case study class 9 science pdf

CBSE Case Study Questions for Class 9 Science - Pdf PDF Download

Cbse case study questions for class 9 science.

Case based questions for Class 9 Science involve exploring a real-world situation through scientific analysis and inquiry. These questions allow students to make connections between science concepts and the world around them, as well as develop critical thinking skills. For example, a case study may involve challenging a student to determine the cause of an illness in a local population by researching the disease, its symptoms, and the local environment. Through this exercise, students learn how to identify a problem, break it down into parts, and come up with a solution that is supported by evidence. This type of question helps students to understand how science is at the centre of solving real-world problems.

Chapter Wise Case Based Questions for Class 9 Science

Chapter-wise case-based questions for Class 9 Science are a set of questions based on specific chapters or topics covered in the science textbook. These questions are designed to help students apply their understanding of scientific concepts to real-world situations and events.

The CBSE Class 9 Case Based Questions can be accessed from Chapetrwise Links provided below:

Chapter 1: Matter In Our Surroundings

Chapter 2: is matter around us pure.

• Case Based Questions: Is Matter Around Us Pure?

Chapter 3: Atoms And Molecules

• Case Based Questions: Atoms And Molecules

Chapter 4: Structure Of The Atom

• Case Based Questions: Structure Of The Atom

Chapter 5: The Fundamental Unit Of Life

• Case Based Questions: The Fundamental Unit Of Life- 1
• Case Based Questions: The Fundamental Unit Of Life- 2

Chapter 6: Tissues

• Case Based Questions: Tissues- 1
• Case Based Questions: Tissues- 2

Chapter 7: Motion

• Case Based Questions: Motion-1 
• Case Based Questions: Motion- 2

Chapter 8: Force And Laws Of Motion

• Case Based Questions: Force And Laws Of Motion

Chapter 9: Gravitation

• Case Based Questions: Gravitation

Chapter 10: Work And Energy

• Case Based Questions: Work And Energy- 1
• Case Based Questions: Work And Energy- 2

Chapter 11: Diversity In Living Organisms

Chapter 12: sound, chapter 13: natural resources, chapter 14: improvement in food resource, chapter 15: why do we fall ill.

• Case Based Questions: Why Do We Fall Ill?

Weightage of Case Based Questions in Class 9 Science

Why are Case Study Questions important in Science Class  9?

• Enhance critical thinking:  Case study questions require students to analyze a real-life scenario and think critically to identify the problem and come up with possible solutions. This enhances their critical thinking and problem-solving skills.
• Apply theoretical concepts:  Case study questions allow students to apply theoretical concepts that they have learned in the classroom to real-life situations. This helps them to understand the practical application of the concepts and reinforces their learning.
• Develop decision-making skills:  Case study questions challenge students to make decisions based on the information provided in the scenario. This helps them to develop their decision-making skills and learn how to make informed decisions.
• Improve communication skills:  Case study questions often require students to present their findings and recommendations in written or oral form. This helps them to improve their communication skills and learn how to present their ideas effectively.
• Enhance teamwork skills:  Case study questions can also be done in groups, which helps students to develop teamwork skills and learn how to work collaboratively to solve problems.

In summary, case study questions are important in Class 9 because they enhance critical thinking, apply theoretical concepts, develop decision-making skills, improve communication skills, and enhance teamwork skills. They provide a practical and engaging way for students to learn and apply their knowledge and skills to real-life situations.

Class 9 Science Curriculum at Glance

The Class 9 Science curriculum in India covers a wide range of topics and concepts. Here is a brief overview of the Science curriculum at a glance:

• Physics:  The Physics section includes topics such as motion, force, work and energy, sound, and light.
• Chemistry:  The Chemistry section includes topics such as matter, atoms and molecules, structure of the atom, and chemical reactions.
• Biology:  The Biology section includes topics such as cell structure and functions, tissues, diversity in living organisms, natural resources, and environmental management.
• Practical Work:  The Science curriculum also includes practical work, where students perform experiments to observe and understand scientific phenomena.

The Class 9 Science curriculum is designed to provide a strong foundation in science and prepare students for higher education in the field. The curriculum is structured to develop critical thinking, problem-solving, and analytical skills, and to promote the application of scientific concepts in real-life situations. The curriculum is also designed to help students prepare for competitive exams and develop a strong scientific base for future academic and professional pursuits.

Students can also access Case Based Questions of all subjects of CBSE Class 9

• Case Based Questions for Class 9 Maths
• Case Based Questions for Class 9 Social Science
• Case Based Questions for Class 9 English
• Case Based Questions for Class 9 Hindi
• Case Based Questions for Class 9 Sanskrit

Frequently Asked Questions (FAQs) on Case Based Questions for Class 9 Science

Are case-based questions on the class 9 science exam.

Yes, case-based questions are often included in science exams at the class 9 level as they test students' ability to apply their scientific knowledge and skills to real-world situations.

How are case-based questions different from traditional science questions?

Traditional science questions typically focus on testing students' knowledge of specific facts, concepts, and theories. Case-based questions, on the other hand, require students to use their knowledge and understanding to analyze and interpret real-world situations and make informed decisions.

How can students prepare for case-based questions in science?

To prepare for case-based questions in science, students should practice analyzing data and interpreting scientific experiments. They should also work on developing their critical thinking and problem-solving skills.

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CBSE Case Study Questions for Class 9 Science - Pdf Free PDF Download

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Class 9 Science NCERT Notes, Sample Papers & Tests

Get the best grades with the help of myCBSEguide where you can access Class 9 Science question papers, revision notes, important questions, NCERT & Exemplar Solutions, and learning videos.

• CBSE Syllabus

CBSE Sample Papers

Cbse last year papers, user submitted papers, case study questions,   mock tests, matter in our surrounding, is matter around us pure, atoms and molecules, structure of the atoms, cell - basic unit of life, forces and laws of motion, gravitation, work and energy, improvement in food resources.

• CBSE Test Papers
• CBSE Revision Notes

CBSE Important Questions

Other useful resourses,   online tests,   learning videos.

At the secondary stage, students are required to understand abstraction and have quantitative reasoning. At this stage, they are introduced to the idea of atoms and molecules being the building blocks of matter. Students come across new experiments and theories, like Newton’s law of gravitation. Student-friendly myCBSEguide app brings to you a complete course for Class 9 Science.

CBSE Class 9 Science Study Material

The best content available for the CBSE course is available on the myCBSEguide mobile app. Our Class 9 Science course includes sample question papers, chapter-based NCERT solutions, revision notes, exemplar solutions, test papers, and learning videos. Our NCERT solutions are available for free download in PDF format, others comprise premium content. The curriculum of Class 9 Science is designed to provide a number of opportunities for the students to engage with the processes of Science like observing, recording observations, drawing, tabulation, plotting graphs, etc. Therefore, students need to have a robust preparation right from this level to have a stronger foundational knowledge for higher education in the subject of science. Here you will find the Class 9 Science CBSE Syllabus, Sample Papers, Last Year Papers (school-based), expert-made Case Study Questions, and Mock Tests. One can use these to plan their preparation.

CBSE Class 9 Science Syllabus

A syllabus is one such thing that is subject to get updated as and when the requirement is. The secondary science syllabus has been designed around seven broad themes. They are as follows:

• The World of The Living
• How Things Work
• Moving Things
• People and Ideas
• Natural Phenomenon and Natural Resources

Students must be super sure about the syllabus and the chapters it includes so that their preparation goes in the right direction. Check the topic details for the Class 9 Science syllabus 2022-23 on this link. The units along with the marks allotted have been reflected in the table given below.

Class 9 Science NCERT Solutions

The importance of NCERT solutions in the CBSE curriculum is known to all. In fact, for all the CBSE subjects, the chapter-end textbook questions have a special position and should not be overlooked at any cost. You can access to Class 9 NCERT Solutions on our students dashboard for all the subjects. The best part is it is available for free and teachers can use them to generate question papers as well.

NCERT Class 9 Science Chapter-wise Solutions

• Atoms and Molecules part 1
• Diversity in Living Organisms part 2
• Forces and Laws of Motion part 1
• Gravitation part 2
• Improvement in Food Resources part 1
• Is Matter around Us Pure part 1

To access the NCERT solutions for other chapters, click on NCERT Solutions for Class 9 Science .

CBE Class 9 Science Revision Notes

Revision Notes have the crux of a chapter, which assures having command over a condensed knowledge of all the important points included in the chapter. On myCBSEguide, you get quick revision notes for all the subjects. You can check our Class 9 Science Revision Notes which will be very helpful for having an overall view of a particular chapter in no time.

The link to the Revision Notes of some of the Class 9 Science Chapters is listed below:

• Diversity in Living Organisms class 9 Notes Science
• Motion class 9 Notes Science
• Forces and Laws of Motion class 9 Notes Science
• Gravitation class 9 Notes Science
• Work and Energy class 9 Notes Science
• Sound class 9 Notes Science

CBSE Class 9 Science Sample Papers & Test Papers

There is no official sample paper for class 9 issued by the CBSE. Taking a cue from class 10 sample papers, we at myCBSEguide prepare such model question papers in all the subjects for class 9. Having good practice with sample papers is absolutely necessary if you aim to score really well in your papers. Here you get to download the Class 9 Science Sample Papers 2022-23 . Be it a school final examination or board sample papers are helpful for both. It not only makes the students more sure but also enables make the best use of their time.

Apart from sample papers, there is a provision for class 9 science test papers, which can be used by students and teachers equally to keep a tab on small targets. Our online tests and mock tests for class 9 science can be used to learn in a play way method.

CBSE Class 09 Science Case Study Questions

To promote competency-based education, CBSE has included new kinds of questions to test the analytical skills of the students. These are the case study questions in which the students require to identify the problems in the given real-life scenario and find solutions in their own way. You can get the best of the case study questions on myCBSEguide .

For the secondary level, the case study questions for science usually are given in the form of a passage having a diagram or a graph based on which a few subjective or objective questions are asked.  It is obvious that the level of difficulty of these case study questions changes with the class. Hence, we bring the case study questions according to the different levels of difficulty. The Class 9 case study questions are designed to engage the students in critical thinking thing and apply problem-solving techniques. Check out Class 9 Case Study Questions for sample questions.

Students who aim to perform well and pass with flying colours must download the myCBSEguide app and prepare with the help of the materials provided here.

Class 9 Science Case Study Questions

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 …

NCERT Solutions for Class 9

NCERT textbook solutions for class 9 are now available in the myCBSEguide App for free. Students can download and access the latest NCERT book chapter-end questions and answers from the myCBSEguide App and student dashboard. We offer NCERT questions and solutions for class 9 Social Science, Science, Mathematics, English Language …

CBSE Class 9 Science Syllabus 2022-23

CBSE Class 9 Science Syllabus 2022-23 includes Matter- Its Nature and Behaviour, Organization in the Living World, Motion, Force and Work, Food Production etc for the session 2022 – 2023. Here is the detailed syllabus. To download class 9 Science CBSE latest sample question papers for the 2023 exams, please …

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JSTSE Previous Year Papers 2009 to 2019

Download JSTSE previous year papers from the year 2009 to 2019 in PDF format for free from myCBSEguide App and website. JSTSE (Junior Science Talent Search Examination) JSTSE Previous Year Question Papers are now available in myCBSEguide App for free. JSTSE past 10 year question papers and solutions are also …

CBSE Syllabus for Class 9 Science 2019-20

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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

(c) Only for 1 particle

(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

(c) No internal forces 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

(c) Become zero

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

(c) Unequal and in the same 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

(c) Both forces acted on different object simultaneously

(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.

Hope the information shed above regarding Case Study and Passage Based Questions for Class 9 Science Chapter 9 Force and Laws of Motion with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 9 Science Force and Laws of Motion Case Study and passage-based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible

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

• Last modified on: 2 years ago
• Reading Time: 6 Minutes

Case Study/Passage Based Questions:

Question 1:

Read the following and answer any four questions from (i) to (v) given below :

In the figure below the card is flicked with a push. It was observed that the card moves ahead while coin falls in glass.

(i) Give reason for the above observation. (a) The coin possesses inertia of rest, it resists the change and hence falls in the glass. (b) The coin possesses inertia of motion; it resists the change and hence falls in the glass. (c) The coin possesses inertia of rest, it accepts the change and hence falls in the glass. (d) The coin possesses inertia of rest, it accepts the change and hence falls in the glass.

(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. (d) Law of conservation of energy

(iii) If the above coin is replaced by a heavy five-rupee coin, what will be your observation. Give reason. (a) Heavy coin will possess more inertia so it will not fall in tumbler. (b) Heavy coin will possess less inertia so it will fall in tumbler. (c) Heavy coin will possess more inertia so it will fall in tumbler. (d) Heavy coin will possess less inertia so it will not fall in tumbler.

(iv) Name the law which provides the definition of force. (a) Law of conservation of mass (b) Newton’s third law. (c) Newton’s first law (d) Newton’s second law.

(v) State Newton’s first law of motion. (a) Energy can neither be created nor be destroyed, it can be converted from one form to another, total amount of energy always remains constant. (b) A body at rest remains at rest or, if in motion, remains in motion at constant velocity unless it is acted upon by an external unbalanced force. (c) For every action in nature there is an equal and opposite reaction. (d) The acceleration in an object is directly related to the net force and inversely related to its mass.

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Last modified on:2 years agoReading Time:4MinutesCase Study Questions for Class 9 Science Chapter 4 Structure of Atom 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…

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Last modified on:2 years agoReading Time:4MinutesCase Study Questions for Class 9 Science Chapter 10 Gravitation 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 questions based…

Case Study and Passage Based Questions for Class 9 Science Chapter 11 Work and Energy

Last modified on:2 years agoReading Time:4MinutesCase Study Questions for Class 9 Science Chapter 11 Work and Energy 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…

Case Study and Passage Based Questions for Class 9 Science Chapter 13 Why Do We Fall Ill

Last modified on:2 years agoReading Time:4MinutesCase Study Questions for Class 9 Science Chapter 13 Why Do We Fall Ill 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,…

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• Important Questions For Class 9 Science

CBSE Important Questions for Class 9 Science

CBSE Important Questions for Class 9 Science play an important role in students’ exam preparation. It gives an idea of what types of questions can be asked in the examination. It also develops skills for miscellaneous questions, which are generally neglected by students. Moreover, practising CBSE Important Questions for Class 9 Science will boost the student’s confidence.

By solving these important questions, the students will be more familiar with the type of questions asked in final exams. Hence, they will be able to face the exams more confidently and without fear. Science is a subject which needs conceptual understanding. Just mugging up the answer will not fulfil the purpose. While solving the CBSE Class 9 important questions of Science, students will easily grasp the concepts. Apart from this, they will also acquire answer writing skills in a better way. They will learn to draw the diagrams and write the formulas wherever it is required while answering. This will help them fetch more marks in exams.

Students must start solving the important questions once they complete the CBSE Class 9 Science Syllabus at least 2 months before the annual exam. Here, we have provided the chapter-wise CBSE important questions for Class 9 Science. These questions are created from an exam perspective and are most likely to be asked in the exam.

CBSE Class 9 Science Marking Scheme

The table below shows the weightage of the types of questions asked in the exam. Go through it to know the Class 9 Science Exam pattern in detail. We have included all these types of questions in the CBSE Important Questions so that students do not miss anything.

Benefits of Important Questions for CBSE Class 9 Science

CBSE Important Questions for Class 9 Science gives a competitive edge to the students. Here are the benefits of studying through these important questions:

• Students will revise everything and cover the entire syllabus.
• The questions are created by the subject experts exclusively for students to prepare better for the exams.
• There is a high chance that some of these questions will be asked in the exam.
• Students will have good practice with different types of questions and will be well prepared for the Science paper.

We hope students have found this information on “CBSE Important Question for Class 9 Science” useful for their studies. Students can also access CBSE Class 9 Science Sample Papers , solutions, exam tips, projects, etc., on BYJU’S website.

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

CBSE Class 9 Science Chapter-9 Important Questions - Free PDF Download

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 covered in this chapter. The chapter Forces and Laws of Motion explains the change in the state of an object in motion or at rest. 

Vedantu provides all the necessary notes and questions on this chapter, including the revision notes , NCERT solutions with step-by-step explanations, etc. to make it easier for students to understand the concepts. Enroll for Class 9 Science tutoring at Vedantu.com to enhance your exam scores. Vedantu offers free CBSE Solutions (NCERT) and additional study resources. Students seeking improved Math solutions can access Class 9 Maths NCERT Solutions, aiding comprehensive syllabus revision and higher exam scores.

Download CBSE Class 9 Science Important Questions 2024-25 PDF

Also, check CBSE Class 9 Science Important Questions for other chapters:

Study Important Questions for Class 9 Science Chapter 9 - Forces and Laws of Motion

Very short answer questions                             (1 mark).

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

The batsman did not hit the ball hard enough.

Velocity is proportional to the force exerted on the ball.

There is a force on the ball opposing the motion.

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

Ans: (c) There is a force on the ball opposing the motion.

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

\[{{(mv)}^{2}}\]

\[m{{v}^{2}}\]

$\dfrac{1}{2}m{{v}^{2}}$

Ans: (d) $mv$

3. Using a horizontal force of $200N$, 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?

Ans: In order to move the cabinet across the floor at a constant velocity, the net force experienced by it must be zero. Thus a frictional force of $200N$ must be exerted on the cabinet to move it across the floor at a constant velocity, against the horizontal force of $200N$.

4. What is the S.I. unit of momentum?

$\dfrac{ms}{kg}$

$kgm{{s}^{-1}}$

$\dfrac{kg}{ms}$

Ans: (c) $kgm{{s}^{-1}}$

5. What is the numerical formula for force?

$\mathsf{F=ma}$

$\mathsf{F=}\dfrac{\mathsf{m}}{\mathsf{a}}$

$\mathsf{F=m}{{\mathsf{a}}^{\mathsf{2}}}$

$F={{m}^{2}}a$

Ans: (a) $\mathrm{F=ma}$

6. If the initial velocity is zero then the force acting is 

Acceleration

Ans: (a) Acceleration

7. What is the S.I. unit of force?

$\dfrac{kgm}{{{s}^{2}}}$

$\dfrac{kgm}{s}$

$\dfrac{kg{{m}^{2}}}{{{s}^{2}}}$

$kg{{m}^{2}}{{s}^{2}}$

Ans: (a) $\dfrac{\mathrm{kgm}}{{{\mathrm{s}}^{\mathrm{2}}}}$

8. Newton’s first law of motion is also called

Law of Inertia

Law of Momentum

Law of Action & Reaction

None of these

Ans: (a) Law of Inertia

9. Which law explains swimming?

Newton’s first law

Newton’s second law

Newton’s third law

All of these

Ans: (c) Newton’s third law

10. The S.I. unit of weight is:

$\dfrac{N}{s}$

$\dfrac{Nm}{s}$

Ans: (a) $\mathrm{N}$

11. Which equation defines Newton’s Second law of motion?

$\mathsf{F=ma=}\dfrac{\mathsf{dp}}{\mathsf{dt}}$

$\mathsf{F=m}\dfrac{\mathsf{da}}{\mathsf{dt}}\mathsf{=P}$

$\dfrac{\mathsf{dF}}{\mathsf{dt}}\mathsf{=ma=P}$

$\mathsf{F=ma=P}$

Ans: (d) $\mathrm{F=ma=P}$

12. The people in the bus are pushed backwards when the bus starts suddenly due to

Inertia due to Rest

Inertia due to Motion

Inertia due to Direction

Ans: (a) Inertia due to Rest

13. If the force acting on the body is zero. Its momentum is

Ans: (b) constant

14. The inability of the body to change its state of rest or motion is

Acceleration.

Ans: (c) Inertia

Short Answer Questions                                          (2 Marks)

1. An object experiences a net zero external unbalanced force. Is it possible for the object to be traveling 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, it is possible for an object to be traveling with a non-zero velocity if it experiences a net zero external unbalanced force. This is based on Newton’s First law of motion, which states that an object will change its state of motion if and only if the net force acting on it is non-zero.  

Thus, in order to change its motion or to bring about motion, some external unbalanced force is required. 

In this case, the object experiences a net zero unbalanced force, which will cause it to move with some non-zero velocity provided that the object was previously moving with a constant velocity.

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

Ans: When a carpet is beaten with a stick, dust comes out of it because of Newton’s First Law of Motion, the law of Inertia. Initially, the dust particles and the carpet are in a state of rest. When the carpet is beaten with a stick, it causes the carpet to move, while the dust particles, due to inertia of rest, will resist the change in motion. Thus, the carpet’s forward motion will exert a backward force on the dust particles, which makes them move in the opposite direction. Therefore the dust comes out of the carpet.

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

It is advised to tie any luggage kept on the roof of a bus with a rope because of Newton’s First Law of Motion, the law of Inertia. 

When the bus moves, the luggage will be in inertia of motion and say the bus suddenly stops, then the luggage tends to resist this change in motion, causing it to move forward and fall off, if not tied up by a rope. 

Similarly, when the bus decelerates or changes its direction while turning, the inertia of motion of the luggage will try to resist this change in motion, causing the luggage to move oppositely and fall off, if not tied up by a rope.

4. A stone of $1kg$ is thrown with a velocity of $20m{{s}^{-1}}$ across the frozen surface of a lake and comes to rest after traveling a distance of $50m$. What is the force of friction between the stone and the ice?

Ans: Given:

Mass of stone: $\mathrm{m=1kg}$

Initial velocity of stone: $\mathrm{u=20m}{{\mathrm{s}}^{\mathrm{-1}}}$

Final velocity of stone:  $\mathrm{v=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it comes to rest)

Distance traveled on ice: $\mathrm{s=50m}$

To find: Force of friction between stone and ice.

First, we need to find the deceleration:

It is known that – ${{\mathrm{v}}^{\mathrm{2}}}\mathrm{=}{{\mathrm{u}}^{\mathrm{2}}}\mathrm{+2as}$

Thus, ${{\mathrm{0}}^{\mathrm{2}}}\mathrm{=(20}{{\mathrm{)}}^{\mathrm{2}}}\mathrm{+2a(50)}$

$\Rightarrow \mathrm{0=400+100a}$

$\Rightarrow -400=100a$

$\Rightarrow a=-4m{{s}^{-2}}$

The negative sign implies deceleration.

Next, finding the frictional force:

$\mathrm{F=ma}$

$\Rightarrow \mathrm{F=(1)(}-4)$

$\Rightarrow \mathrm{F=}-4N$

Thus, the force of friction between stone and ice is $-4N$ .

5. An automobile vehicle has a mass of $1500kg$. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of $1.7m{{s}^{-2}}$?

Mass of vehicle:$\mathrm{m=1500kg}$

Negative acceleration:$\mathrm{a=}-\mathrm{1}\mathrm{.7m}{{\mathrm{s}}^{\mathrm{-2}}}$

To find: Force of friction between road and vehicle.

It is known that - $\mathrm{F=ma}$

$\Rightarrow \mathrm{F=(1500)(}-1.7)$

$\Rightarrow \mathrm{F=}-2550N$

Thus the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of $1.7m{{s}^{-2}}$is $-2550N$.

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

Mass of object:$\mathrm{m=100kg}$

Initial velocity of object:$\mathrm{u=5m}{{\mathrm{s}}^{\mathrm{-1}}}$

Final velocity of object: $\mathrm{v=8m}{{\mathrm{s}}^{\mathrm{-1}}}$

Time duration of acceleration:$t=6s$

To find: 

Initial momentum

Final momentum

Force exerted on the object

It is known that – $momentum=\operatorname{ma}ss\times velocity$

$Initial\_momentum=\operatorname{ma}ss\times initial\_velocity$

$\Rightarrow Initial\_momentum=100\times 5$

$\Rightarrow Initial\_momentum=500kgm{{s}^{-1}}$

$Final\_momentum=\operatorname{ma}ss\times final\_velocity$

$\Rightarrow Final\_momentum=100\times 8$

$\Rightarrow Final\_momentum=800kgm{{s}^{-1}}$

Now, the force – $\mathrm{F=ma}$

$\mathrm{F=m(}\dfrac{v-u}{t})$

$\Rightarrow \mathrm{F=100(}\dfrac{8-5}{6})$

$\Rightarrow \mathrm{F=100(}\dfrac{3}{6})$

$\Rightarrow \mathrm{F=50N}$

Initial momentum:$500kgm{{s}^{-1}}$

Final momentum: $800kgm{{s}^{-1}}$

Force exerted on object: $50N$

7. Akhtar, Kiran, and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with 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.

Kiran’s statement – 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). 

Based on the law of conservation of momentum, the change of momentum experienced by both the insect and car should be equal. The change in velocity of the insect will be greater, due to its small mass, while the change in velocity of the car is insignificant, due to its larger mass. But the change in momentum before and after collision would be the same. Thus, Kiran’s statement is false.

Akhtar’s statement – since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result, the insect died. 

According to Newton’s third law of motion, for every action, there is an equal and opposite reaction. Thus both the car and insect would experience the same force. So, we can say that Akhtar’s statement is also false.

Rahul’s statement – both the motorcar and the insect experienced the same force and a change in their momentum.

Inferring from the law of conservation of momentum and Newton’s third law of motion, we can say that Rahul’s statement is true.

8. State Newton’s second law of motion?

Ans: Newton’s Second law of motion states that when an unbalanced external force acts on an object, its velocity changes, that is, the object gets accelerated. It can also be stated as the time rate of change of the momentum of a body is equal in both magnitude and direction to the force applied on it.

Mathematically – $\mathrm{F=ma}$, where ‘F’ is the force, ‘m’ is the mass of the object, and ‘a’ is the acceleration.

9. What is the momentum of a body of mass $200g$ moving with a velocity of $15m{{s}^{-1}}$?

Ans: Given: 

Mass of body:$\mathrm{m=200g=0}\mathrm{.2kg}$

Velocity of body: $\mathrm{v=15m}{{\mathrm{s}}^{\mathrm{-1}}}$

To find: Momentum of the body.

$\Rightarrow momentum=0.2\times 15$

$\Rightarrow momentum=3kgm{{s}^{-1}}$

Thus, the momentum of the body is $3kgm{{s}^{-1}}$.

10. Define force and what are the various types of forces?

Ans: Force is defined as the push or pulls on an object that produces a change in the state or shape of the object. It can also cause a change in the speed and/or direction of motion of the object.

The various types of force are:

Mechanical force

Gravitational force

Frictional force

Electrostatic force

Electromagnetic force

Nuclear force

11. A force of $25N$ acts on a mass of $500g$ resting on a frictionless surface. What is the acceleration produced?

Mass:$\mathrm{m=500g=0}\mathrm{.5kg}$

Force exerted: $\mathrm{F=25N}$

To find: Acceleration.

It is known that – $\mathrm{F=ma}$

$\Rightarrow a=\dfrac{F}{m}$

$\Rightarrow a=\dfrac{25}{0.5}$

$\Rightarrow a=50m{{s}^{-2}}$

Thus, the acceleration produced is \[50m{{s}^{-2}}\].

12. State Newton’s first law of Motion?

Ans: Newton’s first law of motion is also called the law of Inertia. It states that 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. Or, an object rest will continue to be at rest and an object in motion will continue to be in motion until and unless it is acted upon by an external force.

13. A body of mass $5kg$ starts and rolls down $32m$of an inclined plane in $4s$. Find the force acting on the body?

Mass of body:$\mathrm{m=5kg}$

Initial velocity of body: $\mathrm{u=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it is starting to roll)

Distance travelled on inclined plane: $\mathrm{s=32m}$

Time duration of rolling:$t=4s$

To find: Force acting on the body.

First we need to find the acceleration:

It is known that – $\mathrm{s=ut+}\dfrac{1}{2}\mathrm{a}{{\mathrm{t}}^{\mathrm{2}}}$

Thus, $\mathrm{32=(0}\times \mathrm{4)+}\dfrac{1}{2}(a\times {{4}^{2}})$

$\Rightarrow \mathrm{32=}\dfrac{1}{2}(a\times 16)$

$\Rightarrow \mathrm{32=}(a\times 8)$

$\Rightarrow a=4m{{s}^{-2}}$

Next, finding the force:

$\Rightarrow \mathrm{F=(5}\times 4)$

$\Rightarrow \mathrm{F=20}N$

Thus, the force acting on the body is $20N$ .

14. On a certain planet, a small stone tossed up at $15m{{s}^{-1}}$ vertically upwards takes $7.5s$ to return to the ground. What is the acceleration due to gravity on the planet?

Initial velocity of stone:$\mathrm{u=15m}{{\mathrm{s}}^{\mathrm{-1}}}$

Final velocity of stone: $\mathrm{v=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it becomes zero at the highest point)

Total time duration of flight (tossed up and falling down to the ground):$t=7.5s$

To find: Acceleration due to gravity of the planet.

It is known that – $\mathrm{v=u+at}$

Thus, $\mathrm{0=15+(at)}$

$\mathrm{t=}\dfrac{-15}{a}$  this denotes the time for one-half of the entire flight.

Thus the total duration of the flight is twice this duration.

i.e. $7.5s=2t$

$\Rightarrow 7.5=2(\dfrac{-15}{a})$

Now, the acceleration due to gravity is –

$a=\dfrac{2\times (-15)}{7.5}$

\[\Rightarrow a=-4m{{s}^{-2}}\]

Thus, the acceleration due to gravity of the planet is $-4m{{s}^{-2}}$.

15. Why is the weight of the object more at the poles than at the equator?

The weight of the object is more at the poles than at the equator because the acceleration due to gravity is slightly greater at the poles than at the equator. This is because - $g=\dfrac{GM}{{{r}^{2}}}$, meaning acceleration due to gravity is inversely proportional to the square of the radius. Since the radius of the earth at the equator is greater than at the poles, the acceleration due to gravity is slightly less at the equator, than at the poles.

Also, we know that weight is directly proportional to acceleration due to gravity $(\because w=m\times g\Rightarrow w\alpha g)$.

Using these two implications, we can say that at the equator, where the radius is larger, the acceleration due to gravity is smaller, the weight is lower. And at the poles, where the radius is smaller, the acceleration due to gravity is greater, the weight is higher.

16. Why does the passenger sitting in a moving bus are pushed in the forward direction when the bus stops suddenly?

Ans: The passengers sitting in the moving bus are pushed in the forward direction when the bus stops suddenly due to inertia because the passengers' upper body continues to be in a state of motion, while the lower part of the body that is in contact with the seat remains at rest. As a result, the passenger’s upper body is pushed in the forward direction, in the direction in which the bus was moving before coming to a halt.

17. Why does the boat move backward when the sailor jumps in the forward direction?

Ans: The boat moves backward when the sailor jumps in the forward direction because of Newton’s third law of motion which states that for every action, there is an equal and opposite reaction. Thus, when the sailor jumps in the forward direction he is causing an action force due to which the boat moves backward. In response, the boat exerts an equal and opposite force (the reaction force) on the sailor due to which he is pushed in the forward direction.

18. Derive the law of conservation of momentum from Newton’s third law?

Ans: Newton’s third law of motion states that for every action, there is an equal and opposite reaction, that acts on different bodies.

Say we have two objects A and B of masses ${{m}_{A}}$ and ${{m}_{B}}$ are traveling in the same direction along a straight line at different velocities ${{u}_{A}}$ and ${{u}_{B}}$, respectively. 

Consider that there are no other external unbalanced forces acting on them. 

Let ${{u}_{A}}>{{u}_{B}}$ and the two objects collide with each other.

During collision which lasts for a time $t$, A exerts a force ${{F}_{AB}}$ on B and B exerts a force ${{F}_{BA}}$ on  A. 

Say, ${{v}_{A}}$ and ${{v}_{B}}$ are the velocities of the two A and B after the collision, respectively.

Momentum of A before collision: \[{{m}_{A}}\times {{u}_{A}}\]

Momentum of A after collision: \[{{m}_{A}}\times {{v}_{A}}\]

Momentum of B before collision: \[{{m}_{B}}\times {{u}_{B}}\]

Momentum of B after collision: \[{{m}_{B}}\times {{v}_{B}}\]

It is also known that force can also be defined as the rate of change of momentum, i.e. $F=ma=m(\dfrac{v-u}{t})=\dfrac{mv-mu}{t}$

Now, the rate of change of momentum of A during collision is $\dfrac{{{m}_{A}}{{v}_{A}}-{{m}_{A}}{{u}_{A}}}{t}$, which is the force ${{F}_{AB}}$.

And the rate of change of momentum of B during collision is \[\dfrac{{{m}_{B}}{{v}_{B}}-{{m}_{B}}{{u}_{B}}}{t}\], which is the force ${{F}_{BA}}$.

Based on Newton’s third law of motion, the force ${{F}_{AB}}$  exerted by Aon B and force ${{F}_{BA}}$ exerted by B on A are equal in magnitude but opposite in direction.

i.e.  ${{F}_{AB}}=-{{F}_{BA}}$

Using the formulae – 

${{F}_{AB}}=-{{F}_{BA}}$

\[\Rightarrow \dfrac{{{m}_{A}}{{v}_{A}}-{{m}_{A}}{{u}_{A}}}{t}=-(\dfrac{{{m}_{B}}{{v}_{B}}-{{m}_{B}}{{u}_{B}}}{t})\]

Simplifying,

\[\Rightarrow {{m}_{A}}{{v}_{A}}-{{m}_{A}}{{u}_{A}}=-({{m}_{B}}{{v}_{B}}-{{m}_{B}}{{u}_{B}})\]

\[\Rightarrow {{m}_{A}}{{v}_{A}}-{{m}_{A}}{{u}_{A}}=-{{m}_{B}}{{v}_{B}}+{{m}_{B}}{{u}_{B}}\]

Rearranging,

\[{{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}={{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}\]

Here, \[{{m}_{A}}{{u}_{A}}+{{m}_{B}}{{u}_{B}}\]is the total sum of momentum of A and B before collision and \[{{m}_{A}}{{v}_{A}}+{{m}_{B}}{{v}_{B}}\] is the total sum of momentum of A and B after collision.

This equation implies that the final momentum of the two objects after the collision is equal to the initial momentum of the two objects before the collision.

Thus the law of conservation of momentum, for an isolated system, is that the total initial momentum for an event is equal to total initial momentum.

19. An astronaut has $80kg$ mass on earth.

i) What is his weight on earth?

Mass of astronaut: $\mathrm{m=80kg}$

To find his weight on earth.

It is known that,

Acceleration due to gravity on earth: ${{g}_{e}}=10m{{s}^{-2}}$

Acceleration due to gravity on mars:${{g}_{m}}=3.7m{{s}^{-2}}$

Weight: $w=m\times g$

Weight on earth: ${{w}_{e}}=m\times {{g}_{e}}$

$\Rightarrow {{w}_{e}}=80\times 10$

${{w}_{e}}=800N$

ii) What will be his mass and weight on mars with ${{g}_{m}}=3.7m{{s}^{-2}}$?

Mass of astronaut:$\mathrm{m=80kg}$

To find his mass and weight on mars.

Weight on mars: ${{w}_{m}}=m\times {{g}_{m}}$

$\Rightarrow {{w}_{m}}=80\times 3.7$

${{w}_{m}}=296N$

The mass of astronauts remains the same on mars because it is a constant value. 

Thus, mass on mars is $\mathrm{m=80kg}$.

Short Answer Questions                                     (3 Marks)

1. Which of the following has more inertia:

a. A rubber ball and a stone of the same size?

Ans: Inertia depends on the mass of the object. The larger the mass, the greater will be its inertia and vice-versa. In the case of a rubber ball and a stone of the same size, it is clear that the stone will have greater inertia than the ball. It is because, despite being the same size, the stone weighs more than the rubber ball.

b. A bicycle and a train?

Ans: Inertia depends on the mass of the object. The larger the mass, the greater will be its inertia and vice-versa. In the case of a bicycle and a train, it is clear that the train will have greater inertia than the bicycle because the train weighs more than the bicycle.

c. A five rupees coin and a one-rupee coin?

Ans: Inertia depends on the mass of the object. The larger the mass, the greater will be its inertia and vice-versa. In the case of a five rupees coin and a one-rupee coin, the five rupees coin will have greater inertia than the one-rupee coin because five rupees coin weighs more than a 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: In the example, the number of times the velocity of football changes is four.

(i) The velocity of the football changes first when a player kicks the ball towards another player on his team. Here, the agent supplying the force is the foot of the football player who is kicking the ball.

(ii) The velocity of the football changes second when that another player kicks the ball towards the goal. Here, the agent supplying the force is the foot of the other player who is now kicking the ball towards the goal.

(iii) The velocity of the football changes for the third time when the goalkeeper of the opposite team stops the football by collecting it. Here, the agent supplying the force are the hands of the goalkeeper who collects the ball.

(iv) The velocity of the football changes for the fourth time when the goalkeeper kicks it towards a player of his team. Here, the agent supplying the force is the foot of the goalkeeper who is now kicking the ball towards his teammate.

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

Ans: Some of the leaves may get detached from a tree if we vigorously shake its branch because the branches of the tree will come into motion while the leaves tend to continue in their state of rest. This is due to the inertia of rest of the leaves. The force of shaking will act on the leaves with the change in direction rapidly, which results in the leaves detaching and falling off from the tree.

4. Why do you fall in the forward direction when a moving bus breaks to a stop and fall back when it accelerates from rest?

Similarly, the passengers sitting in the bus are pushed in the backward direction when the bus accelerates from rest due to inertia, because the passengers’ upper body continues to be in a state of rest, while the lower part of the body that is in contact with the seat is set in motion. As a result, the passenger’s upper body is pushed in the backward direction, in the opposite to which the bus starts to move.

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

In this case, we are dealing with unbalanced forces. It is true that the horse exerts an action force on the cart and experiences a reaction force from the cart. But also, the horse creates an action force on the ground over which it is walking, and experiences a reaction force from the ground.

In pulling the cart, the action force of the horse pulling the cart is greater than the reaction force of the cart, resisting the pull. Thus the cart moves in the direction of the pull of the horse.

In stepping on the ground, the horse creates an action force on the ground in the backward direction during each step, while the ground creates a reaction force pushing the horse forward.

In this was a horse can pull a cart.

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

Ans: It is difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity because of Newton’s third law of motion. In this case, the water being ejected out in the forward direction with great force (action) will create a backward force that results in the backward movement (reaction) of the hose pipe. As a result of this backward force and movement, it will be difficult for the fireman to hold the hose properly with stability.

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

Mass of rifle: ${{m}_{1}}=4kg$

Mass of bullet: ${{m}_{2}}=50g=0.05kg$ 

Initial velocity of rifle: ${{u}_{1}}=0m{{s}^{-1}}$ (it is stationary during firing)

Initial velocity of bullet:${{u}_{2}}=0m{{s}^{-1}}$(it starts from rest, inside the barrel of the rifle)

Fired velocity of bullet:${{v}_{2}}=35m{{s}^{-1}}$

To find: Recoil velocity of rifle:${{v}_{1}}$

By using the law of conservation of momentum, for an isolated system, is that the total initial momentum for an event is equal to total initial momentum, we get – 

\[{{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}={{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\]

Here, \[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\]is the total sum of momentum of rifle and bullet before firing and \[{{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\] is the total sum of momentum of rifle and bullet after firing.

Substituting the values in – \[{{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}={{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\]

\[\Rightarrow (4\times {{v}_{1}})+(0.05\times 35)=(4\times 0)+(0.05\times 0)\]

\[\Rightarrow (4\times {{v}_{1}})+(17.5)=0\]

\[\Rightarrow (4\times {{v}_{1}})=-17.5\]

\[\Rightarrow {{v}_{1}}=-4.375m{{s}^{-1}}\](The negative sign indicates the backward direction in which the rifle moves when it recoils)

Thus, the recoil velocity of the rifle is \[4.375m{{s}^{-1}}\].

8. An $8000kg$ engine pulls a train of $5$ wagons, each of $2000kg$, along a horizontal track. If the engine exerts a force of $40000N$ and the track offers a friction force of $5000N$, then calculate:

(a) The net accelerating force

Force exerted by engine on wagons: $F=40000N$

Frictional exerted on wagons:$f=5000N$

Mass of engine:${{m}_{e}}=8000kg$

Mass of each wagon:${{m}_{w}}=2000kg$

Mass of all five wagons:${{m}_{W}}=5\times {{m}_{w}}=5\times 2000=10000kg$

Mass of entire train: ${{m}_{T}}={{m}_{e}}+{{m}_{W}}=8000+10000=18000kg$

To find accelerating force.

Net accelerating force can be found by subtracting the frictional force from the force exerted by the engine on the wagons.

Thus, $NetAcceleratingForce=ForceOfEngine-FrictionalForce$

$\Rightarrow NetAcceleratingForce=F-f$

$\Rightarrow NetAcceleratingForce=40000-5000$

$\Rightarrow NetAcceleratingForce=35000N$

(b) The acceleration of the train

Ans:  Given:

To find the acceleration of the train.

$\Rightarrow a=\dfrac{NetAcceleratingForce}{MassOfTrain}$

$\Rightarrow a=\dfrac{35000}{18000}$

$\Rightarrow a=1.944m{{s}^{-2}}$

(c) The force of wagon $1$ on wagon $2$.

To find the force exerted by wagon $1$ on wagon $2$

Here, wagon $1$ exerts a pulling force on the remaining $4$ wagons

$\Rightarrow {{\mathrm{F}}_{21}}\mathrm{=(4}{{\mathrm{m}}_{w}}\mathrm{)}\times \mathrm{a}$

$\Rightarrow {{\mathrm{F}}_{21}}\mathrm{=(4}\times 2000\mathrm{)}\times 1.944$

$\Rightarrow {{\mathrm{F}}_{21}}\mathrm{=15552}N$

9. Two objects, each of mass $1.5kg$, are moving in the same straight line but in opposite directions. The velocity of each object is $2.5m{{s}^{-1}}$ before the collision during which they stick together. What will be the velocity of the combined object after collision?

Mass of object 1: ${{m}_{1}}=1.5kg$

Mass of object 2: ${{m}_{2}}=1.5kg$ 

Initial velocity of object 1: ${{u}_{1}}=2.5m{{s}^{-1}}$

Initial velocity of object 2:${{u}_{2}}=-2.5m{{s}^{-1}}$(negative sign because it is moving in the opposite direction)

Mass of combined object after collision: $m={{m}_{1}}+{{m}_{2}}=1.5+1.5=3kg$

To find: Final velocity of the combined object after collision:$v$

\[mv={{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\]

Here, \[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\]is the total sum of the momentum of objects before the collision and \[mv\] is the total momentum of the combined objects after the collision.

Substituting the values in – \[mv={{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\]

\[\Rightarrow (3\times v)=(1.5\times 2.5)+(1.5\times -2.5)\]

\[\Rightarrow (3\times v)=(3.75)+(-3.75s)\]

\[\Rightarrow (3\times v)=0\]

$\Rightarrow v=0m{{s}^{-1}}$

Thus, the velocity of the combined object after collision is \[0m{{s}^{-1}}\].

10. 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.

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. This pair of forces is called the action-reaction pair. 

In the case of the massive truck parked alongside the road, the action-reaction pair is the weight of the truck exerting a force on the road in the downward direction (action), and the static friction of the road in the upward direction (reaction), which keeps the truck at rest. These two equal and opposite forces cancel out each other, which is why the truck will not move.

For it to move, we need to apply additional external force to overcome the static friction of the road. 

Thus, as the student explained, the truck does not move because the two opposite and equal forces of the truck and road cancel out each other is valid.

11. A hockey ball of mass $200g$ traveling at $10m{{s}^{-1}}$ is struck by a hockey stick so as to return it along its original path with a velocity at $5m{{s}^{-1}}$. Calculate the change of momentum that occurred in the motion of the hockey ball by the force applied by the hockey stick.

Mass of hockey ball: $m=200g=0.2kg$

Initial velocity of hockey ball: $u=10m{{s}^{-1}}$

Final velocity of hockey ball:$v=-5m{{s}^{-1}}$ (because it moves back in its original direction)

To find: Change in momentum of hockey ball due to the force of hockey stick

\[ChangeOfMomentum=mv-mu\]

Here, \[mu\]is the initial momentum of the hockey ball and \[mv\] is the final momentum of the hockey ball.

Substituting the values in –\[ChangeOfMomentum=mv-mu\]

\[\Rightarrow ChangeOfMomentum=(0.2\times -5)-(0.2\times 10)\]

\[\Rightarrow ChangeOfMomentum=(-1)-(2)\]

\[\Rightarrow ChangeOfMomentum=-3kgm{{s}^{-1}}\]

Thus, the change in momentum of hockey ball due to the force of hockey stick is \[-3kgm{{s}^{-1}}\].

12. A bullet of mass $10g$ traveling horizontally with a velocity of $150m{{s}^{-1}}$ strikes a stationary wooden block and comes to rest in $0.03s$. 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.

Mass of bullet: $m=10g=0.01kg$

Initial velocity of bullet: $u=150m{{s}^{-1}}$

Final velocity of bullet: $\mathrm{v=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it comes to rest after penetration)

Time duration of bullet travel:$t=0.03s$

Distance of penetration of bullet into the block

Force exerted by the block on the bullet

Distance of penetration:

Thus, $\mathrm{0=150+(a}\times 0.03\mathrm{)}$

$\Rightarrow \mathrm{(a}\times 0.03\mathrm{)=}-\mathrm{150}$

$\Rightarrow \mathrm{a=}-5000m{{s}^{-2}}$

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

$\Rightarrow \mathrm{s=(150}\times 0.03\mathrm{)+}\dfrac{1}{2}(-5000\times {{0.03}^{2}})$

$\Rightarrow \mathrm{s=(}4.5\mathrm{)+}(-2.25)$

$\Rightarrow \mathrm{s=}2.25m$

$\Rightarrow \mathrm{F=(0}\mathrm{.01}\times -5000)$

$\Rightarrow \mathrm{F=}-5\mathrm{0}N$

Distance of penetration of bullet into the block is $2.25m$

Force exerted by the block on the bullet is $-50N$

13. An object of mass $1kg$ traveling in a straight line with a velocity of $10m{{s}^{-1}}$ collides with and sticks to, a stationary wooden block of mass $5kg$. 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 object 1: ${{m}_{1}}=1kg$

Mass of object 2: ${{m}_{2}}=5kg$ 

Initial velocity of object 1: ${{u}_{1}}=10m{{s}^{-1}}$

Initial velocity of object 2:${{u}_{2}}=0m{{s}^{-1}}$(because it is stationary)

Mass of combined object after collision: $m={{m}_{1}}+{{m}_{2}}=1+5=6kg$

Momentum before impact

Momentum after impact

The final velocity of the combined object after collision:$v$

Momentum before impact is the Initial momentum:

\[InitialMomentum={{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\]

\[\Rightarrow InitialMomentum=(1\times 10)+(5\times 0)\]

\[\Rightarrow InitialMomentum=10kgm{{s}^{-1}}\]

Momentum after impact is the Final momentum:

\[FinalMomentum=mv\]

Thus we get – \[FinalMomentum=mv={{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}=10kgm{{s}^{-1}}\]

\[\Rightarrow FinalMomentum=10kgm{{s}^{-1}}\]

Finding the final velocity of the combined object:

\[FinalMomentum=10kgm{{s}^{-1}}\]

\[\Rightarrow mv=10\]

\[\Rightarrow (6\times v)=10\]

\[\Rightarrow v=1.67m{{s}^{-1}}\]

Momentum before impact is \[10kgm{{s}^{-1}}\]

Momentum after impact is \[10kgm{{s}^{-1}}\]

Final velocity of the combined object after collision is \[1.67m{{s}^{-1}}\]

14. How much momentum will a dumb-bell of mass $10kg$ transfer to the floor if it falls from a height of $80cm$? Take its downward acceleration to be $10m{{s}^{-1}}$.

Mass of dumbbell: $m=10kg$

Initial velocity of dumbbell: $u=0m{{s}^{-1}}$(as it starts from rest)

Height of fall of dumbbell: $\mathrm{h=80cm=0}\mathrm{.8m}$ 

Acceleration due to gravity: $g=10m{{s}^{-2}}$

To find: Momentum transferred to the ground by dumbbell.

It is known that – ${{v}^{2}}={{u}^{2}}+2gh$

$\Rightarrow {{v}^{2}}={{(0)}^{2}}+(2\times 10\times 0.8)$

$\Rightarrow {{v}^{2}}=16$

$\Rightarrow v=4m{{s}^{-1}}$

Now, \[Momentum=mv\]

\[\Rightarrow Momentum=(10\times 4)\]

\[\Rightarrow Momentum=40kgm{{s}^{-1}}\]

Thus, the momentum transferred to the ground by dumbbell is \[40kgm{{s}^{-1}}\]

15. A force of $15N$acts for $5s$on a body of mass $5kg$ which is initially at rest. Calculate.

(a) Final velocity of the body

Mass of body: $m=5kg$

Initial velocity of body: $u=0m{{s}^{-1}}$(as it starts from rest)

Force acting on the body: $F=15N$

Time: $t=5s$

To find the final velocity of the body.

First we need to find the acceleration produced.

$\Rightarrow \mathrm{a=}\dfrac{F}{m}$

$\Rightarrow \mathrm{a=}\dfrac{15}{5}$

$\Rightarrow a=3m{{s}^{-2}}$

Thus, $\mathrm{v=0+(3}\times 5\mathrm{)}$

$\Rightarrow v\mathrm{=15}m{{s}^{-1}}$

(b) The displacement of the body

To find the displacement of the body.

Next, the distance of penetration:

$\Rightarrow \mathrm{s=(0}\times 5)\mathrm{+}\dfrac{1}{2}(3\times {{5}^{2}})$

$\Rightarrow \mathrm{s=(0)+}(37.5)$

$\Rightarrow \mathrm{s=37}.5m$

16. Differentiate between mass and weight?

Ans: The difference between mass and weight is given below,

17. A scooter is moving with a velocity of $20m{{s}^{-1}}$when brakes are applied. The mass of the scooter and the rider is $180kg$. The constant force applied by the brakes is $500N$.

(a) How long should the brakes be applied to make the scooter comes to a halt?

Mass of scooter and rider: $m=180kg$

Initial velocity of scooter: $u=20m{{s}^{-1}}$

Final velocity of scooter: $v=0m{{s}^{-1}}$(as it halts after applying the brake)

Force of the brake: $F=-500N$(as it opposes the motion)

To find the time duration over which brake should be applied to stop the scooter.

$\Rightarrow -500\mathrm{=180(}\dfrac{0-20}{t})$

$\Rightarrow t\mathrm{=(}\dfrac{180\times (-20)}{-500})$

$\Rightarrow t\mathrm{=(}\dfrac{-3600}{-500})$

$\Rightarrow t\mathrm{=}7.2s$

(b) How far does the scooter travel before it comes to rest?

To find distance travelled by scooter before coming to halt.

Acceleration – $\mathrm{a=(}\dfrac{v-u}{t})$

$\Rightarrow \mathrm{a=(}\dfrac{0-20}{7.2})$

$\Rightarrow \mathrm{a=}-2.78m{{s}^{-2}}$

Acceleration is negative because it is retarding the motion of the scooter.

$\Rightarrow \mathrm{s=(20}\times 7.2)\mathrm{+}\dfrac{1}{2}(-2.78\times {{7.2}^{2}})$

$\Rightarrow \mathrm{s=(144)+}(-72.1)$

$\Rightarrow \mathrm{s=}71.9m$

18. State Newton’s third law of motion and how does it explain the walking of man on the ground?

Ans: Newton’s third law of motion states that for every action, there is an equal and opposite reaction acting on different bodies. This implies the existence of the action-reaction force pair. That is, for every action force created an equal and opposite reaction force will be created.

The walking of a man on the ground can be explained with Newton’s third law of motion. During walking on the ground, the man creates an active force on the ground in the backward direction during each step, while the ground creates a reaction force pushing the man forward enabling him to walk.

19. With what speed must a ball be thrown vertically up in order to rise to a maximum height of $45m$? And for how long will it be in the air?

Final velocity of stone: $v=0m{{s}^{-1}}$(as it attains zero velocity at maximum height)

Height to which stone is to be thrown: $h=45m$

Acceleration due to gravity: $g=-10m{{s}^{-2}}$(it is negative because the stone is thrown against gravity upwards)

The initial velocity with which the stone is to be thrown: $u$

Time duration over which the stone stays in the air

Initial velocity:

$\Rightarrow {{0}^{2}}={{u}^{2}}+(2\times -10\times 45)$

\[\Rightarrow 0={{u}^{2}}+(-900)\]

\[\Rightarrow {{u}^{2}}=900\]

$\Rightarrow u=30m{{s}^{-1}}$

It is known that – $\mathrm{v=u+gt}$

Thus, $\mathrm{0=30+(}-10\times t\mathrm{)}$

$\Rightarrow t=3s$

It takes $3s$ to go up and another $3s$ to come down. So we can say that the total time the stone is air bound is $3s+3s=6s$

The initial velocity with which the stone is to be thrown is $30m{{s}^{-1}}$

Time duration over which the stone stays in the air is $6s$

20. State Newton’s second law of motion and derive it mathematically?

Ans: Newton’s second law of motion states that the rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.

Mathematical derivation:

Say we have an object of mass $m$that is moving along a straight line with an initial velocity, $u$. 

It is then uniformly accelerated to velocity, $v$ in time, $t$ by the application of a constant force, $F$ throughout the time, $t$. 

Initial Momentum of object: \[{{p}_{1}}=m\times u\]

Final Momentum of object: \[{{p}_{2}}=m\times v\]

Now, the change of momentum is the Final momentum subtracted by the Initial momentum

Thus, $\Delta p={{p}_{2}}-{{p}_{1}}=mv-mu=m(v-u)$

$\Rightarrow \Delta p=m(v-u)$

The rate of change of momentum is $\dfrac{\Delta p}{t}$

i.e. $\dfrac{\Delta p}{t}=\dfrac{m(v-u)}{t}$

We know that the applied force is proportional to the rate of change of momentum of the object.

$F=\dfrac{\Delta p}{t}$

$\Rightarrow F=\dfrac{m(v-u)}{t}$

But, acceleration $a=\dfrac{v-u}{t}$

Using these, we get

$\Rightarrow F=ma$

The SI unit of force is Newton ($Kgm{{s}^{-2}}$)

The second law of motion gives a method to measure the force acting on an object as a product of its mass and acceleration.

21. A bullet traveling at $360m{{s}^{-1}}$ strikes a block of softwood. The mass of the bullet is $2.0g$. Does the bullet come to rest after penetrating $10cm$ into the wood?

Find the average deceleration force exerted by the wood.

Mass of bullet: $m=2.0g=0.002kg$

Initial velocity of bullet: $u=360m{{s}^{-1}}$

Distance travelled by the bullet into the block:$s=10cm=0.1m$

To find the average deceleration force exerted by the wood block.

It is known that – ${{v}^{2}}={{u}^{2}}+2as$

$\Rightarrow {{0}^{2}}={{360}^{2}}+(2\times a\times 0.1)$

\[\Rightarrow 0=129600+(0.2a)\]

\[\Rightarrow 0.2a=-129600\]

$\Rightarrow a=-648000m{{s}^{-2}}$

The acceleration is negative because it opposes the motion of bullet.

Next, force

$\Rightarrow \mathrm{F=(0}\mathrm{.002}\times -648000)$

$\Rightarrow \mathrm{F=}-1296N$

(b) Find the time taken by the bullet to come to rest.

Distance travelled by the bullet into the block: $s=10cm=0.1m$

To find the time taken by the bullet to come to rest.

Thus, $\mathrm{0=360+(}-648000\times t\mathrm{)}$

$\Rightarrow -648000t\mathrm{=}-36\mathrm{0}$

$\Rightarrow t=5.56\times {{10}^{-4}}s$

22. Two objects A and B are dropped from a height. The object B being dropped $1s$ after A was dropped. How long after A was dropped will A and B be $10m$apart?

Ans: Given: Object B is dropped one second after object A.

To find: Time at which A and B will be $10m$ apart

We can say that the initial velocity of both A and B as ${{\mathrm{u}}_{A}}={{\mathrm{u}}_{B}}\mathrm{=0m}{{\mathrm{s}}^{\mathrm{-1}}}$, since they are dropped from rest.

For object A – ${{\mathrm{s}}_{A}}\mathrm{=}{{\mathrm{u}}_{A}}\mathrm{t+}\dfrac{1}{2}\mathrm{g}{{\mathrm{t}}_{A}}^{\mathrm{2}}$

For object B –${{\mathrm{s}}_{B}}\mathrm{=}{{\mathrm{u}}_{B}}\mathrm{t+}\dfrac{1}{2}\mathrm{g}{{\mathrm{t}}_{B}}^{\mathrm{2}}$

We take acceleration to be acceleration due to gravity because it is being dropped from a height downwards to the earth.

Since we need to find the time at which A and B will be $10m$ apart

Let’s say - ${{\mathrm{s}}_{A}}-{{\mathrm{s}}_{B}}\mathrm{=10m}$

Also, since object B is dropped one second after object A, we can say that ${{t}_{B}}={{t}_{A}}-1$

Substituting in – ${{\mathrm{s}}_{A}}-{{\mathrm{s}}_{B}}\mathrm{=10m}$

$\Rightarrow {{\mathrm{s}}_{A}}-{{\mathrm{s}}_{B}}\mathrm{=}{{\mathrm{u}}_{A}}\mathrm{t+}\dfrac{1}{2}\mathrm{g}{{\mathrm{t}}_{A}}^{\mathrm{2}}-{{\mathrm{u}}_{B}}\mathrm{t+}\dfrac{1}{2}\mathrm{g}{{\mathrm{t}}_{B}}^{\mathrm{2}}$

$\Rightarrow 10\mathrm{=}{{\mathrm{u}}_{A}}\mathrm{t+}\dfrac{1}{2}\mathrm{g}{{\mathrm{t}}_{A}}^{\mathrm{2}}-\left[ {{\mathrm{u}}_{B}}\mathrm{t+}\dfrac{1}{2}\mathrm{g(}{{\mathrm{t}}_{A}}-1{{)}^{\mathrm{2}}} \right]$

$\Rightarrow 10\mathrm{=(0}\times \mathrm{t)+}\dfrac{1}{2}\mathrm{(10}\times {{\mathrm{t}}_{A}}^{\mathrm{2}})-\left[ \mathrm{(0}\times \mathrm{t)+}\dfrac{1}{2}\mathrm{(10}\times {{\mathrm{(}{{\mathrm{t}}_{A}}-1)}^{\mathrm{2}}}) \right]$

$\Rightarrow 10\mathrm{=(5}\times {{\mathrm{t}}_{A}}^{\mathrm{2}})-\left[ 5\times \mathrm{(}{{\mathrm{t}}_{A}}^{\mathrm{2}}-2{{\mathrm{t}}_{A}}+1) \right]$

$\Rightarrow 10\mathrm{=5}{{\mathrm{t}}_{A}}^{\mathrm{2}}-\mathrm{5}{{\mathrm{t}}_{A}}^{\mathrm{2}}+10{{\mathrm{t}}_{A}}-5$

$\Rightarrow 10\mathrm{=}10{{\mathrm{t}}_{A}}-5$

\[\Rightarrow 10{{\mathrm{t}}_{A}}=15\]

$\Rightarrow {{\mathrm{t}}_{A}}=1.5s$

Thus, the time at which A and B will be $10m$ apart is $1.5s$

23. A boy throws a stone up with a velocity of $60m{{s}^{-1}}$.

(a) How long will it take to reach the maximum height? $(g=-10m{{s}^{-2}})$

Initial velocity of stone: $\mathrm{u=60m}{{\mathrm{s}}^{\mathrm{-1}}}$ 

To find time to reach maximum height.

$\Rightarrow \mathrm{v=u+gt}$

$\Rightarrow 0\mathrm{=60+(}-10\mathrm{t)}$

$\Rightarrow -10\mathrm{t=}-60$

$\Rightarrow t=6s$

(b) What will be the maximum height reached by the stone?

To find maximum height.

$\Rightarrow \mathrm{h=ut+}\dfrac{1}{2}\mathrm{g}{{\mathrm{t}}^{\mathrm{2}}}$

$\Rightarrow \mathrm{h=(60}\times \mathrm{6)+}\dfrac{1}{2}\mathrm{(}-10\times {{6}^{\mathrm{2}}})$

$\Rightarrow \mathrm{h=(360)+(}-180)$

$\Rightarrow \mathrm{h=180m}$

(c) What will be its velocity when it reaches the ground?

To find velocity when it reaches the ground.

Velocity when reaching the ground:

For this, we consider the initial velocity (from its maximum attained height) is zero. And the acceleration due to gravity becomes positive because it is falling down.

i.e. Initial velocity of stone: $\mathrm{u=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ 

Now,  ${{v}^{2}}={{u}^{2}}+2as$

$\Rightarrow {{v}^{2}}={{u}^{2}}+2gs$

$\Rightarrow {{v}^{2}}={{0}^{2}}+(2\times 10\times 180)$

\[\Rightarrow {{v}^{2}}=3600\]

\[\Rightarrow v=\sqrt{3600}\]

$\Rightarrow v=60m{{s}^{-1}}$

24. A certain particle has a weight of $30N$at a place where the acceleration due to gravity is $9.8m{{s}^{-2}}$.

(a) What are its mass and weight at a place where the acceleration due to gravity is $3.5m{{s}^{-2}}$?

Weight of particle:$\mathrm{w=30N}$

Acceleration due to gravity on that planet: ${{g}_{1}}=9.8m{{s}^{-2}}$

Mass of particle: $m$

To find mass and weight of particle on planet with${{g}_{2}}=3.5m{{s}^{-2}}$.

Mass of particle: $m=\dfrac{w}{g}$

$\Rightarrow m=\dfrac{30}{9.8}$

$\Rightarrow m=3.06kg$

Mass and Weight on planet with${{g}_{2}}=3.5m{{s}^{-2}}$

Mass is a constant quantity irrespective of place. 

So, $\Rightarrow m=3.06kg$

Weight: $w=m\times {{g}_{2}}$

$\Rightarrow w=3.06\times 3.5$

(b) What will be its mass and weight at a place where the acceleration due to gravity is zero?

To find mass and weight of particle on planet with${{g}_{3}}=0m{{s}^{-2}}$ s.

Mass and Weight on planet with${{g}_{3}}=0m{{s}^{-2}}$

Weight: $w=m\times {{g}_{3}}$

$\Rightarrow w=3.06\times 0$

25. Why does a person while firing a bullet holds the gun tightly to his shoulders?

Ans: A person while firing a bullet holds the gun tightly to his shoulder because of the recoil of the gun when the bullet is fired. This is in accordance with Newton’s third law of motion, which states that for every action, there is an equal and opposite reaction, that acts on different bodies. 

When a bullet is fired, the forward motion of the bullet (action) creates a recoil or backward motion of the gun (reaction). The action force being much greater will create an equivalent recoil force in the backward direction. 

If the person who holds the gun does not hold it properly to his shoulders that can result in injury. This is because the shoulder absorbs most of the force during recoil that enables the shooter to take a steady shot.

Thus, if not held tightly to his shoulders, the shot will not be precise and this can also cause the gun to fly away from his hands.

26. A car is moving with a velocity of $16m{{s}^{-1}}$ when brakes are applied. The force applied by the brakes is $1000N$. The mass of the car its passengers is $1200kg$.

How long should the brakes be applied to make the car come to a halt?

Mass of car and passengers: $m=1200kg$

Initial velocity of car: $u=16m{{s}^{-1}}$

Final velocity of car: $v=0m{{s}^{-1}}$(as it halts after applying the brake)

Force of the brake: $F=-1000N$(as it opposes the motion)

To find time duration over which brake should be applied to stop the car.

$\Rightarrow -1000\mathrm{=1200(}\dfrac{0-16}{t})$

$\Rightarrow t\mathrm{=(}\dfrac{1200\times (-16)}{-1000})$

$\Rightarrow t\mathrm{=(}\dfrac{-19200}{-1000})$

$\Rightarrow t\mathrm{=19}\mathrm{.2}s$

How far does the car travel before it comes to rest?

To find distance travelled by car before coming to halt.

$\Rightarrow \mathrm{a=(}\dfrac{0-16}{19.2})$

$\Rightarrow \mathrm{a=}-0.83m{{s}^{-2}}$

$\Rightarrow \mathrm{s=(16}\times 19.2)\mathrm{+}\dfrac{1}{2}(-0.83\times {{19.2}^{2}})$

$\Rightarrow \mathrm{s=(307}\mathrm{.2)+}(-152.98)$

$\Rightarrow \mathrm{s=154}\mathrm{.2}m$

Long Answer Questions                       (5 Marks)

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

Mass of object 1: ${{m}_{1}}=100g=0.1kg$

Mass of object 2: ${{m}_{2}}=200g=0.2kg$ 

Initial velocity of object 1: ${{u}_{1}}=2m{{s}^{-1}}$

Initial velocity of object 2:${{u}_{2}}=1m{{s}^{-1}}$

Final velocity of object 1:${{v}_{1}}=1.67m{{s}^{-1}}$

To find: Final velocity of object 2:${{v}_{2}}$

Here, \[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\]is the total sum of momentum of objects before collision and \[{{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\] is the total sum of momentum of objects after collision.

\[\Rightarrow (0.1\times 1.67)+(0.2\times {{v}_{2}})=(0.1\times 2)+(0.2\times 1)\]

\[\Rightarrow (0.167)+(0.2\times {{v}_{2}})=(0.2)+(0.2)\]

\[\Rightarrow (0.167)+(0.2\times {{v}_{2}})=0.4\]

\[\Rightarrow 0.2\times {{v}_{2}}=0.4-0.167\]

\[\Rightarrow 0.2\times {{v}_{2}}=0.233\]

$\Rightarrow {{v}_{2}}=1.165m{{s}^{-1}}$

Thus, the velocity of the second object is \[1.165m{{s}^{-1}}\].

2. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of $400m$ in $20s$. Find its acceleration. Find the force acting on it, if its mass is $7metricTonnes$.( Hint : $1metricTonne=1000kg$)

Mass of truck:$\mathrm{m=7metricTonne=7000kg}$

Initial velocity of truck: $\mathrm{u=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it is stars from rest)

Distance travelled: $\mathrm{s=400m}$

Time duration of travel:$t=20s$

Acceleration of the truck

Force acting on the truck

Thus, $\mathrm{400=(0}\times 20\mathrm{)+}\dfrac{1}{2}(a\times {{20}^{2}})$

$\Rightarrow 400\mathrm{=}\dfrac{1}{2}(a\times 400)$

$\Rightarrow 800\mathrm{=}(a\times 400)$

$\Rightarrow a=2m{{s}^{-2}}$

$\Rightarrow \mathrm{F=(7000}\times 2)$

$\Rightarrow \mathrm{F=14000}N$

Thus, the acceleration of the truck is $2m{{s}^{-2}}$, and the force acting on the truck is $14000N$ .

3. A stone is dropped from a $100m$high tower. How long does it take to fall?

The first $50m$

Initial velocity of stone: $\mathrm{u=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it is starts from rest, before being dropped)

Height of the tower: $\mathrm{s=100m}$ 

Distance travelled in case A: ${{\mathrm{s}}_{1}}\mathrm{=50m}$ (first half distance)

Distance travelled in case B: ${{\mathrm{s}}_{2}}\mathrm{=50m}$ (next half distance)

To find: Time duration of travel:$t$ during the first $50m$.

Since the stone is dropped from a height, we can consider its acceleration to be equal to the acceleration due to gravity.

Acceleration of stone: Acceleration due to gravity  $\Rightarrow a=g=10m{{s}^{-2}}$

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

$\Rightarrow \mathrm{50=(0}\times t\mathrm{)+}\dfrac{1}{2}(10\times {{t}^{2}})$

$\Rightarrow \mathrm{50=5}\times {{t}^{2}}$

$\Rightarrow {{t}^{2}}=10$

$\Rightarrow t=\sqrt{10}=3.16s$

$\Rightarrow {{t}_{1}}=3.16s$

(b) The second $50m$

To find: Time duration of travel:$t$ during the second $50m$.

Time duration for the next $50m$can be found by subtracting time for the first half distance from the time for the total distance of travel.

$\Rightarrow 10\mathrm{0=(0}\times t\mathrm{)+}\dfrac{1}{2}(10\times {{t}^{2}})$

$\Rightarrow 100\mathrm{=5}\times {{t}^{2}}$

$\Rightarrow {{t}^{2}}=20$

$\Rightarrow t=\sqrt{20}=4.47s$

$\Rightarrow t=4.47s$

Thus the time for the second half is – ${{t}_{2}}=t-{{t}_{1}}$

$\Rightarrow {{t}_{2}}=4.47-3.16$

$\Rightarrow {{t}_{2}}=1.31s$

4. A body of mass $10kg$ starts from rest and rolls down an inclined plane. It rolls down $10m$ in $2s$.

What is the acceleration attained by the body?

Mass of body:$\mathrm{m=10kg}$

Distance travelled on inclined plane: $\mathrm{s=10m}$

Time duration of rolling:$t=2s$

To find acceleration attained by the body.

Thus, $\mathrm{10=(0}\times 2\mathrm{)+}\dfrac{1}{2}(a\times {{2}^{2}})$

$\Rightarrow 10\mathrm{=}\dfrac{1}{2}(a\times 4)$

$\Rightarrow 10\mathrm{=}(a\times 2)$

$\Rightarrow a=5m{{s}^{-2}}$

What is the velocity of the body at $2s$?

To find velocity of body at $t=2s$.

Thus, $\mathrm{v=0+(5}\times 2\mathrm{)}$

$\Rightarrow v=10m{{s}^{-1}}$

What is the force acting on the body?

To find force acting on the body.

$\Rightarrow \mathrm{F=(10}\times 5)$

$\Rightarrow \mathrm{F=50}N$

5. A body of mass $2kg$ is at rest at the origin of a frame of reference. A force of $5N$acts on it at $t=0s$. The force acts for $4s$ and then stops.

(a) What is the acceleration produced by the force on the body?

Mass of body:$\mathrm{m=2kg}$

Initial velocity of body: $\mathrm{u=0m}{{\mathrm{s}}^{\mathrm{-1}}}$ (as it is starting from rest)

Force acting on the body: $F=5N$

Time duration for which force is exerted:$t=4s$

To find acceleration produced by the force on the body.

$\Rightarrow a=\dfrac{5}{2}$

$\Rightarrow a=2.5m{{s}^{-2}}$

(b) What is the velocity at $t=4s$?

To find velocity of body at $t=4s$.

Thus, $\mathrm{v=0+(2}\mathrm{.5}\times 4\mathrm{)}$

(c) Draw the v-t graph for the period $t=0s$ to $t=6s$.

To plot the v-t graph.

Plotting the v-t graph

Using the formula – $\mathrm{v=u+at}$

At $t=0s\Rightarrow v=0+(2.5\times 0)=0m{{s}^{-1}}$

At $t=1s\Rightarrow v=0+(2.5\times 1)=2.5m{{s}^{-1}}$

At $t=2s\Rightarrow v=0+(2.5\times 2)=5m{{s}^{-1}}$

At $t=3s\Rightarrow v=0+(2.5\times 3)=7.5m{{s}^{-1}}$

At $t=4s\Rightarrow v=0+(2.5\times 4)=10m{{s}^{-1}}$

At $t=5s\Rightarrow v=0+(2.5\times 5)=12.5m{{s}^{-1}}$

At $t=6s\Rightarrow v=0+(2.5\times 6)=15m{{s}^{-1}}$

(Image will be Uploaded Soon)

(d) Find the distance traveled in $6s$.

To find distance travelled in $t=6s$.

This can be found by calculating the area under the v-t graph.

This is a triangle with base as $6$ and height as $15$

Area of triangle: $\dfrac{1}{2}\times b\times h=\dfrac{1}{2}\times 6\times 15=45$

Thus, the distance travelled in $t=6s$ is $45m$.

Why is Chapter 9 the Forces and Laws of Motion Important for Students?

Forces and Laws of Motion is a crucial chapter. In this chapter, students learn about the Laws of Motion as given by Sir Issac Newton, the various terminologies used for denoting forces of nature and motion of objects, the relation between objects, force, motion, etc. It gives students valuable knowledge about the working of the universe. 

Force and Laws of Motion Class 9 Important Questions with Answers 

Force and Laws of Motion important questions will ensure that students are prepared for all types of questions that might appear in exams. It will help them understand how well they have prepared, how good their time management is, and how they can improve. 

We suggest that students keep practicing the class 9 science chapter 9 important questions to familiarize themselves with the various types of questions, save them precious time in exams, and score good marks. 

The Relevance of Chapter 9 Forces and Laws of Motion

Chapter 9 of Class 9 Physics deals with Forces and Laws of Motion. The chapter talks about the multiple forces in nature, how they act upon objects, how things react, and most importantly, the fundamental Laws of Motion as given by Sir Issac Newton. 

Students need to study this chapter thoroughly and from a young age, so remember them later.  

How to Study for Chapter 9 Forces and Laws of Motion

Given below is the proforma we suggest our students follow for the best possible preparation- 

Study the given class and study notes carefully, and make their notes for better understanding.

Consult online classes on our website.

Watch videos of inertia, forces of motion, and simple examples to understand the concept better.

Practice the Force and Laws of Motion Class 9 Important Questions so that they get to know all the possible questions that might appear.

Practice the questions repeatedly so that they get the best preparation possible and attain good marks in exams. 

Introduction to Forces and Laws of Motion

A force refers to the effort to change an object's state at rest or even at motion. It might also change the object's velocity and direction. The shape of an object can also be changed by force. 

Forces can be Two Types- 

Balanced Forces: Balanced forces do not result in any changes in motion. When it is applied to an object, there will be no such force acting upon the object.

Unbalanced Forces: Unbalanced Forces move in the direction of the force in the highest magnitude. It acts upon an object and can change its speed and direction of motion. 

Types of Forces 

There are different types of forces acting around us. 

Gravitational Force: One of the most commonly known forces, gravitational force, refers to the force that exists due to the attraction between two bodies due to their masses. It is denoted by 

F= G (m1m2/r2) 

Here, G refers to the universal constant while m1 and m2 are the masses of the bodies and r is the distance between them. 

Electromagnetic Force- Electromagnetic Force refers to the force exerted by two charged particles at each other. Two common examples of electromagnetic forces are Friction and Tension. 

Nuclear Force- There are protons and neutrons in every atom. The nuclear force helps to bind neutrons and protons and holds them together in an atom. It is also known as Strong Force or Nuclear Interactions. This force is larger in magnitude than any other force. However, it has a concise range of influence, so in that respect, the other forces are more dominating. 

Weak Force- Weak Forces are responsible for phenomenons known as beta decay. Sometimes a neutron changes itself into a proton, emits an electron, and a particle known as antineutrino. This process is known as Beta Decay. 

The weak force is a force of attraction that works at a concise range of 0.1% of a proton's diameter. These forces differ from gravitational, electromagnetic, and nuclear forces and are known as Weak Forces.

Three Main Laws of Motion

The three primary Laws of Motion are Newton's Laws of Motion. They are as follows-

Newton's First Law of Motion states that an object remains at rest unless an external force acts upon it. Similarly, an object in motion stays in motion in the same direction unless acted upon by an external force. 

Newton's Second Law of Motion states that the force acting on a body will be directly proportional to the rate of change in its momentum.

Newton's Third Law of Motion states that "For every action, there is an equal and opposite reaction." 

The Terminology Used in this Chapter

Net Force: Net force acting on a body refers to the fact that when multiple forces work on one body, it tends to change into one component. 

Frictional Force: The force present between two surfaces in contact and opposes relative motion is known as Frictional Force. 

Inertia: Inertia refers to all bodies' tendency to resist changes in a state of rest or motion. However, all bodies do not have the same inertia. The inertia of a body is directly proportional to its mass.

Momentum: The momentum of an object refers to the product of its mass and velocity. p=mv The impacts that an object or body produces depend on its mass and speed, and this vector quantity has direction and magnitude. 

Inertial and Non-Inertial Frames: Inertial Frame refers to the frame where Newton's Laws hold. On the other hand, Non-Inertial Frames refer to the reference frame where Newton's Law of Motion does not fit. A non-inertial frame undergoes acceleration with respect to an inertial frame. In a non-inertial frame, an accelerometer will detect a non-zero acceleration. 

Significance of Key Questions in CBSE Class 9 Science Chapter 9 - Force and Laws of Motion

The significance of Key Questions for CBSE Class 9 Science Chapter 9 - "Force and Laws of Motion" provided by Vedantu is substantial and cannot be emphasized enough. These curated questions play a crucial role in a student's exam preparation journey. They serve as a targeted resource for revising key concepts and assessing one's understanding of the subject matter. Vedantu's focus on quality education is evident in these questions, which are designed to align with the CBSE curriculum and examination pattern. By practicing these questions, students gain confidence in their problem-solving abilities and enhance their grasp of the fundamental principles of force and motion. These questions are not just aids for academic excellence but also tools for fostering a deeper understanding and appreciation of the laws that govern the physical world.

Why Choose Vedantu to Study?

Vedantu is a one-stop platform for all students who require class notes and solved questions to study. After careful assessment and research, we have provided the best possible class notes, revision notes, and class 9 science chapter 9 important questions for our students. These notes and questions are written by our experts who have immense knowledge of the respective subjects. They carefully go through all the syllabus, notes, and guidelines given by the board and the previous years' question papers before writing the letters and questions available on our website. 

Vedantu also conducts online classes on its website, which can help students understand the study material better. These sessions are also recorded and available on our website if anyone misses the classes and wishes to go through the classes later.

Conclusion 

To conclude, we can say that motion and motion laws are a significant chapter for students in Class 9. Not only does it hold importance in school exams, but it also has considerable weightage in board exams and competitive exams later on. So, we at Vedantu make sure that our students get all the study notes, revision notes, and essential questions of force and motion laws in one place. They are also available in downloadable PDF format so that students can study anytime and anywhere they want. We suggest students practice the critical questions repeatedly for the best preparation possible and fetch good marks. 

Important Related Links for CBSE Class 9 

FAQs on Important Questions for CBSE Class 9 Science Chapter 9 - Force and Laws of Motion

1. When a Carpet is Beaten with a Stick, Dust Comes off it. Explain why?

Answer: When a carpet is beaten with a stick, the carpet is set to motion and it moves back and forth. However, the dust particles on the carpet are not in motion due to the inertia of rest, and they come off the carpet fibers. It is due to this same reason, fruits fall off from the tree on moving the branches vigorously.

2. What are the three Laws of Motion?

Answer: Newton’s first law of motion states that an object at rest remains to be in the state of rest unless acted upon by an unbalanced external force, and an object in motion remains to be in the state of motion in the same direction unless acted upon by an unbalanced external force.

Newton’s second law of motion states that the force acting on a body is directly proportional to the rate of change of its momentum.

Newton’s third law of motion states that for every action there is an equal and opposite reaction.

3. What are the Forces Acting upon a Book Lying on a Table?

Answer: The net force acting upon a book lying on a table is zero. The weight of the book (which is a force) acts in the downward direction. The normal reaction force on the book, which is equal to the weight of the book, acts in the opposite direction to its weight.

4. Are the Important Questions for CBSE Class 9 Science Chapter 9- Force and Laws of Motion Helpful for the Exam Preparation?

Answer: Yes, the important questions for CBSE Class 9 Science Chapter 9- Force and Laws of Motion are definitely helpful for the exam preparation. These important questions and answers from Force and Laws are written and compiled by the subject-matter experts at Vedantu, in a very easy to understand manner. Students can download the important questions for CBSE Class 9 Science Chapter 9- Force and Laws of Motion PDF for free of cost from Vedantu. Since questions from all the essential topics are covered in this PDF, students can rely upon it for their exam preparation and practice purposes. 

5. Which chapter is important for Class 9 Science?

Ans: Class 9 Science textbook contains 15 chapters. Competition and the struggle to fare better than others is increasing and will continue to increase in future. Science can be challenging to study if you don’t have a strong grasp of the basics. You will be forced to cram. You should read all the chapters. You should not start your preparation with the thought of skipping certain chapters. However, you can give more time to study topics such as Motion, Matter, Atoms and Molecules, Work, Energy and Power and Tissue and Natural Resources.

6. What is force according to Chapter 9 of Class 9 Science?

Ans: Force is any movement of push or pull. It is an interaction between two objects that are in contact with each other. If there is no interaction, there will be no force between objects. Depending on the movement and relative force, either the speed of the object increases or it stops moving. Force can also change the shape of an object. Newton (N) is the SI unit of force. We apply force daily when we open a door or when we walk.

7. Why do we fall ill according to Chapter 9 of Class 9 Science?

Ans: Health and disease are complicated terms. Health is a broad area that encompasses mental, physical and social well being. Our body’s immune system protects us from various pathogens that may invade our body. We become ill when we consume contaminated food, water or air. We can even become ill by being in contact with someone who has diseases that can be transmitted. Vectors such as mosquitoes also act as agents carrying pathogens from an infected person to another. To find more about this chapter refer to the study materials provided by Vedantu that can be downloaded absolutely free of cost.

8. Why do we need safety belts when we apply brakes suddenly? Explain scientifically from the information obtained from Chapter 9 of Class 9 Science.

Ans: Newton’s first law of motion is the law of inertia. The law states that an object that is at rest will stay at rest unless force is applied to it. When we are sitting in a car, the car is moving but our body is at rest. When a car stops we move out of the inertia. If the car stops suddenly, our bodies too will move suddenly and abruptly. This can cause injury. Seat belts are a type of safety mechanism that stops our body from moving forward and protects us.

9. What do Newton’s laws of motion say according to Chapter 9 of Class 9 Science?

Ans: There are three laws of motion given by Newton. It helps us understand why an object is at rest, how it interacts with other objects at rest and when it moves. The first law of motion describes the state of inertia. It tells us why we need to apply force to change the state, speed or direction of an object. The second law states that force is equal to the product of mass and acceleration. The third law explains the way two bodies interact. It states that when two bodies interact they apply force on each other which is equal in magnitude but opposite in direction.

CBSE Class 9 Science Important Questions

Cbse study materials.

Class 9 Science Case Study Questions Chapter 3 Atoms and Molecules

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Case study Questions in Class 9 Science Chapter 3  are very important to solve for your exam. Class 9 Science Chapter 3 Class 9 Science 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 3 Atoms and Molecules

Join our Telegram Channel, there you will get various e-books for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.

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.

Atoms and Molecules Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 9 Science  Chapter 3 Atoms and Molecules

Case Study/Passage-Based Questions

Case Study 1: The knowledge of the valencies of various radicals helps us to write the formulae of chemical compounds. The total positive charge on positive ions (cations) is equal to the total negative charge on negative ions (anions) in a molecule. Therefore, in writing the formula of a compound, the positive and negative ions are adjusted in such a way that the total number of positive charges of positive ions (cations) becomes equal to the total number of negative charges of negative ions (anions). There is another simple method for writing the formulae of ionic compounds. In this method, the valencies (or positive or negative charges) of the ions can be ‘crossed over’ to give subscripts. The purpose of crossing over of charges is to find the number of ions required to equalise the number of positive and negative charges.

Element X has two valencies 5 and 3 and Y has valency 2. The elements X and Y are most likely to be respectively (a) copper and sulphur (b) sulphur and iron (c) phosphorus and fluorine (d) nitrogen and iron.

Answer: (d) nitrogen and iron.

The formula of the sulphate of an element X is X 2 (SO 4 ) 3 . The formula of nitride of element X will be (a) X 2 N (b) XN 2 (c) XN (d) X 2 N 3

Answer: (c) XN

The formula of a compound is X 3 Y. The valencies of elements X and Y will be respectively (a) 1 and 3 (b) 3 and 1 (c) 2 and 3 (d) 3 and 2

Answer: (a) 1 and 3

Case Study/Passage Based Questions

Case Study 2: A mole of an atom is a collection of atoms whose total mass is the number of grams equal to the atomic mass. Since an equal number of moles of different elements contain an equal number of atoms it becomes convenient to express the amounts of the elements in terms of moles. A mole represents a definite number of particles viz, atoms, molecules, ions or electrons. This definite number is called the Avogadro number or Avogadro constant which is equal to 6.022 × 1023. Hence a mole represents 6.022 × 1023 particles of the substance. One mole of a substance represents one gram-formula of the substance. One mole of a gas at standard temperature and pressure occupies 22.4 litres.

How many grams of sodium must be taken to get 1 mole of the element? (a) 23 g (b) 35.5 g (c) 63.5 g (d) 46 g

Answer: (a) 23 g

What is the mass in grams of a single atom of chlorine? (Atomic mass of chlorine = 35.5) (a) 6.54 × 10 23 g (b) 5.9 × 10 –23 g (c) 0.0025 g (d) 35.5 g

Answer: (b) 5.9 × 10–23 g

How many number of moles are there in 5.75 g of sodium ? (Atomic mass of sodium = 23) (a) 0.25 (b) 0.5 (c) 1 (d) 2.5

Answer: (a) 0.25

What is the mass in grams of 2.42 mol of zinc? (Atomic mass of Zn = 65.41) (a) 200 g (b) 25 g (c) 85 g (d) 158 g

Answer: (d) 158 g

Case Study 3: According to Dalton’s atomic theory, all matter whether an element, a compound, or a mixture is composed of small particles called atoms which can neither be created nor destroyed during a chemical reaction. Dalton’s theory provides a simple explanation for the laws of chemical combination. He used his theory to explain the law of conservation of masses, the law of constant proportions, and the law of multiple proportions, based on various postulates of the theory. Dalton was the first scientist to use the symbols for the elements in a very specific sense. When he used a symbol for an element he also meant a definite quantity of that element, that is one atom of that element.

Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass? (a) Atoms can neither be created nor destroyed. (b) Each element is composed of extremely small particles called atoms. (c) All the atoms of a given element are identical. (d) During chemical combination, atoms of different elements combine in simple ratios.

Answer: (a) Atoms can neither be created nor destroyed.

Which postulate of Dalton’s atomic theory explains law of definite proportions? (a) Atoms of an element do not change during a chemical reaction. (b) An element consists of atoms having fixed mass and the number and kind of atoms in a given compound is fixed. (c) Different elements have different kind of atoms. (d) Atoms are of various kinds

Answer: (b) An element consists of atoms having fixed mass and the number and kind of atoms in a given compound is fixed.

“If 100 g of calcium carbonate (whether in the form of marble or chalk) is decomposed, 56 g of calcium oxide and 44 g of carbon dioxide are formed.” Which law of chemical combination is illustrated by this statement? (a) Law of constant proportions (b) Law of conservation of mass (c) Law of multiple proportions (d) Law of conservation of energy

Answer: (b) Law of conservation of mass

When 5 g calcium is burnt in 2 g oxygen, 7 g of calcium oxide is produced. When 5 g of calcium is burnt in 20 g of oxygen, then also 7 g of calcium oxide is produced. Which law of chemical combination is being followed? (a) Law of conservation of mass (b) Law of multiple proportions (c) Law of constant proportions (d) No law is being followed.

Answer: (c) Law of constant proportions

Case Study 4: Atoms and molecules are the building blocks of matter. An atom is the smallest unit of an element that retains its chemical properties, while a molecule is a group of two or more atoms held together by chemical bonds. Atoms consist of a positively charged nucleus, which contains protons and neutrons, surrounded by negatively charged electrons in energy levels or shells. The number of protons in an atom determines its atomic number and defines its unique identity as an element. The electrons in an atom occupy specific energy levels, and the outermost shell is known as the valence shell. Atoms gain, lose, or share electrons to achieve a stable electron configuration, forming chemical bonds and giving rise to molecules. Understanding the concept of atoms and molecules is crucial for comprehending various chemical reactions and the composition of substances.

What is the smallest unit of an element that retains its chemical properties? a) Proton b) Electron c) Nucleus d) Atom Answer: d) Atom

What is a group of two or more atoms held together by chemical bonds called? a) Element b) Compound c) Molecule d) Nucleus Answer: c) Molecule

What are the positively charged particles present in the nucleus of an atom called? a) Electrons b) Protons c) Neutrons d) Valence electrons Answer: b) Protons

Which part of an atom contains electrons in energy levels or shells? a) Protons b) Neutrons c) Nucleus d) Valence shell Answer: d) Valence shell

What do atoms do to achieve a stable electron configuration? a) Gain, lose, or share electrons b) Absorb protons c) Increase their atomic number d) Create chemical bonds Answer: a) Gain, lose, or share electrons

Hope the information shed above regarding Case Study and Passage Based Questions for Class 9 Science Chapter 3 Atoms and Molecules with Answers Pdf free download has been useful to an extent. If you have any other queries about the CBSE Class 9 Science Atoms and Molecules Case Study and Passage-Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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CBSE Class 9 Science Syllabus 2024-2025: Download PDF

Cbse class 9 science syllabus 2024-25: download the new cbse syllabus for class 9 science in pdf from here. check the detailed and updated syllabus to know the course content prescribed for 2024-2025..

CBSE Class 9 Science Syllabus 2024-2025: Science being compulsory and one of the most important subjects, requires informed study and practice throughout the year. CBSE syllabus is an essential tool that guides you systematically. It mentions the unit-wise weightage and the topics to be covered throughout the year. Additionally, it provides details regarding the examination pattern and the criteria of internal assessment. Therefore, the CBSE Class 9 Science syllabus is an all-encompassing guide to academic success.

CBSE Class 9   Science Syllabus  2024-25 Highlights

Cbse class 9   science unit-wise weightage 2024-2025, cbse class 9   science syllabus 2024-2025, theme: materials, unit 1: matter-nature and behaviour.

Definition of matter; solid, liquid and gas; characteristics - shape, volume, density; change of statemelting (absorption of heat), freezing, evaporation (cooling by evaporation), condensation, sublimation.

Nature of matter: Elements, compounds and mixtures. Heterogeneous and homogenous mixtures, colloids and suspensions. Physical and chemical changes (excluding separating the components of a mixture).

Particle nature and their basic units:  Atoms and molecules, Law of Chemical Combination, Chemical formula of common compounds, Atomic and molecular masses.

Theme: The World of the Living

Unit 2: 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.

Theme: Moving Things, People and Ideas

Unit 3: 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, 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.

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 (excluding commercial unit of Energy).

Theme: Food

Unit 4: food production, cbse class 9 science practical syllabus, internal assessment for cbse class 9 science, prescribed books for cbse class 9 science:  .

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

NCERT Book for Class 9 Science (Revised)

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CBSE Class 9 Science Practical Syllabus 2024-2025: Download in PDF

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CBSE Class 9 Science   Practical Syllabus: Practical work is an essential part of learning science, allowing students to get hands-on experience and see scientific concepts come to life. Through exciting experiments and activities, students develop important skills like observation, analysis, and problem-solving which are essential to excel in the subject.

CBSE Class 9 Science practical work includes experiments covering all three sub-divisions of Science which are Physics, Chemistry and Biology. Students must be aware of the prescribed list of experiments to Plan and perform the experiments effectively, helping them prepare well for their annual practical exam. Below is the list of Science experiments included in CBSE Class 9 Science Practical Syllabus for 2024-2025.

CBSE Class 9 Science Practical Syllabus 2024-2025

1. Preparation of: Unit-I

a) a true solution of common salt, sugar and alum

b) a suspension of soil, chalk powder and fine sand in water

c) a colloidal solution of starch in water and egg albumin/milk in water and distinguish between these on the basis of

• transparency
• filtration criterion

2. Preparation of Unit-I

a) A mixture

b) A compound

using iron filings and sulphur powder and distinguishing between these on the basis of:

(i) appearance, i.e., homogeneity and heterogeneity

(ii) behaviour towards a magnet

(iii) behaviour towards carbon disulphide as a solvent

(iv) effect of heat

3. Perform the following reactions and classify them as physical or chemical changes: Unit-I

a) Iron with copper sulphate solution in water

b) Burning of magnesium ribbon in air

c) Zinc with dilute sulphuric acid

d) Heating of copper sulphate crystals

e) Sodium sulphate with barium chloride in the form of their solutions in water

4. Preparation of stained temporary mounts of (a) onion peel, (b) human cheek cells & to record observations and draw their labeled diagrams. Unit-II

5. Identification of Parenchyma, Collenchyma and Sclerenchyma tissues in plants, striped, smooth and cardiac muscle fibers and nerve cells in animals, from prepared slides. Draw their labeled diagrams. Unit-II

6. Determination of the melting point of ice and the boiling point of water. Unit-I

7. Verification of the Laws of reflection of sound. Unit-III

8. Determination of the density of solid (denser than water) by using a spring balance and a measuring cylinder. Unit-III

9. Establishing the relation between the loss in weight of a solid when fully immersed in Unit-III

a) Tap water

b) Strongly salty water with the weight of water displaced by it by taking at least two different solids.

10. Determination of the speed of a pulse propagated through a stretched string/slinky (helical spring). Unit-III

11. Verification of the law of conservation of mass in a chemical reaction. Unit-III

You can download the above syllabus in PDF from the link given below:

Prescribed Books

1. Assessment of Practical Skills in Science- Class IX- CBSE Publication

2. Laboratory Manual-Science-Class IX, NCERT Publication

Assessment of Practical Skills in Science outlines the specific experiments and activities included in the practical syllabus for CBSE Class 9 Science. It details the objectives, apparatus required, procedure, observations, and conclusion for each experiment.

Laboratory Manual Science is a practical manual published by the National Council of Educational Research and Training (NCERT). It provides a more detailed explanation of the experiments along with additional information, diagrams, and safety precautions, serving as a valuable resource for understanding the concepts and practising the procedures.

Also Read : CBSE Class 9 Science Syllabus 2024-25 (Theory)

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Case Study Questions Class 9 Science Matter in our Surroundings

Case study questions class 9 science chapter 1 matter in our surroundings.

CBSE Class 9 Case Study Questions Science Matter in our Surroundings. 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 Matter in our Surroundings.

At Case Study Questions there will given a Paragraph. In where some Questions will made on that respective Case Based Study. There will various types of marks will given 1 marks, 2 marks, 3 marks or 4 marks.

CBSE Case Study Questions Class 9 Science – Matter in our Surroundings

Case study 1:.

1.) A matter is anything that has mass and occupies space. Pen, paper, clips, sand, air, ice, etc. are different forms of matter. Every matter is made up of small particles. These particles are so tiny that they can’t be seen with naked eyes. Let’s see about the different characteristics of particles of matter.

• All matter is made up of very small particles.
• .Particles of matter has spaces between them.
• Particles of matter are continuously moving.
• Particles of matter attract each other.

Answer the following questions by referring above paragraph.

i.) Which of following is not matter?

c.) smell of perfume

d.) None of these

ii.) Thoughts coming in our mind are example of matter. True or false

c.) None of these

iii.) Which of the following is true about particles of matter?

a.) Particles of matter has spaces between them

b.) Particles of matter are continuously moving

c.) Particles of matter attract each other

d.) All of these

iv.) Give 5 examples of matter in our surroundings

v.) Enlist all properties of particles of matter

Answer key-1

iv.) pen, pencil, notebook, ice and water

v.) Different characteristics of particles of matter are

Case Study 2:

2.) There are three states of matter – solid, liquid and gas.

Solids have a definite shape, distinct boundaries and fixed volumes, that is, have negligible compressibility. Solids have a tendency to maintain their shape when subjected to outside force. Solids may break under force but it is difficult to change their shape, so they are rigid.

Liquids have no fixed shape but have a fixed volume. They take up the shape of the container in which they are kept. Liquids flow and change shape, so they are not rigid but can be called fluid.

Gas as has indefinite shape, no fixed volume. Gas gets the shape and volume of container.

Gas has very low density hence are light. Gas can flow easily and hence are called fluid.

i.) Which of the following state of matter takes shape of container in which it is filled?

d.) Both b and c

ii.) Distance between particles of matter least in

iii.) Compressibility is least in case of

iv.) Give properties of solids.

v.) Give properties of Gases.

Answer key-2

iv.) properties of solid are given below

• Solid has fixed volume.
• Solid has fixed shape.
• Solid has high density.
• Solids are heavy.
• Solid does not flow.

v.) Properties of gases are

• Gas has indefinite shape
• Gas has no fixed volume.
• Gas gets the shape and volume of container.
• Gas fills the container completely.
• Gas has very low density.
• Because of low density gas are light.
• Gas can flow easily and hence are called fluid.

Case Study 3:

3.) What happens inside the matter during change of state? On increasing the temperature of solids, the kinetic energy of the particles increases. Due to the increase in kinetic energy, the

Particles start vibrating with greater speed. The energy supplied by heat overcomes the forces of attraction between the particles. The particles leave their fixed positions and start moving more freely. A stage is reached when the solid melts and is converted to a liquid. The minimum temperature at which a solid melts to become a liquid at the atmospheric pressure is called its melting point.

The temperature of the system does not change after the melting point is reached, till all the ice melts. This happens even though we continue to heat the beaker, that is, we continue to supply heat. This heat gets used up in changing the state by overcoming the forces of attraction between the particles. The amount of heat energy that is required to change 1 kg of a solid into liquid at atmospheric pressure at its melting point is known as the latent heat of fusion. So, particles in water at 0 0 C (273 K) have more energy as compared to particles in ice at the same temperature.

The temperature at which a liquid starts boiling at the atmospheric pressure is known as its boiling point. Boiling is a bulk phenomenon. Particles from the bulk of the liquid gain enough energy to change into the vapour state. A change of state directly from solid to gas without changing into liquid state is called sublimation and the direct change of gas to solid without changing into liquid is called deposition.

i.) A change of state directly from solid to gas without changing into liquid state is called

a.) Sublimation

b.) Deposition

c.) Boiling point

ii.) The direct change of gas to solid without changing into liquid is called

iii.) The energy supplied by heat to solid is used to overcome the forces of attraction between the particles. True or false

iv.) Define melting point and boiling point

v.) Define latent heat of fusion

Answer key-3

iv.) The minimum temperature at which a solid melts to become a liquid at the atmospheric pressure is called its melting point.

The temperature at which a liquid starts boiling at the atmospheric pressure is known as its boiling point.

v.) The amount of heat energy that is required to change 1 kg of a solid into liquid at atmospheric pressure at its melting point is known as the latent heat of fusion.

Case Study 4:

4 .) Do we always need to heat or change pressure for changing the state of matter? Can you quote some examples from everyday life where change of state from liquid to vapour takes place without the liquid reaching the boiling point? In the case of liquids, a small fraction of particles at the surface, having higher kinetic energy, is able to break away from the forces of attraction of other particles and gets converted into vapour. This phenomenon of change of a liquid into vapors at any temperature below its boiling point is called evaporation.

i.) Evaporation of liquid takes place at

a.) Boiling point

b.) Above boiling point

c.) Below boiling point

ii.) Evaporation takes place at surface of liquid because

a.) They are heavy as compare to other particles

b.) They have sufficient kinetic energy to break the force

c.) They are light weight as compare to other particles

iii.) During evaporation particles of liquid change into vapour

a.) From the surface

b.) From the bottom

c.) From all over the liquid

iv.) Define evaporation.

v.) Explain process of evaporation

Answer key-4

iv.) The phenomenon of change of a liquid into vapors at any temperature below its boiling point is called evaporation.

v.) In the case of liquids, a small fraction of particles at the surface, having higher kinetic energy, is able to break away from the forces of attraction of other particles and gets converted into vapour. This phenomenon of change of a liquid into vapors at any temperature below its boiling point is called evaporation.

Case Study 5:

5.) You must have observed that the rate of evaporation increases with–

• an increase of surface area:
• We know that evaporation is a surface phenomenon. If the surface area is increased, the rate of evaporation increases. For example, while putting clothes for drying up we spread them out.
• an increase of temperature:

With the increase of temperature, more number of particles get enough kinetic energy to go into the vapour state.

In an open vessel, the liquid keeps on evaporating. The particles of liquid absorb energy from the surrounding to regain the energy lost during evaporation. This absorption of energy from the surroundings makes the surroundings cold. What happens when you pour some acetone (nail polish remover) on your palm? The particles gain energy from your palm or surroundings and evaporate causing the palm to feel cool. After a hot sunny day, people sprinkle water on the roof or open ground because the large latent heat of vaporization of water helps to cool the hot surface.

i.) Evaporation is surface phenomenon. True or false

ii.) As temperature increases the rate of evaporation is

a.) increases

b.) decreases

c.) remains constant

iii.) The rate of evaporation increases with

a.) Increase in wind speed

b.) Decrease in wind speed

c.) Does not have any effect from wind speed

iv.) What happens when you pour some acetone (nail polish remover) on your palm?

v.) We are able to sip hot tea from saucer than from cup. Why?

Answer key-5

iv.) The particles gain energy from your palm or surroundings and evaporate causing the palm to feel cool.

v.) We are able to sip hot tea from saucer than from cup. This is because saucer has large surface area, due to large surface area as compare to cut area tea evaporates at faster rate.

Thank you It helped me a lot

Why smell of Perfume is not a matter?

Because there is no particle

Because their are perfume particles suspended in air

These all case study questions are really helpful . Thanks

This is my first I was so nervous but these questions help me alot thank you

Smell of perfume is a matter because it have gas particles means perfume particles

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