Newton's Laws of Motion ICSE Class 9: Forces, Inertia, Momentum and F = ma

What Newton's Laws of Motion is about

Weightage: CISCE publishes no chapter weightage and no question-paper layout for ICSE Class 9 Physics. Class IX is examined by your school: one 80-mark paper of 2 hours plus 20 marks of internal assessment. Newton's Laws of Motion is in the 3. Laws of Motion (i)-(iv) part of the 2028 syllabus.Source: CISCE ICSE 2028 Physics syllabus, Class IX

Key concepts

Contact and non-contact forces

Contact forces act only when bodies touch:

• friction: opposes sliding between surfaces;
• normal reaction: the push of a surface at right angles to it (a table pushing up on a book);
• tension: the pull transmitted by a string or rope.

Non-contact forces act across a distance (field forces): gravitational, electric and magnetic forces.

General properties of non-contact forces: they act without contact, even through a vacuum; gravitational force is always attractive, while electric and magnetic forces can attract or repel; and their strength falls as the distance increases, following an inverse square relation: $F\propto \frac{1}{d^2}$.

Worked example

The force between two charged spheres is $F$ when they are 3 cm apart. What is it when they are 6 cm apart, and when they are 1 cm apart?

Try it first, then show the solutionSolution

1. $F\propto \frac{1}{d^2}$.
2. At 6 cm (distance doubled): $\frac{F}{4}$.
3. At 1 cm (distance one-third): $9F$.

Answer: F/4 and 9F

Units of force

• SI unit: newton (N). 1 N is the force that gives a mass of 1 kg an acceleration of 1 m s⁻².
• cgs unit: dyne. 1 dyne gives 1 g an acceleration of 1 cm s⁻².
• $1\text{N}=1\text{kg}\times 1{\text{m s}}^{-2}=1000\text{g}\times 100{\text{cm s}}^{-2}=10^5$ dyne.

Gravitational units use the Earth's pull:

• 1 kgf (kilogram-force) = the force with which the Earth pulls a 1 kg mass $=9.8$ N.
• 1 gf (gram-force) $=980$ dyne.

Only cgs and SI units of force are used in calculations of $F=ma$.

Worked example

Express 50 gf in newtons and in dynes ($g=9.8$ m s⁻²).

Try it first, then show the solutionSolution

1. $50\text{gf}=0.05\text{kgf}=0.05\times 9.8=0.49$ N.
2. $0.49\times 10^5=49000$ dyne (also $50\times 980=49000$).

Answer: 0.49 N = 49,000 dyne

First law and inertia

Newton's first law: a body continues in its state of rest, or of uniform motion in a straight line, unless an external (unbalanced) force acts on it.

The law tells us two things:

• Inertia: the property of a body to resist any change in its state of rest or motion. More mass means more inertia.
• Force (defined by the law): that which changes, or tends to change, the state of rest or uniform motion of a body.

Two kinds of inertia:

• Inertia of rest: a body at rest tends to stay at rest (passengers fall backwards when a bus starts suddenly; dust leaves a carpet that is beaten).
• Inertia of motion: a moving body tends to keep moving (passengers fall forwards when a bus stops suddenly; an athlete runs on past the finishing line).

Worked example

Why does a passenger standing in a bus fall forwards when the bus stops suddenly?

Try it first, then show the solutionSolution

1. The feet, in contact with the floor, stop with the bus.
2. The upper body, due to inertia of motion, tends to keep moving forwards.
3. So the passenger falls forwards.

Answer: Inertia of motion of the upper body.

Momentum and the second law

Linear momentum $p=mv$ (vector, direction of velocity). SI unit: kg m s⁻¹.

For constant mass, the change of momentum is $\Delta p=m\Delta v=m(v-u)$.

Newton's second law: the rate of change of momentum of a body is directly proportional to the force applied and takes place in the direction of the force.

$F=\frac{\Delta p}{\Delta t}=\frac{m(v-u)}{t}=ma$ (with SI units, the constant of proportionality is 1).

Consequences: the same change of momentum produced over a longer time needs a smaller force. That is why a fielder pulls the hands back while catching, and why vehicles have crumple zones.

Worked example

A 0.2 kg ball hits a wall at 20 m s⁻¹ and rebounds at 15 m s⁻¹. If contact lasts 0.05 s, find the change in momentum and the average force.

Try it first, then show the solutionSolution

1. Take the rebound direction as positive: $u=-20$, $v=+15$ m s⁻¹.
2. $\Delta p=0.2\times (15-(-20))=7$ kg m s⁻¹.
3. $F=\frac{7}{0.05}=140$ N (away from the wall).

Answer: 7 kg m s⁻¹; 140 N

Second law with equations of motion

Combine $F=ma$ with $v=u+at$, $S=ut+\frac{1}{2}a{t}^2$ and $v^2=u^2+2aS$:

• If the force is known, find $a=\frac{F}{m}$, then use the equations for $v$, $S$ or $t$.
• If the motion is known (e.g. a car stopping in a given distance), find $a$ from the equations, then $F=ma$.

A retarding force gives a negative acceleration; quote its size and say it opposes the motion.

Worked example

A 20 g bullet moving at 300 m s⁻¹ is stopped by a wooden block after penetrating 6 cm. Find the average retarding force.

Try it first, then show the solutionSolution

1. $v^2=u^2+2aS$: $0=90000+2a(0.06)$, so $a=-750000$ m s⁻².
2. $F=ma=0.02\times 750000=15000$ N.
3. The block exerts a retarding force of 15,000 N.

Answer: 1.5 × 10⁴ N

Third law: action and reaction

Newton's third law: to every action there is an equal and opposite reaction.

If body A exerts a force $F_{AB}$ on body B, then B exerts $F_{BA}$ on A, equal in size and opposite in direction: $F_{BA}=-F_{AB}$.

Key points:

• action and reaction act on different bodies, so they never cancel each other;
• they act at the same instant and are of the same kind (both contact, or both gravitational).

Examples: when you walk, your foot pushes the ground back and the ground pushes you forward; a gun recoils when it fires a bullet; a rocket pushes gases down and the gases push the rocket up; a swimmer pushes water back to move forward.

Worked example

A book rests on a table. Identify the action–reaction pair involving the book's weight, and explain why the weight and the table's normal reaction are not such a pair.

Try it first, then show the solutionSolution

1. Weight: the Earth pulls the book down; its reaction is the book pulling the Earth up.
2. The table's push on the book is paired with the book's push on the table.
3. Weight and normal reaction both act on the book (same body), so they are not an action–reaction pair; they balance.

Answer: Earth–book pull pair; table–book push pair.

Physics in real life: Newton's Laws of Motion

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Inertia and Newton's First and Third Laws: Reasoning Questions
• Newton's Second Law, Momentum and F = ma

Newton's Laws of Motion formula sheet

Momentum

$p=mv$

kg m s⁻¹; vector.

Change of momentum

$\Delta p=m(v-u)$

Mass constant.

Second law

$F=\frac{\Delta p}{\Delta t}=ma$

SI units.

Units

$1\text{N}=10^5\text{dyne}$

1 kgf = 9.8 N; 1 gf = 980 dyne.

Inverse square

$F\propto \frac{1}{d^2}$

Non-contact forces.

Third law

$F_{BA}=-F_{AB}$

On different bodies.

Newton's Laws of Motion: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Newton's Laws of Motion

The question patterns to expect, and the method that works for each.

Tricks to remember Newton's Laws of Motion

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Lazy matter

  Inertia = matter's laziness: it hates to start and hates to stop.

  Use it when: First law examples.

• Spread the stop

  Same Δp, longer time, smaller force: pull your hands back when catching.

  Use it when: Second law applications.

• Different bodies, never cancel

  Action and reaction live on different bodies, so they cannot cancel.

  Use it when: Third law questions.

• Five zeros

  1 newton = 1 with five zeros of dynes (100,000).

  Use it when: Unit conversion.

Newton's Laws of Motion: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section B style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 9-mark question split into 2 + 3 + 4 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Newton's Laws of Motion

The slips examiners see most often, and what to write instead.

• ✗Mistake: Ignoring direction in momentum change

  When a ball rebounds, the velocities have opposite signs; students subtract the speeds.

  Wrong working

  $\Delta p=0.2(20-15)=1$

  Right working

  $\Delta p=0.2(15-(-20))=7$

  ✓Do this: Choose a positive direction and put signs on both velocities.

  Usually costs: Most of the question.

• ✗Mistake: Saying action and reaction cancel

  They act on different bodies, so they cannot cancel.

  ✓Do this: Name the body each force acts on.

  Usually costs: 1 mark.

• ✗Mistake: Not converting grams and centimetres

  In $F=ma$ with SI units, mass must be in kg and distances in m.

  ✓Do this: Convert 20 g → 0.02 kg and 6 cm → 0.06 m first.

  Usually costs: The answer.

• ✗Mistake: Calling the normal reaction the reaction to weight

  Weight (Earth on book) pairs with the book's pull on the Earth, not with the table's push.

  ✓Do this: Check: same kind of force, different bodies.

  Usually costs: 1 mark.

• ✗Mistake: Confusing kgf with kg

  kg is a unit of mass; kgf is a unit of force (9.8 N).

  ✓Do this: Weight in kgf → multiply by 9.8 to get N.

  Usually costs: 1 mark.

Practice questions on Newton's Laws of Motion

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The SI unit of momentum is:
   (a) kg m s⁻²
   (b) kg m s⁻¹
   (c) N s⁻¹
   (d) kg m

   Hint

   p = mv.

   Answer

   (b)

2. Q2 · 1 mark · MCQExam level

   A blanket is beaten with a stick to remove dust. Which property is used?
   (a) inertia of motion
   (b) inertia of rest
   (c) momentum
   (d) third law

   Hint

   Dust stays where it was.

   Answer

   (b)

3. Q3 · 2 marks · Short answerExam level

   Find the force needed to give a 500 g body an acceleration of 4 m s⁻², in newtons and in dynes.

   Hint

   Mass in kg.

   Answer

   $F=0.5\times 4=2$ N $=2\times 10^5$ dyne.

4. Q4 · 1 mark · Assertion-reasonHarder exam level

   Assertion (A): Action and reaction forces cancel each other, so a horse cannot pull a cart.
   Reason (R): Action and reaction act on different bodies.
   (a) (A) is true and (R) is false.
   (b) (A) is false and (R) is true.
   (c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
   (d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).

   Hint

   Do forces on different bodies cancel?

   Answer

   (b)

5. Q5 · 3 marks · Short answerHarder exam level

   A body of mass 4 kg moving at 6 m s⁻¹ is brought to rest in 2 s. Find the retarding force and the distance covered.

   Hint

   a = −3 m s⁻².

   Answer

   $F=4\times 3=12$ N; $S=\frac{1}{2}(6)(2)=6$ m.

6. Q6 · 3 marks · Short answerHarder exam level

   A truck of mass 5000 kg and a car of mass 1000 kg both move at 20 m s⁻¹. Compare their momenta, and the forces needed to stop each in 10 s.

   Hint

   p = mv.

   Answer

   $p_T=100000$, $p_C=20000$ kg m s⁻¹ (ratio 5 : 1). Forces: $10000$ N and $2000$ N.

7. Q7 · 3 marks · Short answerHarder exam level

   A force acts for 0.1 s on a body of mass 2 kg initially at rest. The force then stops and the body moves 50 cm in the next 1 s. Find the force.

   Hint

   Speed after the push = 0.5 m s⁻¹.

   Answer

   $v=0.5$ m s⁻¹; $a=\frac{0.5}{0.1}=5$ m s⁻²; $F=10$ N.

8. Q8 · 2 marks · Short answerHarder exam level

   Two magnets repel with a force of 0.8 N when 4 cm apart. Assuming the inverse square relation, find the force when they are 2 cm apart and 8 cm apart.

   Hint

   Half distance → ×4.

   Answer

   3.2 N and 0.2 N.

9. Q9 · 3 marks · Short answerHarder exam level

   Explain, using the second law, why a person falling on sand is hurt less than one falling on concrete from the same height.

   Hint

   Same Δp; different stopping time.

   Answer

   The change of momentum on landing is the same. Sand yields, so the stopping time is longer and $F=\frac{\Delta p}{\Delta t}$ is smaller; concrete stops the body in a very short time, giving a large force.

10. Q10 · 3 marks · Short answerHarder exam level

    A 2 kg trolley is pulled by a string with a tension of 6 N against a friction force of 2 N. Find its acceleration and its velocity after 4 s from rest.

    Hint

    Net force = 6 − 2.

    Answer

    $a=\frac{4}{2}=2$ m s⁻²; $v=8$ m s⁻¹.

11. Q11 · 4 marks · Long answerHOTS / competency

    A 50 g ball falls from rest from a height of 5 m onto a floor and rebounds to 1.25 m. Taking $g=10$ m s⁻², find its velocity just before and just after the bounce, and the average force if contact lasts 0.01 s.

    Hint

    v² = 2gh each way.

    Answer

    Before: $\sqrt{100}=10$ m s⁻¹ down. After: $\sqrt{25}=5$ m s⁻¹ up. $\Delta p=0.05\times (5+10)=0.75$ kg m s⁻¹; average net force $F=\frac{0.75}{0.01}=75$ N, upwards.

12. Q12 · 3 marks · Short answerHOTS / competency

    A 1200 kg car moving at 54 km h⁻¹ must stop within 45 m of the point where the driver sees a hazard. (a) Find the least uniform braking force if braking starts at once. (b) If the driver takes 0.6 s to react and the brakes then give a force of 4000 N, does the car stop within 45 m?

    Hint

    Find a from v² = u² + 2aS, then F = ma; add the distance covered during the reaction time.

    Answer

    $u=15$ m s⁻¹. (a) $0=225-2a(45)$, so $a=2.5$ m s⁻² and $F=1200\times 2.5=3000$ N. (b) Reaction distance $=15\times 0.6=9$ m. Retardation $=\frac{4000}{1200}=\frac{10}{3}$ m s⁻²; braking distance $=\frac{225}{2\times 10/3}=33.75$ m. Total $=42.75$ m, so yes, it stops 2.25 m short of the limit.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.SI and cgs units of force?Show answer

   Newton (N) and dyne.

2. 2.1 N in dyne?Show answer

   10⁵ dyne.

3. 3.1 kgf in N?Show answer

   9.8 N.

4. 4.Define momentum.Show answer

   Product of mass and velocity, p = mv.

5. 5.State Newton's second law.Show answer

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

6. 6.Give an example of inertia of rest.Show answer

   Passengers fall backwards when a bus starts suddenly.

7. 7.Do action and reaction act on the same body?Show answer

   No, on different bodies.

8. 8.How does a non-contact force vary with distance?Show answer

   Inversely as the square of the distance.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you classify forces as contact or non-contact?
• ☐Can you convert N, dyne, kgf and gf?
• ☐Can you explain an example of inertia of rest and of motion?
• ☐Do you use signs for velocities in momentum changes?
• ☐Can you combine F = ma with the equations of motion?
• ☐Can you name the bodies in an action–reaction pair?
• ☐Can you explain catching a ball using F = Δp/Δt?

Newton's Laws of Motion: frequently asked questions