NCERT Solutions for Class 9 Science Chapter 9: The NCERT Class 9 chapter is Force and Laws of Motion. In this chapter, students will learn about balanced and unbalanced forces, laws of motion, inertia and mass, conservation of momentum, etc. The NCERT Solutions for Class 9 Science chapter 9 is a crucial reference material, that enables students to quickly and effectively answer the in-text questions. NCERT Solutions for Class 9 Science is prepared in easy-to-understand language, covering all key points.
Students can download the PDF from the link provided on this page. In this chapter, students will learn various phenomenon that are, in one way or the other, based on the Laws of Motion. We recommend students to complete the entire syllabus on time so that they can spend the last few weeks before the exam, revising the concepts and attempting mock tests.
NCERT Solutions for Class 9 Science Chapter 9: Overview
With Embibe’s NCERT Class 9 Science Chapter 9 Solutions, students can quickly prepare for a class test or the final exams. The chapters are organised to allow students to access the study material easily. Even students who are learning the concepts for the first time will find the NCERT Class 9 Science Solutions very helpful as the concepts are explained in a simple and easy-to-remember manner by the Subject experts at Embibe. Moreover, students can also download the NCERT Solutions Science Class 9 Chapter 9 in PDF format and use them as notes or revision purposes.
NCERT Solutions Class 9 Science Chapter 9 PDF
Class 9 Chapter 9 NCERT Solutions provides in-depth answers to all the topic-wise questions. Students can download the the same in PDF format and practice the questions offline.
About Class 9 Science Chapter 9: Force and Laws of Motion
Class 9 Science Chapter 9 Exercise Solutions – Force and Laws of Motion deals with the concept of force and its relation to the motion. It answers questions like why the speed of an object changes with time, whether or not all motions require a cause, etc. Newton’s first, second, and third laws of motion are explained in detail with examples giving a clear idea of what force is and its relationship with motion.
Students will also learn about the concept of Conservation of Momentum with several examples and problems. Plenty of real-life examples are included, which helps students to understand the concepts better.
NCERT Solutions Class 9 Science Chapter 9: Summary
While Newton’s Laws of Motion appears apparent to us today, they were revolutionary centuries ago. The three laws of motion help us understand how objects behave when they are stationary, moving, and subjected to forces.
Newton’s First Law of Motion
Definition: An object continues to be in a state of rest or of uniform motion along a straight line unless acted upon by an unbalanced force.
Important Terms: • The natural tendency of objects to resist a change in their state of rest or uniform motion is called inertia. • The mass of an object is a measure of its inertia. Its SI unit is the kilogram (kg). • Force of friction always opposes the motion of objects.
Newton’s Second Law of Motion
Definition: The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.
• The SI unit of force is kg m s^–2. It is also known as newton and is represented by the symbol N. A force of one newton produces an acceleration of 1 m s^–2 on an object of mass 1 kg. • The momentum of an object is the product of its mass and velocity and has the same direction as that of the velocity. Its SI unit is kg m s^(–1).
Newton’s Third Law of Motion
Definition: There is an equal and opposite reaction for every action, and they act on two different bodies.
Note: The total momentum remains conserved in an isolated system (with no external force).
How to Prepare From NCERT Class 9 Science Chapter 9?
Force and Laws of Motion is an upgrade from the Force and Pressure chapter of Class 8. In this chapter, students are taught about different types of force and their application along with the 3 laws of motion. Students can follow the below-given tips to get their basics clear and score good marks.
1. Brush-up on the Force and Pressure chapter from the previous class. 2. Note down the formulas on a separate sheet. 3. Learn the 3 laws of motion with examples as the probability of a question on these is more than 90%. 4. Keep revising the the previous topics. 5. Make a proper schedule to study the chapter/subject. A thorough knowledge of the CBSE Class 9 Syllabus 2022-23 is essential for students to make a solid preparation plan. 6. Take the help of Chapter 9 Science Class 9 NCERT solutions provided by us. 7. Pay attention to the definitions and important terms mentioned in the previous section as they are important from an exam perspective. 8. Take mock tests on Embibe and clear all doubts immediately.
CBSE Class 9 Science Chapter 9 Important Questions
We have tabulated some of the important questions that students must solve/answer.
Question 1: Define Force with 2 examples. Question 2: State the 3 Laws of Motion. Question 3: What are balanced and unbalanced forces? Question 4: Deduce Newton’s First laws from the second law. Question 5: Explain why some of the leaves may get detached from a tree if we vigorously shake its branch. Question 6: Why do you fall in the forward direction when a moving bus brakes to a stop and fall back when it accelerates from rest? Question 7: Two objects of masses 100 g and 200 g move along the same line and direction with velocities of 2m/s and 1m/s, respectively. They collide, and after the collision, the first object moves at a velocity of 1.67m/s. Determine the velocity of the second object. Question 8: A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 metric tonnes (Hint: 1 metric tonne = 1000 kg.) Question 9: How much momentum will a dumbbell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10m/s^2. Question 10: A force of 15 N acts for 5s on a body of mass 5 Kg, which is initially at rest. Calculate Displacement and Final Velocity.
Advantages of Solving NCERT Solutions Class 9 Chapter 9
Let us look at some of the advantages of solving Embibe’s NCERT Solutions for Class 9 Chapter 9.
Embibe’s NCERT solutions help students understand ideas and assist them in developing strategies to score maximum marks in the final exam.
Preparing from Embibe’s NCERT Solutions Class 9 is an excellent technique for students to grasp the chapter’s subjects.
The NCERT Solutions have been prepared in a highly simplified way that students can easily understand.
Students can use the NCERT solutions for last-minute preparation or revision without difficulty.
FAQs on NCERT Solutions for Class 9 Science Chapter 9
Q.1: Where can I get Class 9 Science Ch 9 NCERT Solutions PDF? Ans: Students can get Class 9 Science Chapter 9 NCERT Solutions in PDF format from this article.
Q.2: What is the momentum of an object of mass m, moving with a velocity v? (a) (mv)2 (b) mv2 (c) ½ mv2 (d) mv Ans: The correct answer is (d) mv.
Q.3: Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 ms-1 before the collision during which they stick together. What will be the velocity of the combined object after the collision? Ans:Mass of the objects (m1 and m2) = 1.5kg (Given) Initial velocity of the first object (u1) = 2.5 m/s The initial velocity of the second object, which is moving in the opposite direction (u2) = -2.5 m/s When the two masses stick together, the resulting object has a mass of 3 kg (m1 + m2). As per the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Total momentum before the collision = m1u1 + m2u2 = (1.5kg) (2.5 m/s) + (1.5 kg) (-2.5 m/s) = 0 Therefore, total momentum after collision = (m1+m2) v = (3kg) v = 0 Therefore v = 0 This implies that the object formed after the collision has a velocity of 0 metres per second.
Q.4: An 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate: (i) the net accelerating force and (ii) the acceleration of the train Ans: (i) Given, force exerted by the train (F) = 40,000 N Force of friction = – 5000 N (the negative sign indicates that the force is applied in the opposite direction) Therefore, the net accelerating force = sum of all forces = 40,000 N + (-5000 N) = 35,000 N (ii) Total mass of the train = mass of engine + mass of each wagon = 8000kg + 5 × 2000kg The total mass of the train is 18000 kg. As per the second law of motion, F = ma (or: a = F/m) Therefore, acceleration of the train = (net accelerating force) / (total mass of the train) = 35,000/18,000 = 1.94 ms-2 The acceleration of the train is 1.94 m.s-2.
Q.5: Explain why some of the leaves may get detached from a tree if we vigorously shake its branch. Ans: When the branch of the tree is shaken, the branch moves in a to-and-fro motion. However, the inertia of the leaves attached to the branch resists the motion of the branch. So, the loosely attached leaves may get detached from the tree.
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