Question 1

Question A1The following is the distance-time table of an object in motion:Time in secondsDistance in meters001128327464512562167343(a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?(b) What do you infer about the forces acting on the object?

Accepted Answer

(a)Time (t) in sDistance (s) in mVelocity (v = s/t) in m/sAcceleration (a = ∆v/∆t) in m/s0000111128433279546416751252596216361173434913We notice from the table above that the acceleration of the object is increasing.(b) We know Force = mass x accelerationAs mass of the object remains same, the increasing acceleration implies that the force acting on the object is increasing as well.

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 8: Force and Laws of MotionWeb Content &mdash; Quick Reference
  
    Chapter 8: Force and Laws of Motion &mdash; Quick Reference

Newton laws
      inertia
      momentum
      F=ma
      conservation of momentum
      action reaction

Quick Revision Points
      
        Balanced forces: net force = 0, no acceleration. Unbalanced: net force ≠ 0, causes acceleration
First Law (Inertia): object stays at rest/uniform motion unless external force acts
Momentum p = mv (kg·m/s); Second Law: F = ma = Δp/Δt
1 N = force giving 1 kg an acceleration of 1 m/s²
Third Law: every action has equal and opposite reaction (on different bodies)
Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (no external force)

Exam Tips for Chapter 8
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch08-force-and-laws-of-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 7: MotionWeb Content &mdash; Quick Reference
  
    Chapter 7: Motion &mdash; Quick Reference

distance
      displacement
      speed
      velocity
      acceleration
      equations of motion
      graphs
      circular motion

Quick Revision Points
      
        Distance (scalar, total path) vs Displacement (vector, shortest path)
Speed = distance/time (scalar); Velocity = displacement/time (vector)
Acceleration a = (v−u)/t; unit: m/s²
Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
Uniform circular motion: constant speed, changing velocity (direction changes)

Exam Tips for Chapter 7
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch07-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 2

Question A3A hammer of mass 500 g, moving at 50 ms-1, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?

Accepted Answer

Given,Mass of the hammer (m) = 500 gConvert g into kg1000 g = 1 kg500 g = 5001000\dfrac{500}{1000}1000500​ = 0.5 kgInitial velocity of the hammer (u) = 50 ms-1Terminal velocity of the hammer (v) = 0Time period (t) = 0.01 sAs per the first equation of motion,a = v - ut\dfrac{	ext{v - u}}{	ext{t}}tv - u​Substituting we get,a = 0−500.01\dfrac{0-50}{0.01}0.010−50​ = -5000 ms-2We know, Force = mass x accelerationSubstituting we get,F = 0.5 x 5000 = -2500 NAccording to the third law of motion, the nail exerts an equal and opposite force on the hammer and as the force exerted on the nail by the hammer is -2500 N, hence, the force exerted on the hammer by the nail will be 2500 N.

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 8: Force and Laws of MotionWeb Content &mdash; Quick Reference
  
    Chapter 8: Force and Laws of Motion &mdash; Quick Reference

Newton laws
      inertia
      momentum
      F=ma
      conservation of momentum
      action reaction

Quick Revision Points
      
        Balanced forces: net force = 0, no acceleration. Unbalanced: net force ≠ 0, causes acceleration
First Law (Inertia): object stays at rest/uniform motion unless external force acts
Momentum p = mv (kg·m/s); Second Law: F = ma = Δp/Δt
1 N = force giving 1 kg an acceleration of 1 m/s²
Third Law: every action has equal and opposite reaction (on different bodies)
Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (no external force)

Exam Tips for Chapter 8
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch08-force-and-laws-of-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Web Content: Motion | Bright Tutorials

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 7: MotionWeb Content &mdash; Quick Reference
  
    Chapter 7: Motion &mdash; Quick Reference

distance
      displacement
      speed
      velocity
      acceleration
      equations of motion
      graphs
      circular motion

Quick Revision Points
      
        Distance (scalar, total path) vs Displacement (vector, shortest path)
Speed = distance/time (scalar); Velocity = displacement/time (vector)
Acceleration a = (v−u)/t; unit: m/s²
Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
Uniform circular motion: constant speed, changing velocity (direction changes)

Exam Tips for Chapter 7
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch07-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 3

Question A4A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.

Accepted Answer

Given,Mass of the car (m) = 1200 kgInitial velocity (u) = 90 kmh-1Convert kmh-1 to ms-1 : multiply by 518\dfrac{5}{18}185​90 x 518\dfrac{5}{18}185​ = 25 ms-1Final velocity (v) = 18 kmh-1Convert kmh-1 to ms-1 : multiply by 518\dfrac{5}{18}185​18 x 518\dfrac{5}{18}185​ = 5 ms-1Time period (t) = 4 secAs per the first equation of motion,a = v - ut\dfrac{	ext{v - u}}{	ext{t}}tv - u​Substituting we get,a = 5−254\dfrac{5-25}{4}45−25​ = -5ms-2Hence, acceleration of the car is -5 ms-2Negative sign shows retardation.Initial momentum of the car = m × u = 1200 × 25 = 30,000 kgms-1Final momentum of the car = m × v = 1200 x 5 = 6,000 kgms-1∴ Change in momentum = mv - mu = 6,000 - 30,000 = -24,000 kgms-1Hence, change in momentum = 24,000 kgms-1We know, Force = mass x accelerationSubstituting we get,F = 1200 × -5 = -6000 NHence, the magnitude of force required to slow down the vehicle = 6000 N

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 8: Force and Laws of MotionWeb Content &mdash; Quick Reference
  
    Chapter 8: Force and Laws of Motion &mdash; Quick Reference

Newton laws
      inertia
      momentum
      F=ma
      conservation of momentum
      action reaction

Quick Revision Points
      
        Balanced forces: net force = 0, no acceleration. Unbalanced: net force ≠ 0, causes acceleration
First Law (Inertia): object stays at rest/uniform motion unless external force acts
Momentum p = mv (kg·m/s); Second Law: F = ma = Δp/Δt
1 N = force giving 1 kg an acceleration of 1 m/s²
Third Law: every action has equal and opposite reaction (on different bodies)
Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (no external force)

Exam Tips for Chapter 8
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch08-force-and-laws-of-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Web Content: Motion | Bright Tutorials

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 7: MotionWeb Content &mdash; Quick Reference
  
    Chapter 7: Motion &mdash; Quick Reference

distance
      displacement
      speed
      velocity
      acceleration
      equations of motion
      graphs
      circular motion

Quick Revision Points
      
        Distance (scalar, total path) vs Displacement (vector, shortest path)
Speed = distance/time (scalar); Velocity = displacement/time (vector)
Acceleration a = (v−u)/t; unit: m/s²
Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
Uniform circular motion: constant speed, changing velocity (direction changes)

Exam Tips for Chapter 7
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch07-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 4

Question 5A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if it's mass is 7 tonnes.(Hint: 1 tonne = 1000 kg)

Accepted Answer

Given,Distance (s) = 400 mTime (t) = 20 secInitial velocity (u) = 0According to the second equation of motion,S = ut + 12\dfrac{1}{2}21​at2Substituting we get,400 = 0 + 12\dfrac{1}{2}21​ x a x 20 x 20a = 400×220×20\dfrac{400 	imes 2}{20 	imes 20}20×20400×2​ = 2 ms-2∴ acceleration of the car = 2 ms-2.Now, mass = 7 tonnesConverting tonne into kg1 tonne = 1000 kg∴ 7 tonnes = 7000 kgForce = mass x accelerationF = 7000 x 2 = 14000 NHence, Force acting = 14000 N

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 8: Force and Laws of MotionWeb Content &mdash; Quick Reference
  
    Chapter 8: Force and Laws of Motion &mdash; Quick Reference

Newton laws
      inertia
      momentum
      F=ma
      conservation of momentum
      action reaction

Quick Revision Points
      
        Balanced forces: net force = 0, no acceleration. Unbalanced: net force ≠ 0, causes acceleration
First Law (Inertia): object stays at rest/uniform motion unless external force acts
Momentum p = mv (kg·m/s); Second Law: F = ma = Δp/Δt
1 N = force giving 1 kg an acceleration of 1 m/s²
Third Law: every action has equal and opposite reaction (on different bodies)
Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (no external force)

Exam Tips for Chapter 8
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch08-force-and-laws-of-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Web Content: Motion | Bright Tutorials

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 7: MotionWeb Content &mdash; Quick Reference
  
    Chapter 7: Motion &mdash; Quick Reference

distance
      displacement
      speed
      velocity
      acceleration
      equations of motion
      graphs
      circular motion

Quick Revision Points
      
        Distance (scalar, total path) vs Displacement (vector, shortest path)
Speed = distance/time (scalar); Velocity = displacement/time (vector)
Acceleration a = (v−u)/t; unit: m/s²
Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
Uniform circular motion: constant speed, changing velocity (direction changes)

Exam Tips for Chapter 7
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch07-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 5

Question 6A stone of 1 kg is thrown with a velocity of 20 ms-1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?

Accepted Answer

Given,Mass (m) = 1 kgInitial velocity (u) = 20 ms-1Final velocity (v) = 0Distance travelled (s) = 50 mAccording to the third equation of motion,2as = v2 - u2ora = v2−u22s\dfrac{	ext{v}^2 - 	ext{u}^2 }{2	ext{s}}2sv2−u2​Substituting we get,a = 0−(20)22×50=−400100\dfrac{0 - (20)^2}{2 	imes 50} = -\dfrac{400}{100}2×500−(20)2​=−100400​ = -4 [retardation]Now, as Force = mass x accelerationSubstituting we get,F = 1 × (-4) = -4 NNegative sign indicates the opposing force which is frictionHence, force of friction between the stone and the ice = -4 N

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 8: Force and Laws of MotionWeb Content &mdash; Quick Reference
  
    Chapter 8: Force and Laws of Motion &mdash; Quick Reference

Newton laws
      inertia
      momentum
      F=ma
      conservation of momentum
      action reaction

Quick Revision Points
      
        Balanced forces: net force = 0, no acceleration. Unbalanced: net force ≠ 0, causes acceleration
First Law (Inertia): object stays at rest/uniform motion unless external force acts
Momentum p = mv (kg·m/s); Second Law: F = ma = Δp/Δt
1 N = force giving 1 kg an acceleration of 1 m/s²
Third Law: every action has equal and opposite reaction (on different bodies)
Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (no external force)

Exam Tips for Chapter 8
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch08-force-and-laws-of-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Web Content: Motion | Bright Tutorials

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    .keyword-tags { display: flex; flex-wrap: wrap; gap: 6px; margin: 10px 0; }
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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 7: MotionWeb Content &mdash; Quick Reference
  
    Chapter 7: Motion &mdash; Quick Reference

distance
      displacement
      speed
      velocity
      acceleration
      equations of motion
      graphs
      circular motion

Quick Revision Points
      
        Distance (scalar, total path) vs Displacement (vector, shortest path)
Speed = distance/time (scalar); Velocity = displacement/time (vector)
Acceleration a = (v−u)/t; unit: m/s²
Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
Uniform circular motion: constant speed, changing velocity (direction changes)

Exam Tips for Chapter 7
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch07-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 6

Question 7An 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate:(a) the net accelerating force and(b) the acceleration of the train

Accepted Answer

(a) Given,Force exerted (F) = 40,000 NForce of friction = -5000 NNet accelerating force = 40,000 + (-5000) = 35,000 NHence, net accelerating force = 35,000 N(b) Total mass of the train = mass of engine + mass of each wagon = 8000 + (5 × 2000) = 8000 + 10000 = 18000 kgNow, as Force = mass x accelerationSubstituting we get,35000 = 18000 x aa = 3500018000\dfrac{35000}{18000}1800035000​ = 1.94 ms-2Hence, acceleration of the train is 1.94 ms-2

Web Content: Force and Laws of Motion | Bright Tutorials

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 8: Force and Laws of MotionWeb Content &mdash; Quick Reference
  
    Chapter 8: Force and Laws of Motion &mdash; Quick Reference

Newton laws
      inertia
      momentum
      F=ma
      conservation of momentum
      action reaction

Quick Revision Points
      
        Balanced forces: net force = 0, no acceleration. Unbalanced: net force ≠ 0, causes acceleration
First Law (Inertia): object stays at rest/uniform motion unless external force acts
Momentum p = mv (kg·m/s); Second Law: F = ma = Δp/Δt
1 N = force giving 1 kg an acceleration of 1 m/s²
Third Law: every action has equal and opposite reaction (on different bodies)
Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (no external force)

Exam Tips for Chapter 8
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch08-force-and-laws-of-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Web Content: Motion | Bright Tutorials

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 7: MotionWeb Content &mdash; Quick Reference
  
    Chapter 7: Motion &mdash; Quick Reference

distance
      displacement
      speed
      velocity
      acceleration
      equations of motion
      graphs
      circular motion

Quick Revision Points
      
        Distance (scalar, total path) vs Displacement (vector, shortest path)
Speed = distance/time (scalar); Velocity = displacement/time (vector)
Acceleration a = (v−u)/t; unit: m/s²
Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
Uniform circular motion: constant speed, changing velocity (direction changes)

Exam Tips for Chapter 7
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch07-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 7

Question 8An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 ms-2?

Accepted Answer

Given,Mass (m) = 1500 kgAcceleration (a) = -1.7 ms-2Now, as Force = mass x accelerationF = 1500 × (-1.7) = -2550 NHence, force required = -2550 N. (Negative sign indicates opposite direction.)

Web Content: Force and Laws of Motion | Bright Tutorials

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 8: Force and Laws of MotionWeb Content &mdash; Quick Reference
  
    Chapter 8: Force and Laws of Motion &mdash; Quick Reference

Newton laws
      inertia
      momentum
      F=ma
      conservation of momentum
      action reaction

Quick Revision Points
      
        Balanced forces: net force = 0, no acceleration. Unbalanced: net force ≠ 0, causes acceleration
First Law (Inertia): object stays at rest/uniform motion unless external force acts
Momentum p = mv (kg·m/s); Second Law: F = ma = Δp/Δt
1 N = force giving 1 kg an acceleration of 1 m/s²
Third Law: every action has equal and opposite reaction (on different bodies)
Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (no external force)

Exam Tips for Chapter 8
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch08-force-and-laws-of-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Web Content: Motion | Bright Tutorials

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    .quick-rev h3 { color: var(--bt-primary); margin-top: 0; }
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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 7: MotionWeb Content &mdash; Quick Reference
  
    Chapter 7: Motion &mdash; Quick Reference

distance
      displacement
      speed
      velocity
      acceleration
      equations of motion
      graphs
      circular motion

Quick Revision Points
      
        Distance (scalar, total path) vs Displacement (vector, shortest path)
Speed = distance/time (scalar); Velocity = displacement/time (vector)
Acceleration a = (v−u)/t; unit: m/s²
Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
Uniform circular motion: constant speed, changing velocity (direction changes)

Exam Tips for Chapter 7
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch07-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 8

Question 9What is the momentum of an object of mass m, moving with a velocity v?(mv)2mv212\dfrac{1}{2}21​mv2mv

Accepted Answer

mvReason — The momentum of a body is defined as the product of its mass and velocity. For a body of mass m, moving with velocity v, linear momentum p is expressed as : p = mv

Web Content: Force and Laws of Motion | Bright Tutorials

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    .quick-rev ul { margin: 0; padding-left: 20px; }
    .quick-rev li { margin-bottom: 6px; font-size: 0.92rem; }
    .keyword-tags { display: flex; flex-wrap: wrap; gap: 6px; margin: 10px 0; }
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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 8: Force and Laws of MotionWeb Content &mdash; Quick Reference
  
    Chapter 8: Force and Laws of Motion &mdash; Quick Reference

Newton laws
      inertia
      momentum
      F=ma
      conservation of momentum
      action reaction

Quick Revision Points
      
        Balanced forces: net force = 0, no acceleration. Unbalanced: net force ≠ 0, causes acceleration
First Law (Inertia): object stays at rest/uniform motion unless external force acts
Momentum p = mv (kg·m/s); Second Law: F = ma = Δp/Δt
1 N = force giving 1 kg an acceleration of 1 m/s²
Third Law: every action has equal and opposite reaction (on different bodies)
Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (no external force)

Exam Tips for Chapter 8
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch08-force-and-laws-of-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Web Content: Motion | Bright Tutorials

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    .quick-rev h3 { color: var(--bt-primary); margin-top: 0; }
    .quick-rev ul { margin: 0; padding-left: 20px; }
    .quick-rev li { margin-bottom: 6px; font-size: 0.92rem; }
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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 7: MotionWeb Content &mdash; Quick Reference
  
    Chapter 7: Motion &mdash; Quick Reference

distance
      displacement
      speed
      velocity
      acceleration
      equations of motion
      graphs
      circular motion

Quick Revision Points
      
        Distance (scalar, total path) vs Displacement (vector, shortest path)
Speed = distance/time (scalar); Velocity = displacement/time (vector)
Acceleration a = (v−u)/t; unit: m/s²
Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
Uniform circular motion: constant speed, changing velocity (direction changes)

Exam Tips for Chapter 7
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch07-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 9

Question 10Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?

Accepted Answer

As the velocity of the cabinet is constant, its acceleration must be zero. Therefore, the effective force acting on it is also zero. Hence, the magnitude of opposing frictional force is equal to the force exerted on the cabinet i.e., 200 N. So, the total friction force is 200 N.

Web Content: Force and Laws of Motion | Bright Tutorials

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    .quick-rev ul { margin: 0; padding-left: 20px; }
    .quick-rev li { margin-bottom: 6px; font-size: 0.92rem; }
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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 8: Force and Laws of MotionWeb Content &mdash; Quick Reference
  
    Chapter 8: Force and Laws of Motion &mdash; Quick Reference

Newton laws
      inertia
      momentum
      F=ma
      conservation of momentum
      action reaction

Quick Revision Points
      
        Balanced forces: net force = 0, no acceleration. Unbalanced: net force ≠ 0, causes acceleration
First Law (Inertia): object stays at rest/uniform motion unless external force acts
Momentum p = mv (kg·m/s); Second Law: F = ma = Δp/Δt
1 N = force giving 1 kg an acceleration of 1 m/s²
Third Law: every action has equal and opposite reaction (on different bodies)
Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (no external force)

Exam Tips for Chapter 8
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch08-force-and-laws-of-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Web Content: Motion | Bright Tutorials

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 7: MotionWeb Content &mdash; Quick Reference
  
    Chapter 7: Motion &mdash; Quick Reference

distance
      displacement
      speed
      velocity
      acceleration
      equations of motion
      graphs
      circular motion

Quick Revision Points
      
        Distance (scalar, total path) vs Displacement (vector, shortest path)
Speed = distance/time (scalar); Velocity = displacement/time (vector)
Acceleration a = (v−u)/t; unit: m/s²
Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
Uniform circular motion: constant speed, changing velocity (direction changes)

Exam Tips for Chapter 7
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch07-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 10

Question 11According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.

Accepted Answer

The student's justification is incorrect since action-reaction force always apply to different bodies and hence cannot cancel each other. When two equal and opposite forces act on the same body, they cancel each other.As, the truck has a very high mass, the static friction between the road and the truck is high. When pushing the truck with a small force, the frictional force cancels out the applied force and the truck does not move. Hence, the truck will only move if a force greater than the friction force is applied to it.

Web Content: Force and Laws of Motion | Bright Tutorials

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 8: Force and Laws of MotionWeb Content &mdash; Quick Reference
  
    Chapter 8: Force and Laws of Motion &mdash; Quick Reference

Newton laws
      inertia
      momentum
      F=ma
      conservation of momentum
      action reaction

Quick Revision Points
      
        Balanced forces: net force = 0, no acceleration. Unbalanced: net force ≠ 0, causes acceleration
First Law (Inertia): object stays at rest/uniform motion unless external force acts
Momentum p = mv (kg·m/s); Second Law: F = ma = Δp/Δt
1 N = force giving 1 kg an acceleration of 1 m/s²
Third Law: every action has equal and opposite reaction (on different bodies)
Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (no external force)

Exam Tips for Chapter 8
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch08-force-and-laws-of-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Web Content: Motion | Bright Tutorials

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    .quick-rev ul { margin: 0; padding-left: 20px; }
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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 7: MotionWeb Content &mdash; Quick Reference
  
    Chapter 7: Motion &mdash; Quick Reference

distance
      displacement
      speed
      velocity
      acceleration
      equations of motion
      graphs
      circular motion

Quick Revision Points
      
        Distance (scalar, total path) vs Displacement (vector, shortest path)
Speed = distance/time (scalar); Velocity = displacement/time (vector)
Acceleration a = (v−u)/t; unit: m/s²
Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
Uniform circular motion: constant speed, changing velocity (direction changes)

Exam Tips for Chapter 7
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch07-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 11

Question 12A hockey ball of mass 200 g travelling at 10 ms-1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms-1. Calculate the magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

Accepted Answer

Given,Mass (m) = 200 gInitial velocity (u) = 10 ms-1Final velocity (v) = -5 ms-1Initial momentum of the ball = mu = 200 × 10 = 2000 gms-1Final momentum of the ball = mv = 200 × -5 = -1000 gms-1∴ The change in momentum (mv - mu) = -1000 - 2000 = -3000 gms-1Convert gms-1 to kgms-1 :1000 gms-1 = 1 kgms-1So, 3000 gms-1 = 30001000\dfrac{3000}{1000}10003000​ = -3 kgms-1Hence, momentum of the ball reduces by 3 kgms-1 after being struck by the hockey stick.

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 8: Force and Laws of MotionWeb Content &mdash; Quick Reference
  
    Chapter 8: Force and Laws of Motion &mdash; Quick Reference

Newton laws
      inertia
      momentum
      F=ma
      conservation of momentum
      action reaction

Quick Revision Points
      
        Balanced forces: net force = 0, no acceleration. Unbalanced: net force ≠ 0, causes acceleration
First Law (Inertia): object stays at rest/uniform motion unless external force acts
Momentum p = mv (kg·m/s); Second Law: F = ma = Δp/Δt
1 N = force giving 1 kg an acceleration of 1 m/s²
Third Law: every action has equal and opposite reaction (on different bodies)
Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (no external force)

Exam Tips for Chapter 8
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch08-force-and-laws-of-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Web Content: Motion | Bright Tutorials

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 7: MotionWeb Content &mdash; Quick Reference
  
    Chapter 7: Motion &mdash; Quick Reference

distance
      displacement
      speed
      velocity
      acceleration
      equations of motion
      graphs
      circular motion

Quick Revision Points
      
        Distance (scalar, total path) vs Displacement (vector, shortest path)
Speed = distance/time (scalar); Velocity = displacement/time (vector)
Acceleration a = (v−u)/t; unit: m/s²
Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
Uniform circular motion: constant speed, changing velocity (direction changes)

Exam Tips for Chapter 7
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch07-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 12

Question 13A bullet of mass 10 g travelling horizontally with a velocity of 150 ms-1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.

Accepted Answer

Given,Mass of the bullet (m) = 10 gConvert g into kg1000 g = 1 kgSo, 10 g = 101000\dfrac{10}{1000}100010​ = 0.01 kgInitial velocity of the bullet (u) = 150 ms-1Terminal velocity of the bullet (v) = 0Time period (t) = 0.03 sTo find the distance of penetration, the acceleration of the bullet must be calculatedLet the distance of penetration be sAccording to the first equation of motion,v = u + atora = v - ut\dfrac{	ext {v - u}}{	ext{t}}tv - u​Substituting we get,a = 0−1500.03=1500.03=−150003\dfrac{0-150}{0.03} = \dfrac{150}{0.03} =\dfrac{-15000}{3}0.030−150​=0.03150​=3−15000​ = -5000 ms-2∴ Acceleration = -5000 ms-2.According to the third equation of motion,2as = v2 - u2ora = v2−u22s\dfrac{	ext{v}^2 - 	ext{u}^2 }{2	ext{s}}2sv2−u2​Substituting we get,s = 0−(150)22×−5000=−22500−10000=\dfrac{0 - (150)^2}{2 	imes -5000} = \dfrac{-22500}{-10000} =2×−50000−(150)2​=−10000−22500​= = 2.25 m.Hence, the total distance travelled is 2.25 mNow, as Force = Mass x AccelerationSubstituting we get,F = 0.01 × -5000 = -50 NNegative sign indicates direction of recoil of bullet is opposite to the direction of bullet.Hence, magnitude of the force = 50 N

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 8: Force and Laws of MotionWeb Content &mdash; Quick Reference
  
    Chapter 8: Force and Laws of Motion &mdash; Quick Reference

Newton laws
      inertia
      momentum
      F=ma
      conservation of momentum
      action reaction

Quick Revision Points
      
        Balanced forces: net force = 0, no acceleration. Unbalanced: net force ≠ 0, causes acceleration
First Law (Inertia): object stays at rest/uniform motion unless external force acts
Momentum p = mv (kg·m/s); Second Law: F = ma = Δp/Δt
1 N = force giving 1 kg an acceleration of 1 m/s²
Third Law: every action has equal and opposite reaction (on different bodies)
Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (no external force)

Exam Tips for Chapter 8
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch08-force-and-laws-of-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Web Content: Motion | Bright Tutorials

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 7: MotionWeb Content &mdash; Quick Reference
  
    Chapter 7: Motion &mdash; Quick Reference

distance
      displacement
      speed
      velocity
      acceleration
      equations of motion
      graphs
      circular motion

Quick Revision Points
      
        Distance (scalar, total path) vs Displacement (vector, shortest path)
Speed = distance/time (scalar); Velocity = displacement/time (vector)
Acceleration a = (v−u)/t; unit: m/s²
Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
Uniform circular motion: constant speed, changing velocity (direction changes)

Exam Tips for Chapter 7
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch07-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 13

Question 14An object of mass 1 kg travelling in a straight line with a velocity of 10 ms-1 collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.

Accepted Answer

Given,Mass of the object (m1) = 1 kgMass of the block (m2) = 5 kgInitial velocity of the object (u1) = 10 ms-1Initial velocity of the block (u2) = 0Mass of the resulting object = m1 + m2 = 6 kgVelocity of the resulting object (v) = ?Total momentum before the collision = m1u1 + m2u2 = (1 × 10) + 0 = 10 kgms-1According to the law of conservation of momentum,total momentum before the collision = total momentum after the collision.∴ the total momentum post the collision is also 10 kgms-1Hence, total momentum just before the impact = 10 kgms-1 and just after the impact = 10 kgms-1Now, (m1 + m2) × v = 10 kgms-16 x v = 10v = 106\dfrac{10}{6}610​ = 53\dfrac{5}{3}35​ ms-1Hence, velocity of the combined object = 53\dfrac{5}{3}35​ ms-1

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 8: Force and Laws of MotionWeb Content &mdash; Quick Reference
  
    Chapter 8: Force and Laws of Motion &mdash; Quick Reference

Newton laws
      inertia
      momentum
      F=ma
      conservation of momentum
      action reaction

Quick Revision Points
      
        Balanced forces: net force = 0, no acceleration. Unbalanced: net force ≠ 0, causes acceleration
First Law (Inertia): object stays at rest/uniform motion unless external force acts
Momentum p = mv (kg·m/s); Second Law: F = ma = Δp/Δt
1 N = force giving 1 kg an acceleration of 1 m/s²
Third Law: every action has equal and opposite reaction (on different bodies)
Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (no external force)

Exam Tips for Chapter 8
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch08-force-and-laws-of-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Web Content: Motion | Bright Tutorials

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 7: MotionWeb Content &mdash; Quick Reference
  
    Chapter 7: Motion &mdash; Quick Reference

distance
      displacement
      speed
      velocity
      acceleration
      equations of motion
      graphs
      circular motion

Quick Revision Points
      
        Distance (scalar, total path) vs Displacement (vector, shortest path)
Speed = distance/time (scalar); Velocity = displacement/time (vector)
Acceleration a = (v−u)/t; unit: m/s²
Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
Uniform circular motion: constant speed, changing velocity (direction changes)

Exam Tips for Chapter 7
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch07-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 14

Question 15An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms-1 to 8 ms-1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.

Accepted Answer

Given,Mass of the object (m) = 100 kgInitial velocity (u) = 5 ms-1Final velocity (v) = 8 ms-1Time period (t) = 6 sNow, initial momentum (m × u) = 100 × 5 = 500 kgms-1Final momentum (m × v) = 100 × 8 = 800 kgms-1Hence, Initial momentum = 500 kgms-1 and final momentum = 800 kgms-1As per the first equation of motion,a = v - ut\dfrac{	ext{v - u}}{	ext{t}}tv - u​Substituting we get,a = 8−56\dfrac{8-5}{6}68−5​ = 0.5 ms-2We know, Force = mass x accelerationSubstituting we get,F = 100 x 0.5 = 50 NHence, magnitude of the force = 50 N

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 8: Force and Laws of MotionWeb Content &mdash; Quick Reference
  
    Chapter 8: Force and Laws of Motion &mdash; Quick Reference

Newton laws
      inertia
      momentum
      F=ma
      conservation of momentum
      action reaction

Quick Revision Points
      
        Balanced forces: net force = 0, no acceleration. Unbalanced: net force ≠ 0, causes acceleration
First Law (Inertia): object stays at rest/uniform motion unless external force acts
Momentum p = mv (kg·m/s); Second Law: F = ma = Δp/Δt
1 N = force giving 1 kg an acceleration of 1 m/s²
Third Law: every action has equal and opposite reaction (on different bodies)
Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (no external force)

Exam Tips for Chapter 8
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch08-force-and-laws-of-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Web Content: Motion | Bright Tutorials

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 7: MotionWeb Content &mdash; Quick Reference
  
    Chapter 7: Motion &mdash; Quick Reference

distance
      displacement
      speed
      velocity
      acceleration
      equations of motion
      graphs
      circular motion

Quick Revision Points
      
        Distance (scalar, total path) vs Displacement (vector, shortest path)
Speed = distance/time (scalar); Velocity = displacement/time (vector)
Acceleration a = (v−u)/t; unit: m/s²
Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
Uniform circular motion: constant speed, changing velocity (direction changes)

Exam Tips for Chapter 7
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch07-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 15

Question 16Akhtar, Kiran, and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.

Accepted Answer

Kiran's statement is false. The change in momentum of the insect and the motorcar is equal by conservation of momentum.Akhtar's statement is false. Force exerted is directly proportional to the mass and change in momentum. It does not depend on velocity. Secondly, according to Newton's third law of motion car and insect would apply equal and opposite force on each other.Rahul's statement is correct. As per the Newton's third law of motion, the force exerted by the insect on the car is equal and opposite to the force exerted by the car on the insect. Hence, insect and car experience same force.The change in momentum, however, depends on the mass and resulting acceleration. The motorcar, with its larger mass, experiences less acceleration and, therefore, less change in velocity. On the other hand, the mass of the insect is very small compared to the motorcar, it suffers a huge change of velocity as compared to the motorcar. Due to this, insect dies.

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 8: Force and Laws of MotionWeb Content &mdash; Quick Reference
  
    Chapter 8: Force and Laws of Motion &mdash; Quick Reference

Newton laws
      inertia
      momentum
      F=ma
      conservation of momentum
      action reaction

Quick Revision Points
      
        Balanced forces: net force = 0, no acceleration. Unbalanced: net force ≠ 0, causes acceleration
First Law (Inertia): object stays at rest/uniform motion unless external force acts
Momentum p = mv (kg·m/s); Second Law: F = ma = Δp/Δt
1 N = force giving 1 kg an acceleration of 1 m/s²
Third Law: every action has equal and opposite reaction (on different bodies)
Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (no external force)

Exam Tips for Chapter 8
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch08-force-and-laws-of-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

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BRIGHT TUTORIALS
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  Science | Chapter 7: MotionWeb Content &mdash; Quick Reference
  
    Chapter 7: Motion &mdash; Quick Reference

distance
      displacement
      speed
      velocity
      acceleration
      equations of motion
      graphs
      circular motion

Quick Revision Points
      
        Distance (scalar, total path) vs Displacement (vector, shortest path)
Speed = distance/time (scalar); Velocity = displacement/time (vector)
Acceleration a = (v−u)/t; unit: m/s²
Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
Uniform circular motion: constant speed, changing velocity (direction changes)

Exam Tips for Chapter 7
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch07-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 16

Question 17How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 ms-2.

Accepted Answer

Given,Mass of the dumb-bell (m) = 10 kgDistance covered (s) = 80 cmConvert cm to m100 cm = 1 mSo, 80 cm = 80100\dfrac{80}{100}10080​ = 0.8 mInitial velocity (u) = 0Acceleration (a) = 10 ms-2Terminal velocity (v) = ?Momentum of the dumb-bell when it hits the ground = mvAs per the third law of motion,v2 - u2 = 2asSubstituting we get,v2 - 0 = 2 x 10 x 0.8v2 = 16v = 16\sqrt{16}16​ ms-1v = 4 ms-1Hence, dumb-bell strikes the ground with a velocity of 4 ms-1The momentum transferred by the dumb-bell to the floor = mv = 10 × 4 = 40 kgms-1Hence, momentum transferred = 40 kgms-1

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 8: Force and Laws of MotionWeb Content &mdash; Quick Reference
  
    Chapter 8: Force and Laws of Motion &mdash; Quick Reference

Newton laws
      inertia
      momentum
      F=ma
      conservation of momentum
      action reaction

Quick Revision Points
      
        Balanced forces: net force = 0, no acceleration. Unbalanced: net force ≠ 0, causes acceleration
First Law (Inertia): object stays at rest/uniform motion unless external force acts
Momentum p = mv (kg·m/s); Second Law: F = ma = Δp/Δt
1 N = force giving 1 kg an acceleration of 1 m/s²
Third Law: every action has equal and opposite reaction (on different bodies)
Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (no external force)

Exam Tips for Chapter 8
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch08-force-and-laws-of-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Web Content: Motion | Bright Tutorials

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 7: MotionWeb Content &mdash; Quick Reference
  
    Chapter 7: Motion &mdash; Quick Reference

distance
      displacement
      speed
      velocity
      acceleration
      equations of motion
      graphs
      circular motion

Quick Revision Points
      
        Distance (scalar, total path) vs Displacement (vector, shortest path)
Speed = distance/time (scalar); Velocity = displacement/time (vector)
Acceleration a = (v−u)/t; unit: m/s²
Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
Uniform circular motion: constant speed, changing velocity (direction changes)

Exam Tips for Chapter 7
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch07-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 17

Question 2In the following example, try to identify the number of times the velocity of the ball changes:'A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team'.Also identify the agent supplying the force in each case.

Accepted Answer

The velocity of the ball changes :When the football player kicks a football to another player,When that player kicks the football towards the goal.When the goalkeeper stops the football.When the goalkeeper kicks the football towards his team player.Hence, the velocity of the football changes four times.Agent supplying the force:(a) The First case is the First player(b) The Second case is the Second player(c) The Third case is Goalkeeper(d) The Fourth case is Goalkeeper

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 8: Force and Laws of MotionWeb Content &mdash; Quick Reference
  
    Chapter 8: Force and Laws of Motion &mdash; Quick Reference

Newton laws
      inertia
      momentum
      F=ma
      conservation of momentum
      action reaction

Quick Revision Points
      
        Balanced forces: net force = 0, no acceleration. Unbalanced: net force ≠ 0, causes acceleration
First Law (Inertia): object stays at rest/uniform motion unless external force acts
Momentum p = mv (kg·m/s); Second Law: F = ma = Δp/Δt
1 N = force giving 1 kg an acceleration of 1 m/s²
Third Law: every action has equal and opposite reaction (on different bodies)
Conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (no external force)

Exam Tips for Chapter 8
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch08-force-and-laws-of-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Web Content: Motion | Bright Tutorials

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 7: MotionWeb Content &mdash; Quick Reference
  
    Chapter 7: Motion &mdash; Quick Reference

distance
      displacement
      speed
      velocity
      acceleration
      equations of motion
      graphs
      circular motion

Quick Revision Points
      
        Distance (scalar, total path) vs Displacement (vector, shortest path)
Speed = distance/time (scalar); Velocity = displacement/time (vector)
Acceleration a = (v−u)/t; unit: m/s²
Equations: v = u + at; s = ut + ½at²; v² = u² + 2as
d-t graph slope = speed; v-t graph slope = acceleration; area under v-t = distance
1 km/h = 5/18 m/s; Free fall: u = 0, a = g = 9.8 m/s²
Uniform circular motion: constant speed, changing velocity (direction changes)

Exam Tips for Chapter 7
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch07-motion.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Force and Laws of Motion

Additional Exercises

Exercises

Intext Questions 1