NCERT Solutions for Class 9 Science Chapter 8 - Force and Laws of Motion

NCERT Solutions for Class 9 Science Chapter 8 - Force and Laws of Motion

Vishal kumarUpdated on 16 Sep 2025, 02:04 AM IST

Have you ever wondered why a football would run faster when you kick it, why it would run slower gradually or how a car can alter its direction when you turn the steering wheel? All these observations in daily life are discussed in Chapter 8 of Science Class 9 - Force and Laws of Motion, which is the first to present students with the basics of force, motion, and laws of Newton. The chapter provides the background of mechanics and assists the students in correlating the classroom material with real-life experiences.

This Story also Contains

  1. Class 9 Science Chapter 8 - Force and Laws of Motion Question Answers: Download Solution PDF
  2. Class 9 Science Chapter 8 - Force and Laws of Motion: Solved In-text Questions
  3. Class 9 Science Chapter 8 - Force and Laws of Motion: Solved Exercise Questions
  4. Force and Laws of Motion NCERT Solutions: Solved Additional Exercises
  5. Class 9 Science Chapter 8 - Force and Laws of Motion: Higher Order Thinking Skills (HOTS) Questions
  6. NCERT Solutions for Class 9 Science Chapter 8: Key Topics
  7. NCERT Solutions for Class 9 Science Chapter 8: Important Formulas
  8. Approach to Solve Questions of Class 9 NCERT Chapter 8:Force and Laws of Motion
  9. NCERT Solutions for Class 9 Science Chapter-wise
NCERT Solutions for Class 9 Science Chapter 8 - Force and Laws of Motion
Force and Laws of Motion

NCERT Solutions for Class 9 Science Chapter 8 - Force and Laws of Motion offer well-structured, correct and step-by-step solutions that are set out by subject experts. The solutions simplify, make learning fast and exam-oriented, so that students can enhance their theoretical and numerical problem-solving ability. In-text Question, Exercise Question, HOTS (Higher Order Thinking Skills) and problem-based on application will also be covered and all the answers will be in accordance with the latest CBSE syllabus and exam pattern (2025-26). Using these NCERT Solutions for Class 9 Science Chapter 8 - Force and Laws of Motion, the students will be able to conceptualize the First, Second and Third Laws of Motion as stated by Newton with examples, learn good problem solving skills, doing derivations and numericals, make fast and brief revisions using brief explanations and diagrams, use science to solve real world problems to memorize better. With Force and Laws of Motion NCERT Solutions, students would be able to study the subject well, prevent errors and score more points in exams, as well as establish a good foundation to further education and competitive examinations.

Class 9 Science Chapter 8 - Force and Laws of Motion Question Answers: Download Solution PDF

The Class 9 Science Chapter 8 - Force and Laws of Motion question answers provide a well-organised step-by-step solution to all questions of the textbook, and students find it easy to understand the laws of Newton and how to apply them in real-life situations. These solutions are created by the same experts in the subject as in accordance with the current CBSE syllabus, which makes them very useful in preparing exams, in doing daily homework and in a quick revision. Through these solutions, students would be in a position to reinforce their knowledge of the vital concepts of force, inertia, momentum, and the three laws of motion, thus performing better on exams. To be able to study at any time, anywhere, you will find it more convenient to download the free PDF of Force and Laws of Motion Class 9 question answers.

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Class 9 Science Chapter 8 - Force and Laws of Motion: Solved In-text Questions

The Force and Laws of Motion NCERT Solutions (In-text Questions) will give proper and in-depth responses to all questions inquired in the textbook. These are solutions to make students grasp vital ideas such as the laws of motion, force, inertia and momentum of Newton in a straightforward and ordered manner. These in-text question answers allow the students to enhance their conceptual clarity, enhance their skills of solving problems and be in a good position to pass exams.

Topic 8.3(Inertia and Mass)

Q1. Which of the following has more inertia:

(a) a rubber ball and a stone of the same size?

(b) a bicycle and a train?

(c) a five-rupee coin and a one-rupee coin?

Answer:

We know that inertia is defined by the mass of the body. So the body has more mass/weight will have more inertia.

(a) In the case of a rubber ball and a stone of the same size, the stone will have greater mass, so its inertia will be greater compared to a rubber ball.

(b) The train has much more mass as compared to the bicycle, so the inertia of the train is larger.

(c) A five-rupee coin has more weight than a rupee coin, so the inertia of five five-rupee coins is more.

Q2. In 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.

Answer:

Velocity has both magnitude and direction. Thus change in any of the magnitude or direction results in a change in velocity.

In the given case, when the player kicks the ball to another player, the direction of the ball is changed, thus the velocity is changed.

The second player then pushes the ball in a forward direction, sothe velocity changes. Then the goalkeeper brings the ball to rest, hence velocity changes (magnitude is changed). After that, he kicks the ball, thus increasing the speed and direction is opposite, hence the velocity changes again.

The agent of the supplying force is the 1st player, the 2nd player, goalkeeper, respectively.

Q3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.

Answer:

This can be explained with the help of the concept of inertia. When the tree is at rest, its leaves are also at rest. But when we bring the tree in motion by means of shaking it, due to the inertia of the leaves, they still tend to be at rest. Thus force is acting on leaves vigorously with a changing direction rapidly. This results in the detachment of leaves from the tree.

Q4 . Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?

Answer:

In the situation of a moving bus, the passengers are also in motion along with the bus. Now, when the brakes are applied, the bus comes to rest, but due to the inertia of the passengers, they still tend to move in the forward direction. Thus, they move in the forward direction after applying the brakes.

Now that the bus is resting, the passengers are also at rest. The inertia tends them to be at rest. That's why when the bus accelerates, they move in a backwards direction.

Additional questions

Q1. If action is always equal to the reaction, explain how a horse can pull a cart.

Answer:

For motion, the horse applies force on the ground, and thus the frictional force of the ground pushes the horse forward (action-reaction pair). Thus horse applies force on the cart and the cart applies backwards force on the hors,e but due to an unbalanced force, the horse moves in the forward direction.

Q2. Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.

Answer:

This can be explained with the help of Newton's third law of motion (action-reaction pair). When the velocity of water is more i.e., water is coming out with a higher force thus the reaction force on the hose is also large. This makes it difficult for a fireman to hold the hose.

Q3. From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m s-1. Calculate the initial recoil velocity of the rifle.

Answer:

Here we will use conservation of momentum :

Initial momentum of system ( rifle + bullet ) = Final momentum of system

m1u1 + m2u2 = m1v1 + m2v2

or 4×0 + 0.05×0 = 4×v1 + 0.05×35

or 4×v1 = 1.75

or v1 = 0.4375 m/s

The negative sign shows that the velocity is opposite to that of the bullet. This is because the force on the rifle will be opposite to the bullet (action-reaction pair).

Q4. Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 m s-1 and 1 m s-1 , respectively. They collide and after the collision, the first object moves at a velocity of 1.67 m s-1. Determine the velocity of the second object.

Answer:

In this question, we will use conservation of momentum :

Initial momentum = Final momentum

m1u1 + m2u2 = m1v1 + m2v2

or 0.1×2 + 0.2×1 = 0.1×1.67 + 0.2×v2

or v2 = 0.2330.2

or v2 = 1.165 m/s

Hence, the required velocity is 1.165 m/s.

Class 9 Science Chapter 8 - Force and Laws of Motion: Solved Exercise Questions

The exercise questions that are answered at the end of this chapter facilitate students in realising how the laws of Newton's motion, conservation of momentum, and effects of force can be applied in a single clear way. These solutions empower the problem-solving capabilities and also adequately prepare students towards examinations.

Q 1. An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.

Answer:

When a net zero external unbalanced force is applied and the object is in motion, then there will be a condition in which its motion is unopposed by any external force. Then, as a result object will continue to remain in motion. The velocity, in this case, will be constant and in a particular direction.

Q 2. When a carpet is beaten with a stick, dust comes out of it. Explain.

Answer:

During beating the carpet, we bring it into motion. But the dust particles have a tendency to be at rest, so they resist the motion. This is why they come out of the carpet. The factor responsible for this is inertia.

Q 3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?

Answer:

This is because the luggage has some mass, so the inertia. When the bus is in motion, the luggage is also in motion. But when the bus stops suddenly, the luggage still tends to move with the same velocity due to inertia. Thus, it may fall down. Similarly, when the bus starts from rest, the luggage tends to be at rest thus it is pushed in a backward direction.

Q 5. A 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 its mass is 7 tonnes (Hint: 1 tonne = 1000 kg.)

Answer:

From the equations of motion, we can write :

s = ut + 12at2

or 400 = 0×20 + 12×a×202

or a = 400200 = 2 m/s2

And the force acting on the truck is given by :

F=ma=7000×2

F = 14000 N

Q 6. A stone of 1 kg is thrown with a velocity of 20 m s-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?

Answer:

By the equation of motion, we know that,

v2 = u2 + 2as

or 02 = 202 + 2a×50 (Since the ball comes to rest, thus final velocity is zero.)

or a = 400100 = 4 m/s2

The acceleration is negative. This implies that the force is opposing the motion.

The force is given by :

F = ma

or F = 1× (4) = 4 N

Hence, the frictional force between the stone and the ice is 4 N.

Q 7. (a) A 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 the net accelerating force.

Answer:

The net accelerating force is given by :

Net accelerating force = Force by engine - Frictional force

= 40000 5000 = 35000 N

Q 7.(b) An 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 the acceleration of the train

Answer:

We know that the force is given by :

F = ma

or 35000 = (8000 + 10000)a

or a = 3500018000 = 1.944 m/s2

Thus acceleration of the train is 1.944 m/s2 .

Q 8. An 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 m s-2 ?

Answer:

By Newton's law of motion, we have :

F = ma

and F= 1500×1.7

or F = 2550 N

Thus the retarding force required is 2550 N in the direction opposite to the motion of the vehicle

Q 9. What is the momentum of an object of mass m, moving with a velocity v?

(a) (mv)2 (b) mv2 (c) 12mv2 (d) mv

Answer:

We know that the momentum of a body is given by:-

Momentum = Mass × Velocity

Thus we have : p = mv

Hence option (d) is correct.

Q 10. Using 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?

Answer:

By Newton's third law, it is clear that the frictional force acting on the cabinet will be 200 N. (Note that the limit of frictional force depends upon the mass of the body and the coefficient of friction between the bodies. So it might be possible that if the frictional force acting is less than 200 N.

Q 11. According 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.

Answer :

The student is correct. The force applied by him is less than the static friction present between the tyres of the truck and the road (as the mass of the truck is large, thus the frictional force is also high). Thus the force applied by the student is balanced by the static frictional force, which is the action-reaction pair.

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

Answer:

The change in the momentum is given by :

Δp = mΔv

or Δp= 0.2×(5 10)

or Δp= 3 Kg m/s

In the above equation, we have taken v = 5 m/s as its direction is opposite.

Q 13. A bullet of mass 10 g travelling horizontally with a velocity of 150 m s-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.

Answer:

By the equation of motion we know that:-

v = u + at

Putting values of v, u and t in the equation :

0 = 150 + a×0.03

or a = 1500.03

or a = 5000 m/s2

Now we have :

s = ut + 12at2

or s= 150×0.03 + 12×(5000)×0.032

or s= 2.25 m

Hence the distance of penetration of the bullet in the block is 2.25 m.

Now, for the force we have :

F = ma

or F= 0.01×(5000) = 50 N

Thus the retarding force acting on the bullet is 50 N.

Q 14. An object of mass 1 kg travelling in a straight line with a velocity of 10 m s-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 .

Answer:

For calculating the velocity of the combined mass, we need to use the law of conservation of momentum.

m1u1 + m2u2 = (m1 + m2)v

or 1×10 + 5×0 = (1 + 5)v

or v = 106

or v = 1.66 m/s

Now, the momentum before the collision is :

pi = m1u1 + m2u2

or pi= 1×10 + 5×0 = 10 Kg m/s

By momentum conservation, the final momentum is also 10 Kg m/s .

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

Answer:

The momentum is given by :

p = mv

Initial momentum is : pi = mu = 100×5 = 500 Kg m/s

And the final momentum is :

pf = mv = 100×8 = 800 Kg m/s

Also, the force is defined as rate of change of momentum.

Thus F = pf pit

or F= 800 5006 = 50 N

Hence the force exerted on the body is 50 N.

Q 16. Akhtar, 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 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

Answer:

Kiran’s observation:- The insect experiences a greater change in its velocity as the mass of the insect is much less compared to the motorcar, so it experiences a greater change in its momentum. Thus Kiran's observation is correct.

Akhtar’s observation:- We know that a motorcar has a larger velocity and has a bigger mass as compared to the insect. Also, the motorcar continues to move in the same direction even after the collision. This suggests that the change in momentum of a motorcar is very less, whereas the insect experiences a great change in its momentum.

Rahul’s observation:- The momentum gained by the insect is equal to the momentum lost by the motorcar (law of conservation of momentum). Thus his observation is also correct.

Q 17. How 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 m s-2 .

Answer :

For calculating momentum, we need the final velocity of the bell.

By equation of motion we can write :

v2 = u2 + 2as

or v2 = 02 + 2×10×0.8

or v = 4 m/s

Thus the momentum is : p = mv

or p= 10×4 = 40 Kg m/s

Force and Laws of Motion NCERT Solutions: Solved Additional Exercises

The additional exercises that have been solved offer additional practice with Newton's laws, inertia, momentum, and balanced-unbalanced forces, and students find additional conceptual understanding. The solutions improve with the application of analytical thinking and confidence to solve higher-level questions in the exams.

Q A1. (a) The following is the distance-time table of an object in motion: What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?

Time in secondsDistance in metre
00
11
28
327
464
5125
6216
7343

Answer:

From the table, the relation between time and distance can be seen.

s = t3

Thus the velocity of the particle is increasing with time.

v = dsdt = 3t2

and a = dvdt = 6t

Hence acceleration increases linearly with time.

Q A1. (b) The following is the distance-time table of an object in motion: What do you infer about the forces acting on the object?

Time in secondsDistance in metre
00
11
28
327
464
5125
6216
7343

Answer:

i)Velocity is the change in distance with time. So the velocity is increasing non uniformly, so the acceleration increases with time.

F = ma

ii) Thus, the net force acting on the body is also increasing.

Q A2. Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 m s-2 . With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.)

Answer:

From the question, it is clear that two people push the car with constant velocity. But when the third person pushes, it has some acceleration. Thus, the third person is responsible for the acceleration generated.

Force by the third person is given by :

F = ma

F= 1200×0.2 = 240 N

Hence the force from each man is 240 N.

Q A3. A hammer of mass 500 g, moving at 50 m s-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?

Answer:

Thus, the force is given by :

F=mvut=500×103(0500.01)=2500N

or F= 0.5×(5000) = 2500 N

The negative sign implies that the force of the nail on the hammer is in the opposite direction.

Q A4. A 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.

Answer:

The initial and final velocities are :

u = 90×518 = 25 m/s

and v = 18×518 = 5 m/s

Using the equation of motion, we can write :

v = u + at

or 5 = 25 + a×4

or a = 5 m/s2

Now the force is given by :

F = ma

or F= 1200×(5) = 6000 N

And the change in momentum is: Δp = mΔv

or Δp= 1200(20) = 24000 Kg m/s

Class 9 Science Chapter 8 - Force and Laws of Motion: Higher Order Thinking Skills (HOTS) Questions

The Force and Laws of Motion NCERT Solutions (HOTS Questions) is created to improve the level of analysis and reasoning and is not limited to the simplistic textbook-problems. These questions of Higher Order Thinking Skills will ask students to use the laws of Newton and the concept of force in real-life scenarios and complex cases. Such HOTS questions will assist the learners to think critically, understand the chapter better, and score better during the exams.

Q1:

The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle θ should be

Answer:

For block A, Tmg=m(o)
For block B, T-mg=m(o)
(Since both the A and B blocks are similar and the tension in the strings is mg. It has zero acceleration, as the system is in equilibrium.

So,2 Tcosθ=2mg2mgcosθ=2mgθ=45


Q2:

The figure shows the displacement of a particle going along the X−axis as a function of time. The force acting on the particle is zero in the region

Answer:

In the given displacement-time graph, in regions AB and CD, the graph is linear i.e. velocity (slope) is uniform i.e., constant, and we know that acceleration = change in velocity/time. Since velocity is constant, so change in velocity is 0, and acceleration is 0.

From Newton's second law of motion,

And F = m × a = 0


Q3:

Two particles of mass m each are tied at the ends of a light string of length 2a. The whole system is kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance a from the centre P (as shown in the figure). Now, the midpoint of the string is pulled vertically upwards with a small but constant force F. As a result, the particles move towards each other on the surface. Neglecting the earth's gravitation, the magnitude of acceleration, when the separation between them becomes 2x, is :

Answer:

The acceleration of mass m is due to the force Tcosθ

Tcosθ=maa=Tcosθm(i)Also, F=2TsinθT=F2sinθ(ii)
From(i)and(ii)

a=(F2sinθ)cosθm=F2mtanθ=F2mxa2x2[tanθa2x2x]


Q4:

There are cars with masses of 4 kg and 10 kg, respectively, that are at rest. A car having a mass of 10 kg moves towards the east with a velocity of 5 m/s. Find the velocity of the car with a mass of 4 kg with respect to the ground.

Answer:

Given,

m1 = 4 kg

m2 = 10 kg

v1 = ?

v2 = 5 m.s-1

We know from the law of conservation of momentum that,

pinitial = 0, as the cars are at rest

pfinal = p1 + p2

pfinal = m1.v1 + m2.v2

pfinal = 4 kg.v1 + 10 kg.5 m.s-1

pi = pf

0=4 kg.v1+50 kg.m.s-1

v1 = 12.5 m.s-1


Q5:

A body of mass 300g kept at rest breaks into two parts due to internal forces. One part of mass 200 g is found to move at a speed of 12 m/s towards the east. What will be the velocity of the other part?

Answer: Before it broke, the body was at rest. The linear momentum of the body was thus p=mv=0. The body breaks due to internal forces. As the external force acting on it is zero, its linear momentum will remain constant, i.e., zero. The linear momentum of the first part is p1=m1v1=(200 g)×(12 m/s), towards the east. For the total momentum to remain zero, the linear momentum of the other part must have the same magnitude and should be opposite in direction. Let its speed be v2 then its linear momentum is p2=m2v2=(100 g)×v2 Now, pi=pf(200 g)×(12 m/s)=(100 g)×v2v2=24 m/s

NCERT Solutions for Class 9 Science Chapter 8: Key Topics

Force and Laws of Motion cover all the key topics that explain how forces affect the motion of objects, including Newton’s laws, inertia, and momentum. These topics form the foundation of classical mechanics and are essential for understanding real-world motion.

8.1 Balanced And Unbalanced Forces
8.2 First Law of Motion
8.3 Inertia And Mass
8.4 Second Law of Motion
8.4.1 Mathematical Formulation of the Second Law of Motion
8.5 Third Law of Motion

NCERT Solutions for Class 9 Science Chapter 8: Important Formulas

This chapter highlights the key formulas based on Force and Newton’s Laws of Motion, making it easier for students to solve numerical problems accurately. These formulas act as quick revision tools and help in building a strong foundation for higher-level physics concepts.

1. Force

Force = Mass × Acceleration F=ma
Where:
F= force (in newtons, N )
m= mass (in kg)
a= acceleration (in m/s2 )

2. Momentum

Momentum = Mass × Velocity p=mv
Where:
p= momentum (in kgm/s )
m= mass (in kg)
v= velocity (in m/s)

3. Newton’s Second Law (in terms of momentum change)

F=ΔpΔt=mvmut
Where:
Δp= change in momentum
t= time interval
u= initial velocity, v= final velocity

4. Law of Conservation of Momentum

m1u1+m2u2=m1v1+m2v2

Approach to Solve Questions of Class 9 NCERT Chapter 8:Force and Laws of Motion

Class 9 Science Chapter 8 - Force and Laws of Motion describes the action of forces on the movement of objects as per Newton's laws of motion. The key elements that students should work on to effectively solve problems in this chapter include the development of a solid conceptual foundation, knowledge of real-life applications, and proper application of formulas. A systematic approach guarantees that both theory-based and numerical problems are clear in assisting students to perform well in exams.

  • Learn Newton's Three Laws:- Get to know the statement, meaning, and examples of each law (especially how they are used in real life).
  • Draw Free-Body Diagrams:- Use simple diagrams to illustrate the different forces acting on a body.
  • Find Out Given and Needed Values:- Label what is being given in the question (mass, force, acceleration) and what is to be found out.
  • Use the Right Formula for Force and Change in Momentum
  • Use Units Carefully:- Always report values in SI units (kg, m/s, N, etc.).
  • Verify Direction of Forces:- It is necessary to determine if the motion is one-way or includes counteracting forces, like friction. Revise Examples and Numericals. Solve the examples given in the NCERT and try similar problems.

NCERT Solutions for Class 9 Science Chapter-wise

NCERT Solutions of Class 9 Science have been organised and divided answers in chapters and assist the students in enhancing their knowledge of the key concepts in Physics, Chemistry, and Biology. These solutions, created according to the most recent CBSE syllabus, will include in-text questions, exercises, and numericals, as well as HOTS, which will allow for making the exam preparation more organised and efficient. They can be used as a sure source of revision, homework, and competitive exam preparation with detailed descriptions and step-by-step solutions.

NCERT Solutions for Class 9 - Subject Wise

Also Read,

Also, check NCERT Books and NCERT Syllabus here:

Frequently Asked Questions (FAQs)

Q: Why is the study of force significant in physics?
A:

The study of force enables us to understand why and how things move or remain stationary. It's the basis for the study of motion and mechanics.

Q: Why is it difficult to push a heavy stationary object?
A:

Due to its high inertia. Heavy bodies resist motion changes more than light bodies.

Q: Why are balanced forces not the same as unbalanced forces?
A:

Equilibrium in forces results in no change in the movement of an object, but unbalanced forces result in a change in movement or direction.

Q: Why do passengers fall forward when a moving bus suddenly stops?
A:

Due to inertia, the lower body which is in contact with the bus does not move anymore, but the upper body still shows signs of movement, and the person finds himself falling forward.

Q: Is it possible for an object to travel under a balanced force?
A:

Yes, if it was already uniformly moving, it will keep on moving with the same velocity. Equilibrium forces do not produce acceleration.

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