Chapter 8 of 12

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CBSE Class 9 Science★ 8–10 marks · CBSE Class 9 board (Unit III — Motion, Force and Work)

Force and Laws of Motion

A push, a pull, a kick — every change in motion comes from a force. Class 9 Force and Laws of Motion covers inertia, balanced vs unbalanced forces, Newton's three laws derived from everyday experience, momentum, conservation of momentum, and 25+ exam-style numericals with full step-by-step solutions.

⏱ 45 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Define force; state its effects on motion, shape and size
2Distinguish balanced vs unbalanced forces and predict the consequence on a body's motion
3State Newton's three laws of motion and apply each to identify forces in real situations
4Define and compute momentum and impulse with correct SI units
5Apply F = ma to numerical problems involving acceleration, mass and net force
6Apply conservation of momentum to two-body collisions and recoil problems
7Identify and explain Newton's third law action-reaction pairs in everyday situations

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Why this chapter matters

Newton's three laws are the single most useful set of equations in all of school physics. Every subsequent mechanics chapter — gravitation, work and energy, sound, fluids, even electromagnetism — uses F = ma somewhere. Master this and you have the working tool of mechanics.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Motion (Class 9 Ch7)

Equations of motion, units of acceleration.

Force and Laws of Motion — Class 9 (CBSE)

Galileo started physics. Isaac Newton finished it — or rather, finished the first chapter. In 1687 he published Philosophiae Naturalis Principia Mathematica, in which three laws explain why every object on Earth (and in the solar system) moves the way it does. This chapter is those three laws, derived, proven, and applied.

1. The story — overturning 2000 years of Aristotle

Aristotle (~350 BCE) said: "Heavy objects fall faster than light ones. Motion requires a continuous push; without push, objects naturally come to rest." For 2000 years, this was the standard view in Europe.

Galileo dropped balls from the Leaning Tower of Pisa (~1590) and showed they hit the ground at the same time regardless of mass. He rolled balls down inclined planes and observed: without friction, an object in motion stays in motion indefinitely. The "natural state" isn't rest — it's UNIFORM MOTION.

Newton built on Galileo and turned it into three universal laws:

First law (law of inertia): an object stays at rest, or in uniform motion, unless acted upon by a net external force.
Second law: $F = ma$ . Force = mass × acceleration. Quantitative.
Third law: every action has an equal and opposite reaction.

Three sentences. They describe the motion of a falling apple, a moon orbiting Earth, a rocket lifting off, a bullet recoiling, a fish swimming, a car turning. Master them and the universe becomes predictable.

2. What is a force?

A force is a push or a pull that:

Can change the state of motion of an object (start, stop, speed up, slow down, change direction).
Can change the shape of an object (squish, stretch, bend).
Can change the size of an object (compress a spring).

SI unit: newton (N). One newton is the force that accelerates a 1 kg mass at 1 m/s².

$1 N = 1 kg \cdot m/s^{2}$

Force is a vector — has magnitude and direction.

Types of forces (preview)

Contact forces: friction, tension, normal force, applied force, spring force.
Non-contact forces: gravity, magnetism, electric force.

3. Balanced and unbalanced forces

When multiple forces act on a body, what matters is the net force (vector sum of all forces).

Balanced forces

Net force = 0. The body's motion does NOT change.

A book on a table: gravity pulls down ( $m g$ ), the table pushes up (normal force). They cancel. Book stays put.
A tug-of-war with no winner: equal pulls from both sides.

Balanced forces can CHANGE the shape of an object (a soft ball squished from both sides), but not its motion.

Unbalanced forces

Net force ≠ 0. The body accelerates (changes speed or direction).

Push a stationary ball; it starts moving.
Brake a moving car; it slows down.

The unbalanced force causes the change in motion. Equal magnitudes from both sides would have made the ball stay still.

4. Newton's first law — inertia

An object continues in its state of rest or of uniform motion in a straight line, unless acted upon by a net external force.

This is the formal statement. The first law has two parts:

A body at rest stays at rest (no force, no movement).
A body in uniform motion stays in uniform motion (no force, no stopping or speeding up).

Inertia

Inertia is the property of a body to resist a change in its state of motion. Higher mass = higher inertia. A truck has more inertia than a bicycle.

Mass IS a measure of inertia. A heavier object is harder to start (or stop) moving.

Three kinds of inertia

Inertia of rest — a body at rest tends to stay at rest. Example: when a bus suddenly starts, passengers jerk backward (their bodies were at rest and resist starting motion).
Inertia of motion — a body in motion tends to stay in motion. Example: when a moving bus suddenly stops, passengers jerk forward (their bodies were in motion and resist stopping).
Inertia of direction — a body moving in a straight line tends to keep moving in a straight line. Example: when a car takes a sharp turn, passengers feel thrown to the outside (their bodies want to keep going straight).

Seat belts are designed around these — they apply the necessary force to overcome the passenger's inertia during sudden acceleration, braking, or collision.

5. Newton's second law — F = ma

The most-used equation in mechanics.

Momentum — a new quantity

Newton actually phrased the second law in terms of momentum ( $p$ ), defined as:

$p = m v$

Where $m$ is mass (kg) and $v$ is velocity (m/s). Momentum is a vector. SI unit: kg·m/s.

A heavy slow truck and a light fast bullet can have the same momentum — momentum captures "quantity of motion" combining both mass and speed.

The law — formally

The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction of the force.

Symbolically:

$F \propto \frac{Δ p}{Δ t}$

With the constant of proportionality set to 1 by the choice of SI units:

$F = \frac{Δ p}{Δ t} = \frac{m Δ v}{Δ t} = m \cdot \frac{Δ v}{Δ t} = ma$

The famous result:

$F = ma$

What this means

A bigger force → bigger acceleration (for a given mass).
A bigger mass → smaller acceleration (for a given force).
Force and acceleration are in the SAME direction (both vectors, parallel).

Examples

Hitting a cricket ball: ball comes at you at 30 m/s, leaves your bat at 50 m/s. If the ball is 0.16 kg and contact lasts 0.001 s, force = m(v-u)/t = 0.16 × (50-(-30))/0.001 = 12800 N. (Note: take incoming as negative since direction reverses.)

Stopping a moving car: brakes apply force, mass × decel = stopping force. Heavy car + same braking system → longer stopping distance. That's why trucks need bigger brakes than cars.

6. Newton's third law — action and reaction

To every action, there is an equal and opposite reaction.

Forces always come in pairs. When body A pushes body B with force $F$ , body B pushes back on A with $- F$ (same magnitude, opposite direction).

Important: action and reaction act on DIFFERENT bodies

You push a wall (action on wall); wall pushes back on you (reaction on YOU).
A rocket pushes hot gas downward (action on gas); the gas pushes the rocket upward (reaction on rocket) → rocket lifts off.
You walk forward by pushing the ground backward (action on ground); the ground pushes you forward (reaction on you).
A swimmer pushes water backward (action on water); water pushes swimmer forward (reaction on swimmer).

Why action and reaction don't cancel

They act on DIFFERENT bodies, so they don't cancel each other. (Forces on the SAME body would cancel.)

This is the most common Class 9 question conceptual error. If a horse pulls a cart with force $F$ and the cart pulls back with $- F$ , why does the cart move? Because:

$F$ on the cart causes the cart to accelerate (it's an unbalanced force on the cart, given its weight + friction).
$- F$ on the horse does NOT prevent the horse from accelerating — the horse separately pushes the ground backward, and the ground reaction pushes the horse forward.

7. Law of conservation of momentum

A direct consequence of Newton's third law: in a closed system (no external forces), the total momentum is conserved.

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$

This applies before and after any collision or explosion — for any two-body interaction in an isolated system.

Derivation (brief)

By Newton's 3rd law, the force on body 1 from body 2 equals minus the force on body 2 from body 1. By Newton's 2nd law, force = rate of change of momentum. So $Δ p_{1} = - Δ p_{2}$ , meaning $Δ p_{1} + Δ p_{2} = 0$ . Total momentum is constant.

Applications

Rocket propulsion: rocket ejects hot gas backward (gas gains momentum in one direction); rocket gains equal and opposite momentum (forward). $0 = m_{gas} v_{gas} + m_{rocket} v_{rocket}$ → $v_{rocket} = - m_{gas} v_{gas} / m_{rocket}$ .
Recoil of a gun: gun + bullet at rest; bullet fires forward with high velocity → gun recoils backward. Momentum conservation: $0 = m_{b} v_{b} + m_{g} v_{g}$ , so $v_{g} = - m_{b} v_{b} / m_{g}$ . Small mass × big velocity for the bullet = big mass × small velocity for the gun.
Collision of two cars: total momentum before = total momentum after, regardless of whether the collision is elastic, inelastic, or partial. (Energy may or may not be conserved; momentum always is.)

8. Impulse — change in momentum

When a force acts on a body for a short time, the resulting change in momentum is called impulse.

$Impulse = F \times t = Δ p = m (v - u)$

SI unit: N·s (or equivalently kg·m/s, same as momentum).

Examples

A catcher pulls his hands back when catching a cricket ball: extends the contact time $t$ → reduces the force experienced (impulse is fixed).
Air bags in cars: extend the time of impact during collision → reduce the force on the passenger.
Boxer rolling with a punch: extends contact time → reduces force.

For a fixed change in momentum, a longer time means a smaller force. Useful in many safety designs.

9. Worked example — recoil of a gun

A gun of mass 5 kg fires a bullet of mass 50 g with a velocity of 200 m/s. Find the recoil velocity of the gun.

Step 1 — Identify before and after.

Before: gun + bullet at rest. Total momentum = 0.
After: bullet moves at 200 m/s forward; gun recoils at velocity $v$ (find this).

Step 2 — Convert units. Bullet mass = 50 g = 0.05 kg. Gun mass = 5 kg.

Step 3 — Apply conservation of momentum. $0 = (0.05) (200) + (5) (v)$ $0 = 10 + 5 v$ $v = - 2 m/s$

The negative sign means the gun recoils in the direction opposite to the bullet's motion.

Answer: The gun recoils at 2 m/s backward.

10. Closing thought

You started this chapter knowing forces push and pull. You're ending it with the three laws that govern the motion of every object in every direction at every scale below light-speed.

Newton's laws built the modern world:

Civil engineering uses them for bridges and buildings.
Mechanical engineering uses them for engines and turbines.
Aerospace engineering uses them for planes and rockets.
Sports science uses them to optimise human performance.
Medical engineering uses them for prosthetics, cardiac modeling, MRI design.

In 1687 they were three sentences. In 2026, they're the working tools of millions of engineers worldwide. That's how good physics travels through time.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Force (definition)

F = ma

Newton's 2nd law. SI unit: 1 N = 1 kg·m/s².

Force from momentum

F = Δp/Δt = m(v − u)/t

Newton's original phrasing. F = rate of change of momentum.

Momentum

p = mv

Vector. SI unit: kg·m/s. Equivalent to N·s.

Impulse

Impulse = F × t = Δp = m(v − u)

When applied for time t, force changes momentum by impulse. SI unit: N·s.

Conservation of momentum

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Total momentum before = after, for isolated systems.

Newton's 1st law

If F_net = 0, then v = constant (or 0)

Body in motion stays in motion; body at rest stays at rest.

Newton's 2nd law

F_net = m × a

Quantitative; force and acceleration are vectors, same direction.

Newton's 3rd law

F_AB = − F_BA

Action-reaction pair. Equal magnitude, opposite direction, on DIFFERENT bodies.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Saying balanced forces produce zero motion

✓ Balanced forces produce zero ACCELERATION (no change in motion). A body in uniform motion with balanced forces keeps moving uniformly forever (no friction case).

WATCH OUT

✗ Saying action and reaction cancel each other out

✓ They act on DIFFERENT BODIES, so they don't cancel. Newton's 3rd law pairs never appear on the same body simultaneously — that's why they don't cancel and why motion can occur.

WATCH OUT

✗ Confusing mass and weight

✓ Mass is a scalar measure of inertia (in kg). Weight is the force gravity exerts on a mass (in newtons, W = mg). Same mass, different weight on Earth vs Moon.

WATCH OUT

✗ Forgetting to convert grams to kilograms in F = ma

✓ SI F = ma uses kg and m/s² to give N. A 200 g bullet needs to be 0.2 kg. Otherwise the force will be 1000× too large.

WATCH OUT

✗ Using positive sign for both initial and final velocity in a collision

✓ If a body reverses direction, one velocity is positive and the other negative. Pick a convention (e.g., right = positive) and stick with it.

WATCH OUT

✗ Saying inertia depends on speed

✓ Inertia depends on MASS, not on speed. A speeding bullet has high momentum but the same inertia as the bullet at rest.

WATCH OUT

✗ Saying impulse is the same as force

✓ Force is in newtons. Impulse = F × t, in N·s. Force is a rate; impulse is an accumulated quantity.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Section 8.1 (in-text)

Balanced vs unbalanced forces; Galileo and inertia

Questions

Section 8.2 (in-text)

Newton's 1st law of motion + inertia and mass

Questions

Section 8.3 (in-text)

Newton's 2nd law: momentum, F = ma, numericals

Questions

Section 8.4 (in-text)

Newton's 3rd law: action-reaction, conservation of momentum

Questions

End-of-chapter

Mixed: force calculations, momentum problems, recoil and collision numericals

Questions

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Definition

Define one newton.

▶ Show solution

Step 1 — From F = ma.
  Force = mass × acceleration.

Step 2 — Set both to 1 in SI.
  1 N = 1 kg × 1 m/s².

✦ Answer: 1 newton is the force that produces an acceleration of 1 m/s² in a body of mass 1 kg.

Q2EASY· F=ma

What force is needed to accelerate a 2 kg object at 5 m/s²?

▶ Show solution

Step 1 — Use F = ma.
  F = 2 × 5 = 10 N.

✦ Answer: 10 N.

Q3EASY· Momentum

Find the momentum of a 60 kg cyclist moving at 5 m/s.

▶ Show solution

Step 1 — p = mv = 60 × 5 = 300 kg·m/s.

✦ Answer: 300 kg·m/s.

Q4EASY· Inertia

Why do passengers fall forward when a moving bus suddenly stops?

▶ Show solution

Step 1 — Identify the type of inertia.
  Inertia of MOTION — bodies in motion tend to stay in motion.

Step 2 — Apply to the situation.
  Before braking, both the bus and the passenger are moving forward at the same velocity.
  When the bus suddenly stops (large deceleration on the bus), passengers' bodies — having inertia of motion — tend to continue moving forward.
  The lower body (in contact with the seat) decelerates with the bus, but the upper body lurches forward.

✦ Answer: Passengers' bodies have inertia of motion — they continue moving forward when the bus suddenly stops (which decelerates only the bus, not the loose passengers). They jerk forward until the seat belt or other force stops them.

Q5EASY· 3rd law

When you walk on the ground, identify the action-reaction force pair.

▶ Show solution

Step 1 — When walking, your feet push the ground backward.
  This is the ACTION force (your foot on the ground, directed backward).

Step 2 — By Newton's 3rd law, the ground pushes back on your foot.
  This is the REACTION force (the ground on your foot, directed forward).
  The forward reaction force is what propels you forward.

✦ Answer: Action: your foot pushes ground backward. Reaction: ground pushes your foot forward. You move forward thanks to the reaction force.

Q6MEDIUM· F=ma

A car of mass 1000 kg accelerates from rest to 20 m/s in 10 s. Find the net force on it.

▶ Show solution

Step 1 — Find acceleration.
  a = (v − u)/t = (20 − 0)/10 = 2 m/s².

Step 2 — Apply F = ma.
  F = 1000 × 2 = 2000 N = 2 kN.

✦ Answer: Net force = 2000 N (2 kN), in the direction of motion.

Note: This is the NET force. The actual engine force is larger (counters air drag + friction), but the net is what gives the acceleration.

Q7MEDIUM· Momentum

A body of mass 5 kg moving with velocity 10 m/s is brought to rest in 2 s. Find (a) the impulse, (b) the average force.

▶ Show solution

Step 1 — Find change in momentum.
  Initial p = mv = 5 × 10 = 50 kg·m/s.
  Final p = 0.
  Change Δp = 0 − 50 = −50 kg·m/s.

Step 2 (a) — Impulse = Δp.
  Impulse = −50 kg·m/s = 50 N·s (magnitude).

Step 3 (b) — Average force.
  F = Δp/Δt = −50/2 = −25 N (magnitude 25 N, directed opposite to motion).

✦ Answer: (a) Impulse = 50 N·s (in the direction of stopping force); (b) Average force = 25 N (decelerating). The negative sign reflects that the force opposes the original motion.

Q8MEDIUM· Recoil

A gun of mass 4 kg fires a bullet of mass 20 g with a muzzle velocity of 300 m/s. Find the recoil velocity of the gun.

▶ Show solution

Step 1 — Conservation of momentum.
  Initial total momentum = 0 (both at rest).
  Final total momentum = m_bullet × v_bullet + m_gun × v_gun = 0.

Step 2 — Substitute.
  (0.020)(300) + (4)(v_gun) = 0
  6 + 4 × v_gun = 0
  v_gun = −1.5 m/s.

Step 3 — Interpret sign.
  Negative means gun recoils in the opposite direction to the bullet.

✦ Answer: Recoil velocity of the gun = 1.5 m/s, in the direction opposite to the bullet's motion.

This is why a heavy rifle is easier to fire: same momentum, more mass → smaller recoil velocity (less kick on your shoulder).

Q9MEDIUM· Collision

A 50 kg boy running at 4 m/s jumps onto a 200 kg cart moving in the same direction at 1 m/s. Find their common velocity after he lands (assume he sticks to the cart).

▶ Show solution

Step 1 — Conservation of momentum (inelastic collision, two bodies become one).
  m_boy × v_boy + m_cart × v_cart = (m_boy + m_cart) × v_common

Step 2 — Substitute.
  50 × 4 + 200 × 1 = (50 + 200) × v
  200 + 200 = 250 × v
  400 = 250v
  v = 400/250 = 1.6 m/s.

✦ Answer: Common velocity = 1.6 m/s (in the direction of original motion).

Observation: the boy slows down (4 → 1.6), the cart speeds up (1 → 1.6). Total momentum 400 kg·m/s is unchanged.

Q10MEDIUM· Concept

A book is at rest on a table. Identify the action-reaction pair acting between the book and the table.

▶ Show solution

Step 1 — Recognise: action and reaction are between TWO BODIES, not on the same body.

Step 2 — Identify the pair.
  Action: the book pushes DOWN on the table due to its weight (book on table).
  Reaction: the table pushes UP on the book (table on book) — this is the 'normal force'.

Step 3 — NOTE: gravity (the Earth pulling the book down) is NOT the action paired with the normal force. The reaction to gravity is the book pulling the Earth up (negligible but real, Newton's 3rd law).

✦ Answer: Action: book pushes the table downward (force = weight). Reaction: table pushes the book upward (normal force). These act on different bodies; that's why they don't 'cancel' — they're a Newton's 3rd-law pair.

The weight (mg) and the normal force happen to balance on the book — but balanced forces and action-reaction pairs are different concepts.

Q11HARD· F=ma + decel

A bus of mass 5000 kg moves with velocity 72 km/h. The driver applies brakes and the bus stops in 5 s. Find: (a) the deceleration, (b) the braking force.

▶ Show solution

Step 1 — Convert speed.
  72 km/h = 72 × 5/18 = 20 m/s.

Step 2 (a) — Deceleration.
  a = (v − u)/t = (0 − 20)/5 = −4 m/s².
  Magnitude 4 m/s² (deceleration).

Step 3 (b) — Braking force.
  F = ma = 5000 × (−4) = −20,000 N.
  Magnitude 20,000 N = 20 kN, directed opposite to motion.

✦ Answer: Deceleration = 4 m/s². Braking force = 20,000 N (20 kN).

Real-world calibration: heavy braking systems for buses produce 10–25 kN. The number is plausible for an emergency stop.

Q12HARD· HOTS

Explain why a fast-moving cricket ball hurts more if caught with a stiff hand than if the hand is moved back during the catch.

▶ Show solution

Step 1 — Impulse = F × t = change in momentum.
  When a cricket ball is caught, Δp (change in ball's momentum) is fixed — it goes from high momentum to zero.

Step 2 — Same Δp, different time.
  Stiff hand: ball stops in a very short time t. Impulse formula → F × t = Δp → F = Δp/t (small t → large F → painful).
  Hand moved back: stopping time t is increased. Same Δp → smaller force F. Less pain.

Step 3 — Practical principle.
  Spreading impact over time reduces peak force. Same idea behind:
  • Air bags (slow stop in 50 ms instead of 5 ms).
  • Crumple zones in cars.
  • Padded gloves in boxing.
  • Springs in a parachute landing.

✦ Answer: For a given change in momentum (Δp), extending the contact time t reduces the average force (F = Δp/t). Moving the hand back stretches the stopping time → smaller force → less pain. Same principle as airbags and crumple zones.

Q13HARD· Combined

Two blocks of masses 3 kg and 2 kg are connected by a string and placed on a frictionless horizontal surface. A force of 10 N is applied to the 3 kg block, pulling both forward. Find (a) the acceleration of the system, (b) the tension in the string.

▶ Show solution

Step 1 — Analyse the system as a whole.
  Total mass = 3 + 2 = 5 kg.
  Net force = 10 N (only applied force, no friction).
  Acceleration: a = F/m = 10/5 = 2 m/s².

Step 2 — Find the tension by analysing just the 2 kg block.
  Only the string pulls the 2 kg block forward with tension T.
  T = m × a = 2 × 2 = 4 N.

Step 3 — Verify by analysing the 3 kg block.
  Forces on 3 kg block: +10 N applied, −T (string pulls back).
  Net force = 10 − T = 3 × 2 = 6 N (mass × acceleration).
  T = 10 − 6 = 4 N. ✓

✦ Answer: (a) Acceleration of system = 2 m/s²; (b) Tension in string = 4 N.

Q14HARD· Collision

An object of mass 1 kg moving at 10 m/s collides head-on with another stationary object of mass 4 kg. After collision, the 1 kg object moves backward at 2 m/s. Find the velocity of the 4 kg object after collision.

▶ Show solution

Step 1 — Pick a positive direction (rightward = positive).
  Before: m₁ = 1 kg at v₁ = +10 m/s; m₂ = 4 kg at v₂ = 0.
  After: 1 kg moves BACKWARD at 2 m/s → v₁' = −2 m/s. v₂' = ?

Step 2 — Conservation of momentum.
  m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
  1 × 10 + 4 × 0 = 1 × (−2) + 4 × v₂'
  10 = −2 + 4 × v₂'
  4 × v₂' = 12
  v₂' = 3 m/s.

Step 3 — Interpret.
  Positive sign → 4 kg object moves forward (in the original direction of the 1 kg).

✦ Answer: The 4 kg object moves forward at 3 m/s after the collision.

Sanity check on momentum: Before = +10 kg·m/s. After = 1×(−2) + 4×3 = −2 + 12 = +10 kg·m/s. ✓

Q15HARD· HOTS

A rocket of mass 1000 kg is to be lifted vertically. To lift, the rocket must overcome gravity AND accelerate upward. If the upward acceleration is 5 m/s², find the thrust (force exerted by the engines). (g = 10 m/s²)

▶ Show solution

Step 1 — Forces on the rocket (vertical only).
  Weight (downward) = mg = 1000 × 10 = 10,000 N.
  Thrust (upward) = T.
  Net force = T − mg (upward positive).

Step 2 — Apply Newton's 2nd law.
  Net force = ma.
  T − mg = m × a.
  T = m(g + a) = 1000 × (10 + 5) = 15,000 N.

Step 3 — Interpret.
  Of the 15 kN thrust, 10 kN cancels gravity, and 5 kN provides the upward acceleration.

✦ Answer: Thrust required = 15,000 N (15 kN).

The rocket's thrust must always exceed its weight (mg) for any net upward motion. Just hovering needs T = mg; accelerating up needs T > mg.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Force is a push/pull. Changes motion, shape, size. Vector. SI unit: newton (N) = kg·m/s².
•Balanced force → no change in motion. Unbalanced force → acceleration.
•Newton's 1st law (inertia): in absence of net force, body stays at rest or moves uniformly.
•Mass = quantitative measure of inertia. Higher mass → more inertia → harder to start/stop.
•Newton's 2nd law: F = ma (or F = Δp/Δt). Force ∝ rate of change of momentum.
•Momentum p = mv. Vector. SI: kg·m/s.
•Impulse = F × t = Δp. SI: N·s. Used in safety design (airbags, crumple zones).
•Newton's 3rd law: action and reaction equal and opposite, on DIFFERENT bodies → don't cancel.
•Conservation of momentum: total momentum before = after, for an isolated system.
•Recoil: m_b × v_b = m_g × v_g (in magnitude, opposite directions).
•Three kinds of inertia: rest, motion, direction (passengers jerk back, jerk forward, thrown to the side respectively).

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Because they act on DIFFERENT bodies. Forces on the same body would cancel. The horse-cart paradox is resolved by remembering: F on the cart causes the cart to accelerate; the reaction force is on the horse, not the cart.

Mass is the AMOUNT of matter / measure of inertia (kg, scalar). Weight is the FORCE of gravity on that mass (N = kg × m/s², vector). Your mass is the same on Earth and Moon; your weight is 6× less on the Moon (because g_moon ≈ g_earth/6).

A newton is calibrated for 1 kg accelerating at 1 m/s², which is gentle. Earth's gravity is 9.8 m/s², so a 1 kg object weighs 9.8 N. A 5 kg bag of rice weighs about 50 N. Designers chose this unit so g comes out to a familiar single-digit number.

It defines the framework: it tells you what INERTIAL frames are (frames where the laws apply). In a accelerating car (non-inertial), things accelerate without any obvious force — and you need pseudo-forces or to switch frames. The 1st law gives you the rulebook for when F = ma is valid.

Yes — a body moving at constant velocity has zero acceleration. Velocity stays constant, no change → a = 0. Net force = 0 (balanced).

F_brake required = m × a (for the same deceleration). Heavy truck = bigger m → bigger braking force needed. Heavy vehicles also have larger momentum (p = mv) for the same speed, so stopping takes longer or harder force.

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Matter in Our Surroundings

From the air you breathe to the ice in your glass — everything around you is matter. Class 9 Matter in Our Surroundings with the particle theory, the three classical states, the curious fourth and fifth states (plasma and Bose-Einstein condensate), latent heat, evaporation, and 25+ exam-style problems with full step-by-step solutions.

Is Matter Around Us Pure?

Almost nothing in everyday life is chemically "pure" — your air is a mixture, your milk is a colloid, your salt water is a solution. Class 9 Is Matter Around Us Pure? distinguishes pure substances, mixtures, solutions, suspensions, colloids; the Tyndall effect; concentration calculations; and separation techniques with 20+ exam-style problems.

Atoms and Molecules

The atom is the indivisible building block — except when it isn't. Class 9 Atoms and Molecules with Dalton's atomic theory, laws of chemical combination, atomic and molecular masses, the mole concept, Avogadro's number, formula writing, and 25+ exam-style numericals with full step-by-step solutions.

Structure of the Atom

Dalton thought atoms were indivisible. He was wrong. Class 9 Structure of the Atom covers Thomson's plum-pudding model, Rutherford's gold-foil experiment, Bohr's planetary atom, the discovery of the neutron, electron-configuration rules, isotopes & isobars, valency and 25+ exam-style problems with full step-by-step solutions.