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

Newton’s laws explain why objects at rest stay at rest and why every force comes with an equal, opposite reaction. This set of Class 11 Physics Chapter 4 questions tests momentum, self-adjusting static friction, and the forces behind circular motion.

Last Updated: September 23, 2026

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.

Quick visual: a worked diagram from the full Solutions page, for reference.

Frequently Asked Questions

A 5 kg block is pushed with a force of 20 N on a frictionless surface, how would you find its acceleration?
Using F = ma, a = F over m = 20 over 5 = 4 m per second squared.

If a rocket of mass 1000 kg ejects gas at 500 m/s to gain a thrust of 50000 N, how would you find the rate of mass ejection?
Using thrust = velocity times rate of mass loss, rate = 50000 over 500 = 100 kg per second.

Written by Satish

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