: 10 kg body stopped by 50 N in 2 s. Verify a.

Solution: a = −50/10 = −5 m/s²

Answer: Consistent with Ex 3.4

Advanced

Q: 5 kg block, μ_k=0.1, pulled by 10 N horizontally.

Solution: f_k=4.9 N; F_net=5.1; a=1.02

Answer: a = 1.02 m/s²

Exam

Q: Is F=ma valid for a rocket losing fuel?

Solution: Mass changes → use F=Δp/Δt

Answer: F=ma only for constant m

F₁₂ = −F₂₁

Definition: Interaction forces between two bodies are equal in magnitude and opposite in direction.

Derivation

Newton's Third Law — fundamental postulate from mutual interaction of matter.

Variables

F₁₂ = force on 1 by 2 · F₂₁ = force on 2 by 1 (both in N)

Why it works

Every push is mutual. You push Earth down when jumping; Earth pushes you up — that is the reaction.

Historical context

Newton's Third Law corrected the idea that only active objects exert force. Both partners always interact.

Deep understanding

Forces act on DIFFERENT bodies — they never cancel on a single object. Common NIOS trap!

2. Diagrams & Visuals

Third Law — action & reaction Book F₁₂↓ F₂₁↑ Different bodies!

Color-coded visual · step-by-step breakdown below

  1. Identify the two interacting bodies
  2. Name force on A due to B
  3. Write equal opposite force on B due to A
  4. Never add them on one FBD

3. Solved Examples

Basic

Q: 2 kg book on table. Force on table by book?

Solution: F = mg = 2×9.8

Answer: 19.6 N downward

Intermediate

Q: Name the reaction pair for book on table.

Solution: Book on table ↓; table on book ↑

Answer: F₁₂ = −F₂₁

Advanced

Q: Student says action-reaction cancel on book. Correct?

Solution: Forces on different bodies

Answer: WRONG — Intext 3.3 trap

Exam

Q: Man kicks football. Identify action-reaction.

Solution: Foot on ball; ball on foot

Answer: Equal magnitude, opposite

Impulse = F·Δt = Δp

Definition: Impulse is force applied over a time interval; it equals momentum change.

Derivation

Multiply F = Δp/Δt by Δt on both sides.

Variables

Impulse (N·s) = F (N) × Δt (s) = Δp (kg·m·s⁻¹)

Why it works

Same momentum change spread over longer time needs smaller force — why fielders soften hands.

Historical context

Impulse concept links Newton's mechanics to collision analysis used in engineering and sports science.

Deep understanding

Impulse is a vector. Area under F–t graph equals Δp.

2. Diagrams & Visuals

Impulse — area under F-t graph F·Δt = Δp wider base (Δt) → lower peak F

Color-coded visual · step-by-step breakdown below

  1. Find Δp from initial and final velocity
  2. OR find F and Δt
  3. Set F·Δt = Δp
  4. Solve for unknown

3. Solved Examples

Basic

Q: 10 N acts for 3 s. Impulse?

Solution: J = 10×3

Answer: 30 N·s

Intermediate

Q: Ball Δp = 5 kg·m·s⁻¹. Force averaged over 0.5 s?

Solution: F = Δp/Δt = 5/0.5

Answer: F = 10 N

Advanced

Q: Catch: Δp=3.2, stiff hands 0.04 s vs soft 0.20 s.

Solution: F = 80 N vs 16 N

Answer: Soft hands reduce force 5×

Exam

Q: Why draw hands back when catching?

Solution: Increases Δt, reduces F

Answer: Same Δp, smaller peak force

p_total = constant

Definition: Total momentum of an isolated system remains constant.

Derivation

From F_AB = −F_BA → Δp_A + Δp_B = 0 → total p unchanged.

Variables

Isolated system: no net external force

Why it works

Internal forces cancel in pairs; only external forces change total system momentum.

Historical context

Conservation laws are cornerstones of physics — momentum conservation predates energy conservation in formal use.

Deep understanding

Applies to collisions, explosions, gun recoil, rockets. Individual parts can change; sum cannot.

2. Diagrams & Visuals

Momentum conservation p_before = p_after Σp_i (isolated) = constant

Color-coded visual · step-by-step breakdown below

  1. Check system is isolated
  2. Write total p before event
  3. Write total p after event
  4. Equate and solve

3. Solved Examples

Basic

Q: Two skaters at rest push apart. Total p after?

Solution: Started at 0

Answer: Still 0

Intermediate

Q: 2 trolleys (m) at v hit 3 at rest. Find v′.

Solution: 2mv = 5mv′

Answer: v′ = 2v/5

Advanced

Q: Explosion: bomb at rest splits into two pieces.

Solution: p₁ + p₂ = 0

Answer: Pieces move opposite ways

Exam

Q: When is momentum NOT conserved?

Solution: Net external force present

Answer: Friction from outside counts

mv₁ + Mv₂ = 0

Definition: Gun + bullet system momentum balance when starting from rest.

Derivation

Special case of conservation: 0 = mv₁ + Mv₂.

Variables

m, v₁ = bullet · M, v₂ = gun recoil

Why it works

Forward bullet momentum is balanced by backward gun momentum so total stays zero.

Historical context

Recoil observations helped artillery engineers and led to rocket propulsion principles.

Deep understanding

v₂ = −(m/M)v₁ — lighter bullet or heavier gun means smaller recoil speed.

2. Diagrams & Visuals

Gun M v₁→ v₂←

Color-coded visual · step-by-step breakdown below

  1. Set total initial momentum (often 0)
  2. Write mv₁ + Mv₂ = 0
  3. Solve for unknown velocity
  4. Check direction signs

3. Solved Examples

Basic

Q: 5 kg gun, 0.01 kg bullet at 300 m/s. Recoil?

Solution: v₂ = −(0.01/5)×300

Answer: v₂ = −0.6 m/s

Intermediate

Q: 4 kg gun fires 8 g at 250 m/s.

Solution: v₂ = −(0.008/4)×250

Answer: v₂ = −0.5 m/s

Advanced

Q: Why is recoil much slower than bullet?

Solution: m ≪ M

Answer: |v₂| ≪ |v₁|

Exam

Q: NIOS gun recoil: state formula used.

Solution: mv₁ + Mv₂ = 0

Answer: Conservation of momentum

F₁ + F₂ + F₃ = 0

Definition: Concurrent forces in equilibrium have zero vector resultant.

Derivation

From ΣF = 0 for static equilibrium of a point mass.

Variables

F₁, F₂, F₃ = concurrent forces (N)

Why it works

If forces balance at a point, the object does not accelerate — essential for bridges, cranes, ropes.

Historical context

Static equilibrium analysis dates to ancient architecture; vector methods modernized it.

Deep understanding

Graphically: triangle of forces closes. Any two sum to oppose the third.

2. Diagrams & Visuals

Concurrent forces in equilibrium F₁ F₂ F₃ F₁+F₂+F₃=0

Color-coded visual · step-by-step breakdown below

  1. Draw all forces from the point
  2. Resolve into x and y components
  3. Set ΣF_x = 0 and ΣF_y = 0
  4. Solve simultaneous equations

3. Solved Examples

Basic

Q: Two forces 3 N and 4 N perpendicular, equilibrium. Third force?

Solution: |F₃| = √(9+16)

Answer: F₃ = 5 N

Intermediate

Q: Three ropes on ring in equilibrium. Given two, find third.

Solution: Vector triangle closes

Answer: Use component method

Advanced

Q: Lami's theorem application with 3 concurrent forces.

Solution: F/sinα = constant

Answer: Useful for angled ropes

Exam

Q: When is a body in static equilibrium?

Solution: ΣF = 0 on point mass

Answer: No net force

f_s(max) = μ_s F_N

Definition: Maximum static friction before sliding starts.

Derivation

Empirical — from experiments; μ_s depends on surface pair.

Variables

f_s(max) (N) · μ_s (no unit) · F_N (N)

Why it works

Surfaces interlock microscopically until push exceeds limit — then sliding begins.

Historical context

Coulomb studied friction in 18th century; μ_s and μ_k distinction is key to his work.

Deep understanding

While f_s < f_s(max), static friction self-adjusts: f_s = F_ext. Independent of contact area.

2. Diagrams & Visuals

block push f_s max at μ_s F_N

Color-coded visual · step-by-step breakdown below

  1. Find normal force F_N (often mg on flat floor)
  2. Look up or use given μ_s
  3. f_s(max) = μ_s F_N
  4. Compare with applied force

3. Solved Examples

Basic

Q: 2 kg block, μ_s=0.25, horizontal floor.

Solution: F_N=19.6; f_s(max)=0.25×19.6

Answer: 4.9 N

Intermediate

Q: Push 5 N on above block. Does it move?

Solution: 5 > 4.9

Answer: Yes — sliding starts

Advanced

Q: Block on 30° incline. Find f_s(max).

Solution: F_N = mg cosθ; then μ_s F_N

Answer: Depends on angle

Exam

Q: Terminal Q12 style: min push to start sliding?

Solution: Beat f_s(max)

Answer: F > μ_s mg on horizontal

f_k = μ_k F_N

Definition: Kinetic friction opposes sliding motion.

Derivation

Empirical; μ_k usually less than μ_s.

Variables

f_k (N) · μ_k · F_N (N)

Why it works

Sliding surfaces ride over bumps — steady opposing force while moving.

Historical context

Engineering applications (brakes, bearings) rely on controlling μ_k.

Deep understanding

Rolling friction ≪ f_k (~1/100 for steel on steel). Wheels save energy.

2. Diagrams & Visuals

pull f_k

Color-coded visual · step-by-step breakdown below

  1. Confirm body is sliding
  2. Find F_N
  3. f_k = μ_k F_N (opposite motion)
  4. Use in ΣF = ma

3. Solved Examples

Basic

Q: 5 kg block, μ_k=0.1. Find f_k on horizontal floor.

Solution: f_k = 0.1×49

Answer: 4.9 N

Intermediate

Q: 5 kg, μ_k=0.1, pulled by 10 N. Find a.

Solution: F_net=5.1; a=1.02

Answer: a = 1.02 m/s²

Advanced

Q: Why harder to start than keep moving?

Solution: μ_s > μ_k

Answer: Beat f_s(max) first

Exam

Q: MCQ: rolling friction vs kinetic?

Solution: Rolling ~ 1/100 of sliding

Answer: Option C in Q10

ΣF = ma

Definition: Vector sum of external forces on one body equals ma.

Derivation

Newton's Second Law applied to a free body.

Variables

ΣF (N) · m (kg) · a (m/s²)

Why it works

FBD + ΣF = ma is the universal recipe for mechanics numericals.

Historical context

Free-body analysis became standard in 20th-century physics education.

Deep understanding

Use components: ΣF_x = ma_x, ΣF_y = ma_y. Choose axes wisely (along incline helps).

2. Diagrams & Visuals

FBD → ΣF = ma mg N f

Color-coded visual · step-by-step breakdown below

  1. Isolate ONE body
  2. Draw all external forces
  3. Choose coordinate axes
  4. ΣF = ma (component-wise)
  5. Solve

3. Solved Examples

Basic

Q: 2 kg block, net 6 N horizontal. a?

Solution: a = 6/2

Answer: a = 3 m/s²

Intermediate

Q: Block on incline: use mg sinθ − f = ma.

Solution: Component along plane

Answer: Standard incline setup

Advanced

Q: Two bodies connected — system + individual FBDs.

Solution: System for a; one body for T

Answer: Ex 3.8 method

Exam

Q: First step in every numerical?

Solution: Draw FBD

Answer: MCQ Q9 answer: ΣF = ma

a = F / (m₁ + m₂)

Definition: Common acceleration of two blocks tied on smooth table.

Derivation

System approach: ΣF = (m₁+m₂)a.

Variables

F = pull · m₁, m₂ = masses

Why it works

They move together — treat as single mass for acceleration.

Historical context

Classic Atwood/connected-mass problems appear in every physics curriculum since Newtonian mechanics.

Deep understanding

Smooth table means no friction on system. String assumed massless and inextensible.

2. Diagrams & Visuals

m₂ m₁ F T

Color-coded visual · step-by-step breakdown below

  1. Draw both blocks + string
  2. Consider system m₁+m₂
  3. ΣF = (m₁+m₂)a
  4. a = F/(m₁+m₂)

3. Solved Examples

Basic

Q: m₁=2, m₂=3 kg, F=10 N. Find a.

Solution: a = 10/5

Answer: a = 2 m/s²

Intermediate

Q: m₁=3, m₂=2, F=10. Find a and T.

Solution: a=2; T=m₁a=6

Answer: a=2, T=6 N

Advanced

Q: Add friction on table — modify F_net.

Solution: Subtract f_k from F

Answer: Harder variant

Exam

Q: NIOS Ex 3.8: state system equation.

Solution: F = (m₁+m₂)a

Answer: Then T = m₁a

T = m₁F / (m₁ + m₂)

Definition: Tension on leading block (pulled system).

Derivation

On m₁: T = m₁a = m₁F/(m₁+m₂).

Variables

T = tension (N)

Why it works

Only tension accelerates the leading block — it transmits force through string.

Historical context

String tension problems connect to engineering cable and bridge analysis.

Deep understanding

Tension is same throughout massless string. Direction pulls leading block forward.

2. Diagrams & Visuals

m₁ T → T = m₁a

Color-coded visual · step-by-step breakdown below

  1. First find a = F/(m₁+m₂)
  2. Isolate m₁ in FBD
  3. T = m₁a
  4. Substitute a

3. Solved Examples

Basic

Q: m₁=4, m₂=1, F=10 N. Find T.

Solution: a=2; T=4×2

Answer: T = 8 N

Intermediate

Q: m₁=3, m₂=2, F=10.

Solution: a=2; T=6

Answer: T = 6 N

Advanced

Q: Prove T < F always.

Solution: T = m₁F/(m₁+m₂) < F

Answer: Only fraction of F on string

Exam

Q: Which block to use for T = m₁a?

Solution: Leading block (far from pull)

Answer: Ex 3.8 logic

a = (m₁ − m₂)g / (m₁ + m₂)

Definition: Atwood machine acceleration (m₁ > m₂).

Derivation

m₁g − T = m₁a; T − m₂g = m₂a → add equations.

Variables

m₁ down, m₂ up · g = 9.8 m/s²

Why it works

Weight difference drives motion; total inertia is m₁+m₂.

Historical context

Atwood machine (1784) let Galileo-style slow-motion free fall be measured accurately.

Deep understanding

If m₁ = m₂, a = 0 — balanced. Pulley assumed frictionless, string massless.

2. Diagrams & Visuals

m₁ m₂

Color-coded visual · step-by-step breakdown below

  1. Draw FBD for m₁ and m₂
  2. Write m₁g − T = m₁a (down +)
  3. Write T − m₂g = m₂a (up +)
  4. Add → solve for a

3. Solved Examples

Basic

Q: m₁=5, m₂=3 kg. Find a.

Solution: a = 2×9.8/8

Answer: a = 2.45 m/s²

Intermediate

Q: m₁=4, m₂=4. Find a.

Solution: m₁−m₂=0

Answer: a = 0

Advanced

Q: m₁=6, m₂=2. Find a and T.

Solution: a=4.9; T=2×(9.8+4.9)

Answer: T = 39.2 N

Exam

Q: NIOS pulley: which mass goes down?

Solution: Heavier m₁

Answer: a uses (m₁−m₂)g

T = 2m₁m₂g / (m₁ + m₂)

Definition: String tension in Atwood machine.

Derivation

Substitute a into T = m₂(g+a) or T = m₁(g−a).

Variables

T in N

Why it works

Tension is between m₁g and m₂g — supports both while they accelerate.

Historical context

Precise tension measurement in lab Atwood setups verified g experimentally.

Deep understanding

Symmetric form: same formula if you swap labels (check with m₁(g−a) = m₂(g+a)).

2. Diagrams & Visuals

T = 2m₁m₂g / (m₁+m₂) between m₁g and m₂g

Color-coded visual · step-by-step breakdown below

  1. Find a first
  2. Use T = m₂(g+a) or m₁(g−a)
  3. Substitute a
  4. Simplify to standard form

3. Solved Examples

Basic

Q: m₁=3, m₂=1. Find T (g=10).

Solution: a=5; T=1×15

Answer: T = 15 N

Intermediate

Q: m₁=5, m₂=3, g=9.8.

Solution: a=2.45; T=36.75

Answer: T = 36.75 N

Advanced

Q: Show T < m₁g and T > m₂g when m₁>m₂.

Solution: From force balance

Answer: Physically sensible

Exam

Q: Two formulas for Atwood — name both.

Solution: a and T formulas

Answer: Sec 3.7 examples

5. Special Features & Extras

Complete study guide — exam tips, common mistakes, memory aids, and quick reference.

Exam Tips & Tricks

  • Always draw a FBD before writing ΣF = ma — NIOS MCQ Q9 tests this.
  • Use F = Δp/Δt for catches, collisions, and changing mass; F = ma when mass is constant.
  • Third law pairs act on different bodies — never cancel on one object.
  • For friction: check if object is at rest (μ_s) or sliding (μ_k).
  • Connected blocks: find a from system first, then T from one block.
  • Atwood: heavier mass goes down; use (m₁−m₂)g in numerator.

Common Student Mistakes

  • Thinking action-reaction cancel on a single body
  • Using F = ma when mass changes (rockets, sandbags)
  • Assuming velocity direction = force direction
  • Forgetting friction opposes motion or impending motion
  • Skipping vector signs in momentum problems
  • Using f_k when block is still at rest

Memory Aids & Mnemonics

IF PET: Impulse = FΔt = Δp · Equilibrium: ΣF = 0 · Third law: pairs on different bodies
Friction ladder: Static → Kinetic → Rolling (each step ~100× smaller)
Atwood: "Heavy minus light over heavy plus light" → a = (m₁−m₂)g/(m₁+m₂)

Which Formula When?

  • Stopping / catching? → Impulse or F = Δp/Δt
  • Find acceleration? → F = ma or ΣF = ma
  • Two bodies interact? → Third law or momentum conservation
  • About to slide? → f_s(max) = μ_s F_N
  • Already sliding? → f_k = μ_k F_N
  • Blocks tied? → a = F/(m₁+m₂), T = m₁F/(m₁+m₂)
  • Pulley? → Atwood a and T formulas

QUICK REFERENCE — Ch 3 Laws of Motion

p = mvF = Δp/ΔtF = maF₁₂ = −F₂₁ Impulse = FΔtΣp = constmv₁+Mv₂=0 f_s=μ_s F_Nf_k=μ_k F_NΣF=ma a=F/(m₁+m₂)T=m₁F/(m₁+m₂) a=(m₁−m₂)g/(m₁+m₂)T=2m₁m₂g/(m₁+m₂)

Units: N = kg·m·s⁻² · p in kg·m·s⁻¹ · impulse in N·s · μ has no unit

Tip: Draw the FBD first — it saves half your mistakes on the NIOS exam.

PYQ — Previous Year Questions

Extracted from NIOS Physics (312) board exam papers in your PDF. Chapter L3 — Laws of Motion only. Use Model Answer for marking points; Explanation for concept clarity.

Chapter 3 — Laws of Motion (L3)

20 questions · Section A (MCQ & objective) + Section B (short/long) · Sources: 312/TUS/104A, 312/MAY/204A–C, 68/ESS/1-312-A

Section A — Multiple Choice (1 mark)

PYQ1. When a constant net external force acts on a body, which of the following physical quantities may not change?

1 mark · Section A Q3 · 312/TUS/104A

Model Answer

Answer: (D) Acceleration

From Newton's second law F = ma, a constant net force gives constant acceleration. Position, speed and velocity change with time; acceleration remains unchanged.

Explanation

Under constant F_net, a = F/m is fixed. Velocity v = u + at changes; position changes. Acceleration is the quantity that stays constant (not zero — unchanged in magnitude and direction).

PYQ2. A force F acts on a body of mass m for t seconds. The change in its linear momentum will be — (A) Ft   (B) Fm   (C) F/t   (D) F/m

1 mark · Section A Q3 (OR) · 312/TUS/104A

Model Answer

Answer: (A) Ft

Explanation

Impulse–momentum theorem: Δp = FΔt when F is constant. Hence change in linear momentum = Ft. Units: N·s = kg·m·s⁻¹.

PYQ3. The mass of a body is 2 kg. Its weight is — (A) 19.6 N   (B) 9.8 N   (C) 10 N   (D) 5 N

1 mark · Section A Q2(i) · Marking Scheme · g = 9.8 m·s⁻²

Model Answer

Answer: (A) 19.6 N

W = mg = 2 × 9.8 = 19.6 N

Explanation

Weight is gravitational force W = mg near Earth's surface. Distinct from mass (kg).

PYQ4. A body of mass 200 g falls through air with acceleration 6 m·s⁻². The air drag on the body is — (A) 1200 N   (B) 1.2 N   (C) 1.96 N   (D) 0.76 N

1 mark · Section A Q2(ii) OR · Marking Scheme

Model Answer

Answer: (D) 0.76 N

Mg − F_a = ma → F_a = Mg − ma = 0.20×9.8 − 0.20×6 = 1.96 − 1.2 = 0.76 N

Explanation

Newton's second law with air resistance upward: net force = weight − drag = ma. Drag reduces acceleration below g.

PYQ5. A passenger in a moving bus is thrown forward when the bus suddenly stops. This is explained by — (A) Newton's first law   (B) Newton's second law   (C) Newton's third law   (D) conservation of mass

1 mark · Section A Q3(i) · Marking Scheme

Model Answer

Answer: (A) Newton's first law

Explanation

Inertia (first law): body tends to continue its state of motion. Bus stops but passenger's upper body keeps moving forward until seat belt or friction acts.

PYQ6. The need of banking of roads is — (A) additional gravitational force   (B) additional centrifugal force   (C) additional centripetal force   (D) additional electrostatic force

1 mark · Section A Q3(ii) OR · Marking Scheme

Model Answer

Answer: (C) To provide additional centripetal force for higher velocity

Explanation

Banking tilts normal reaction so its horizontal component supplies centripetal force mv²/r, allowing safe turning at higher speed without excessive friction.

PYQ7. A body of mass m just starts sliding on an incline when the plane makes 30° with the vertical. Coefficient of friction μ is — (A) 1/√3   (B) √3   (C) mg/√3   (D) √3·mg

1 mark · Section A Q10 · 68/ESS/1-312-A

Model Answer

Answer: (B) √3

Angle with horizontal θ = 90° − 30° = 60°. At limiting equilibrium: μ = tan θ = tan 60° = √3.

Explanation

Along incline: mg sin θ = μ mg cos θ at limiting friction → μ = tan θ. Read angle carefully — 30° with vertical means 60° with horizontal.

PYQ8. A 2 kg body moves at constant velocity 5 m·s⁻¹ under constant force 3 N. Power loss due to friction is — (A) Zero   (B) 15 W   (C) −15 W   (D) 30 W

1 mark · Section A Q13 · 68/ESS/1-312-A

Model Answer

Answer: (B) 15 W

Constant v ⇒ net force = 0, so friction = 3 N opposite motion. P = F·v = 3 × 5 = 15 W.

Explanation

At uniform velocity, applied force balances friction. Power dissipated against friction P = f_k × v (L6 link: P = Fv).

PYQ9. Passage — Friction: F = μR. (i) Max static friction independent of — ? (ii) Unit of μ? (iii) Arrange μr, μk, μms ascending? (iv) Static friction on body at rest under 5 N applied force?

2 marks (1×2) · Section A Q17 · Marking Scheme: (i)(c) (ii)(d) (iii)(b) (iv)(b)

Model Answer

  • (i) (c) area of contact
  • (ii) (d) unitless
  • (iii) (b) μr < μk < μms
  • (iv) (b) 5 N (static friction equals applied force until limiting value)

Explanation

Limiting friction F = μs F_N depends on normal reaction and μ, not contact area. μ has no dimensions. Rolling friction is smallest; static max exceeds kinetic. Static friction is self-adjusting up to μs F_N.

PYQ10. Complete using [more, force, linear momentum, inertia, isolated, less]: (i) Total linear momentum of _______ system is conserved. (ii) Rate of change of momentum is higher when force is _______.

2 marks (1×2) · Section A Q18 · Marking Scheme

Model Answer

  • (i) isolated
  • (ii) large
  • (iii) force (also accepted in scheme)
  • (iv) linear momentum

Explanation

Conservation of momentum holds for isolated system (no net external force). Newton's second law: F = dp/dt — larger force ⇒ faster momentum change.

PYQ11. Passage — Newton's third law: (a) Why ignore intermolecular forces when finding net force? (b) Equilibrium under two forces? (c) Equilibrium under three forces? (d) Forces on a book on a table?

2 marks (1×2) · Section A Q19 · 312/TUS/104A

Model Answer

  • (a) Internal forces occur in equal and opposite pairs (3rd law) — cancel in pairs.
  • (b) Two forces: equal magnitude, opposite direction, same line of action.
  • (c) Three forces: vector sum zero (closed polygon / Lami's theorem).
  • (d) Weight mg downward; normal reaction N upward; both act on book.

Explanation

Only external forces change centre-of-mass motion. Translational equilibrium: ΣF_ext = 0. Weight and normal on the same body are not an action–reaction pair.

PYQ12. Match Column I with II: (i) Law of conservation of linear momentum — (ii) Expression for friction force — Options: (a) F = μR   (b) F = ma   (c) Ptotal = constant

2 marks · Section A Q21 · Marking Scheme: (i)–R → (c), (ii)–P → (a)

Model Answer

(i) Conservation of linear momentum ↔ (c) Ptotal = constant

(ii) Friction force ↔ (a) F = μR

Explanation

F = ma is Newton's second law, not friction. Momentum conservation: total p unchanged in isolated collision/interaction.

PYQ13. A boy throws a ball vertically upward with velocity v₀ and catches it on return. What is the change in linear momentum of the ball?

2 marks · Section B Q29 · 312/TUS/104A

Model Answer

Initial momentum p_i = mv₀ (upward). On return, velocity = −v₀ (downward before catch). Final p_f = −mv₀.

Δp = p_f − p_i = −mv₀ − mv₀ = −2mv₀

Magnitude of change = 2mv₀.

Explanation

Momentum is vector. Same speed but reversed direction ⇒ change is not zero. Sign shows downward impulse delivered by hands when catching.

PYQ14. Give any two methods of reducing friction between two surfaces.

2 marks · Section B Q29 (OR) / Q32 · Multiple papers

Model Answer

Any two from L3 notes:

  • Applying lubricants (oil/grease) between surfaces
  • Using ball bearings or rollers (rolling friction < sliding)
  • Polishing/smoothing surfaces
  • Streamlining (for fluids)
  • Using wheels instead of dragging

Explanation

Friction arises from surface irregularities and adhesion. Lubrication separates surfaces; rollers convert sliding to rolling with smaller μ.

PYQ15. Give any two examples of conservation of linear momentum.

2 marks · Section B (OR) · 312/MAY/204A, 204B, 204C

Model Answer

  • Recoil of a gun when bullet is fired
  • Two ice skaters pushing apart and moving in opposite directions
  • Collision of two billiard balls on a smooth table
  • Rocket propulsion (ejecting gases backward)

Explanation

If net external force is zero, total momentum before = total momentum after. Internal forces cancel in pairs.

PYQ16. Identify the action–reaction forces on a book lying on a table, explaining how these forces are developed.

2 marks · Section B Q31/Q33/Q29 · 312/MAY/204A, 204B, 204C

Model Answer

Pair 1: Book presses table downward → table presses book upward (normal contact forces).

Pair 2: Earth attracts book (weight) → book attracts Earth with equal and opposite gravitational force.

Forces develop from deformation at contact and gravitational interaction.

Explanation

Action and reaction act on different bodies, same line, equal magnitude, opposite direction. Weight and normal on the book are NOT a third-law pair (both on book).

PYQ17. A boy throws a ball of mass m upward with speed v; it returns to his hands at the same speed. Find the change in momentum.

2 marks · Section B (OR) · 312/MAY/204A, 204B, 204C

Model Answer

Same as PYQ13: |Δp| = 2mv (direction of momentum reverses).

Explanation

Take upward positive: Δp = m(−v) − m(v) = −2mv.

PYQ18. Calculate limiting friction: (a) 2 kg block, μs = 0.3, g = 10 m·s⁻²   (b) 5 kg block, μ = 0.1   (c) 3 kg block, μs = 0.4

2 marks each variant · Section B Q35 · 312/MAY/204A/B/C

Model Answer

flimit = μs F_N = μs mg

  • (a) 0.3 × 2 × 10 = 6 N
  • (b) 0.1 × 5 × 10 = 5 N
  • (c) 0.4 × 3 × 10 = 12 N

Explanation

On horizontal surface F_N = mg. Limiting (maximum static) friction is threshold before sliding begins: fs,max = μs mg.

PYQ19. A 1 kg body at rest explodes into three fragments of mass ratio 1 : 1 : 3. The two equal fragments fly off perpendicular to each other at 30 m·s⁻¹ each. Find the velocity of the heavier fragment.

3 marks · Section B Q38 · 312/TUS/104A (OR alternative)

Model Answer

Masses: m₁ = m₂ = 0.2 kg, m₃ = 0.6 kg. Conservation of momentum (initial = 0):

0.2×30 î + 0.2×30 ĵ + 0.6 v₃ = 0

|v₃| = √(6² + 6²)/0.6 = 6√2/0.6 = 10√2 ≈ 14.1 m·s⁻¹

Direction: opposite to resultant of the two lighter fragments (135° from each).

Explanation

Explosion — internal forces only ⇒ total momentum conserved. Vector addition of perpendicular momenta gives magnitude 6√2 kg·m·s⁻¹; divide by 0.6 kg for heavy piece speed.

PYQ20. Explain how to determine impulse of a force when the force is (a) constant and (b) variable.

3 marks · Section B Q38 (OR) · 312/TUS/104A

Model Answer

(a) Constant F: Impulse J = FΔt (area of rectangle under F–t graph). Equals change in momentum Δp.

(b) Variable F: J = ∫ F dt from t₁ to t₂ (area under F–t curve). Measure F at intervals, sum FΔt or integrate if F(t) known.

Explanation

Impulse–momentum theorem J = Δp links force applied over time to momentum change. Graphical method works for any F(t).

Problem Solving — L3 Laws of Motion

Six problems spanning this chapter’s NIOS syllabus. Every question is built from the notes and formula sheet: solve with equations first, then read the formal textbook-style write-up, the easy explanation, and the topic in depth (formulas, meaning, exam tips). Explanations open by default.

Question 1 of 6InertiaNewton I

Draw a free-body diagram of a book resting on a table. Identify all forces on the book and explain, using Newton’s first law, why the book does not accelerate. If the table is suddenly jerked sideways with large acceleration, what happens to the book (qualitatively) and which property of matter is responsible?

If ΣF = 0 ⇒ v = constant (including zero)

Pencil sketch (labelled)

table book mg N FBD — book on table ΣF = 0 (N = mg)
Pencil sketch: free-body diagram of book (N and mg labelled)

Solution — step by step with formulas

  1. Draw free-body diagram: weight mg downward; normal force N upward from table (see pencil sketch).
  2. Horizontal forces on a stationary book on a fixed table: none (or friction = 0 if no horizontal push).
  3. Vertical equilibrium: N = mg ⇒ ΣF = 0 ⇒ acceleration a = 0 (first law).
  4. When table jerks sideways, book tends to remain at rest relative to ground due to inertia until friction accelerates it.

Final answer: Book in equilibrium under N and mg (ΣF = 0). On sudden jerk, book lags due to inertia (mass).

Formulas used in this problem

If ΣF = 0 ⇒ v = constant (including zero)

Textbook formal language

Newton’s first law asserts that a body continues in its state of rest or of uniform motion in a straight line unless compelled by a net external force to change that state. The property of matter by virtue of which it resists change in its state of rest or uniform motion is inertia; mass is the quantitative measure of inertia. For the book on a stationary table, the vector sum of gravitational force and normal reaction vanishes; hence acceleration is zero. When the table is accelerated suddenly, the external horizontal force on the book is only friction, which is finite; for a sufficiently large acceleration of the table, the book does not instantly acquire the table’s velocity—manifestation of inertia.

Formulae applied: If ΣF = 0 ⇒ v = constant (including zero).

Easy language (same idea, plain words)

The book sits still because two forces cancel: gravity pulls down, the table pushes up the same amount. No leftover force means no speeding up or slowing down. If someone yanks the table sideways, the book “wants” to stay where it was for a moment—that laziness of matter is inertia. Friction may drag it along later, but it doesn’t jump instantly with the table.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Inertia and Newton’s first law

Inertia appears in everyday life: passengers lean back when a bus starts and lurch forward when it stops. The first law defines an inertial frame as one in which free particles move with constant velocity. Always begin motion problems by asking whether net force is zero; if yes, velocity is constant even if non-zero (uniform motion).

Link to chapter notes (L3 — Inertia and Newton’s first law): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: If ΣF = 0 ⇒ v = constant (including zero). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write If ΣF = 0 ⇒ v = constant (including zero) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 2 of 6Momentump = mv

Draw a sketch showing a ball hitting a wall and rebounding (label pᵢ and p_f). A ball of mass 0.20 kg moving east at 10 m·s⁻¹ hits a wall and rebounds west at 8.0 m·s⁻¹. Taking east as positive, calculate (i) initial momentum, (ii) final momentum, (iii) change in momentum. What does the large change in momentum imply about the wall’s force?

p = mv
Δp = p_f − p_i

Pencil sketch (labelled)

Ball hits wall — momenta wall pᵢ = +mv (east) p_f (west)
Pencil sketch: incident and rebound momentum labelled

Solution — step by step with formulas

  1. m = 0.20 kg; v_i = +10 m·s⁻¹; v_f = −8.0 m·s⁻¹.
  2. p_i = mv_i = 0.20 × 10 = +2.0 kg·m·s⁻¹.
  3. p_f = mv_f = 0.20 × (−8) = −1.6 kg·m·s⁻¹.
  4. Δp = p_f − p_i = −1.6 − 2.0 = −3.6 kg·m·s⁻¹.
  5. By Newton II, F_avg = Δp/Δt; large |Δp| in short contact time ⇒ large average force by the wall on the ball.

Final answer: p_i = +2.0 kg·m·s⁻¹; p_f = −1.6 kg·m·s⁻¹; Δp = −3.6 kg·m·s⁻¹ (large wall force).

Formulas used in this problem

p = mv
Δp = p_f − p_i

Textbook formal language

Linear momentum of a particle is the product of its mass and velocity, a vector quantity collinear with velocity. The change in momentum is the vector difference of final and initial momenta. Newton’s second law in momentum form states that the net external force equals the time rate of change of momentum; hence a large momentum change in a small interval implies a large average force.

Formulae applied: p = mv; Δp = p_f − p_i.

Easy language (same idea, plain words)

Momentum is “mass times how fast and which way.” East was +, so start with +2.0. After bounce it goes west, so momentum is negative (−1.6). The change is final minus initial: −1.6 − 2.0 = −3.6. That big flip in a short hit means the wall pushed hard on the ball.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Linear momentum

Momentum is conserved only for isolated systems; a single ball hitting a wall is not isolated—the Earth–wall system supplies external force. Always fix a positive direction before assigning signs to velocity and momentum.

Link to chapter notes (L3 — Linear momentum): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: p = mv; Δp = p_f − p_i. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write p = mv; Δp = p_f − p_i before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 3 of 6Newton IIF = ma

A constant net force of 12 N acts on a 3.0 kg body initially at rest. Find (i) acceleration, (ii) velocity after 4.0 s, (iii) momentum after 4.0 s. Verify that F = Δp/Δt gives the same force.

F = ma
F = Δp/Δt

Solution — step by step with formulas

  1. a = F/m = 12/3.0 = 4.0 m·s⁻².
  2. From rest: v = u + at = 0 + 4.0×4.0 = 16 m·s⁻¹.
  3. p = mv = 3.0×16 = 48 kg·m·s⁻¹; p_i = 0 ⇒ Δp = 48.
  4. Δp/Δt = 48/4.0 = 12 N = F (consistent).

Final answer: a = 4.0 m·s⁻²; v = 16 m·s⁻¹; p = 48 kg·m·s⁻¹; F = Δp/Δt checks.

Formulas used in this problem

F = ma
F = Δp/Δt

Textbook formal language

For constant mass, Newton’s second law reduces to F = ma, where F is the net external force and a the acceleration of the centre of mass. Equivalently F = dp/dt. With constant F and u = 0, v = at and p = mat, so Δp/Δt = ma = F.

Working formula set for this problem: F = ma; F = Δp/Δt. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Force 12 N on 3 kg means it speeds up by 4 m/s every second. After 4 s it moves at 16 m/s, momentum 48. Force is also “how fast momentum changes”: 48 in 4 s is 12 N—same number, two views of one law.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Newton’s second law

Use F = ma only for constant mass. For rockets (mass varying) retain F_ext + v_rel(dm/dt) forms. Net force means vector sum after free-body diagram; forgotten friction or components cause most exam errors.

Link to chapter notes (L3 — Newton’s second law): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: F = ma; F = Δp/Δt. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write F = ma; F = Δp/Δt before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 4 of 6Impulse

A 0.15 kg cricket ball arrives at 20 m·s⁻¹ and is stopped by a fielder in 0.040 s. Find the magnitude of average force exerted by the fielder. Why does drawing the hands backward while catching reduce injury risk?

J = F_avg Δt = Δp
F_avg = Δp/Δt

Solution — step by step with formulas

  1. Δp = m(v_f − v_i) = 0.15(0 − 20) = −3.0 kg·m·s⁻¹; |Δp| = 3.0.
  2. F_avg = |Δp|/Δt = 3.0/0.040 = 75 N.
  3. If hands move back, Δt increases; same |Δp| ⇒ smaller |F_avg|.

Final answer: F_avg = 75 N; larger Δt (drawing hands back) reduces average force.

Formulas used in this problem

J = F_avg Δt = Δp
F_avg = Δp/Δt

Textbook formal language

Impulse delivered by a force in time Δt is J = ∫F dt = F_avg Δt and equals the change in momentum of the body. For fixed Δp, increasing interaction time decreases the average force. Fielders increase Δt by drawing the hands backward, thereby reducing peak force on hands and ball.

Working formula set for this problem: J = F_avg Δt = Δp. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Stopping the ball means removing all its forward momentum. That change spread over 0.04 s needs about 75 N on average. If you pull hands back, you take longer to stop the ball, so the push on your hands is gentler—same “momentum cancel,” longer time.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Impulse–momentum theorem

Airbags, crumple zones, and soft landings all trade longer collision time for smaller peak force. Graphically, impulse is the area under the F–t curve.

Link to chapter notes (L3 — Impulse–momentum theorem): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: J = F_avg Δt = Δp. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Impulse–momentum is the integrated form of Newton’s second law. Same Δp with larger Δt means smaller average force — classic catching / airbag idea in NIOS notes.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 5 of 6Momentum conservationRecoil

Draw a labelled diagram of gun recoil (bullet forward, gun backward). A gun of mass 4.0 kg fires a 20 g bullet at 300 m·s⁻¹. If the system is initially at rest and recoils freely, find the recoil speed of the gun. State the principle used and one condition for its validity.

m_g v_g + m_b v_b = 0 (initially at rest)
v_g = −(m_b/m_g) v_b

Pencil sketch (labelled)

Gun recoil — momentum conservation Gun M v_g ← bullet m → m v_b + M v_g = 0
Pencil sketch: gun and bullet momenta labelled

Solution — step by step with formulas

  1. m_b = 0.020 kg; m_g = 4.0 kg; v_b = +300 m·s⁻¹ (forward).
  2. Initial total p = 0.
  3. m_g v_g + m_b v_b = 0 ⇒ v_g = −(m_b/m_g)v_b = −(0.020/4.0)×300 = −1.5 m·s⁻¹.
  4. Recoil speed = 1.5 m·s⁻¹ opposite to bullet.

Final answer: Recoil speed 1.5 m·s⁻¹ backward; conservation of momentum (isolated system).

Formulas used in this problem

m_g v_g + m_b v_b = 0 (initially at rest)
v_g = −(m_b/m_g) v_b

Textbook formal language

In the absence of external force in a given direction, the component of total linear momentum of an isolated system along that direction remains constant. For gun + bullet initially at rest, total momentum is zero; after firing, momenta are equal in magnitude and opposite in direction: m_g v_g = −m_b v_b.

Working formula set for this problem: m_g v_g + m_b v_b = 0 (initially at rest); v_g = −(m_b/m_g) v_b. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Before the shot, nothing is moving, so total momentum is zero. The bullet goes one way with small mass but high speed; the gun must go the other way so the two momenta cancel. Light bullet × big speed = heavy gun × small speed → about 1.5 m/s kick.

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Conservation of linear momentum

External forces (shoulder, ground friction) can reduce observed recoil; the ideal free-gun model assumes no external horizontal force. Rockets use the same principle with continuous exhaust of mass.

Link to chapter notes (L3 — Conservation of linear momentum): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: m_g v_g + m_b v_b = 0 (initially at rest); v_g = −(m_b/m_g) v_b. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write m_g v_g + m_b v_b = 0 (initially at rest); v_g = −(m_b/m_g) v_b before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Question 6 of 6Frictionμ

Draw the free-body diagram of a block on a rough horizontal surface (label F, f, N, mg). A 5.0 kg block rests on a horizontal surface with μ_s = 0.40 and μ_k = 0.30. Take g = 10 m·s⁻². (i) Find maximum static friction. (ii) What happens if a horizontal force of 15 N is applied? (iii) If 25 N is applied and the block moves, find kinetic friction and acceleration.

f_s(max) = μ_s N
f_k = μ_k N
N = mg (horizontal)

Pencil sketch (labelled)

Block on rough surface — FBD surface m F f N mg
Pencil sketch: FBD with F, f, N, mg labelled

Solution — step by step with formulas

  1. N = mg = 5.0×10 = 50 N.
  2. f_s(max) = μ_s N = 0.40×50 = 20 N.
  3. 15 N < 20 N ⇒ block does not start; f_s = 15 N (balances applied force).
  4. 25 N > 20 N ⇒ motion begins; f_k = μ_k N = 0.30×50 = 15 N.
  5. a = (F − f_k)/m = (25 − 15)/5.0 = 2.0 m·s⁻².

Final answer: (i) 20 N (ii) remains at rest, f_s=15 N (iii) f_k=15 N, a=2.0 m·s⁻²

Formulas used in this problem

f_s(max) = μ_s N
f_k = μ_k N
N = mg (horizontal)

Textbook formal language

Limiting static friction is μ_s N; static friction adjusts up to this limit to prevent relative motion. Once sliding occurs, kinetic friction is approximately μ_k N, usually with μ_k < μ_s. Net force along the surface equals ma.

Working formula set for this problem: f_s(max) = μ_s N; f_k = μ_k N; N = mg (horizontal). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.

Easy language (same idea, plain words)

Max “grip” before sliding is 20 N. A 15 N push is weaker than grip, so the block stays put and friction equals 15 N. A 25 N push breaks the grip; sliding friction drops to 15 N, leaving 10 N net force, so it accelerates at 2 m/s².

Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.

Topic in depth — Static and kinetic friction

μ depends on surface pair, not area (ideal Amontons–Coulomb model). On an incline, N = mg cos θ and the component mg sin θ competes with friction—standard banking and ladder problems use the same laws.

Link to chapter notes (L3 — Static and kinetic friction): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: f_s(max) = μ_s N; f_k = μ_k N; N = mg (horizontal). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.

Exam tip

Quote the law in one line, then write f_s(max) = μ_s N; f_k = μ_k N; N = mg (horizontal) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.

Common mistakes

  • Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
  • Using the wrong sign convention for work/heat/force direction.
  • Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
  • Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).