CBSE Class 11 Physics Chapter 4 Laws of Motion Worksheet with Answers

Quick Chapter Recap: Laws of Motion

• Newton's First Law: Inertia and the concept of force; Galileo's experiments.
• Newton's Second Law: F = ma; momentum form F = dp/dt; impulse.
• Newton's Third Law: Action-reaction pairs; thrust and recoil.
• Friction: Static (fs ≤ μsN), kinetic (fk = μkN), rolling; angle of friction and angle of repose.
• Circular motion: Centripetal acceleration a = v²/r; banking of roads; conical pendulum.

Section A: Multiple Choice Questions (1 mark each)

• Q1. A body of mass 2 kg is at rest on a frictionless surface. A force of 10 N acts on it for 3 seconds. The final velocity of the body is: (a) 5 m/s (b) 10 m/s (c) 15 m/s (d) 20 m/s
• Q2. Newton's first law is valid in: (a) all frames (b) inertial frames only (c) non-inertial frames only (d) rotating frames only
• Q3. A book rests on a table. The reaction force to the weight of the book is: (a) normal force by table on book (b) gravitational pull by book on Earth (c) normal force by book on table (d) friction by table on book
• Q4. Coefficient of static friction is always: (a) less than kinetic friction (b) greater than or equal to kinetic friction (c) equal to kinetic friction (d) unrelated to kinetic friction
• Q5. A car moves in a horizontal circular track of radius 50 m at 10 m/s. The centripetal acceleration is: (a) 1 m/s² (b) 2 m/s² (c) 5 m/s² (d) 10 m/s²
• Q6. A 5 kg block is pulled horizontally by a force of 20 N. If friction force is 5 N, the acceleration is: (a) 1 m/s² (b) 3 m/s² (c) 4 m/s² (d) 5 m/s²

Section B: Fill in the Blanks (1 mark each)

• Q7. Newton's second law in terms of momentum is F = ________.
• Q8. The maximum static friction is given by fs,max = ________.
• Q9. The SI unit of coefficient of friction is ________.
• Q10. For a particle moving in a circle of radius r with speed v, centripetal acceleration is ________.
• Q11. The minimum force required to move a body on a horizontal surface is called ________ friction.
• Q12. When a car negotiates a banked curve at the optimum speed, friction is ________.

Section C: True or False (1 mark each)

• Q13. Inertia is a force that keeps objects at rest.
• Q14. Kinetic friction is always less than or equal to the maximum static friction for the same pair of surfaces.
• Q15. In Newton's third law, action and reaction forces act on the same body.
• Q16. A body can have acceleration even if the net force on it is zero in an inertial frame.
• Q17. The centripetal force is always directed towards the centre of the circular path.
• Q18. Rolling friction is generally greater than kinetic friction for the same materials.

Section D: Short Answer Questions (2-3 marks each)

• Q19. (2 marks) A force of 50 N acts on a body of mass 10 kg at rest. Calculate the acceleration and the distance covered in 4 seconds.
• Q20. (3 marks) Explain why it is easier to pull a lawn roller than to push it, using the concept of normal force and friction.
• Q21. (2 marks) A 2 kg block on a table has μs = 0.4 and μk = 0.3. Find the maximum static friction and the kinetic friction. (Take g = 10 m/s²)
• Q22. (3 marks) A stone of mass 0.5 kg is tied to a string and whirled in a horizontal circle of radius 1 m with speed 4 m/s. Calculate the tension in the string.
• Q23. (3 marks) State Newton's third law. Give two examples from daily life where action-reaction pairs are clearly observable.

Section E: Long Answer and HOTS Questions (4-5 marks each)

• Q24. (5 marks) Derive the expression for the angle of banking of a road for a vehicle moving with speed v on a curve of radius r, assuming no friction. Discuss what happens if the vehicle moves faster than the optimum speed.
• Q25. (4 marks) A block of mass 5 kg rests on a rough inclined plane making an angle of 30° with the horizontal. The coefficient of static friction is 0.4. Will the block slide down? If an upward force of 20 N parallel to the plane is applied, find the net force on the block. (Take g = 10 m/s²)
• Q26. (5 marks) State and explain Newton's second law of motion. Show that F = dp/dt is the general form. How does this form help in understanding rocket propulsion?

Section F: Case-Study Question (4 marks)

Answer Key with Explanations

• A1. (c) 15 m/s. Using F = ma, a = 10/2 = 5 m/s². v = u + at = 0 + 5×3 = 15 m/s.
• A2. (b) inertial frames only. Newton's first law defines inertial frames; it does not hold in accelerating or rotating frames without pseudo-forces.
• A3. (b) gravitational pull by book on Earth. The reaction to Earth pulling the book down (weight) is the book pulling Earth up, not the normal force.
• A4. (b) greater than or equal to kinetic friction. Typically μs ≥ μk; static friction can sustain larger forces before motion begins.
• A5. (b) 2 m/s². a = v²/r = 10²/50 = 100/50 = 2 m/s².
• A6. (b) 3 m/s². Net force F = 20 − 5 = 15 N. a = F/m = 15/5 = 3 m/s².
• A7. dp/dt (rate of change of momentum).
• A8. μsN (coefficient of static friction times normal force).
• A9. No unit (dimensionless). Coefficient of friction is a ratio of forces.
• A10. v²/r or ω²r. Both forms are acceptable.
• A11. Limiting or maximum static friction. This is the threshold to overcome static friction.
• A12. Zero (or not required). At optimum speed on a banked road, friction is not needed; centripetal force is provided entirely by the horizontal component of normal force.
• A13. False. Inertia is the property of mass; it is not a force. No force is needed to maintain constant velocity in an inertial frame.
• A14. True. Once motion starts, kinetic friction fk = μkN is generally less than the maximum static friction fs,max = μsN, and μk ≤ μs.
• A15. False. Action and reaction forces in Newton's third law always act on different bodies, never on the same body.
• A16. False. In an inertial frame, if net force is zero, acceleration must be zero (Newton's second law).
• A17. True. By definition, centripetal acceleration and centripetal force point toward the centre of the circular path.
• A18. False. Rolling friction is typically much smaller than kinetic (sliding) friction for the same materials, which is why wheels are used.
• A19. a = F/m = 50/10 = 5 m/s²; s = ½at² = ½×5×16 = 40 m. (2 marks: 1 for acceleration, 1 for distance with working.)
• A20. When pulling at an angle, the vertical component reduces the normal force N, hence friction f = μN decreases. When pushing, the vertical component adds to N, increasing friction. Thus pulling is easier. (3 marks: concept of normal force 1, friction dependence 1, conclusion 1.)
• A21. N = mg = 2×10 = 20 N. fs,max = μsN = 0.4×20 = 8 N; fk = μkN = 0.3×20 = 6 N. (2 marks: correct values with units.)
• A22. Centripetal force = mv²/r = 0.5×4²/1 = 8 N. Since the circle is horizontal, tension T provides the centripetal force. T = 8 N. (3 marks: formula 1, substitution 1, answer with unit 1.)
• A23. Newton's third law: To every action there is an equal and opposite reaction acting on different bodies. Examples: (i) A book on a table exerts a downward force; the table exerts an equal upward normal force. (ii) A swimmer pushes water backward; water pushes the swimmer forward. (3 marks: statement 1, two valid examples 1 each.)
• A24. Free-body diagram: weight mg down, normal N perpendicular to road surface. Resolve N into N cosθ (vertical) and N sinθ (horizontal centripetal). Vertical equilibrium: N cosθ = mg. Centripetal: N sinθ = mv²/r. Dividing: tanθ = v²/(rg), so θ = arctan(v²/(rg)). If v > vopt, the required centripetal force exceeds N sinθ, so friction must act down the slope toward the centre. If v < vopt, friction acts up the slope. (5 marks: diagram 1, resolution 2, derivation 1, discussion 1.)
• A25. Weight component down the plane = mg sinθ = 5×10×sin30° = 25 N. Normal force N = mg cosθ = 5×10×cos30° = 43.3 N. fs,max = μsN = 0.4×43.3 ≈ 17.3 N. Since 25 N > 17.3 N, the block will slide down without external force. With 20 N upward parallel force, net down-slope force = 25 − 20 − 17.3 (friction now acts up) = not valid—recalculate: friction opposes relative motion. If 20 N applied upward, net applied up = 20 N, component down = 25 N, static friction can be up to 17.3 N up. Net = 25 − 20 − 17.3 = −12.3 N (block will not move or move down?). Re-assess: 25 N down > 20 N up, difference 5 N, which is less than fs,max 17.3 N, so block remains stationary; net force = 0. (4 marks: component calculations 2, friction analysis 1, conclusion 1.) [Note: nuanced question; careful sign and direction needed.]
• A26. Newton's second law: The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction of the force. Mathematically, F = dp/dt. For constant mass, dp/dt = d(mv)/dt = m dv/dt = ma, recovering F = ma. In rocket propulsion, mass changes as fuel is ejected. Using F = dp/dt, thrust = (dm/dt)v_exhaust, explaining how rockets accelerate by expelling mass. (5 marks: statement 1, derivation from momentum 2, F=ma 1, rocket application 1.)
• Case (a): N = mg = 80×10 = 800 N. fmax = μsN = 0.6×800 = 480 N. (2 marks)
• Case (b): Centripetal force ≤ fmax, so mv²/r ≤ 480. v² ≤ 480×20/80 = 120. v ≤ √120 ≈ 10.95 m/s ≈ 11 m/s. (2 marks)

How to Use This Worksheet Effectively

• Attempt the worksheet in exam conditions: quiet room, 90-minute timer, no textbook or notes.
• Use rough paper for calculations; write final answers clearly on the worksheet.
• Do not peek at the answer key until you have answered every question, even if you are unsure.
• After self-assessment, calculate your percentage: (marks scored / total marks) × 100.
• Identify weak areas (e.g. friction numericals, circular motion conceptual questions) and revisit NCERT Class 11 Physics explanations.
• Redo incorrect questions on a separate sheet after studying the concept; check your work again.
• Share your score and doubts with your physics teacher or peers; collaborative learning deepens understanding.

Why Laws of Motion Matters for CBSE Board Exams

• Chapter 4 carries approximately 8-12 marks in CBSE Class 11 Physics annual exams.
• Newton's laws underpin mechanics; weak foundation here affects Chapters 5 (Work, Energy, Power), 6 (Gravitation), and 7 (Rotational Motion).
• Free-body diagrams and force resolution are tested in board practicals and viva questions.
• Case-study and HOTS questions now form 20-25 percent of the theory paper, demanding application skills.
• Friction problems often combine with inclined planes, pulleys, and circular motion, requiring multi-concept integration.
• Strong performance in Laws of Motion boosts overall Physics percentage and builds problem-solving stamina for JEE/NEET aspirants.

Beyond Worksheets: Building Mastery in Laws of Motion

• Read NCERT Class 11 Physics Chapter 4 thoroughly; annotate margins with your own examples.
• Solve all in-text questions and additional exercises; these are the source pool for many board exam questions.
• Watch animations of free-body diagrams and circular motion to build spatial intuition.
• Maintain a personal error log: every mistake in worksheets or tests goes in, with the correct method written beside it.
• Use flashcards for definitions (inertia, momentum, impulse) and key formulae (F=ma, fk=μkN, a=v²/r).
• Join online doubt-clearing platforms like CBSETUTOR.ai for 24×7 AI tutor support, photo-upload solving, and personalised practice at a flat ₹999/month across all classes.