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Case Study Questions Class 9 Science Force and Laws of Motion

Case study questions class 9 science chapter 9 force and laws of motion.

CBSE Class 9 Case Study Questions Science Force and Laws of Motion. Important Case Study Questions for Class 9 Exam. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Force and Laws of Motion.

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CBSE Case Study Questions Class 9 Science – Force and Laws of Motion

(1 ) Newton’s first law of motion states that a body at rest will remain at rest position only and a body which is in motion continues to be in motion unless otherwise they are acted upon by an external force. In other words, all objects resist a changein their state of motion. In a qualitative way, the tendency of undisturbed objects to stayat rest or to keep moving with the same velocity is called inertia. This is why, the firstlaw of motion is also known as the law of inertia. Answer the following questions.

(i) The first law of motion is also known as

(a)law of inertia

(b)law of thermodynamics

(c)both a and b

(d)none of these

(ii) If no external force acts on object which is at rest. it will

(a)remain at rest

(b)start to move

(c)both a and b can possible

(iii) If no external force acts on moving object. it will

(a)stop moving

(b)continue to move with same speed in same direction

(c)changes its direction of motion

(iv) State Newton’s first law of motion.

(v) why Newton’s first law of motion is called law of inertia

Answer key -1

(iv) Newton’s first law of motion states that a body at rest will remain at rest position only and a body which is in motion continues to be in motion unless otherwise they are acted upon by an external force.

(v) All objects resist a change in their state of motion. In a qualitative way, the tendency of undisturbed objects to stay at rest or to keep moving with the same velocity is called inertia. This is why, the first law of motion is also known as the law of inertia.

(2) Two strings X and Y are tied to the two opposite faces of the block as shown in figure. If we apply a force by pulling the string X, the block begins to move to the right. Similarly, if we pull the string Y, the block moves to the left. But, if the block is pulled from both the sides with equal forces, the block will not move. Such forces are called balanced forces and do not change the state of rest or of motion of an object. Now, let us consider a situation in which two opposite forces of different magnitudes pull the block. In this case, the block would begin to move in the direction ofthe greater force. Thus, the two forces are not balanced and the unbalanced force acts in the direction the block moves. This suggests that an unbalanced force acting on an object brings it in motion. Force is push or pull.

(i) Force is nothing but

(c)both push or pulls

(d)none of the above

(ii) When balanced forces acting on moving object then

(a) Object continue to move with same speed

(b) Object will change its direction of motion

(d) None of the above

(iii) When unbalanced force acts on moving object opposite to direction of motion then

(b) Object will come to rest

(iv) Differentiate between balanced and unbalanced force. give 3 points each .

(v) From above diagram if one person pull from Y rope with 10N force and another person pull from X rope with 5N force. In which direction box will move? Is this a case of unbalanced force or balance force?

Answer key -2

(iv) Difference between balanced and unbalanced force

(3) The second law of motion is quantitative expression of force and it states that the rate of change of momentum of an object isproportional to the applied unbalanced force in the direction of force. Mathematically, F = ma, the unit of force is kg-m/s 2 or Newton,which has the symbol N. The second law ofmotion gives us a method to measure the force acting on an object as a product of its mass and acceleration.Answer the following questions.

(i) SI unit of force is

(b) Kg-m/s 2

(ii) The quantitative expression of force is given by

(a) First law of motion

(b) Second law of motion

(iii) Force is directly proportional to

(a) Acceleration of object

(b) Time for which force acts on object

(iv) State second law of motion. State whether it is scalar or vector quantity

(v) Differentiate between first law and second law of motion.(give 3 points)

Answer key -3

(iv) The second law of motion is quantitative expression of force and it states that the rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of force. Mathematically, F = ma, the unit of force is kg-m/s 2 or Newton.

(v) Difference between first law and second law of motion is given by

(4) The third law of motion states that when one object exerts a force on another object, the second object instantaneously exerts a force back on the first. These two forces are always equal in magnitude but opposite in direction. These forces act on different objects and neveron the same object. It is important to note that even though the action and reaction forces are always equal in magnitude; these forces may not produce accelerations of equal magnitudes, this is because each force acts on a different object that may have a different mass.The two opposing forces are also known as action and reaction forces. Answer the following questions.

(i) Action reaction forces are always

(a) Equal and in same direction

(b) Equal and in opposite direction

(ii) Which of the following is correct about action reaction forces?

(a) They act on different objects

(b) They are equal in magnitude and opposite in direction

(d) All the above

(iii) State third law of motion

(iv) Give 5 examples of third law of motion

(v) Even though the action and reaction forces are always equal in magnitude; these forces may not produce accelerations of equal magnitudes. Give your justification on this statement

Answer key -4

(iii) The third law of motion states that when one object exerts a force on another object, the second object instantaneously exerts a force back on the first. These two forces are always equal in magnitude but opposite in direction. These forces act on different objects and neveron the same object.

(iv) Examples of third law of motion are

Swimming or rowing a boat. • Static friction while pushing an object. • Walking. • Standing on the ground or sitting on a chair. • The upward thrust of a rocket. • Resting against a wall or tree.

(v) Even though the action and reaction forces are always equal in magnitude; these forces may not produce accelerations of equal magnitudes because these action reaction forces are acting on two different objects having different masses that’s why they are acceleration with different magnitude.

(5) The sum of momentum of the two objects before collision is equal to the sum of momentum after the collision provided there is no external unbalanced force acting on them. This is known as the law of conservation of momentum. This statement can alternatively be given as the total momentum of the two objects is unchanged or conserved by the collision. Law of conservation of momentum is applicable to system of particle. Answer the following questions.

(i)Law of conservation of momentum is applicable to

(a) A system of particles

(b) Only for 2 particles

(ii) Law of conservation of momentum holds good provided that

(a) There should be external unbalanced force acting on particles

(b) There should not be any external unbalanced force acting on particles

(iii)The total momentum of the two objects when collision occurs is

(a) Changed

(b) Remains conserved

(iv) State law of conservation of momentum.

(v) If action and Reaction are equal in magnitude and opposite in direction then why they do not cancel each other?

Answer key -5

(iv) The sum of momentum of the two objects before collision is equal to the sum of momentum after the collision provided there is no external unbalanced force acting on them. This is known as the law of conservation of momentum. This statement can alternatively be given as the total momentum of the two objects is unchanged or conserved by the collision.

(v) Action and Reaction are equal in magnitude and opposite in direction but they do not cancel each other because they are not action on sane object. As these forces are acting on different object hence produces different acceleration and does not cancel each other.

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Case Study Questions of Chapter 9 Force and Laws of Motion PDF Download

Case study Questions on Class 9 Science Chapter 9 are very important to solve for your exam. Class 9 Science Chapter 9 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based questions for Class 9 Science Chapter 9 Force and Laws of Motion

In CBSE Class 9 Science Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Force and Laws of Motion Case Study Questions With answers

Here, we have provided case-based/passage-based questions for Class 9 Science Chapter 9 Force and Laws of Motion

Case Study/Passage-Based Questions

Question 1:

The sum of the momentum of the two objects before the collision is equal to the sum of momentum after the collision provided there is no external unbalanced force acting on them. This is known as the law of conservation of momentum. This statement can alternatively be given as the total momentum of the two objects is unchanged or conserved by the collision. The Law of conservation of momentum is applicable to the system of particles. Answer the following questions.

(i)Law of conservation of momentum is applicable to

(a) A system of particles

(b) Only for 2 particles

(d) None of the above

Answer: (a) A system of particles

(ii) Law of conservation of momentum holds good provided that

(a) There should be external unbalanced force acting on particles

(b) There should not be any external unbalanced force acting on particles

Answer: (b) There should not be any external unbalanced force acting on particles

(iii)The total momentum of the two objects when collision occurs is

(a) Changed

(b) Remains conserved

Answer: (b) Remains conserved

(iv) State law of conservation of momentum.

Answer: The sum of momentum of the two objects before collision is equal to the sum of momentum after the collision provided there is no external unbalanced force acting on them. This is known as the law of conservation of momentum. This statement can alternatively be given as the total momentum of the two objects is unchanged or conserved by the collision.

(v) If action and Reaction are equal in magnitude and opposite in direction then why they do not cancel each other?

Answer: Action and Reaction are equal in magnitude and opposite in direction but they do not cancel each other because they are not action on sane object. As these forces are acting on different object hence produces different acceleration and does not cancel each other.

Question 2:

The third law of motion states that when one object exerts a force on another object, the second object instantaneously exerts a force back on the first. These two forces are always equal in magnitude but opposite in direction. These forces act on different objects and never on the same object. It is important to note that even though the action and reaction forces are always equal in magnitude; these forces may not produce accelerations of equal magnitudes, this is because each force acts on a different object that may have a different mass. The two opposing forces are also known as action and reaction forces. Answer the following questions.

(i) Action reaction forces are always

(a) Equal and in the same direction

(b) Equal and in the opposite direction

Answer: (b) Equal and in the opposite direction

(ii) Which of the following is correct about action reaction forces?

(a) They act on different objects

(b) They are equal in magnitude and opposite in direction

(d) All the above

Answer: (d) All the above

(iii) State third law of motion

Answer: The third law of motion states that when one object exerts a force on another object, the second object instantaneously exerts a force back on the first. These two forces are always equal in magnitude but opposite in direction. These forces act on different objects and neveron the same object.

(iv) Give 5 examples of third law of motion

Answer: Examples of third law of motion are Swimming or rowing a boat. •Static friction while pushing an object. •Walking. •Standing on the ground or sitting on a chair. •The upward thrust of a rocket. •Resting against a wall or tree.

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case study questions on laws of motion class 9

9th Standard CBSE

Class 9th Science - Motion Case Study Questions and Answers 2022 - 2023

Class 9th Science - Motion Case Study Questions and Answers 2022 - 2023 Study Materials Sep-09 , 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 9th Science Subject - Motion, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

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Motion case study questions with answer key.

Final Semester - June 2015

(ii) Find the Velocity of the boy.

(iii) A boy is sitting on a merry-go-round which is moving with a constant speed of 10m/s. This means that the boy is :

(iv) In which of the following cases of motion, the distance moved and the magnitude of displacement are equal ?

(ii) How far does it travel in 1 second ?

(iii) How far does it travel in 6 seconds ?

(iv) How long does it take to travel 240 m ?

(v) Which of the following statement is correct regarding velocity and speed of a moving body? (a) velocity of a moving body is always higher than its speed (b) speed of a moving body is always higher than its velocity (c) speed of a moving body is its velocity in a given direction (d) velocity of a moving body is its speed in a given direction

(ii) What type of motion is represented by BC ?

(iii) Find out the acceleration of the body.

(iv) Calculate the retardation of the body.

(v) Find out the distance travelled by the body from A to B.

*****************************************

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Class 9 Science Case Study Questions Chapter 9 Force and Laws of Motion

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Case study Questions in Class 9 Science Chapter 9 are very important to solve for your exam. Class 9 Science Chapter 9 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving Class 9 Science Case Study Questions Chapter 9 Force and Laws of Motion

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Force and Laws of Motion Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 9 Science Chapter 9 Force and Laws of Motion

Case Study/Passage-Based Questions

Case Study 1: The sum of the momentum of the two objects before the collision is equal to the sum of momentum after the collision provided there is no external unbalanced force acting on them. This is known as the law of conservation of momentum. This statement can alternatively be given as the total momentum of the two objects is unchanged or conserved by the collision. The Law of conservation of momentum is applicable to the system of particles. Answer the following questions.

(i)Law of conservation of momentum is applicable to

(a) A system of particles

(b) Only for 2 particles

(d) None of the above

Answer: (a) A system of particles

(ii) Law of conservation of momentum holds good provided that

(a) There should be external unbalanced force acting on particles

(b) There should not be any external unbalanced force acting on particles

Answer: (b) There should not be any external unbalanced force acting on particles

(iii)The total momentum of the two objects when collision occurs is

(a) Changed

(b) Remains conserved

Answer: (b) Remains conserved

(iv) State law of conservation of momentum.

(v) If action and Reaction are equal in magnitude and opposite in direction then why they do not cancel each other?

Case Study 2: The third law of motion states that when one object exerts a force on another object, the second object instantaneously exerts a force back on the first. These two forces are always equal in magnitude but opposite in direction. These forces act on different objects and never on the same object. It is important to note that even though the action and reaction forces are always equal in magnitude; these forces may not produce accelerations of equal magnitudes, this is because each force acts on a different object that may have a different mass. The two opposing forces are also known as action and reaction forces. Answer the following questions.

(i) Action reaction forces are always

(a) Equal and in the same direction

(b) Equal and in the opposite direction

Answer: (b) Equal and in the opposite direction

(ii) Which of the following is correct about action reaction forces?

(a) They act on different objects

(b) They are equal in magnitude and opposite in direction

(d) All the above

Answer: (d) All the above

(iii) State third law of motion

(iv) Give 5 examples of third law of motion

Case Study 3:

Force is a push or pull that can change the state of motion of an object. According to Newton’s first law of motion, an object at rest will remain at rest, and an object in motion will continue moving with a constant velocity in a straight line unless acted upon by an external force. This is known as the law of inertia. Newton’s second law of motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, this can be expressed as F = ma, where F is the force, m is the mass of the object, and a is the acceleration produced. Newton’s third law of motion states that for every action, there is an equal and opposite reaction. This means that whenever an object exerts a force on another object, the second object exerts an equal and opposite force on the first object. Understanding the concepts of force and the laws of motion helps us explain the behavior of objects and the factors that influence their motion.

What is force? a) A change in the state of motion of an object b) A push or pull that can change the state of motion of an object c) The mass of an object d) The velocity of an object Answer: b) A push or pull that can change the state of motion of an object

What does Newton’s first law of motion state? a) An object at rest will remain at rest, and an object in motion will continue moving with a constant velocity in a straight line unless acted upon by an external force. b) The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. c) For every action, there is an equal and opposite reaction. d) The force exerted by an object is equal to its mass multiplied by its acceleration. Answer: a) An object at rest will remain at rest, and an object in motion will continue moving with a constant velocity in a straight line unless acted upon by an external force.

What is Newton’s second law of motion? a) An object at rest will remain at rest, and an object in motion will continue moving with a constant velocity in a straight line unless acted upon by an external force. b) The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. c) For every action, there is an equal and opposite reaction. d) The force exerted by an object is equal to its mass multiplied by its acceleration. Answer: b) The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

What is Newton’s third law of motion? a) An object at rest will remain at rest, and an object in motion will continue moving with a constant velocity in a straight line unless acted upon by an external force. b) The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. c) For every action, there is an equal and opposite reaction. d) The force exerted by an object is equal to its mass multiplied by its acceleration. Answer: c) For every action, there is an equal and opposite reaction.

How do the concepts of force and the laws of motion help us? a) Explain the behavior of objects and the factors that influence their motion. b) Calculate the speed of objects. c) Classify objects into different categories. d) Determine the position of objects. Answer: a) Explain the behavior of objects and the factors that influence their motion.

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NCERT Solutions for Class 9 Science Chapter 9 - Force And Laws Of Motion
NCERT Solutions

NCERT Solutions for Class 9 Science Chapter 9 - Free PDF Download

Delve into the realm of physics with NCERT Solutions for Class 9 Science Chapter 9 - 'Force and Laws of Motion.' This chapter serves as a gateway to understanding the fundamental principles that govern motion and the forces behind it. Here, we offer free PDF downloads of comprehensive NCERT solutions to assist your learning journey. These solutions are thoughtfully curated to help you grasp concepts like force, inertia, and momentum, setting a strong foundation in physics. Beyond exam readiness, this resource nurtures a deeper understanding of the laws that shape the dynamic world. Explore and empower yourself with our Class 9 Science Chapter 9 PDF downloads.

A Glance About the Topic Force and Laws of Motion

Newton's first law of motion states that, An object in the rest of the object in motion will always remain constant unless some external unbalanced force is applied on it.

Newton’s second law of motion states that the acceleration of the object completely depends on the force acting upon it and the mass of an object.

Newton’s third law of motion states that, when the two objects interact with each other, they both apply a force of equal magnitude in opposite directions.

According to the Conservation of Momentum, the total momentum of the closed system remains constant until the external force is applied to it.

NCERT Solutions Class 9 Science answers are like an all-time available solution to the problems of students. With these solutions, you can prepare easily for a class test or the final exam. These NCERT Solutions present the entire chapter in an organized way to make students confident about the chapter. Even if a student has never read the chapter, by going through these NCERT solutions, they will be ready to face the exams. NCERT Solutions Science Class 9 Chapter 9 is prepared in an easy and understandable language by subject experts at Vedantu and is also easily accessible.

Access NCERT Solutions For Class 9 Science Chapter 9 – Force and Laws of Motion

INTEXT EXERCISE 1

1. Which of the following has more inertia:

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

Ans: The inertia of an object is measured by its mass. Heavier or more massive objects offer larger inertia.

Stone is heavier than the rubber ball of the same size. e. Hence, inertia of the stone is greater than that of a rubber ball.

(b) a bicycle and a train?

Ans: Train is heavier than bicycle. Hence, inertia of the train is greater than that of the

Ans: A five rupee coin is heavier than a one rupee coin. . Hence, inertia of the five rupee coin is greater than that of the one-rupee coin.

2. 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. Ans: The ball's velocity changes four times.

First change: The ball's speed changes from 0 to a specific amount as the football player kicks it. value. As a result, the ball's velocity is altered.

Second change:Another player is kicking the ball to the goal post in the second change. As a result of this, the direction of the ball is changed. As a result, its speed varies. In this case, the player used force. to change the velocity of the ball.

Third change: The ball is being collected by the goalie in the third change. The ball finally comes to a halt. As a result, its speed is lowered to zero from a specific value The pace of the ball has changed. In this situation, the goalie utilised a counter-force to slow down or modify the pace of the ball.

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

Ans : Because of the inertia of rest, when the branch is quickly moved, the leaves attached to it tend to stay in their resting position. The leaves and branch junctions are put under a lot of stress as a result of this. This strain causes some leaves to detach off the branch .

Fourth change-The goalkeeper kicks the ball to his teammates. As a result, the ball's velocity increases from zero to a certain number. As a result, its velocity shifts once more. In this case, the goalkeeper used force to change the ball's velocity.

4. Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest? Ans: We move in the forward direction when a moving bus is braking because our upper portion of the body and the bus are both in motion when the bus is moving, and when the bus is breaking our body is trying to be in motion due to inertia of motion and thereby we experience a forward push. Similarly, when the bus accelerates from the rest, the passenger tends to fall backwards. This is because the passenger's inertia tends to oppose the bus's forward motion when the bus accelerates. Therefore, when the bus accelerates, the passenger tends to fall backwards.

INTEXT EXERCISE 2

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

Ans: With his foot, a horse pushes the earth in a rearward way. According to Newton's third law of motion, the Earth exerts a reaction force on the horse in the forward direction. As a result, the cart advances.

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

Ans: When a significant volume of water is discharged from a hose at a high velocity, Newton's Third Law of Motion states that the water pushes the hose backwards with the same force. As a result, gripping a hose that ejects a significant volume of water at a rapid rate is difficult for a firefighter.

3. 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.\[\]

Mass of the rifle, ${m_1} = 4$ kg

Mass of the bullet, ${m_2} = 50$ g $ = 0.05$ kg

Recoil velocity of the rifle $ = {v_1}$

Initial velocity of bullet, ${v_2} = 35$ m/s

Ans: As, the riffle is at rest, its initial velocity, $v = 0$

Total initial momentum of the rifle and bullet system $ = \left( {{m_1} + {m_2}} \right)v = 0$

Total momentum of the rifle and bullet system after firing:

$ = {m_1}{v_1} + {m_2}{v_2}$

$ = 4\left( {{v_1}} \right) + 0.05 \times 35$

$ = 4{v_1} + 1.75$

According to the law of conservation of momentum,

Total momentum after the firing = Total momentum before the firing

$4{v_1} + 1.75 = 0$

$4{v_1} = - 1.75$

${v_1} = \dfrac{{ - 1.75}}{4}$

${v_1} = - 0.4375$ m/s

The negative sign indicates that the rifle recoils backwards with a velocity ${v_1} = - 0.4375$ m/s

4. 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.

Mass of one of the objects, ${m_1} = 100$ g $ = 0.1$ kg

Mass of the other object, ${m_2} = 200$ g $ = 0.2$ kg

Velocity of m 1 before collision, ${v_1} = 2$ m/s

Velocity of m 2 before collision, ${v_2} = 1$ m/s

Velocity of m 1 after collision, ${v_3} = 1.67$ m/s

Velocity of m 2 after collision $ = {v_4}$

According to the law of conservation of momentum:

Total momentum before collision $ = $ Total momentum after collision

${m_1}{v_1} + {m_2}{v_2} = {m_3}{v_3} + {m_4}{v_4}$

$\left( {0.1} \right)2 + \left( {0.2} \right)1 = \left( {0.1} \right)1.67 + \left( {0.2} \right){v_4}$

$0.2 + 0.2 = 0.167 + 0.2{v_4}$

$0.4 = 0.167 + 0.2{v_4}$

$0.4 - 0.167 = 0.2{v_4}$

$0.233 = 0.2{v_4}$

${v_4} = \dfrac{{0.233}}{{0.2}}$

${v_4} = 1.165$ m/s

Hence, the velocity of the second object becomes $1.165$ m/s after the collision.

NCERT EXERCISE

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.

Ans: Yes. An object can travel at a non-zero velocity even if it has a net zero external unbalanced force. This is only possible if the item moves at a consistent speed in a specified direction. As a result, the body is not subjected to any net imbalanced forces. The item will continue to travel at a velocity greater than zero. A net non-zero external unbalanced force must be supplied to the item to change its state of motion.

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

Ans: Using a stick to beat a carpet; causing the carpet to move quickly, while dust particles trapped in the carpet's pores prefer to stay still, since inertia of an item resists any change in its state of rest or motion. The dust particles, according to Newton's first rule of motion, remain at rest as the carpet moves. As a result, dust particles emerge from the carpet.

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

Ans: According to Newton's First Law of Motion, luggage on a bus' roof tends to maintain its condition of rest when the bus is at rest and retain its state of motion when the bus is in motion. When the bus starts moving again after a period of rest, luggage on the roof may fall down to maintain the resting spot. Similarly, owing to inertia of motion, luggage on the roof top of a moving bus will tumble forward when it arrives in the rest state. To avoid this, any luggage kept on a bus's roof should be tied with a rope.

4. A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because

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

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

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

Ans: Option(c). When the ball moves on the ground, the force of friction opposes its movement and after some time ball comes to a state of rest.

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 $7$ metric tonnes (Hint: $1$ metric tonne $ = 1000$ kg).

Given:

Initial velocity of the truck , $u = 0$ (since the truck is initially at rest)

Distance travelled, s $ = 400$ m

Time taken, t $ = 20$ s

According to the second equation of motion:

$s = ut + \dfrac{1}{2}a{t^2}$

$400 = 0 + \dfrac{1}{2}a{\left( {20} \right)^2}$

$400 = \dfrac{1}{2}a\left( {400} \right)$

$400 = a\left( {200} \right)$

\[a = \dfrac{{400}}{{200}}\]

\[a = 2\] m/s 2

\[1\] metric tonne \[ = 1000\] kg

\[\therefore 7\] metric tonnes \[ = 7000\] kg

Mass of truck, \[m = 7000\] kg

From Newton’s second law of motion:

Force, F = Mass × Acceleration

F = ma

F= \[ = 7000 \times 2\]

F \[ = 14000\] N

Hence, the acceleration of the truck is \[2\] m/s 2 and the force acting on the truck F \[ = 14000\] N

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?

Initial velocity of the stone, u \[ = 20\] m/s

Final velocity of the stone, v \[ = 0\] (finally the stone comes to rest)

Distance covered by the stone, s \[ = 50\] m

According to the third equation of motion: \[\]

${v^2} = {u^2} + 2as$

${0^2} = {\left( {20} \right)^2} + 2 \times a \times 50$

$0 = 400 + 100a$

$100a = - 400$

$a = - \dfrac{{400}}{{100}}$

$a = - 4$

a = −4 $\dfrac{m}{{{s^2}}}$

The negative sign indicates that acceleration is acting against the motion of the stone.

Mass of the stone, m $ = 1$ kg

F $ = 1 \times - 4$

F $ = - 4$ N

Hence, the force of friction between the stone and the ice F $ = - 4$ N .

7. 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:

(a) the net accelerating force;

Force exerted by the engine, F $ = 40000$ N

Frictional force offered by the track, ${F_{fraction}} = 5000$ N

Net accelerating force,

${F_{net}} = F - {F_{friction}}$

\[{F_{net}} = 40000 - 5000\]

\[{F_{net}} = 35000\] N

Hence, the net accelerating force \[{F_{net}} = 35000\] N

(b) the acceleration of the train; and

The engine exerts a force of \[40000\] N on all the five wagons.

Net accelerating force on the wagons, \[{F_{net}} = 35000\] N

Mass of a wagon \[ = 2000\] kg

Number of wagons \[ = 5\]

Total Mass of the wagons,

m = Mass of a wagon × Number of wagons

m \[ = 2000 \times 5\]

m \[ = 10000\] kg

Mass of the engine, m′ \[ = 8000\] kg

Total mass, M = m + m′

\[ = 10000 + 8000\]

\[ = 18000\] kg

\[Fa = Ma\]

\[a = \dfrac{{Fa}}{m}\]

\[a = \dfrac{{35000}}{{18000}}\]

\[a = 1.944\] m/s 2

Hence, the acceleration of the wagons and the train \[a = 1.944\] m/s 2

Ans:

The force of wagon 1 on wagon 2 = mass of four wagons x acceleration

Mass of 4 wagons

\[ = 4 \times 2000\]

\[ = 8000\] kg

F \[ = 8000\] kg \[ \times 1.944\] m/s 2

F \[ = 1552\] N

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 \[1.7\] \[m{s^{ - 2}}\]?

Mass of the automobile vehicle, m \[ = 1500\] kg

Final velocity, \[v = 0\]

Acceleration of the automobile, a \[ = - 1.7\] \[m{s^{ - 2}}\]

From Newton’s second law of motion,

Force = Mass × Acceleration

\[ = 1500 \times \left( { - 1.7} \right)\]

\[ = - 2550\] N

Hence, the force between the automobile and the road \[ = - 2550\] N.

Negative sign shows that the force is acting in the opposite direction of the vehicle.

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

Mass of the object \[ = m\]

Velocity \[ = v\]

Momentum = Mass × Velocity

Momentum \[ = mv\]

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?

A same amount of force will act in the opposite direction, according to Newton's third law of motion.

Friction is the name of this force. As a result, the cabinet is subjected to a \[200\] N frictional force.

11. Two objects, each of mass \[1.5\] kg are moving in the same straight line but in opposite directions. The velocity of each object is \[2.5\] m s−1 before the collision during which they stick together. What will be the velocity of the combined object after collision?

Mass of first object , m 1 \[ = 1.5\] kg

Mass of second object , m 2 \[ = 1.5\] kg

Velocity of m 1 before collision, v 1 \[ = 2.5\] m/s

Velocity of m 2 , (moving in opposite direction ) before collision, v 2 \[ = - 2.5\] m/s

After collision, the two objects stick together.

Total mass of the combined object \[ = {m_1} + {m_2}\]

\[ = 1.5\] kg \[ + 1.5\] kg

\[ = 3\] kg

Velocity of the combined object \[ = v\]

Total momentum before collision = Total momentum after collision

\[{m_1}{v_1} + {m_2}{v_2} = ({m_1} + {m_2})v\]

\[ \Rightarrow 1.5 \times 2.5 + 1.5\left( { - 2.5} \right) = \left( {1.5 + 1.5} \right)v\]

\[ \Rightarrow 3.75 - 3.75 = 3v\]

\[ \Rightarrow v = 0\]

Hence, the velocity of the combined object after collision \[v = 0\] m/s.

12. 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.

The static friction force is quite strong due to the truck's massive bulk. Because the student's effort is insufficient to overcome the static friction, the truck cannot be moved. In this circumstance, the net imbalanced force in either direction is zero, which explains why there is no movement. The force exerted by the learner and the force exerted owing to static friction cancel each other out.

As a result, the student is correct in claiming that the two equal and opposing forces cancel each other out.

13. 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 change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

Ans: Mass of the hockey ball, m \[ = 200\] g \[ = 0.2\] kg

velocity of the ball , \[{v_1} = 10\] m/s

Initial momentum \[ = m{v_1}\]

velocity of the ball after struck by the stick, \[{v_2} = - 5\] m/s

Final momentum \[ = m{v_2}\]

Change in momentum

\[ = m{v_1} - m{v_2}\]

\[ = m\left( {{v_1} - {v_2}} \right)\]

\[ = 0.2\left( {10 - \left( { - 5} \right)} \right)\]

\[ = 0.2 \times 15\]

\[ = 3\] kg \[m{s^{ - 1}}\]

Hence, the change in momentum of the hockey ball \[ = 3\] kg \[m{s^{ - 1}}\]

14. 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.

Initial velocity of the bullet , u \[ = 150\] m/s

Final velocity, \[v = 0\] Time, \[t = 0.03\] s

According to the first equation of motion, \[v = u + at\]

Acceleration of the bullet, a

\[0 = 150 + \left( {a \times 0.03s} \right)\]

\[a = - \dfrac{{150}}{{0.03}}\]

\[a = - 5000\] m/s 2

(Negative sign indicates that the velocity of the bullet is decreasing.)

According to the third equation of motion:

\[{v^2} = {u^2} + 2as\]

\[{0^2} = {\left( {150} \right)^2} + 2\left( { - 5000} \right)s\]

\[0 = 22,500 - 10000s\]

\[10000s = 22,500\]

\[s = \dfrac{{22,500}}{{10000}}\]

\[s = 2.25\] m

Hence, the distance of penetration of the bullet into the block \[s = 2.25\] m

Mass of the bullet, m \[ = 10\] g \[ = 0.01\] kg

Acceleration of the bullet, a \[ = - 5000\] \[\dfrac{m}{{{s^2}}}\]

\[ = 0.01 \times - 5000\]

\[ = - 50\] N

Hence, the magnitude of force exerted by the wooden block on the bullet \[ = - 50\] N

but it acts in opposite direction.

15. 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.

Mass of the object, \[{m_1} = 1\] kg

Velocity of the object before collision, \[{v_1} = 10\] m/s

Mass of the wooden block, \[{m_2} = 5\] kg

Velocity of the wooden block before collision, \[{v_2} = 0\] m/s

∴ Total momentum before collision

$ = {m_1}{v_1} + {m_2}{v_2}$

\[ = 1\left( {10} \right) + 5(0)\]

\[ = 10\] kg \[m{s^{ - 1}}\]

It is given that after collision, the object and the wooden block stick together.

Total mass of the combined system,

\[m = {m_1} + {m_2}\]

\[ = 1\] kg \[ + 5\] kg

\[ = 6\] kg

Total momentum before collision \[ = \] Total momentum after collision

$ \Rightarrow {m_1}{v_1} + {m_2}{v_2}$ $ = \left( {{m_1} + {m_2}} \right)v$

$ \Rightarrow 1\left( {10} \right) + 5\left( 0 \right) = \left( {1 + 5} \right)v$

$ \Rightarrow 10 = 6v$

$ \Rightarrow v = \dfrac{{10}}{6}$

$ \Rightarrow v = \dfrac{5}{3}$ m/s

$v = 1.66$ m/s

Total momentum after collision

\[{m_1}v + {m_2}v\]

\[ = v\left( {{m_1} + {m_2}} \right)\]

\[ = 10\left( {6 \times 6} \right)\]

\[ = 10\] kg m/s

The total momentum after collision is also \[10\] kg m/s.

Total momentum just before the impact \[ = 10\] kg m/s .

Total momentum just after the impact \[ = 10\] kg m/s .

Hence, velocity of the combined object after collision \[ = \dfrac{5}{3}\] m/s .

16. 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.

Initial velocity of the object, u \[ = 5\] m/s

Final velocity of the object, v \[ = 8\] m/s

Mass of the object, m \[ = 100\] kg

Time taken by the object to accelerate, t \[ = 6\] s

Initial momentum \[ = \] mu

\[ = 100 \times 5\]

\[ = 500\] kg \[m{s^{ - 1}}\]

Final momentum \[ = \] mv

\[ = 100 \times 8\]

\[ = 800\] kg \[m{s^{ - 1}}\]

Force exerted on the object,

F \[ = \] mv-mu/t

F \[ = \left( {\dfrac{{800 - 500}}{6}} \right)\]

F \[ = \dfrac{{300}}{6}\]

F \[ = 50\] N

Initial momentum of the object is \[500\] kg \[m{s^{ - 1}}\] .

Final momentum of the object is \[800\] kg \[m{s^{ - 1}}\] .

Force exerted on the object is \[50\] N.

17. Akhtar, Kiran and Rahul were riding in a motor car 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 a 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.

As a result, the vehicle and insect systems have no change in momentum.

In this case, the insect experiences a bigger change in velocity, which results in a greater shift in momentum. Kiran's assessment is correct from this perspective.

The motorcar travels at a faster speed and has a bigger mass than the insect.

Furthermore, the motorcar continues to travel in the same direction after the collision, indicating that the motorcar has the least amount of momentum change, whilst the insect has the most. As a result, Akhtar's statement is likewise correct.

Because the momentum acquired by the bug is equal to the momentum lost by the motorcar, Rahul's observation is likewise true. This is also in agreement with the conservation of momentum law. However, he committed an error since the system suffers from a flaw. Because the momentum before the collision is identical to the momentum after the impact, there is no change in momentum following the accident.

18. How much momentum will a dumbbell of mass \[10\] kg transfer to the floor if it falls from a height of \[80\] cm? Take its downward acceleration to be \[10\] \[m{s^{ - 2}}\] .

Mass of the dumbbell, m \[ = 10\] kg

Distance covered by the dumbbell, s \[ = 80\] cm \[ = 0.8\] m

Acceleration in the downward direction, a \[ = 10\] \[\dfrac{m}{{{s^2}}}\]

Initial velocity of the dumbbell, u \[ = 0\]

Final velocity of the dumbbell v = ?

\[{v^2} = {u_2} + 2as\]

\[{v^2} = 0 + 2\left( {10} \right)0.8\]

\[{v^2} = 20 \times 0.8\]

\[{v^2} = 16\]

\[v = \sqrt {16} \]

\[v = 4\] m/s

Hence, the momentum with which the dumbbell hits the floor is

\[ = 10 \times 4\]

\[ = 40\] kg \[m{s^{ - 1}}\]

ADDITIONAL EXERCISE:

1. The following is the distance-time table of an object in motion:

(a) What conclusion can you draw about the acceleration? Is it constant, increasing,

decreasing, or zero?

From the table, we can see that the distance changes unequally in equal intervals of time. Thus the object is said to be in non- uniform motion. Since, velocity of the object is increasing with time, the acceleration is also increasing.

(b)What do you infer about the forces acting on the object?

According to Newton’s second law of motion, \[F = mat\] . In the given case, acceleration is increasing , which indicates that the force is also increasing.

2. 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)

Mass of the motor car \[ = 1200\] kg

Acceleration produced by the car, when it is pushed by the third person, a \[ = 0.2\] \[\dfrac{m}{{{s^2}}}\]

Let the force applied by the third person be F.

Force = Mass × Acceleration

F \[ = 1200 \times 0.2\]

F \[ = 240\] N

Thus, the third person applies a force of magnitude \[240\] N.

Hence, each person applies a force of \[240\] N to push the motor car.

3. 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?

Mass of the hammer, m \[ = 500\] g \[ = 0.5\] kg

Initial velocity of the hammer, u \[ = 50\] m/s

Time taken by the nail to the stop the hammer, t \[ = 0.01\] s

Velocity of the hammer, v \[ = 0\]

Force, F =m(v-u)/t

F \[ = \dfrac{{0.5\left( {0 - 50} \right)}}{{0.01}}\]

F \[ = - 2500\] N

The hammer strikes the nail with a force F \[ = - 2500\] N.

Hence, from Newton’s third law of motion, the force of the nail on the hammer is equal and opposite, i.e., \[ + 2500\] N.

4. 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.

Mass of the motor car, m \[ = 1200\] kg

Initial velocity of the motor car, u \[ = 90\] km/h \[ = 25\] m/s

Final velocity of the motor car, v \[ = 18\] km/h \[ = 5\] m/s

Time taken, t \[ = 4\] s

According to the first equation of motion:

\[v = u + at\]

\[5 = 25 + a\left( 4 \right)\]

\[5 - 25 = a\left( 4 \right)\]

\[20 = a\left( 4 \right)\]

\[a = \dfrac{{20}}{4}\]

\[a = - 5\] m/s 2

= mv − mu

\[ = 1200\left( {5 - 25} \right)\]

\[ = 1200\left( { - 20} \right)\]

\[ = - 24000\] kg \[m{s^{ - 1}}\]

Force \[ = 1200 \times - 5\]

Force \[ = - 6000\] N

Acceleration of the motor car \[ = - 5\] m/s 2

Change in momentum of the motor car \[ = - 24000\] kg \[m{s^{ - 1}}\]

Hence, the force required to decrease the velocity \[ = - 6000\] N.

5. A large truck and a car, both moving with a velocity of magnitude v, have a head-on collision and both of them come to a halt after that. If the collision lasts for \[1\] s:

Let the mass of the truck be M and that of the car be m.

Thus, M > m

Initial velocity of both vehicles, v

Final velocity of both vehicles, v’ = 0 (since the vehicles come to rest after collision)

Time of impact, t \[ = 1\] s

(a) Which vehicle experiences the greater force of impact?

From Newton’s second law of motion, the net force experienced by each vehicle is given by the relation:

\[{F_{car}} = m\left( {v' - v} \right)/t = - mv\]

\[{F_{truck}} = m\left( {v' - v} \right)/t = - Mv\]

Since the mass of the truck is greater than that of the car, it will experience a greater force of impact.

(b) Which vehicle experiences the greater change in momentum?

Initial momentum of the car = mv

Final momentum of the car = 0

Change in momentum = mv - 0

Initial momentum of the truck = Mv

Final momentum of the truck = 0

Change in momentum = Mv -0

Since the mass of the truck is greater than that of the car, it will experience a greater change in momentum.

By Newton's third law of motion, for every action there is an equal and opposite reaction that acts on different bodies. Since the truck experiences a greater force of impact (action), this larger impact force is also experienced by the car (reaction). Thus, the car is likely to suffer more damage than the truck.

(d) Why is the car likely to suffer more damage than the truck?

Truck experiences a greater force of impact ( action), this larger impact force is also

experienced by the car ( reaction).Thus, the car is likely to suffer more damage than the truck.

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

You can opt for Chapter 9 - Force and Laws of Motion NCERT Solutions for Class 9 Science PDF for Upcoming Exams and also You can Find the Solutions of All the Maths Chapters below.

NCERT Solutions For Class 9 Science- Free PDF Download

Technology doesn't act as a barrier with NCERT Solutions if you are worried about internet connectivity. Our NCERT Solutions Class 9 is available in pdf format and is easy to download. Get them from our website or app and they can be accessed anytime and anywhere upon download. NCERT Solutions Class 9 is entirely free of cost. So if you are going to have a test or exam near, NCERT Solutions Class 9 is there for you. Solutions are made in such a way that all students, whether bright or average can rely on them.

An Overview of Class 9 Science Chapters 9- Force And Laws Of Motion

In the curriculum of Class 9 Science, Chapter 9 is the Force and Laws of Motion This chapter belongs to Unit II- Motion, Force, and Work. If you are a student of Class 9 then you might be well aware of the chapter. This chapter is all about Force, Types of Forces, and Laws of motion given by Sir Isaac Newton. The topics of this chapter are Force and its types, First law of motion, Inertia and mass, Second law of motion, Mathematical formulation of the second law of motion, Third law of motion, Conservation of momentum, etc. All these concepts are explained in a simple language combined with diagrams, activities/experiments involved, and an explanation of the numerical problems if any.

Our subject matter experts have prepared these NCERT Solutions Class 9 Chapter 9 in an efficient manner which not only makes the study interesting but also builds a strong foundation for students.

Class 9 Science Chapter 9 Force and Laws of Motion Weightage

Chapter 9 belongs to Unit II of the Class 9 curriculum and this unit has a weightage of 27 marks. Many questions of the Physics section are formed from this Chapter. Preparing with these NCERT Solutions will help the student to score better in their exams.

Here is More Detail About The Contents of Chapter 9

9.1 Balanced and Unbalanced forces

9.2 First Law of Motion

9.3 Inertia and mass

9.4 Second law of motion

9.4.1 Mathematical formulation of the second law of motion

9.5 Third law of motion

9.6 Conservation of motion

Benefits of NCERT Solutions Class 9 Chapter 9

Preparing from our NCERT Solutions Class 9 is a great way for students through which they have a strong grip on the topics of the chapter

These solutions not only build concepts but also help in strategy formation for students to excel in exams.

Detailed analysis of topics with weightage is given which helps the students in better preparation.

Highly simplified language is used by our experts to prepare these NCERT Solutions which makes it understandable for the students.

Students without any hesitation can rely upon these NCERT solutions for last-minute preparation or revision starting from the zero levels.

NCERT Solutions for Class 9 Science

Chapter 1 - Matter in Our Surroundings

Chapter 2 - Is Matter Around us Pure

Chapter 3 - Atoms and Molecules

Chapter 4 - Structure of Atom

Chapter 5 - The Fundamental Unit of Life

Chapter 6 - Tissues

Chapter 7 - Diversity in Living Organisms

Chapter 8 - Motion

Chapter 9 - Force and Laws of Motion

Chapter 10 - Gravitation

Chapter 11 - Work and Energy

Chapter 12 - Sound

Chapter 13 - Why do We Fall ill

Chapter 14 - Natural Resources

Chapter 15 - Improvement in Food Resources

Along with this, students can also view additional study materials provided by Vedantu, for Class 9

NCERT Solutions For Class 9

Revision Notes for Class 9

In this article, you will find all the NCERT study material which is required to be studied by the students while they prepare for their Class 9 CBSE exam . We also have covered the discussion on important topics from this chapter - Motion.

Students must also take care of the numerical present in this chapter, and revise the formulae regularly.

Vedantu’s NCERT Solutions for Class 9 Science Chapter 9 - 'Force and Laws of Motion' offers students a valuable resource for understanding the fundamental principles that govern motion in the physical world. These solutions, available as free PDF downloads, not only aid in exam preparation but also foster a deeper appreciation for the laws that dictate the behavior of objects in motion. Equipped with these insights, students are empowered to explore and comprehend the dynamic forces at play in their surroundings. We encourage learners to make the most of these resources to enhance their knowledge and excel in their science studies.

FAQs on NCERT Solutions for Class 9 Science Chapter 9 - Force And Laws Of Motion

1. What are force and its effects?

A push or a pull on anybody is called Force . The direction in which a body is pushed or pulled is called the direction of the force. For example, if a horse cart is pulled by a horse in the east direction then that ‘pull’ is the force and east is the direction of the force.

Effects of Force

We cannot see the force but through its effect, we can identify the force. There are various effects of force as explained below-

Making a stationary body move. For example, Kicking a ball at rest.

A force can stop a moving body. For example, Brakes applied on a moving cycle.

2. What are the 3 Laws of Motion?

Newton gave the 3 laws of motion that describe the motion of moving bodies.

First Law of Motion:- A body at rest will remain at rest, and a body in motion will continue in motion with uniform speed unless an external force is applied on the body to change its state of rest or uniform motion.

Second Law of Motion:- The rate of change of momentum is directly proportional to the applied force, and takes place in the direction in which the force acts.

Third Law of motion: To every action, there is an equal and opposite reaction. Example: Firing of Gun.

3. Which concepts in the NCERT Solutions for Class 9 Science Chapter 9 are important from the exam perspective?

Class 9 Science Chapter 9 Force and Laws of Motion is a practical chapter that carries high weightage in the exam. This chapter carries 27 marks, hence you need to know the important topics that you should prepare well. The following are some of the important topics from this chapter that you should prepare thoroughly:

Balanced and Unbalanced forces

First Law of Motion

Inertia and mass

Second Law of Motion

Mathematical Formulation of the second law of motion

Third Law of Motion

Conservation of motion.

4. What is Force Class 9th NCERT?

Force is referred to as the frequency of action to change the motion of any object or person. You apply force to change the motion of an object from the resting stage to motion or vice versa. Several characteristics, such as the weight of the object, the height at which the object is placed, and the slope of the path, determine the force needed to be applied on an object. Force is applied to accelerate or develop the motion in an object or to decline the already induced motion of the object.

5. How many laws of motion are there and what do they imply?

There are three laws of motion described by Newton. These are:

First Law of Motion - If an object is at rest, it will stay at rest unless a net force is applied to it. If an object is in motion, it will stay in motion unless a net force is applied to it.

Second Law of Motion - More force applied, more acceleration.

Third Law of Motion - For every action, there is an equal and opposite reaction.

6. Where can I find the downloadable solutions for Class 9 NCERT Chapter 9?

To find the downloadable solutions for NCERT Class 9 chapter 9 , follow these steps -

Click on the link NCERT Solutions for Class 9 Science (Physics) Chapter 9

You will land on the Vedantu Solutions page for NCERT Class 9 Chapter 9 “ Force and Laws of Motions ”.

At the top of the page, you will see an option to download the PDF of the Solutions for NCERT Chapter 9.

You can also get important questions here to practice more questions for the exam.

7. What are the key points to choose NCERT Solutions for Class 9 Science Chapter 9?

In NCERT Class 9 Science Chapter 9 , Force and Laws of Motion , you will find many identities and formulae that you need to keep in mind while solving the numericals. You should have a guide with yourself to understand the tricks to solve these questions faster. NCERT Solutions for Class 9 Chapter 9 are prepared by subject specialists, and they are highly accurate. You will get many tricks to solve your question even faster than before. You can have a deep study about these on the Vedantu Mobile app and for free of cost.

NCERT Solutions for Class 9

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Unit 2: Force & laws of motion

About this unit.

Is a force needed to keep things in motion? What does a force do? Do things always move in the direction of the push? Answers to these questions might seem pretty common sense at first, but they aren't. In this chapter, we will explore answers to these questions rediscover Newton's laws of motion.

Newton's first law & unbalanced forces

Newton's first law intro (forces causes motion?) (Opens a modal)
Balanced & unbalanced forces (Opens a modal)
Newton's first law of motion (Opens a modal)
Net force and acceleration exercise 4 questions Practice

Newton's second law & Inertia

Newton's second law of motion (Opens a modal)
Finding force - Newton's second law (Solved example) (Opens a modal)

Newton's third law

Newton's third law of motion (Opens a modal)
More on Newton's third law (Opens a modal)
Identifying equal and opposite forces 4 questions Practice
Newton's third law of motion 7 questions Practice

Momentum and force

Intro to momentum (& it's meaning) (Opens a modal)
Newton's second law & momentum (Opens a modal)
Calculating momentum changes - Solved example (Opens a modal)
Calculating linear momentum and change in momentum 4 questions Practice

Conservation of momentum

Conservation of momentum (Opens a modal)
Momentum conservation - Solved example (Opens a modal)
Momentum: Ice skater throws a ball (Opens a modal)
Momentum conservation derivation (Opens a modal)
Calculating speed and mass using conservation of momentum 4 questions Practice

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Class 9 Science Case...

Class 9 Science Case Study Questions

Table of Contents

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Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

If you are wondering how to solve class 9 science case study questions, then myCBSEguide is the best platform to choose. With the help of our well-trained and experienced faculty, we provide solved examples and detailed explanations for the recently added Class 9 Science case study questions.

You can find a wide range of solved case studies on myCBSEguide, covering various topics and concepts. Class 9 Science case studies are designed to help you understand the application of various concepts in real-life situations.

The rationale behind Science

Science is crucial for Class 9 students’ cognitive, emotional, and psychomotor development. It encourages curiosity, inventiveness, objectivity, and aesthetic sense.

In the upper primary stage, students should be given a variety of opportunities to engage with scientific processes such as observing, recording observations, drawing, tabulating, plotting graphs, and so on, whereas in the secondary stage, abstraction and quantitative reasoning should take a more prominent role in science teaching and learning. As a result, the concept of atoms and molecules as matter’s building units, as well as Newton’s law of gravitation, emerges.

Science is important because it allows Class 9 Science students to understand the world around us. It helps to find out how things work and to find solutions to problems at the Class 9 Science level. Science is also a source of enjoyment for many people. It can be a hobby, a career, or a source of intellectual stimulation.

Case study questions in Class 9 Science

The inclusion of case study questions in Class 9 science CBSE is a great way to engage students in critical thinking and problem-solving. By working through real-world scenarios, Class 9 Science students will be better prepared to tackle challenges they may face in their future studies and careers. Class 9 Science Case study questions also promote higher-order thinking skills, such as analysis and synthesis. In addition, case study questions can help to foster creativity and innovation in students. As per the recent pattern of the Class 9 Science examination, a few questions based on case studies/passages will be included in the CBSE Class 9 Science Paper. There will be a paragraph presented, followed by questions based on it.

Examples of Class 9 science class case study questions

Class 9 science case study questions have been prepared by myCBSEguide’s qualified teachers. Class 9 case study questions are meant to evaluate students’ knowledge and comprehension of the material. They are not intended to be difficult, but they will require you to think critically about the material. We hope you find Class 9 science case study questions beneficial and that they assist you in your exam preparation.

The following are a few examples of Class 9 science case study questions.

Class 9 science case study question 1

due to its high compressibility
large volumes of a gas can be compressed into a small cylinder
transported easily
all of these
shape, volume
volume, shape
shape, size
size, shape
the presence of dissolved carbon dioxide in water
the presence of dissolved oxygen in the water
the presence of dissolved Nitrogen in the water
liquid particles move freely
liquid have greater space between each other
both (a) and (b)
none of these
Only gases behave like fluids
Gases and solids behave like fluids
Gases and liquids behave like fluids
Only liquids are fluids

Answer Key:

(d) all of these
(a) shape, volume
(b) the presence of dissolved oxygen in the water
(c) both (a) and (b)
(c) Gases and liquids behave like fluids

Class 9 science case study question 2

12/32 times
18 g of O 2
18 g of CO 2
18 g of CH 4
1 g of CO 2
1 g of CH 4 CH 4
2 moles of H2O
20 moles of water
6.022 × 1023 molecules of water
1.2044 × 1025 molecules of water
(I) and (IV)
(II) and (III)
(II) and (IV)
Sulphate molecule
Ozone molecule
Phosphorus molecule
Methane molecule
(c) 8/3 times
(d) 18g of CH 4
(c) 1g of H 2
(d) (II) and (IV)
(c) phosphorus molecule

Class 9 science case study question 3

collenchyma
chlorenchyma
It performs photosynthesis
It helps the aquatic plant to float
It provides mechanical support
Sclerenchyma
Collenchyma
Epithelial tissue
Parenchyma tissues have intercellular spaces.
Collenchymatous tissues are irregularly thickened at corners.
Apical and intercalary meristems are permanent tissues.
Meristematic tissues, in its early stage, lack vacuoles, muscles
(I) and (II)
(III) and (I)
Transpiration
Provides mechanical support
Provides strength to the plant parts
None of these
(a) Collenchyma
(b) help aquatic plant to float
(b) Sclerenchyma
(d) Only (III)
(c) provide strength to plant parts

Cracking Class 9 Science Case Study Questions

There is no one definitive answer to Class 9 Science case study questions. Every case study is unique and will necessitate a unique strategy. There are, nevertheless, certain general guidelines to follow while answering case study questions.

To begin, double-check that you understand the Class 9 science case study questions. Make sure you understand what is being asked by reading it carefully. If you’re unclear, seek clarification from your teacher or tutor.
It’s critical to read the Class 9 Science case study material thoroughly once you’ve grasped the question. This will provide you with a thorough understanding of the problem as well as the various potential solutions.
Brainstorming potential solutions with classmates or other students might also be beneficial. This might provide you with multiple viewpoints on the situation and assist you in determining the best solution.
Finally, make sure your answer is presented simply and concisely. Make sure you clarify your rationale and back up your claim with evidence.

A look at the Class 9 Science Syllabus

The CBSE class 9 science syllabus provides a strong foundation for students who want to pursue a career in science. The topics are chosen in such a way that they build on the concepts learned in the previous classes and provide a strong foundation for further studies in science. The table below lists the topics covered in the Class 9 Science syllabus of the Central Board of Secondary Education (CBSE). As can be seen, the Class 9 science syllabus is divided into three sections: Physics, Chemistry and Biology. Each section contains a number of topics that Class 9 science students must study during the course.

CBSE Class 9 Science (Code No. 086)

Theme: Materials Unit I: Matter-Nature and Behaviour Definition of matter; solid, liquid and gas; characteristics – shape, volume, density; change of state-melting (absorption of heat), freezing, evaporation (cooling by evaporation), condensation, sublimation. Nature of matter: Elements, compounds and mixtures. Heterogeneous and homogenous mixtures, colloids and suspensions. Particle nature and their basic units: Atoms and molecules, Law of constant proportions, Atomic and molecular masses. Mole concept: Relationship of mole to mass of the particles and numbers. Structure of atoms: Electrons, protons and neutrons, valency, the chemical formula of common compounds. Isotopes and Isobars.

Theme: The World of the Living Unit II: Organization in the Living World Cell – Basic Unit of life: Cell as a basic unit of life; prokaryotic and eukaryotic cells, multicellular organisms; cell membrane and cell wall, cell organelles and cell inclusions; chloroplast, mitochondria, vacuoles, endoplasmic reticulum, Golgi apparatus; nucleus, chromosomes – basic structure, number. Tissues, Organs, Organ System, Organism: Structure and functions of animal and plant tissues (only four types of tissues in animals; Meristematic and Permanent tissues in plants).

Theme: Moving Things, People and Ideas Unit III: Motion, Force and Work Motion: Distance and displacement, velocity; uniform and non-uniform motion along a straight line; acceleration, distance-time and velocity-time graphs for uniform motion and uniformly accelerated motion, derivation of equations of motion by graphical method; elementary idea of uniform circular motion. Force and Newton’s laws: Force and Motion, Newton’s Laws of Motion, Action and Reaction forces, Inertia of a body, Inertia and mass, Momentum, Force and Acceleration. Elementary idea of conservation of Momentum. Gravitation: Gravitation; Universal Law of Gravitation, Force of Gravitation of the earth (gravity), Acceleration due to Gravity; Mass and Weight; Free fall. Floatation: Thrust and Pressure. Archimedes’ Principle; Buoyancy. Work, energy and power: Work done by a Force, Energy, power; Kinetic and Potential energy; Law of conservation of energy. Sound: Nature of sound and its propagation in various media, speed of sound, range of hearing in humans; ultrasound; reflection of sound; echo.

Theme: Food Unit IV: Food Production Plant and animal breeding and selection for quality improvement and management; Use of fertilizers and manures; Protection from pests and diseases; Organic farming.

PRESCRIBED BOOKS:

Science-Textbook for class IX-NCERT Publication
Assessment of Practical Skills in Science-Class IX – CBSE Publication
Laboratory Manual-Science-Class IX, NCERT Publication
Exemplar Problems Class IX – NCERT Publication

myCBSEguide: A true helper

There are numerous advantages to using myCBSEguide to achieve the highest results in Class 9 Science.

myCBSEguide offers high-quality study materials that cover all of the topics in the Class 9 Science curriculum.
myCBSEguide provides practice questions and mock examinations to assist students in the best possible preparation for their exams.
On our myCBSEguide app, you’ll find a variety of solved Class 9 Science case study questions covering a variety of topics and concepts. These case studies are intended to help you understand how certain principles are applied in real-world settings
myCBSEguide is that the study material and practice problems are developed by a team of specialists who are always accessible to assist students with any questions they may have. As a result, students may be confident that they will receive the finest possible assistance and support when studying for their exams.

So, if you’re seeking the most effective strategy to study for your Class 9 Science examinations, myCBSEguide is the place to go!

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Important Questions

Force and Laws of Motion

Ncert revision notes for chapter 9 force and laws of motion class 9 science.

→ Force can make a stationary body in object. For example: a football can be set to move by kicking it i.e. by applying a force.

→ Force can stop a moving body. For example, by applying brakes, a running cycle or a running vehicle can be stopped.

→ Force can change the direction of a moving object. For example: by applying force i.e. by moving handle, the direction of a running bicycle can be changed. Similarly by moving steering, the direction of a running vehicle is changed.

→ Force can change the speed of a moving body. By accelerating, the speed of a running vehicle can be increased or by applying brakes the speed of a running vehicle can be decreased.

→ Force can change the shape and size of an object. For example: by hammering, a block of metal can be turned into a thin sheet. By hammering, a stone can be broken into pieces.

• Forces are mainly of two types:

(i) Balanced forces

(ii) Unbalanced forces

→ If the resultant of applied forces is equal to zero, it is called balanced forces. Example: In the tug of war if both the team apply similar magnitude of forces in opposite directions, rope does not move in either side. This happens because of balanced forces in which resultant of applied forces become zero.

→ Balanced forces do not cause any change of state of an object. Balanced forces are equal in magnitude and opposite in direction.

→ Balanced forces can change the shape and size of an object. For example: When forces are applied from both sides over a balloon, the size and shape of balloon is changed.

→ If the resultant of applied forces are greater than zero, the forces are called unbalanced forces.

→ An object in rest can be moved because of applying balanced forces.

• Unbalanced forces can do the following:

→ Move a stationary object

→ Increase the speed of a moving object

→ Decrease the speed of a moving object

→ Stop a moving object

→ Change the shape and size of an object

• Galileo Galilei: Galileo first of all said that object move with a constant speed when no foces act on them.

→ This means if an object is moving on a frictionless path and no other force is acting upon it, the object would be moving forever. That is, there is no unbalanced force working on the object.

→ But practically it is not possible for any object. Because to attain the condition of zero, unbalanced force is impossible.

→ Force of friction, force of air and many other forces are always acting upon an object.

→ Newton studied the ideas of Galileo and gave the three laws of motion. These laws are known as Newton’s laws of motion.

Newton’s First Law of Motion (Law of Inertia)

→ Any object remains in the state of rest or in uniform motion along a straight line, until it is compelled to change the state by applying external force.

Explanation: If any object is in the state of rest, then it will remain in rest until a external force is applied to change its state. Similarly, an object will remain in motion until any external force is applied over it to change its state. This means all objects resist to in changing their state. The state of any object can be changed by applying external forces only.

• Newton’s First Law of Motion in Everyday Life

(i) A person standing in a bus falls backward when bus starts moving suddenly.

→ This happens because the person and bus both are in rest while bus is not moving, but as the bus starts moving, the legs of the person start moving along with bus but rest portion of his body has the tendency to remain in rest. Because of this, the person falls backward; if he is not alert.

(ii) A person standing in a moving bus falls forward if driver applies brakes suddenly.

→ This happens because when bus is moving, the person standing in it is also in motion along with bus. But when driver applies brakes the speed of bus decreases suddenly or bus comes in the state of rest suddenly, in this condition the legs of the person which are in contact with the bus come in rest while the rest part of his body have the tendency to remain in motion. Because of this person falls forward if he is not alert.

(iii) Before hanging the wet clothes over laundry line, usually many jerks are given to the clothes to get them dried quickly. Because of jerks, droplets of water from the pores of the cloth falls on the ground and reduced amount of water in clothes dries them quickly.

→ This happens because when suddenly clothes are made in motion by giving jerks, the water droplets in it have the tendency to remain in rest and they are separated from clothes and fall on the ground.

(iv) When the pile of coin on the carom-board is hit by a striker, coin only at the bottom moves away leaving rest of the pile of coin at same place.

→ This happens because when the pile is struck with a striker, the coin at the bottom comes in motion while rest of the coin in the pile has the tendency to remain in the rest and they vertically falls the carom-board and remain at same place.

→ The property of an object because of which it resists to get disturb its state is called inertia.

→ Inertia of an object is measured by its mass. Inertia is directly proportional to the mass. This means inertia increases with increase in mass and decreases with decrease in mass.

→ A heavy object will have more inertia than the lighter one. In other words, the natural tendency of an object that resists the change in state of motion or rest of the object is called inertia.

→ Since a heavy object has more inertia, thus it is difficult to push or pull a heavy box over the ground than the lighter one.

→ Momentum is the power of motion of an object.

→ The product of velocity and mass is called the momentum. Momentum is denoted by ‘p’.

→ Therefore, Momentum of the object = Mass × Velocity (p = m × v), where, p = momentum, m = mass of the object and v = velocity of the object.

• Some explanations to understand the momentum:

→ A person get injured in the case of hitting by a moving object, such as stone, pebbles or anything because of momentum of the object.

→ Even a small bullet is able to kill a person when it is fired from a gun because of its momentum due to great velocity.

→ A person get injured severely when hit by a moving vehicle because of momentum of vehicle due to mass and velocity.

Momentum and Mass

→ Since momentum is the product of mass and velocity (p = m × v) of an object. This means momentum is directly proportional to mass and velocity. Momentum increases with increase of either mass or velocity of an object.

→ This means if a lighter and a heavier object is moving with same velocity, then heavier object will have more momentum than the lighter one.

→ If a small object is moving with great velocity, it has tremendous momentum. And because of momentum, it can harm an object more severely. Example: a small bullet having a little mass even kills a person when it is fired from a gun.

→ Usually, road accidents prove more fatal because of high speed than in slower speed. This happens because vehicles running with high speed have greater momentum compared to a vehicle running with slower speed. Momentum of an object which is in the state of rest.

Let an object with mass ‘m’ is in the rest.

Since, object is in rest, therefore, its velocity, v = 0

Now we know that, Momentum = mass × velocity

⇒ p = m × 0 = 0

Thus, the momentum of an object in the rest i.e. non-moving, is equal to zero.

Unit of momentum

→ SI unit of mass = kg

→ SI unit of velocity = m/s

We know that, Momentum (p) = m × v

∴ p = kg × m/s ⇒ p = kgm/s

Numerical Problems Based on Momentum

Type I: Calculation of Momentum

Example 1: What will be the momentum of a stone having mass of 10 kg when it is thrown with a velocity of 2 m/s?

Mass (m) = 10 kg

Velocity (v) = 2 m/s

We know that, Momentum (p) = Mass (m) × Velocity (v)

∴ p = 10 kg × 2 m/s = 20 kg m/s

Thus, the momentum of the stone = 20 kg m/s.

Example 2: Calculate the momentum of a bullet of 25 g when it is fired from a gun with a velocity of 100 m/s.

Velocity of the bullet (v) = 100 m/s

Mass of the bullet (m) = 25 g = 25/1000 kg = 0.025 kg

Since, p = m × v

∴ p = 0.025 × 100 = 2.5 kg m/s

Momentum of the bullet = 2.5 kg m/s.

Example 3: Calculate the momentum of a bullet having mass of 25 g is thrown using hand with a velocity of 0.1 m/s.

Velocity of the bullet (v) = 0.1 m/s

Momentum (p) = Mass (m) × Velocity (v)

∴ p = 0.025 kg × 0.1 m/s

⇒ p = 0.0025 kg m/s

Momentum of the bullet = 0.0025 kg m/s.

Example 4: The mass of a goods lorry is 4000 kg and the mass of goods loaded on it is 20000 kg. If the lorry is moving with a velocity of 2 m/s, what will be its momentum?

Mass of lorry = 4000 kg

Mass of goods on the lorry = 20000 kg

∴ Total mass (m) on the lorry = 4000 kg + 20000 kg = 24000 kg

∴ p = 24000 kg × 2 m/s

⇒ p = 48000 kg m/s

Momentum of the lorry = 48000 kg m/s.

Example 5: A car having mass of 1000 kg is moving with a velocity of 0.5 m/s. What will be its momentum?

Velocity of the car (v) = 0.5 m/s

Mass of the car (m) = 1000 kg

∴ p = 1000 kg × 0.5 m/s = 500 kg m/s

Momentum of the car = 500 kg m/s.

→ Rate of change of momentum of an object is proportional to applied unbalanced force in the direction of force.

Mathematical expression

Mass of an object = m kg

Initial velocity of an object = u m/s

Final velocity of an object = v m/s

∴ Initial momentum, p 1 = mu

Final momentum, p 2 = mv

∴ Change in momentum = Final momentum – Initial momentum

∴ Rate of change of momentum = Change in momentum/Time taken

• According to 2nd law, this rate of change is momentum is directly proportional to force.

We know that, a = (v-u)/t (From 1st equation of motion)

where, k is a constant. Its value can be assumed as 1.

∴ F = 1 × m × a = ma

• SI unit = kg m/s 2 or Newton

1 Newton: When an acceleration of 1 m/s 2 is seen in a body of mass 1 kg, then the force applied on the body is said to be 1 Newton.

Proof of Newton’s First Law of Motion from Second Law

→ First law states that if external force F = 0, then a moving body keeps moving with the same velocity, or a body at rest continues to be at rest.

We know, F = m(v-u)/t

(i) A body is moving with initial velocity u then,

m(v-u)/t = 0 ⇒ v – u = 0

Thus, final velocity is also same.

(ii) A body is at rest i.e., u = 0

Therefore, from above u = v = 0

So, the body will continue to be at rest.

→ For every action there is an equal an opposite reaction.

Applications

(i) Walking is enabled by 3rd law.

(ii) A boat moves back when we deboard it.

(iii) A gun recoils.

(iv) Rowing of a boat.

Law of Conservation of Momentum

→ When two (or more) bodies act upon one another, their total momentum remains constant (or conserved) provided no external forces are acting.

• Initial momentum = Final momentum

Suppose, two objects A and B each of mass m 1 and mass m 2 are moving initially with velocities u 1 and u 2 , strike each other after time t and start moving with velocities v 1 and v 2 respectively.

Initial momentum of object A = m 1 u 1

Initial momentum of object B = m 2 u 2

Final momentum of object A = m 1 v 1

Final momentum of object B = m 2 v 2

So, Rate of change of momentum in A,

F1 = (m 1 v 1 - m 1 u 1 )t = m 1 (v 1 - u 1 )/t ....(i)

Rate of change of momentum in B,

F2 = (m 2 v 2 - m 2 u 2 )t = m 2 (v 2 - u 2 )/t ....(ii)

We know from 3rd law of motion, F 1 = −F 2

So, m 1 (v 1 - u 1 )/t = -m 2 (v 2 - u 2 )/t

⇒ m 1 v 1 – m 2 u 2 = −m 2 v 2 + m 1 u 1

⇒ m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2

Thus, Initial momentum = Final momentum

Example 1: A bullet of mass 20 g is fired horizontally with a velocity of 150 m/s from a pistol of mass 2 kg. Find the recoil velocity of the pistol.

Mass (m 1 ) of bullet = 20 g = 0.02 kg

Mass (m 2 ) of pistol = 2 kg

Initially bullet is inside the gun and it is not moving.

∵ Mass = m 1 +m 2 = (0.02 + 2) kg = 2.02 kg and u 1 = 0

∴ Initial momentum = 2.02 × 0 = 0 ....(i)

Let the velocity of pistol be v 2 and v 1 for bullet = 150

∴ Final momentum = m 1 v 1 + m 2 v 2 = 0.02×150 + 2v 2 ...(ii)

We know that, Initial momentum = Final momentum

∴ (0.02×150)/100 + 2v 2 = 0 [From equations (i) and (ii)]

⇒ 3 + 2v 2 = 0

⇒ 2v 2 = −3

⇒ v 2 = −1.5 m/s

(−)ve sign indicates that gun recoils in direction opposite to that of the bullet.

Example 2: Two hockey players viz A of mass 50 kg is moving with a velocity of 4 m/s and another one B belonging to opposite team with mass 60 kg is moving with 3 m/s, get entangled while chasing and fall down. Find the velocity with which they fall down and in which direction?

m A = 50 kg, u A = 4 m/s

m B = 60 kg, u B = 3 m/s

Initial momentum A = m A u A = 50 × 4 = 200 kgm/s

Initial momentum B = m B u B = 60 × 3 = 180 kgm/s

∴ Total initial momentum = 200 + 180 = 380 kgm/s ....(i)

Final momentum = (m A + m B )v = (50 + 60)v = 110v ....(ii)

According to the law of conservation of momentum,

⇒ v = 380/110 = 3.454 m/s

NCERT Solutions for Chapter 4 The Age of Industrialisation Class 10 History

Related chapters.

Matter in Our Surroundings
Is Matter Around Us Pure
Atoms and Molecules
Structure of the Atom
The Fundamental Unit of Life

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9th Class Science Force and laws of motion Question Bank

Done case based mcqs - force and laws of motion total questions - 25.

A) same inertia done clear

B) same momentum done clear

C) different inertia done clear

D) different momentum done clear

question_answer 2) The inertia of a moving object depends on

A) mass of object done clear

B) momentum of object done clear

C) speed of object done clear

D) shape of object done clear

A) balanced forces act on ball done clear

B) unbalanced forces act on ball done clear

C) frictional forces act on ball done clear

D) gravitational force acts on ball done clear

A) A only done clear

B) B only done clear

C) Both A and B done clear

D) None of the above done clear

question_answer 5) A water tanker filled upto \[\frac{1}{3}\]of its height is moving with a uniform speed. On a sudden application of brakes, the water in the tank would

A) move backward done clear

B) move forward done clear

C) be unaffected done clear

D) rise upwards done clear

A) Due to large force done clear

B) Due to large velocity done clear

C) Both [a] and [b] done clear

question_answer 7) Momentum of a bullet of mass 0.2 kg moving at 400 m/s will be

A) 4 kg m/s done clear

B) 80 kg m/s done clear

C) 8 kg m/s done clear

D) 40 kg m/s done clear

question_answer 8) The unit of measuring momentum of moving body is

A) \[{{\operatorname{ms}}^{-1}}\] done clear

B) kg-\[{{\operatorname{ms}}^{-1}}\] done clear

C) kg-\[{{\operatorname{ms}}^{-2}}\] done clear

D) \[N-{{m}^{2}}k{{g}^{-2}}\] done clear

question_answer 9) If the mass of a body and the force acting on it are both doubled, what happens to the acceleration?

A) doubled done clear

B) halved done clear

C) remains same done clear

D) becomes zero done clear

question_answer 10) If the force acting on the body is zero, its momentum will be

A) zero done clear

B) constant done clear

C) infinite done clear

D) variable done clear

A) 2.25\[{{\operatorname{ms}}^{-2}}\] done clear

B) 6.25\[{{\operatorname{ms}}^{-2}}\] done clear

C) 7\[{{\operatorname{ms}}^{-2}}\] done clear

D) 7.85\[{{\operatorname{ms}}^{-2}}\] done clear

question_answer 12) What is the magnitude of braking force, if the mass of the car is 900 kg?

A) 5625 N done clear

B) 6625 N done clear

C) 7025 N done clear

D) 7625 N done clear

question_answer 13) Calculate the braking distance travelled by the car.

A) 36 m done clear

B) 60 m done clear

C) 72 m done clear

D) 86 m done clear

question_answer 14) A constant force (F) is applied on a stationary particle of mass (m). The velocity attained by the particle after a certain displacement will be proportional to

A) \[\frac{1}{\sqrt{m}}\] done clear

B) m done clear

C) \[\frac{1}{m}\] done clear

D) \[\sqrt{m}\] done clear

question_answer 15) The same net force is applied to object A and B. The observed accelerations of the two objects are not the same; object A has an acceleration three times that of object B. Which of the following is correct?

A) Object A has three times the mass of object B. done clear

B) Object A has one-third mass of object B. done clear

C) Object A has a different, less streamlined shape than object B. done clear

D) Object A has more friction than object B. done clear

A) I, II done clear

B) II, IV done clear

C) II, III done clear

D) I, IV done clear

A) P done clear

B) P, R done clear

C) P, Q done clear

D) S done clear

A) the velocity of the body is variable done clear

B) the velocity of the body is constant done clear

C) the mass of the body is constant done clear

A) 0 done clear

B) 4 N done clear

C) 8 N done clear

D) 10 N done clear

question_answer 20) 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 \[10c{{m}^{-2}}\] ]

A) 10kg-\[{{\operatorname{ms}}^{-1}}\] done clear

B) 20kg-\[{{\operatorname{ms}}^{-1}}\] done clear

C) 3ukg-\[{{\operatorname{ms}}^{-1}}\] done clear

D) 40kg-\[{{\operatorname{ms}}^{-1}}\] done clear

A) Mass of rocket is greater than mass of gases it expels. done clear

B) Mass of rocket is less than mass of gases it expels. done clear

C) Mass of rocket is equal to mass of gases it expels. done clear

D) Mass of rocket is not related to the mass of gases it expels. done clear

question_answer 22) The fuel in the rocket is burnt and exhaust gases are made to escape in

A) parallel to rocket done clear

B) upward direction done clear

C) perpendicular to rocket done clear

D) downward direction done clear

question_answer 23) The statement "For every action there is an equal and opposite reaction", is related to which of the following?

A) Newton's first law done clear

B) Newton's second law done clear

C) Newton's third law done clear

D) Law of conservation of momentum done clear

question_answer 24) What type of motion is described by the rocket?

A) Uniform, done clear

B) Accelerated done clear

C) Both [a] and [b] done clear

D) None of these done clear

question_answer 25) A horse pulling a tanga moves forward due to the force exerted by

A) The horse on the ground with his feet done clear

B) The horse on the tanga done clear

C) The ground on the horse's feet done clear

D) The tanga on the horse done clear

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Case Based MCQs - Force and Laws of Motion

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Important Questions for Class 9 Science Chapter 9 Force and Laws of Motion

Get important questions for Class 9 Science Chapter 9 Force and Laws of Motion with PDF. Our subject expert prepared these important questions and answers as per the latest NCERT textbook. These important questions will be helpful to revise the important topics and concepts. You can easily download all the questions and answers in PDF format from our app.

Force and Laws of Motion Class 9 Science Important Questions with Answers

Important Questions for Class 9 Science Chapter 9 Force and Laws of Motion 00001

Extra Questions for Class 9th: Ch 9 Force and Laws of Motion Science

Extra questions for class 9th: ch 9 force and laws of motion (science) important questions answer included.

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Case Study Questions for Class 9 Science Chapter 1 Matter in Our Surroundings

Last modified on: 5 days ago
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Here we are providing case study questions for class 9 science chapter 12 sound. Students are suggested to go through each and every case study questions for better understanding of the chapter.

Case Study/Passage Based Questions:

Question 1:

Read the following passage and answer the questions given below.

Every matter is made up of tiny particles. These particles are so tiny that they can’t be seen with naked eyes.

The three characteristics shown by particles of matter are as follows:

(i) There are small voids between particles in a matter. This characteristic is the concept behind the solubility of a substance in other substances.

(ii) Particles of matter show continuous random movements, that is they possess kinetic energy. The spreading of ink in a beaker of glass, smell of agarbattis, etc. are few illustrations that show the movement of particles of a substance.

(iii) The particles of matter attract each other with a force called interparticle force of attraction. Read the given passage carefully and give the answer of the following questions:

Q 1. Spreading of fragrance of a burning incense stick in a room shows that:

a. particles of matter have spaces between them.

b. particles of matter attract each other.

c. particles of matter are constantly moving.

d. None of the above

Q 2. What happens when we add sugar to water?

a. Volume of water doubles.

b. Volume of water decreases

c. Volume of water remains the same.

Q 3. A stream of water cannot be cut by fingers. Which property of matter does this observation show?

a. Particles of matter attract each other.

b. Particles of matter have spaces between them.

c. Particles of matter are continuously moving.

Q 4. When we put some crystals of potassium permanganate in a beaker containing water, we observe that after some time, the whole water turns pink. This intermixing of particles of two different types of matter on their own is called:

a. Brownian motion

c. sublimation

d. diffusion

Q 5. Why is the rate of diffusion of liquids higher than that of solids?

a. In the liquid state, particles are tightly packed as compared to solids.

b. In the liquid state, particles move freely as compared to solids.

c. In solid state, particles have least force of attraction between the particles.

d. In solid state, particles cannot be compressed easily.

(c) particles of matter are constantly moving.
(c) Volume of water remains the same.
(a) Particles of matter attract each other.
(d) diffusion
(b) In the liquid state, particles move freely as compared to solids

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Case Study Questions Class 9 Science Force and Laws of Motion
CBSE Case Study Questions Class 9 Science - Force and Laws of Motion. Case 1: (1) Newton's first law of motion states that a body at rest will remain at rest position only and a body which is in motion continues to be in motion unless otherwise they are acted upon by an external force. In other words, all objects resist a changein their ...
Case Study and Passage Based Questions for Class 9 Science Chapter 9
(ii) Name the law involved in this case. (a) Newton's second law of motion. (b) Newton's first law of motion. (c) Newton's third law of motion. ... Last modified on:2 years agoReading Time:5MinutesCase Study Questions for Class 9 Science Chapter 3 Atoms and Molecules In CBSE Class 9 Science Paper, Students will have to answer some ...
Case Study Questions of Chapter 9 Force and Laws of Motion PDF Download
Answer the following questions. (i)Law of conservation of momentum is applicable to. (a) A system of particles. (b) Only for 2 particles. (c) Only for 1 particle. (d) None of the above. Show Answer. (ii) Law of conservation of momentum holds good provided that. (a) There should be external unbalanced force acting on particles.
Class 9 Science Case Study Questions Chapter 8 Motion
Case Study 1: Answer the following questions by observing following diagram. (i)What is distance and displacement when the particle moves from point A to B? (a)distance is equal to the displacement. (b)distance is greater than and equal to the displacement. (c)distance is lesser than and equal to the displacement.
Class 9th Science
Answers. <. Force and Laws of Motion Case Study Questions With Answer Key Answer Keys. Case Study. (i) (a) The coin possesses inertia of rest, it resists the change and hence falls in the glass. (ii) (b) Newton's first law of motion. (iii) (c) Heavy coin will possess more inertia so it will fall in tumbler.
9th Science Motion Case Study Questions and Answers 2022-23
Class 9th Science - Motion Case Study Questions and Answers 2022 - 2023. QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 9th Science Subject - Motion, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.
Class 9th Science
Study Materials. Sep-09 , 2022. QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 9th Science Subject - Motion, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks. QB365 Mobile App For Practice Question Papers.
CBSE Class 9 Science Chapter 9 Force and Laws of Motion ...
Vedantu now offers Chapter 9's crucial science questions for Class 9 in a downloadable PDF format. Students can download the PDF and study anytime and anywhere in offline mode. Force and Laws of Motion important questions along with their detailed answers are prepared by our subject experts to help students get a clear idea of the concepts ...
Class 9 Science Case Study Questions Chapter 9 Force and Laws of Motion
Case Study/Passage-Based Questions. Case Study 1: The sum of the momentum of the two objects before the collision is equal to the sum of momentum after the collision provided there is no external unbalanced force acting on them. This is known as the law of conservation of momentum. This statement can alternatively be given as the total momentum ...
NCERT Solutions for Class 9 Science Chapter 9
Class 9 Science Chapter 9 Force and Laws of Motion Weightage. Chapter 9 belongs to Unit II of the Class 9 curriculum and this unit has a weightage of 27 marks. Many questions of the Physics section are formed from this Chapter. Preparing with these NCERT Solutions will help the student to score better in their exams.
Force & laws of motion
Answers to these questions might seem pretty common sense at first, but they aren't. In this chapter, we will explore answers to these questions rediscover Newton's laws of motion. If you're seeing this message, it means we're having trouble loading external resources on our website.
Class 9 Science Case Study Questions
Class 9 case study questions are meant to evaluate students' knowledge and comprehension of the material. They are not intended to be difficult, but they will require you to think critically about the material. ... Force and Newton's laws: Force and Motion, Newton's Laws of Motion, Action and Reaction forces, Inertia of a body, Inertia ...
Case Study and Passage Based Questions for Class 9 Science Chapter 8 Motion
Case Study/Passage Based Questions: Question 1: Read the following paragraph and any four questions from (i) to (v). Distance is the length of the actual path covered by an object, irrespective of its direction of motion. Displacement is the shortest distance between the initial and final positions of an object in a given direction.
NCERT Revision Notes for Chapter 9 Force and Laws of Motion Class 9
Momentum (p) = Mass (m) × Velocity (v) ∴ p = 1000 kg × 0.5 m/s = 500 kg m/s. Momentum of the car = 500 kg m/s. 9. Second Law of Motion. Answer. → Rate of change of momentum of an object is proportional to applied unbalanced force in the direction of force. Mathematical expression.
9th Class Science Force and laws of motion Question Bank
The first law of motion is stated as: An object remains in a state of rest or of uniform motion in a straight line unless compelled to change that state by an applied force. In other words, all objects resist a change in their state of motion. In a qualitative way, the tendency of undisturbed objects to stay at rest or to keep moving with the ...
NCERT Solutions for Class 9 Science Chapter 9 Force and Laws of Motion
Question 1: If action is always equal to the reaction, explain how a horse can pull a cart. Answer: A horse pushes the ground in the backward direction. According to Newton's third law of motion, a reaction force is exerted by the Earth on the horse in the forward direction. As a result, the cart moves forward.
Extra Questions for Class 9 Science Chapter 9 Force and Laws of Motion
Short Answer Type Questions. 1: State the difference in balanced and unbalanced force. Answer: Balanced force. Unbalanced force. Forces acting on a body from the opposite directions are same. Forces acting on a body from two opposite directions are not same. It does not change the state of rest or of motion of an object.
Force and Laws of Motion Class 9 Extra Questions Science Chapter 9
Question 9. Define force of friction. Answer: The force acting between any two surfaces in contact and tending to oppose motion is called force of friction. Question 10. Define electrostatic force. Answer: The force exerted by an electrically charged body is called electrostatic force. Question 11.
Important Questions for Class 9 Science Chapter 9 Force and Laws of Motion
Important Questions for Class 9 Science Chapter 9 Force and Laws of Motion. Get important questions for Class 9 Science Chapter 9 Force and Laws of Motion with PDF. Our subject expert prepared these important questions and answers as per the latest NCERT textbook. These important questions will be helpful to revise the important topics and ...
NCERT Solutions for Class 9 Science Chapter 9 Force and Laws of Motion
Final velocity, v = 0 (finally the automobile stops) Acceleration of the automobile, a = −1.7 ms −2. From Newton's second law of motion: Force = Mass x Acceleration = 1500 x (−1.7) = −2550 N. Hence, the force between the automobile and the road is −2550 N, in the direction opposite to the motion of the automobile. 9.
CBSE Class 9 Case Study Questions
TopperLearning provides a complete collection of case studies for CBSE Class 9 students. Improve your understanding of biological concepts and develop problem-solving skills with expert advice.
Extra Questions for Class 9th: Ch 9 Force and Laws of Motion Science
Mention any two kinds of changes that can be brought about in a body by force. Answer. Change in speed/change of direction/change of shape. Q8. State two effects, a force can produce in a non rigid body fixed at a position. Answer. Change its shape and size. Short Answer Questions-I (SAQs-I) : 2 Marks.
Case Study Questions for Class 9 Science Chapter 1 Matter in Our
Here we are providing case study questions for class 9 science chapter 12 sound. Students are suggested to go through each and every case study questions for better understanding of the chapter. Case Study/Passage Based Questions:

Case Study Questions Class 9 Science Force and Laws of Motion

CBSE Case Study Questions Class 9 Science – Force and Laws of Motion

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