Class 11 Physics Chapter 4 Laws of Motion – Extra Questions with Answers

Extra practice questions for Class 11 Physics Chapter 4 (Laws of Motion), beyond the textbook. These Class 11 Physics Chapter 4 important questions are handy for last-minute exam practice.

Very Short Answer Questions (1 mark)

Q1. State Newton’s first law of motion.
Ans: A body continues in its state of rest or uniform motion in a straight line unless acted upon by an external unbalanced force.

Q2. What is the SI unit of force?
Ans: Newton (N).

Q3. Define momentum.
Ans: Momentum is the product of mass and velocity, p=mv.

Q4. Is friction a self-adjusting force? Explain in one line.
Ans: Yes, static friction adjusts its magnitude (up to a limit) to exactly balance the applied force to prevent motion.

Q5. What provides the centripetal force for a car turning on a level road?
Ans: Friction between the tyres and the road.

Short Answer Questions (2–3 marks)

Q6. A force of 10 N acts on a mass of 2 kg. Find the acceleration produced.
Ans: a=F/m=10/2=5 m/s².

Q7. A bullet of mass 0.02 kg moving at 300 m/s embeds in a stationary block of mass 5.98 kg. Find the common velocity after collision using momentum conservation.
Ans: m1u1=(m1+m2)v; 0.02(300)=(6)v; v=6/6=1 m/s.

Q8. A block of mass 4 kg rests on a surface with coefficient of static friction 0.5. Find the maximum static friction force (g=10 m/s²).
Ans: f_s(max)=μ_sN=0.5(4)(10)=20 N.

Higher-Order Thinking / Application Questions

Q9. A body of mass 5 kg is pulled by a horizontal force of 20 N on a rough surface with μ_k=0.2 (g=10 m/s²). Calculate the net acceleration of the body, showing how friction opposes the applied force.
Ans: Normal force N=mg=5(10)=50 N. Kinetic friction f_k=μ_kN=0.2(50)=10 N, opposing motion. Net force = 20−10=10 N. Acceleration a=F_net/m=10/5=2 m/s².

Q10. Two blocks of masses 3 kg and 2 kg are connected by a string over a frictionless pulley, with the 3 kg block on a table and the 2 kg block hanging. Explain the concept used and calculate the system’s acceleration (g=10 m/s², ignore friction on the table).
Ans: This uses Newton’s second law applied to a connected system: the tension is the same throughout an ideal string, and both blocks share the same magnitude of acceleration. For the hanging block: m2g−T=m2a. For the block on table: T=m1a. Adding: m2g=(m1+m2)a, so a=m2g/(m1+m2)=2(10)/5=4 m/s².

Written by Satish

NCERTBooks.org is an independent educational resource run by a small team focused on making official NCERT textbooks easy to find, read, and download for students, parents, and teachers across India. We are not affiliated with NCERT or the Ministry of Education -- we organise publicly available NCERT content by class and subject, verify links against official sources, and build tools (like our in-browser reader) that make studying more convenient. Every guide we publish is written and reviewed by our team based on the actual NCERT curriculum and syllabus.