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

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

Edited By Vishal kumar | Updated on Sep 08, 2023 03:19 PM IST

## NCERT Solutions Class 9 Science Chapter 9 – Access and Download Free PDF

NCERT Solutions for Class 9 Science Chapter 9 Force And Laws Of Motion: Solution for force and laws of motion is part of NCERT Solutions for Class 9 Science. This Solution for class 9 chapter 9 science force and laws of motion are designed by the highly experienced faculty of Careeas360 in a very easy and systematic manner. By utilizing these NCERT solutions, students can prepare for exams even if they have not previously studied the material covered in the chapter. Students can download force and laws of motion class 9 pdf and use them according to their convenience which are free of cost.

This chapter will help you to understand the concepts of force and laws of motion. Have you ever thought about what causes a body to move? How can we stop a moving body? How can we change the direction of a body? The answer to all these three questions is by applying a force, With the help of chapter 9 science class 9, you will be able to understand the concepts of the chapter. Along with this chapter's solutions, you can also check NCERT solutions for other chapters and subjects.

CBSE class 9 chapter 9 science question answer Force and Laws of Motion, will help you to understand Newton's 3 laws of motion and another main topic like inertia. The class 9 science chapter 9 question answer given here as per the latest syllabus & exam pattern for the questions mentioned at the end of the chapter Force And Laws Of Motion will help you to revise the chapter.

### NCERT Solutions for Class 9 Science: Important Formulas and Diagrams + eBook link

Formulas for chapter 9 class 9 science are essential for understanding and solving problems involving the laws of motion and forces. They provide an easy-to-understand explanation of the mathematical relationships behind these topics. Here are some important formulas from the chapter on the laws of motion:

• Force (F) = mass (m) × acceleration (a) [SI unit of force = N]

• Momentum (p) = mass (m) × velocity (v) [SI unit of momentum=kg.m/s]

• Law of Conservation of Momentum

Total initial momentum = Total final momentum

Mathematically,

(m1*u1) + (m2*u2) = (m1*v1) + (m2*v2)

where:

m1 and m2 are the masses of the objects involved,u1 and u2 are the initial velocities of the objects, v1 and v2 are the final velocities of the objects.

The above-mentioned formulae are some of the more essential ones from Chapter 9 class 9 Science Force and Laws of Motion. Careers360 experts have compiled all of the crucial science formulas chapter by chapter. These formulas are extremely beneficial for exam preparation, revision, and answering difficult questions. Students can download the Force and Laws of Motion Class 9 pdf of chapter by chapter formula by clicking on the given below link and use them online/ offline according to their comfort for free of cost.

### NCERT Solutions for Class 9 Science Chapter 9 Force and Laws Of Motion - Important Topics

• Force: Understanding the concept of force, including its definition and effects on objects.
• Inertia: Studying the concept of inertia and its function in the resistance of objects to changes in their state of motion.

Newton's Laws of Motion: Studying Newton's three laws of motion:

a. First Law: Law of Inertia, which states that an object at rest or in motion will continue in that state unless acted upon by an external force.

b. Second Law: Relationship between force, mass, and acceleration, as described by the equation F = ma.

c. Third Law: Law of Action and Reaction, which states that every action has an equal and opposite reaction.

• Momentum: understanding momentum as the product of an object's mass and velocity, as well as its importance in terms of motion.
• Conservation of Momentum: Exploring the concept that the total momentum of a system remains constant unless external forces act upon it.
• Friction: Examining the force of friction and its different types, such as static friction and kinetic friction, and their effects on motion.
• Applications of Force and Laws of Motion: Understanding how the principles of force and laws of motion apply to real-life situations, such as the working of vehicles, sports activities, and safety measures.

This chapter has been renumbered as Chapter 8 in accordance with the CBSE Syllabus 2023–24.

## NCERT Solutions for Class 9 Science Chapter 9 Force and Laws of Motion: Solved In-text Questions

Class 9 Science Chapter 9 Solutions: Topic 9.3(Inertia and Mass)

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

(b) a bicycle and a train?

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

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

(a) In the case of rubber ball and stone of the same size, the stone will have greater mass so its inertia will be more as 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 rupees coin has more weight than one rupee coin so the inertia of five rupees coin is more.

“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.

Velocity has both magnitude and direction. Thus change in any of 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 velocity is changed.

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

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

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 leaves they still tend to be in rest. Thus force is acting on leaves vigorously with changing direction rapidly. This results in detaching of leaves from the tree.

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

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

Force And Laws of Motion Class 9 Topic 9.5

For motion, 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 cart applies backward force on the horse but due to unbalanced force, the horse moves in the forward direction.

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 higher force thus the reaction force on the hose is also large. This makes it difficult for a fireman to hold the hose.

Here we will use conservation of momentum :

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

$m_1u_1\ +\ m_2u_2\ =\ m_1v_1\ +\ m_2v_2$

or $4\times 0\ +\ 0.05\times 0\ =\ 4\times v_1\ +\ 0.05\times 35$

or $-4 \times v_1\ =\ 1.75$

or $v_1 \ =\ -\ 0.4375\ m/s$

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

In this question we will use conservation of momentum :

Initial momentum = Final momentum

$m_1u_1\ +\ m_2u_2\ =\ m_1v_1\ +\ m_2v_2$

or $0.1\times 2\ +\ 0.2\times 1\ =\ 0.1\times 1.67\ +\ 0.2\times v_2$

or $v_2\ =\ \frac{0.233}{0.2}$

or $v_2\ =\ 1.165\ m/s$

Hence required velocity is 1.165 m/s.

## Force and Laws of Motion: Solved Exercise Questions-

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.

During beating the carpet we bring it in the 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 .

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 fell down. Similarly, when the bus starts from the rest, the luggage tends to be in rest thus it is pushed in a backward direction.

(a) the batsman did not hit the ball hard enough.

(b) velocity is proportional to the force exerted on the ball.

(c) there is a force on the ball opposing the motion.

(d) there is no unbalanced force on the ball, so the ball would want to come to rest

The frictional force present between the ground and the ball acts in the opposite direction of the motion of the ball. Thus balls stop after travelling a few distances.

Hence option (c) is correct.

From the equations of motion we can write :

$s\ =\ ut\ +\ \frac{1}{2}at^2$

or $400\ =\ 0\times 20\ +\ \frac{1}{2}\times a\times 20^2$

or $a\ =\ \frac{400}{200}\ =\ 2\ m/s^2$

And the force acting on the truck is given by :

$F\ =\ ma$

$=\ 7000\times 2$

$F\ =\ 14000\ N$

By the equation of motion, we know that,

$v^2\ =\ u^2\ +\ 2as$

or $0^2\ =\ 20^2\ +\ 2a\times 50$ (Since ball comes to rest thus final velocity is zero.)

or $a\ =\ \frac{-400}{100}\ =\ -4\ m/s^2$

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

The force is given by :

$F\ =\ ma$

or $F\ =\ 1\times \ (-4)\ =\ -4\ N$

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

• the net accelerating force and

The net accelerating force is given by :

Net accelerating force = Force by engine - Frictional force

$=\ 40000\ -\ 5000\ =\ 35000\ N$

• the acceleration of the train

We know that the force is given by :

$F\ =\ ma$

or $35000\ =\ (8000\ +\ 10000)a$

or $a\ =\ \frac{35000}{18000}\ =\ 1.944\ m/s^2$

Thus acceleration of the train is $1.944\ m/s^2$ .

By Newton's law of motion, we have :

$F\ =\ ma$

and $=\ 1500\times 1.7$

or $F\ =\ -2550\ N$

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

(a) $(mv)^{2}$ (b) $mv^{2}$ (c) $\frac{1}{2}mv^{2}$ (d) $mv$

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

Momentum = Mass $\times$ Velocity

Thus we have : $p\ =\ mv$

Hence option (D) is correct.

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 frictional force acting is less than 200 N).

The velocity of the combined mass after the collision will be zero.

This can be found using the law of conservation of the momentum :

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v$

or $mu\ +\ m(-u)\ =\ (2 m)v$

or $2mv\ =\ 0$

Thus $v\ =\ 0$

The student is correct. The force applied by him is less than the static friction present between tyres of the truck and the road (as the mass of 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.

The change in the momentum is given by :

$\Delta p\ =\ m\Delta v$

or $=\ 0.2\times (-5\ -\ 10)$

or $=\ -3\ Kg\ m/s$

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

By the equation of motion we know that:-

$v\ =\ u\ +\ at$

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

$0\ =\ 150\ +\ a\times 0.03$

or $a\ =\ \frac{-150}{0.03}$

or $a\ =\ -5000\ m/s^2$

Now we have :

$s\ =\ ut\ +\ \frac{1}{2}at^2$

or $=\ 150\times 0.03\ +\ \frac{1}{2}\times (-5000)\times 0.03^2$

or $=\ 2.25\ m$

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

Now, for the force we have :

$F\ =\ ma$

or $=\ 0.01\times (-5000)\ =\ -50\ N$

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

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

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v$

or $1\times 10\ +\ 5\times 0\ =\ (1\ +\ 5)v$

or $v\ =\ \frac{10}{6}$

or $v\ =\ 1.66\ m/s$

Now, the momentum before the collision is :

$p_i\ =\ m_1u_1\ +\ m_2u_2$

or $=\ 1\times 10\ +\ 5\times 0\ =\ 10\ Kg\ m/s$

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

The momentum is given by :

$p\ =\ mv$

Initial momentum is : $p_i\ =\ mu\ =\ 100\times 5\ =\ 500\ Kg\ m/s$

And the final momentum is :

$p_f\ =\ mv\ =\ 100\times 8 \ =\ 800\ Kg\ m/s$

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

Thus $F\ =\ \frac{p_f\ -\ p_i}{t}$

or $=\ \frac{800\ -\ 500}{6}\ =\ 50\ N$

Hence the force exerted on the body is 50 N.

Kiran’s observation:- The insect experiences a greater change in its velocity as the mass of insects is much less as 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 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.

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

By equation of motion we can write :

$v^2\ =\ u^2\ +\ 2as$

or $v^2\ =\ 0^2\ +\ 2\times 10\times 0.8$

or $v\ =\ 4\ m/s$

Thus the momentum is : $p\ =\ mv$

or $=\ 10\times 4\ =\ 40\ Kg\ m/s$

## NCERT Class 9 Science Chapter 9 Exercise Solutions: Solved Additional Exercises

• What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?
 Time in seconds Distance in metre 0 0 1 1 2 8 3 27 4 64 5 125 6 216 7 343

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

$s\ =\ t^3$

Thus the velocity of the particle is increasing with time.

$v\ =\ \frac{ds}{dt}\ =\ 3t^2$

and $a\ =\ \frac{dv}{dt}\ =\ 6t$

Hence acceleration increases linearly with time.

• What do you infer about the forces acting on the object?
 Time in seconds Distance in metre 0 0 1 1 2 8 3 27 4 64 5 125 6 216 7 343

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

$F\ =\ ma$

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

From the question, it is clear that two-person pushes 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 :

$F\ =\ ma$

$=\ 1200\times 0.2\ =\ 240\ N$

Hence the force from each man is 240 N.

Thus the force is given by :

$\\F = m\frac{v-u}{t}\\=500\times10^{-3}(\frac{0-50}{0.01})=-2500N$

or $=\ 0.5\times (-5000)\ =\ -2500\ N$

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

The initial and final velocities are :

$u\ =\ 90\times \frac{5}{18}\ =\ 25\ m/s$

and $v\ =\ 18\times \frac{5}{18}\ =\ 5\ m/s$

Using the equation of motion we can write :

$v\ =\ u\ +\ at$

or $5\ =\ 25\ +\ a\times 4$

or $a\ =\ -5\ m/s^2$

Now the force is given by :

$F\ =\ ma$

or $=\ 1200\times (-5)\ =\ -\ 6000\ N$

And the change in momentum is : $\Delta p\ =\ m\Delta v$

or $=\ 1200(-20)\ =\ - 24000\ Kg\ m/s$

## Force And Laws of Motion Class 9 Topics:

topics for ncert solutions for class 9 science chapter 9 exercise questions are listed below:

9.1 Balanced and Unbalanced Forces

9.2 First Law of Motion

9.3 Inertia and Mass

9.4 Second Law of Motion

9.5 Third Law of Motion

9.6 Conservation of Momentum

## Key feathers Of NCERT solutions For Class 9 Science Chapter 9 Force And Laws of Motion

• Expertly Created Solutions: The NCERT solutions for Class 9 Science Chapter 9 are written by subject matter experts who have a thorough understanding of the subjects covered in the chapter. The solutions are correct, dependable, and adhere to the NCERT curriculum.
• Comprehensive Coverage: The class 9 science chapter 9 solutions cover all the important topics and questions mentioned in the chapter. They provide step-by-step explanations and solutions for better understanding and clarity.
• Easy to Understand: The class 9 science chapter 9 question answer are designed in a simple and concise manner, making them easier for students to understand. The language used is student-friendly, ensuring that you can effectively grasp the topics.
• Doubt Clarification: If you have any doubts or queries about the force and laws of motion class 9 numericals, you can seek assistance from subject matter experts via chat support. They are available 24 hours a day, seven days a week to provide clarifications and support, ensuring that your questions are immediately handled.
• Enhances Learning Experience: By utilizing the NCERT class 9 science chapter 9 exercise solutions, you can enhance your learning experience and develop a strong foundation in the concepts of force and laws of motion. The solutions provide additional practice and reinforcement of the chapter's content.
• Accessible Anytime: The ncert solutions for class 9 science chapter 9 exercise questions are easily accessible online, allowing you to study and revise at your own convenience. You can access them on various devices like smartphones, tablets, or computers.

Also Check -

NCERT Books and NCERT Syllabus here:

## NCERT Solutions For Class 9 Science- Chapter Wise

 Chapter No. Chapter Name Chapter 1 Matter in Our Surroundings Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Motion Chapter 9 Force and Laws of Motion Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15

## NCERT Science Exemplar Solutions Class 9 - Chapter Wise

 Chapter No. Chapter Name Chapter 1 Chapter 1 Matter in our Surroundings Chapter 2 Chapter 2 Is Matter Around Us Pure? Chapter 3 Chapter 3 Atoms and Molecules Chapter 4 Chapter 4 Structure of the Atom Chapter 5 Chapter 5 The Fundamental Unit of Life Chapter 6 Chapter 6 Tissues Chapter 7 Chapter 7 Diversity in Living Organisms Chapter 8 Chapter 8 Motion Chapter 9 Chapter 9 Forces and Laws of Motion Chapter 10 Chapter 10 Gravitation Chapter 11 Chapter 11 Work and Energy Chapter 12 Chapter 12 Sound Chapter 13 Chapter 13 Why do We Fall ill? Chapter 14 Chapter 14 Natural Resources Chapter 15 Chapter 15 Improvement in Food Resources

1. What are the important topics of the chapter Force And Laws Of Motion

All the NCERT book topics are important from Force And Laws Of Motion Class 9 Science chapter . The topics are

• Balanced and Unbalanced Forces
• First Law of Motion
• Inertia and Mass
• Second Law of Motion
• Third Law of Motion
• Conservation of Momentum
2. What is the acceleration of a body with constant velocity along a straight line?

Constant velocity means the change in velocity is zero. There for the acceleration is zero.

3. What is the name of Chapter 9 in Class 9 science?

The name of chapter 9 in class 9 science is "Force And Laws Of Motion".

4. What is the SI unit of Force?

The SI unit of force is Newton (N).

5. What are the 3 laws of motion Class 9?

The three laws of motion, as stated by Sir Isaac Newton, are:

• Newton's First Law: An object at rest will remain at rest, and an object in motion will remain in motion with a constant velocity (that is, it will continue moving at the same speed and in the same direction) unless acted upon by an external force.

• Newton's Second Law: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. In other words, F = ma, where F is the net force, m is the mass of the object, and a is the acceleration produced.

• Newton's Third Law: For every action, there is an equal and opposite reaction. This means that whenever one object exerts a force on another object, the second object exerts an equal and opposite force on the first object.

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