Laws of Motion — Class 11 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Physics chapter "Laws of Motion" — 7 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 11 Physics chapter "Laws of Motion" — 7 important questions with detailed answers for CBSE board exam prep…
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NCERT Solutions for Class 11 Physics — Laws of Motion. Step-by-step answers to all 7 textbook questions from this chapter, written for CBSE board preparation and free to use.
Laws of Motion — All 7 Questions Solved
Q1. State Newton's three laws of motion and explain the second law with a mathematical equation.
Newton's Three Laws of Motion:
First Law (Law of Inertia):
Every object continues to be in a state of rest or uniform motion in a straight line unless compelled by an external force to change that state.
Key points:
- An object at rest tends to remain at rest
- An object in motion tends to remain in motion
- This tendency to resist change in motion is called inertia
- Inertia is directly proportional to mass
Second Law (Law of Acceleration):
The rate of change of momentum of a body is directly proportional to the net external force acting on it, and this change takes place in the direction of the applied force.
Mathematical form:
F = dp/dt = d(mv)/dt
For constant mass m:
F = m(dv/dt) = ma
Where:
F = net force (in N)
m = mass (in kg)
a = acceleration (in m/s²)
p = momentum (in kg⋅m/s)
This can also be written as:
F_net = Σ F (sum of all forces)
Third Law (Law of Action and Reaction):
For every action, there is an equal and opposite reaction. The forces always occur in pairs acting on different bodies.
Mathematically: F₁₂ = -F₂₁
Key points:
- Action and reaction forces are equal in magnitude
- They are opposite in direction
- They act on different bodies
- They act simultaneously
- They are of the same type (e.g., both gravitational or both frictional)
Example of Third Law:
When a book rests on a table:
- Book exerts downward force on table (action)
- Table exerts upward normal force on book (reaction)
- These forces are equal in magnitude but opposite in direction
- Yet the book doesn't accelerate because these forces act on different bodies
Q2. A block of mass 5 kg is placed on a horizontal surface. If a horizontal force of 20 N is applied and the block accelerates at 3 m/s², find the friction force acting on the block.
Given:
Mass m = 5 kg
Applied force F = 20 N (horizontal)
Acceleration a = 3 m/s²
Find: Friction force f
Step 1: Apply Newton's second law
The net force on the block is:
F_net = ma = 5 × 3 = 15 N
Step 2: Identify forces in horizontal direction
In the horizontal direction:
- Applied force F = 20 N (forward)
- Friction force f (backward, opposing motion)
Step 3: Apply Newton's second law in horizontal direction
F_net = F - f
15 = 20 - f
f = 20 - 15 = 5 N
Step 4: Verify using free body diagram
Forces on block:
- Applied force: 20 N (→)
- Friction force: 5 N (←)
- Net force: 20 - 5 = 15 N (→)
- This produces acceleration: a = F_net/m = 15/5 = 3 m/s² ✓
Step 5: Find coefficient of friction
Normal force N = mg = 5 × 10 = 50 N (taking g = 10 m/s²)
Coefficient of kinetic friction μ_k = f/N = 5/50 = 0.1
Answer: The friction force acting on the block is 5 N, opposing the direction of motion. The coefficient of kinetic friction is 0.1.
Q3. Two blocks of masses 4 kg and 6 kg are connected by a light string and are pulled horizontally by a force of 50 N. Find (a) acceleration of the system (b) tension in the string connecting the two blocks (take g = 10 m/s²).
Given:
Mass m₁ = 4 kg (leading block)
Mass m₂ = 6 kg (trailing block)
Applied force F = 50 N
Assuming frictionless surface and g = 10 m/s²
Part (a): Acceleration of the system
Step 1: Consider the entire system as one body
Total mass M = m₁ + m₂ = 4 + 6 = 10 kg
Step 2: Apply Newton's second law to the system
F_net = Ma
50 = 10 × a
a = 50/10 = 5 m/s²
Part (b): Tension in the string
Step 1: Consider only the second block (6 kg)
Forces on the 6 kg block:
- Tension T (pulling forward)
- No other horizontal forces (assuming frictionless)
Step 2: Apply Newton's second law to the second block
T = m₂ × a
T = 6 × 5 = 30 N
Step 3: Verification using the first block
Forces on the 4 kg block:
- Applied force F = 50 N (forward)
- Tension T (backward, pulling the string)
Applying Newton's second law:
F - T = m₁ × a
50 - T = 4 × 5
50 - T = 20
T = 30 N ✓
The tension is confirmed.
Alternative approach for tension:
The 6 kg block is accelerated by tension alone:
T = m₂a = 6 × 5 = 30 N
Or, the 4 kg block experiences net force:
F_net on m₁ = F - T = 50 - 30 = 20 N
Which gives: a = 20/4 = 5 m/s² ✓
Answers:
(a) Acceleration of the system = 5 m/s²
(b) Tension in the string = 30 N
Q4. A lift with a person inside accelerates upward at 2 m/s². If the person has a mass of 60 kg, calculate the normal force exerted by the floor of the lift on the person. (Take g = 10 m/s²)
Given:
Mass of person m = 60 kg
Acceleration of lift a = 2 m/s² (upward)
g = 10 m/s²
Find: Normal force N exerted by floor on person
Step 1: Identify forces acting on the person
- Weight W = mg (downward)
- Normal force N (upward, from floor)
Step 2: Set up coordinate system
Let upward direction be positive (+)
Downward direction be negative (-)
Step 3: Calculate weight
W = mg = 60 × 10 = 600 N
Step 4: Apply Newton's second law in vertical direction
F_net = ma (upward)
N - W = ma
N - 600 = 60 × 2
N - 600 = 120
N = 720 N
Step 5: Physical interpretation
The normal force (720 N) is greater than the weight (600 N) because the person experiences an upward acceleration. The extra force (120 N) is needed to produce this upward acceleration.
Step 6: Find apparent weight
Apparent weight = Normal force = 720 N
Increase in apparent weight = N - W = 720 - 600 = 120 N
Alternative form:
N = m(g + a) = 60(10 + 2) = 60 × 12 = 720 N
Step 7: Special cases
If lift moves with constant velocity (a = 0): N = mg = 600 N
If lift accelerates downward at 2 m/s²: N = m(g - a) = 60(10 - 2) = 480 N
If lift is in free fall (a = g): N = 0 (weightlessness)
Answer: The normal force exerted by the floor on the person is 720 N. The person experiences an apparent weight of 720 N, which is 120 N more than their true weight.
Q5. Define static friction and kinetic friction. A block of mass 10 kg is on a horizontal surface with coefficient of static friction μ_s = 0.5 and coefficient of kinetic friction μ_k = 0.3. (g = 10 m/s²). What is the minimum force needed to start moving the block, and what force is needed to keep it moving at constant velocity?
Friction Definitions:
Static Friction (f_s):
Static friction is the friction force that acts between two surfaces when there is no relative motion between them. It prevents an object from moving when an external force is applied.
Properties:
- Acts when there is a tendency to move but no actual motion
- Can have any value from 0 to μ_s × N
- Maximum static friction f_s(max) = μ_s × N
- Direction: opposite to the applied force
- μ_s is usually greater than μ_k
Kinetic Friction (f_k):
Kinetic friction is the friction force that acts between two surfaces when they are moving relative to each other. It opposes the motion.
Properties:
- Acts when there is actual motion between surfaces
- Has a constant value f_k = μ_k × N
- Always opposite to the direction of motion
- μ_k is usually less than μ_s
- Independent of the relative speed
Given:
Mass m = 10 kg
Coefficient of static friction μ_s = 0.5
Coefficient of kinetic friction μ_k = 0.3
g = 10 m/s²
Step 1: Calculate normal force
N = mg = 10 × 10 = 100 N
Step 2: Minimum force to start moving (overcome static friction)
Maximum static friction:
f_s(max) = μ_s × N = 0.5 × 100 = 50 N
The minimum force needed to start moving = f_s(max) = 50 N
This is the threshold force. Any force greater than 50 N will cause the block to start moving.
Step 3: Force to keep moving at constant velocity
When moving at constant velocity, acceleration = 0
Net force = 0
Kinetic friction:
f_k = μ_k × N = 0.3 × 100 = 30 N
For constant velocity motion:
Applied force = Kinetic friction
F = 30 N
This force of 30 N just balances the kinetic friction, resulting in zero net force and constant velocity.
Step 4: Comparison
Force to start moving = 50 N
Force to keep moving = 30 N
The force needed to start moving (50 N) is greater than the force needed to keep moving (30 N) because:
- Initial static friction (50 N) > kinetic friction (30 N)
- Once in motion, only kinetic friction needs to be overcome
Answers:
Minimum force to start moving the block = 50 N
Force to keep moving at constant velocity = 30 N
Q6. A car of mass 1000 kg is moving on a horizontal road with coefficient of friction 0.8. The car suddenly applies brakes. Calculate (a) maximum deceleration possible (b) minimum distance to stop from 20 m/s (c) time taken to stop.
Given:
Mass m = 1000 kg
Coefficient of friction μ = 0.8 (kinetic, during braking)
Initial velocity u = 20 m/s
Final velocity v = 0 m/s (car stops)
g = 10 m/s²
Part (a): Maximum deceleration possible
Step 1: Calculate normal force
N = mg = 1000 × 10 = 10,000 N
Step 2: Calculate maximum friction force
When braking, kinetic friction acts:
f_max = μ × N = 0.8 × 10,000 = 8000 N
Step 3: Apply Newton's second law
F = ma
8000 = 1000 × a_max
a_max = 8000/1000 = 8 m/s²
Maximum deceleration = 8 m/s² (magnitude)
Part (b): Minimum distance to stop from 20 m/s
Using v² = u² - 2as (deceleration, so -a)
0² = (20)² - 2(8)s
0 = 400 - 16s
16s = 400
s = 25 m
Alternative using work-energy theorem:
Work done by friction = Change in kinetic energy
-f × s = 0 - (1/2)mu²
-8000 × s = -(1/2) × 1000 × (20)²
8000s = 500 × 400
8000s = 200,000
s = 25 m
Part (c): Time taken to stop
Using v = u - at
0 = 20 - 8 × t
8t = 20
t = 2.5 s
Alternative using s = ut - (1/2)at²
25 = 20t - (1/2)(8)t²
25 = 20t - 4t²
4t² - 20t + 25 = 0
Using quadratic formula or factoring:
(2t - 5)² = 0
t = 2.5 s
Answers:
(a) Maximum deceleration = 8 m/s²
(b) Minimum distance to stop = 25 m
(c) Time taken to stop = 2.5 s
Q7. A sphere of mass 2 kg is hanging from a light string of length 1 m. If the string breaks, how long does the sphere take to reach the ground after falling from rest? Also calculate the velocity with which it hits the ground. (Take g = 10 m/s²)
Given:
Mass m = 2 kg
Height h = 1 m (length of string = height from ground)
Initial velocity u = 0 m/s (falls from rest)
g = 10 m/s²
Part 1: Time to reach the ground
Using equation of motion:
h = ut + (1/2)gt²
1 = 0 + (1/2)(10)t²
1 = 5t²
t² = 1/5 = 0.2
t = √0.2 = √(1/5) = 1/√5 ≈ 0.447 s
Or in simpler form:
t = √(2h/g) = √(2 × 1/10) = √(2/10) = √(1/5) = 1/√5 s
Rationalizing:
t = 1/√5 × √5/√5 = √5/5 s ≈ 0.447 s
Alternative (rationalizing differently):
t = √(2/10) = √(1/5) ≈ 0.447 s
More practical form:
t = √(0.2) ≈ 0.447 s or approximately 0.45 s
Part 2: Velocity with which it hits the ground
Method 1: Using v = u + gt
v = 0 + 10 × (√(1/5))
v = 10/√5 = 10√5/5 = 2√5 m/s ≈ 4.47 m/s
Method 2: Using v² = u² + 2gh
v² = 0 + 2(10)(1)
v² = 20
v = √20 = 2√5 m/s ≈ 4.47 m/s
Method 3: Using energy conservation
Potential energy at height h = Kinetic energy at ground
mgh = (1/2)mv²
gh = (1/2)v²
v² = 2gh = 2 × 10 × 1 = 20
v = √20 = 2√5 m/s ≈ 4.47 m/s
Note: Mass of the sphere (2 kg) does not affect the time or velocity of fall in free fall, only gravity matters.
Answers:
Time to reach ground = √(1/5) s ≈ 0.447 s (or √(2/10) s)
Velocity at impact = 2√5 m/s ≈ 4.47 m/s (or √20 m/s)
Laws of Motion — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. What is the fundamental property of matter by which it resists any change in its state of rest or uniform motion in a straight line?
- A. Force
- B. Momentum
- C. Inertia
- D. Impulse
Answer: C. Inertia — This is the definition of inertia, which is an object's inherent resistance to changes in its state of motion.
MCQ 2. A force of 20 N acts on an object of mass 4 kg. What is the acceleration produced?
- A. 5 m/s²
- B. 0.2 m/s²
- C. 80 m/s²
- D. 24 m/s²
Answer: A. 5 m/s² — According to Newton's Second Law, force equals mass times acceleration (F=ma), which can be rearranged to find acceleration.
MCQ 3. A car of mass 1200 kg is moving with a velocity of 72 km/h. Calculate its momentum.
- A. 24000 kg m/s
- B. 86400 kg m/s
- C. 240000 kg m/s
- D. 864000 kg m/s
Answer: A. 24000 kg m/s — Momentum is the product of mass and velocity. Ensure consistent units (SI units) before calculation.
MCQ 4. When a bullet is fired from a gun, the gun recoils. This is an example of which of Newton's Laws?
- A. First Law of Motion
- B. Second Law of Motion
- C. Third Law of Motion
- D. Law of Conservation of Energy
Answer: C. Third Law of Motion — The gun recoiling is a direct consequence of the bullet exerting a forward force on the gun (action) and the gun exerting an equal and opposite force on the bullet (reaction), as described by Newton's Third Law.
MCQ 5. A man of mass 60 kg stands on a weighing machine in a lift. If the lift is accelerating upwards at 2 m/s², what is the reading on the weighing machine? (Take g = 10 m/s²)
- A. 600 N
- B. 480 N
- C. 720 N
- D. 500 N
Answer: C. 720 N — When the lift accelerates upwards, the apparent weight (normal force) increases and is given by N = m(g + a).
MCQ 6. A bullet of mass 50 g is fired horizontally into a wooden block of mass 1.95 kg resting on a rough horizontal surface. The bullet gets embedded in the block, and the combined system moves a distance of 5 m before coming to rest. If the coefficient of kinetic friction between the block and the surface is 0.25, what was the initial speed of the bullet? (Take g = 10 m/s²)
- A. 100 m/s
- B. 200 m/s
- C. 300 m/s
- D. 400 m/s
Answer: B. 200 m/s — This problem involves two parts: first, use kinematics and friction to find the velocity after impact; then, use conservation of momentum to find the initial bullet speed.
MCQ 7. A block of mass 10 kg is placed on a horizontal surface. The coefficient of static friction is 0.5, and the coefficient of kinetic friction is 0.4. A horizontal force of 40 N is applied to the block. What is the frictional force acting on the block? (Take g = 10 m/s²)
- A. 40 N
- B. 50 N
- C. 0 N
- D. 30 N
Answer: A. 40 N — The maximum static friction is 50 N. Since the applied force (40 N) is less than this, the block will not move, and the static friction will equal the applied force.
MCQ 8. A 2 kg ball moving at 5 m/s hits a wall and rebounds at 3 m/s in the opposite direction. What is the impulse imparted to the ball by the wall?
- A. 16 Ns
- B. 4 Ns
- C. 10 Ns
- D. 6 Ns
Answer: A. 16 Ns — Impulse is defined as the change in momentum (mass times change in velocity). Be careful with the direction of velocities.
MCQ 9. Two masses m₁ = 5 kg and m₂ = 3 kg are connected by a light string passing over a frictionless pulley. What is the acceleration of the system? (Take g = 10 m/s²)
- A. 2.5 m/s²
- B. 2 m/s²
- C. 5 m/s²
- D. 10 m/s²
Answer: A. 2.5 m/s² — For an Atwood machine, the acceleration is found by dividing the net force (difference in weights) by the total mass of the system.
MCQ 10. Which of the following is considered an inertial frame of reference?
- A. A car accelerating
- B. A lift moving upwards with constant acceleration
- C. A train turning a corner
- D. A laboratory at rest relative to the Earth's surface (ignoring Earth's rotation)
Answer: D. A laboratory at rest relative to the Earth's surface (ignoring Earth's rotation) — An inertial frame is one that is either at rest or moving with constant velocity (zero acceleration). All other options describe accelerating frames.
MCQ 11. A 2 kg block is placed inside a truck. The coefficient of static friction between the block and the truck floor is 0.6. What is the maximum acceleration of the truck for which the block will not slide? (Take g = 10 m/s²)
- A. 6 m/s²
- B. 4 m/s²
- C. 10 m/s²
- D. 0.6 m/s²
Answer: A. 6 m/s² — In the truck's non-inertial frame, a pseudo force opposes the truck's acceleration. For no sliding, this pseudo force must be balanced by the maximum static friction.
MCQ 12. A car of mass 1000 kg is negotiating a curve of radius 50 m at a speed of 10 m/s. What is the necessary centripetal force provided by friction?
- A. 2000 N
- B. 1000 N
- C. 500 N
- D. 200 N
Answer: A. 2000 N — The centripetal force required to keep an object in circular motion is calculated using the formula F_c = mv²/r.
MCQ 13. For a road banked at an angle θ, what is the optimum speed (the speed at which there is no reliance on friction) for negotiating the curve of radius r?
- A. √(rg tanθ)
- B. √(rg / tanθ)
- C. rg tanθ
- D. (rg tanθ)²
Answer: A. √(rg tanθ) — At the optimum speed on a banked road, the horizontal component of the normal force provides the necessary centripetal force, leading to the formula v = √(rg tanθ).
MCQ 14. What is the physical quantity that is a measure of the inertia of a body?
- A. Volume
- B. Density
- C. Mass
- D. Weight
Answer: C. Mass — Mass is the intrinsic property of a body that quantifies its inertia, which is its resistance to changes in motion.
MCQ 15. A particle of mass 2 kg moves with a velocity given by $\vec{v} = (3t\hat{i} + 2t^2\hat{j}) $ m/s. What is the magnitude of the force acting on the particle at t = 1 s?
- A. 10 N
- B. 12 N
- C. 16 N
- D. 20 N
Answer: A. 10 N — First find the acceleration vector by differentiating the velocity vector with respect to time, then calculate the force vector using F=ma, and finally find its magnitude.
MCQ 16. A block of mass 5 kg is placed on an inclined plane with an angle of inclination 30°. The coefficient of kinetic friction between the block and the plane is 0.2. What is the acceleration of the block down the incline? (Take g = 10 m/s²)
- A. 3.3 m/s²
- B. 3.46 m/s²
- C. 4.0 m/s²
- D. 4.33 m/s²
Answer: A. 3.3 m/s² — Resolve forces along and perpendicular to the incline. The net force down the incline is the component of gravity minus the kinetic friction, leading to a = g(sinθ - μ_k cosθ).
MCQ 17. An object is moving with a constant velocity. Which of the following statements about the forces acting on it is true?
- A. A net force acts on it in the direction of motion
- B. No forces are acting on it
- C. The net force acting on it is zero
- D. The force of gravity is zero
Answer: C. The net force acting on it is zero — According to Newton's First Law, an object in uniform motion (constant velocity) has zero net force acting on it.
MCQ 18. A 5 kg mass hangs from a string which passes over a frictionless pulley and is attached to a 3 kg mass resting on a frictionless horizontal table. What is the tension in the string? (Take g = 10 m/s²)
- A. 18.75 N
- B. 30 N
- C. 50 N
- D. 12.5 N
Answer: A. 18.75 N — First, calculate the acceleration of the system using the net force (due to the hanging mass) and total mass. Then, use this acceleration to find the tension by considering the block on the table.
MCQ 19. A 100 g cricket ball moving at 30 m/s is caught by a fielder. If the ball is brought to rest in 0.1 s, what is the average force exerted by the fielder on the ball?
- A. 30 N
- B. 3 N
- C. 0.3 N
- D. 300 N
Answer: A. 30 N — Impulse (change in momentum) is equal to the product of average force and time interval (F_avg Δt). Equate these two to find the average force.
MCQ 20. If an object is experiencing forces of 5 N to the east and 3 N to the west, what is the net force acting on the object?
- A. 8 N East
- B. 2 N West
- C. 2 N East
- D. 8 N West
Answer: C. 2 N East — The net force is the vector sum of all individual forces. Since the forces are in opposite directions, subtract their magnitudes to find the resultant force, which acts in the direction of the larger force.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.