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CBSE Worksheets for Class 9 Physics

CBSE Worksheets for Class 9 Physics: One of the best teaching strategies employed in most classrooms today is Worksheets. CBSE Class 9 Physics Worksheet for students has been used by teachers & students to develop logical, lingual, analytical, and problem-solving capabilities. So in order to help you with that, we at WorksheetsBuddy have come up with Kendriya Vidyalaya Class 9 Physics Worksheets for the students of Class 9. All our CBSE NCERT Class 9 Physics practice worksheets are designed for helping students to understand various topics, practice skills and improve their subject knowledge which in turn helps students to improve their academic performance. These chapter wise test papers for Class 9 Physics will be useful to test your conceptual understanding.

Board: Central Board of Secondary Education(www.cbse.nic.in) Subject: Class 9 Physics Number of Worksheets: 25

CBSE Class 9 Physics Worksheets PDF

All the CBSE Worksheets for Class 9 Physics provided in this page are provided for free which can be downloaded by students, teachers as well as by parents. We have covered all the Class 9 Physics important questions and answers in the worksheets which are included in CBSE NCERT Syllabus. Just click on the following link and download the CBSE Class 9 Physics Worksheet. CBSE Worksheets for Class 9 Physics can also use like assignments for Class 9 Physics students.

• CBSE Worksheets for Class 9 Physics All Chapters Assignments 
• CBSE Worksheets for Class 9 Physics Force and Laws of Motion Assignment 1
• CBSE Worksheets for Class 9 Physics Force and Laws of Motion Assignment 2
• CBSE Worksheets for Class 9 Physics Force and Laws of Motion Assignment 3
• CBSE Worksheets for Class 9 Physics Gravitation Assignment 1
• CBSE Worksheets for Class 9 Physics Gravitation Assignment 2
• CBSE Worksheets for Class 9 Physics Gravitation Assignment 3
• CBSE Worksheets for Class 9 Physics Motion Assignment 1
• CBSE Worksheets for Class 9 Physics Motion Assignment 2
• CBSE Worksheets for Class 9 Physics Sound Assignment
• CBSE Worksheets for Class 9 Physics Work & Energy Assignment 1
• CBSE Worksheets for Class 9 Physics Work & Energy Assignment 2
• CBSE Worksheets for Class 9 Physics Assignment 1
• CBSE Worksheets for Class 9 Physics Assignment 2
• CBSE Worksheets for Class 9 Physics Assignment 3
• CBSE Worksheets for Class 9 Physics Assignment 4
• CBSE Worksheets for Class 9 Physics Assignment 5
• CBSE Worksheets for Class 9 Physics Assignment 6
• CBSE Worksheets for Class 9 Physics Assignment 7
• CBSE Worksheets for Class 9 Physics Assignment 8
• CBSE Worksheets for Class 9 Physics Assignment 9
• CBSE Worksheets for Class 9 Physics Assignment 10
• CBSE Worksheets for Class 9 Physics Assignment 11
• CBSE Worksheets for Class 9 Physics Assignment 12
• CBSE Worksheets for Class 9 Physics Assignment 13

Advantages of CBSE Class 9 Physics Worksheets

• By practising NCERT CBSE Class 9 Physics Worksheet , students can improve their problem solving skills.
• Helps to develop the subject knowledge in a simple, fun and interactive way.
• No need for tuition or attend extra classes if students practise on worksheets daily.
• Working on CBSE worksheets are time-saving.
• Helps students to promote hands-on learning.
• One of the helpful resources used in classroom revision.
• CBSE Class 9 Physics Workbook Helps to improve subject-knowledge.
• CBSE Class 9 Physics Worksheets encourages classroom activities.

Worksheets of CBSE Class 9 Physics are devised by experts of WorksheetsBuddy experts who have great experience and expertise in teaching Maths. So practising these worksheets will promote students problem-solving skills and subject knowledge in an interactive method. Students can also download CBSE Class 9 Physics Chapter wise question bank pdf and access it anytime, anywhere for free. Browse further to download free CBSE Class 9 Physics Worksheets PDF .

Now that you are provided all the necessary information regarding CBSE Class 9 Physics Worksheet and we hope this detailed article is helpful. So Students who are preparing for the exams must need to have great solving skills. And in order to have these skills, one must practice enough of Class 9 Physics revision worksheets . And more importantly, students should need to follow through the worksheets after completing their syllabus.  Working on CBSE Class 9 Physics Worksheets will be a great help to secure good marks in the examination. So start working on Class 9 Physics Worksheets to secure good score.

CBSE Worksheets For Class 9

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NCERT Solutions for Class 9 Science Chapter 8 Motion

NCERT Solutions for Class 9 Science (physics) Chapter 8 Motion are given below. In these solutions, we have answered all the intext and exercise questions provided in NCERT class 9 science textbook. Class 9 NCERT Solutions Science Chapter 8 provided in this article are strictly based on the CBSE syllabus and curriculum. Students can easily download these solutions in PDF format for free from our app.

Class 9 Science Chapter 8 Textbook Questions and Answers

INTEXT QUESTIONS

PAGE NO. 100

Question 1: An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

Answer: Yes, zero displacement is possible if an object has moved through a distance. Suppose a body is moving in a circular path and starts moving from point A and it returns back at same point A after completing one revolution, then the distance will be equal to its circumference while displacement will be zero.

Question 2: A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Given, side of the square field = 10 m Therefore, perimeter = 10 m × 4 = 40 m Farmer moves along the boundary in 40 second Time = 2 minutes 20 second = 2 × 60 + 20 = 140 s Since, in 40 second farmer moves 40 m

Therefore, in 1 second distance covered by farmer = 40 ÷ 40 = 1m.

Therefore, in 140 second distance covered by farmer = 1 × 140 m = 140 m

Thus, after 2 minute 20 second the displacement of farmer will be equal to 14.1 m north east from initial position.

Question 3: Which of the following is true for displacement? (a) It cannot be zero. (b) Its magnitude is greater than the distance travelled by the object.

Answer: (a) Not true

Displacement can become zero when the initial and final position of the object is the same.

(b) Not true

Displacement is the shortest measurable distance between the initial and final positions of an object. It cannot be greater than the magnitude of the distance travelled by an object. However, sometimes, it may be equal to the distance travelled by the object.

PAGE NO. 102

Question 1: Distinguish between speed and velocity

Answer: Speed has only magnitude while velocity has both magnitude and direction. So, speed is a scalar quantity but velocity is a vector quantity.

Question 2: Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

Answer: The magnitude of average velocity of an object will be equal to its average speed in the condition of uniform velocity in a straight line motion.

Question 3: What does the odometer of an automobile measure?

Answer: In automobiles, odometer is used to measure the distance.

Question 4: What does the path of an object look like when it is in uniform motion?

Answer: In the case of uniform motion, the path of an object will look like a straight line.

Question 5: During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3×10 8 ms -1 .

Answer: Here we have, speed = 3 × 10 8 m/s

Time = 5 minute = 5 × 60 s = 300 s

Using, Distance = Speed × Time ⇒ Distance = 3 × 10 8 × 300 m = 900 × 10 8 m = 9.0 × 10 10 m

PAGE NO 103

Question 1: When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?

Answer: (i) A body is said in uniform acceleration when its motion is along a straight line and its velocity changes by equal magnitude in equal interval of time.

(ii) A body is said in non-uniform acceleration when its motion is along a straight line and its velocity changes by unequal magnitude in equal interval of time.

Question 2: A bus decreases its speed from 80 km/h to 60 km/h in 5 s. Find the acceleration of the bus.

Question 3: A train starting from a railway station and moving with uniform acceleration attains a speed of 40 km/h in 10 minutes. Find its acceleration.

PAGE NO. 107

Question 1: What is the nature of the distance – time graphs for uniform and non-uniform motion of an object?

Answer: (a) The slope of the distance-time graph for an object in uniform motion is a straight line. (b) The slope of the distance-time graph for an object in non-uniform motion is not a straight line.

Question 2: What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?

Answer: When the slope of distance-time graph is a straight line parallel to time axis, the object is stationary.

Question 3: What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?

Answer: When the graph of a speed time graph is a straight line parallel to the time axis, the object is moving with constant speed.

Question 4: What is the quantity which is measured by the area occupied below the velocity- time graph?

Answer: The quantity of distance is measured by the area occupied below the velocity time graph.

PAGE NO 109-110

Question 1: A bus starting from rest moves with a uniform acceleration of 0.1ms –2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.

Answer:  Here we have, Initial velocity (u) = 0 m/s Acceleration (a) = 0.1ms –2   Time (t) = 2 minute = 120 seconds 

(a) The speed acquired: We know that, v = u + at ⇒ v = 0 + 0.1 × 120 m/s ⇒ v = 12 m/s Thus, the bus will acquire a speed of 12 m/s after 2 minute with the given acceleration.

(b) The distance travelled:

Thus, bus will travel a distance of 720 m in the given time of 2 minute.  

Question 2: A train is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of – 0.5 m/s 2 . Find how far the train will go before it is brought to rest.

Question 3: A trolley, while going down an inclined plane, has an acceleration of 2 cm/s 2 . What will be its velocity 3 s after the start?

Answer: Here we have, Initial velocity, u = 0 m/s Acceleration (a) = 2 cm/s 2 = 0.02 m/s 2 Time (t) = 3 s Final velocity, v = ?

We know that, v = u + at            Therefore, v = 0 + 0.02 × 3 m/s           ⇒ v = 0.06 m/s Therefore, the final velocity will be 0.06 m/s after start.

Question 4:  A racing car has a uniform acceleration of 4 m/s 2 . What distance will it cover in 10 s after start?

Thus, racing car will cover a distance of 200 m after start in 10 s with given acceleration.

Question 5: A stone is thrown in a vertically upward direction with a velocity of 5 m/s. If the acceleration of the stone during its motion is 10 m/s 2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Thus, stone will attain a height of 1.25 m and time taken to attain the height is 0.5 s.

Question 1: An athlete completes one round of circular track of diameter 200 m in 40 sec.  What will be the distance covered and the displacement at the end of 2 minutes 20 sec? 

Answer: Time taken = 2 min 20 sec = 140 sec. Radius, r = 100 m. In 40 sec the athlete complete one round.    

So, in 140 sec the athlete will complete = 140 ÷ 40 = 3.5 round.

At the end of his motion, the athlete will be in the diametrically opposite position.

Displacement = diameter = 200 m.

Question 2: Joseph jogs from one end A to another end B of a straight 300 m road in 2 minutes and 30 sec and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging.

(a) from A to B (b) from A to C?

Answer: For motion from A to B: Distance covered = 300 m = Displacement = 300 m.

Time taken = 150 sec.

(a) We know that,   Average speed  = Total distance covered ÷ Total time taken  = 300 m ÷ 150 sec = 2 ms -1  

Average velocity = Net displacement ÷ time taken = 300 m ÷ 150 sec = 2 ms -1

(b) For motion from A to C:

Distance covered = 300 + 100 = 400 m.

Displacement = AB – CB = 300 – 100 = 200 m.

Time taken = 2.5 min + 1 min = 3.5 min = 3.5 × 60 min = 210 sec.

Therefore,        Average speed = Total distance covered ÷ Total time taken  = 400 ÷ 210 = 1.90 ms -1 .

Average velocity = Net displacement ÷ time taken  = 200 m ÷ 210 sec = 0.952ms -1 . 

Question 3: Abdul, while driving to school, computes the average speed for his trip to be 20 kmh -1 . On his return trip along the same route, there is less traffic and the average speed is 30 km/h. What is the average speed of Abdul’s trip? 

Question 4: A motor boat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms -2 for 8.0 s. How far does the boat travel during this time?

Answer:  Here,      u = 0 m/s                  a = 3 ms -2                t = 8 s

Using,  s = ut + ½ at 2   ⇒ s = 0 × 8 + ½ × 3×8 2 ⇒ s = 96 m.

Question 5: A driver of a car travelling at 52 kmh -1 applies the brakes and accelerates uniformly in the opposite direction. The car stops after 5 s. Another driver going at 34 kmh -1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for two cars. Which of the two cars travelled farther after the brakes were applied?  

Answer: In in the following graph, AB and CD are the time graphs for the two cars whose initial speeds are 52 km/h(14.4 m/s) and 34 km/h(8.9 m/s), respectively.

Distance covered by the first car before coming to rest = Area of triangle AOB = ½ × AO × BO = ½ × 52 kmh -1 × 5 s = ½  (52 × 1000 × 1/3600) ms -1 × 5 s = 36.1 m 

Distance covered by the second car before coming to rest = Area of triangle COD = ½ × CO × DO = ½ × 34 km h -1 × 10 s = ½ × (34 × 1000 × 1/3600) ms -1 ×10 s = 47.2 m

Thus, the second car travels farther than the first car after they applied the brakes.

Question 6: Fig 8.11 shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:

(a) Which of the three is travelling the fastest? (b) Are all three ever at the same point on the road? (c) How far has C travelled when B passes A? (d) How far has B travelled by the time it passes C?

Answer: (a) B is travelling fastest as he is taking less time to cover more distance. (b) All three are never at the same point on the road. (c) Approximately 6 kms.   [as 8 – 2 = 6] (d) Approximately 7 kms. [as 7 – 0 = 7] 

Question 7: A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10ms -2 , with what velocity will it strike the ground? After what time will it strike the ground?

Answer: Given, initial velocity of the ball (u) = 0 (since it began at the rest position) Distance travelled by the ball (s) = 20m Acceleration (a) = 10 ms -2

As per the third motion equation, v 2 = u 2 +2as ⇒ v 2 = 2 × (10ms ‒2 ) × (20m) + 0 ⇒ v 2  = 400m 2 s ‒2 ⇒ v = 20ms -1

The ball hits the ground with a velocity of 20 meters per second.

As per the first motion equation, t = (v-u)/a = (20-0)ms ‒1  / 10ms ‒2 = 2 seconds

Therefore, the ball reaches the ground after 2 seconds.

Question 8: The speed – time graph for a car is shown in Figure:

(a) Find out how far the car travels in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.

(b) Which part of the graph represents uniform motion of the car?

Answer: Shaded area representing the distance travelled is as follows:

(a) The shaded area represents the displacement of the car over a time period of 4 seconds. It can be calculated as:

Distance = 1/2 × 4 × 6 = 12 m

Therefore, the car travels a total of 12 meters in the first four seconds.

(b) Since the speed of the car does not change from the points (x=6) and (x=10), the car is said to be in uniform motion from the 6 th  to the 10 th  second.

Question 9: State which of the following situations are possible and give an example of each of the following:

(a) an object with a constant acceleration but with zero velocity

(b) an object moving with an acceleration but with uniform speed.

(c) an object moving in a certain direction with an acceleration in the perpendicular direction.

Answer: (a) It is possible; an object thrown up into the air has a constant acceleration due to gravity acting on it. However, when it reaches its maximum height, its velocity is zero.

(b) it is impossible; acceleration implies an increase or decrease in speed, and uniform speed implies that the speed does not change over time

(c) It is possible; for an object accelerating in a circular trajectory, the acceleration is perpendicular to the direction followed by the object.

Question 10: An artificial is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hrs to revolve around the earth.   

Answer: Here, 

Given, radius of the orbit = 42250 km

Therefore, circumference of the orbit = 2 × π × 42250 km = 265571.42 km

Time taken for the orbit = 24 hours

Therefore, speed of the satellite = 265571.42 ÷ 24 = 11065.4 km.h -1

The satellite orbits the Earth at a speed of 11065.4 kilometers per hour.

Class 9 Science NCERT Solutions Chapter 8 Motion

CBSE Class 9 Science NCERT Solutions Chapter 8 helps students to clear their doubts and to score good marks in the board exam. All the questions are solved by experts with a detailed explanation that will help students complete their assignments & homework. Having a good grasp over CBSE NCERT Solutions for Class 9 Science will further help the students in their preparation for board exams and other competitive exams such as NTSE, Olympiad, etc.

NCERT Solutions for Class 9 Science Chapter 8 PDF

Below we have listed the topics discussed in NCERT Solutions for Class 9 Science Chapter 8. The list gives you a quick look at the different topics and subtopics of this chapter.

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

• NCERT Solutions
• Chapter 9 Force And Laws Of Motion

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

The updated NCERT Solutions for Chapter 8 force and laws of Motion Class 9 is now available on Vedantu. Our subject experts prepare these NCERT solutions with close reference to the latest NCERT Class 9 Science textbook edition.

Students can download the PDF and refer to these solutions for free from our website. All the important topics and sub-topics Laws of motion, Inertia, Mass, Types of forces, etc covered in Class 9 science chapter 8 have been included in these solutions according to the latest Class 9 CBSE Science Syllabus . Therefore, students can rely upon these NCERT Solutions for Class 9 Science for their exam preparation.

Glance on NCERT Solutions for Class 9 Chapter 8 Force and Laws of Motion

Class 9 force and laws of motion will embark on a fascinating journey to understand different laws. Newton's first law of motion states that An object in the rest of the object in motion will always remain constant unless some external unbalanced force is applied to it.

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

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

Ch 8 Science class 9 belongs to Unit II- Motion, Force, and Work. This chapter is all about Force, Types of Forces, and Laws of motion given by Sir Isaac Newton.

 The topics of this chapter are Force and its types, the First law of motion, Inertia and mass, the Second law of motion, the Mathematical formulation of the second law of motion, the Third law of motion, etc.

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

INTEXT EXERCISE 1

1. Which of the following has more inertia:

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

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

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

(b) a bicycle and a train?

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

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

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

2. In the following example, try to identify the number of times the velocity of the ball changes:“A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team”. Also identify the agent supplying the force in each case. Ans:   The ball's velocity changes four times.

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

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

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

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

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

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

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

INTEXT EXERCISE 2

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

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

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

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

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

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

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

Recoil velocity of the rifle $ = {v_1}$

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

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

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

Total momentum of the rifle and bullet system after firing:

$ = {m_1}{v_1} + {m_2}{v_2}$  

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

$ = 4{v_1} + 1.75$

According to the law of conservation of momentum,

Total momentum after the firing = Total momentum before the firing

$4{v_1} + 1.75 = 0$

$4{v_1} =  - 1.75$

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

${v_1} =  - 0.4375$ m/s

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

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

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

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

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

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

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

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

According to the law of conservation of momentum:

Total momentum before collision $ = $ Total momentum after collision

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

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

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

$0.4 = 0.167 + 0.2{v_4}$

$0.4 - 0.167 = 0.2{v_4}$

$0.233 = 0.2{v_4}$

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

${v_4} = 1.165$ m/s

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

NCERT EXERCISE

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

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

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

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

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

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

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

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

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

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

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

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

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

Given:  

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

Distance travelled, s $ = 400$ m

Time taken, t $ = 20$ s

According to the second equation of motion:

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

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

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

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

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

\[a = 2\] m/s 2

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

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

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

From Newton’s second law of motion:

Force, F = Mass × Acceleration

F = ma 

F= \[ = 7000 \times 2\]

F \[ = 14000\] N

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

6. A stone of \[1\] kg is thrown with a velocity of \[20\]m s \[ - 1\] across the frozen surface of a lake and comes to rest after travelling a distance of \[50\] m. What is the force of friction between the stone and the ice?

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

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

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

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

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

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

$0 = 400 + 100a$

$100a =  - 400$

$a =  - \dfrac{{400}}{{100}}$

$a =  - 4$

a = −4 $\dfrac{m}{{{s^2}}}$

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

Mass of the stone, m $ = 1$ kg

F $ = 1 \times  - 4$  

F $ =  - 4$ N

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

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

(a) the net accelerating force;

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

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

Net accelerating force,

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

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

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

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

(b) the acceleration of the train; and

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

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

Mass of a wagon \[ = 2000\] kg

Number of wagons \[ = 5\]

Total Mass of the wagons,

m = Mass of a wagon × Number of wagons

m \[ = 2000 \times 5\]

m \[ = 10000\] kg

Mass of the engine, m′ \[ = 8000\] kg

Total mass, M = m + m′ 

\[ = 10000 + 8000\]

\[ = 18000\] kg

\[Fa = Ma\]

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

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

\[a = 1.944\] m/s 2

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

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

Ans:  

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

Mass of 4 wagons 

\[ = 4 \times 2000\]

\[ = 8000\] kg

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

F \[ = 1552\] N

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

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

Final velocity, \[v = 0\]

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

From Newton’s second law of motion, 

Force = Mass × Acceleration 

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

\[ =  - 2550\] N

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

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

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

(c)1/2 mv 2

Mass of the object \[ = m\]

Velocity \[ = v\]

Momentum = Mass × Velocity

Momentum \[ = mv\]

10. Using a horizontal force of \[200\] N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?

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

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

11. According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.

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

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

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

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

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

Initial momentum \[ = m{v_1}\]

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

Final momentum \[ = m{v_2}\]

Change in momentum 

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

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

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

\[ = 0.2 \times 15\]

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

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

13. A bullet of mass \[10\] g travelling horizontally with a velocity of \[150\] \[m{s^{ - 1}}\] strikes a stationary wooden block and comes to rest in \[0.03\] s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.

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

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

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

Acceleration of the bullet, a

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

\[a =  - \dfrac{{150}}{{0.03}}\]

\[a =  - 5000\] m/s 2

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

According to the third equation of motion:

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

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

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

\[10000s = 22,500\]

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

\[s = 2.25\] m

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

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

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

\[ = 0.01 \times  - 5000\]

\[ =  - 50\] N

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

but it acts in opposite direction.

14. An object of mass \[1\] kg travelling in a straight line with a velocity of \[10\] \[m{s^{ - 1}}\]collides with, and sticks to, a stationary wooden block of mass \[5\] kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.

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

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

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

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

∴ Total momentum before collision

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

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

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

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

Total mass of the combined system, 

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

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

\[ = 6\] kg

Velocity of the combined object \[ = v\]

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

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

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

$ \Rightarrow 10 = 6v$

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

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

$v = 1.66$ m/s

Total momentum after collision

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

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

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

\[ = 10\] kg m/s

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

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

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

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

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

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

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

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

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

Initial momentum \[ = \] mu

\[ = 100 \times 5\]

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

Final momentum \[ = \] mv 

\[ = 100 \times 8\]

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

Force exerted on the object, 

F \[ = \] mv-mu/t

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

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

F \[ = 50\] N

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

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

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

16. Akhtar, Kiran and Rahul were riding in a motor car that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.

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

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

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

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

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

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

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

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

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

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

Final velocity of the dumbbell v = ?

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

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

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

\[{v^2} = 16\]

\[v = \sqrt {16} \]

\[v = 4\] m/s

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

\[ = 10 \times 4\]

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

ADDITIONAL EXERCISE:

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

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

decreasing, or zero?

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

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

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

2. Two persons manage to push a motorcar of mass \[1200\]kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of \[0.2\] \[m{s^{ - 2}}\]. With what force does each person push the motorcar?

(Assume that all persons push the motorcar with the same muscular effort)

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

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

Let the force applied by the third person be F.

Force = Mass × Acceleration

F \[ = 1200 \times 0.2\]

F \[ = 240\] N

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

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

3. A hammer of mass \[500\]g, moving at \[50\] \[m{s^{ - 1}}\], strikes a nail. The nail stops the hammer in a very short time of \[0.01\] s. What is the force of the nail on the hammer?

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

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

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

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

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

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

F \[ =  - 2500\] N

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

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

4. A motorcar of mass \[1200\] kg is moving along a straight line with a uniform velocity of \[90\] km/h. Its velocity is slowed down to \[18\] km/h in \[4\] s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.

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

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

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

Time taken, t \[ = 4\] s

According to the first equation of motion:

\[v = u + at\]

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

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

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

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

\[a =  - 5\] m/s 2

= mv − mu 

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

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

\[ =  - 24000\] kg \[m{s^{ - 1}}\]  

Force \[ = 1200 \times  - 5\]

Force \[ =  - 6000\] N

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

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

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

Topics Covered in Class 9 Science Chapter 8 - Force And Laws Of Motion

Deleted Topics in Class 9 Chapter 8 Force and Laws of Motion

Conservation of Momentum

Conservation Laws

Benefits of NCERT Solutions Class 9 Chapter 8 Force and Laws of Motion

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

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

Detailed analysis of topics with weightage is given, which helps the students in better preparation. Laws of motion class 9 exercise answers PDF provides in depth solutions and explanations.

Highly simplified language is used by our experts to prepare these NCERT Solutions for class 9 force and laws of motion, which makes it understandable for the students.

Students without any hesitation can rely upon these NCERT solutions and laws of motion class 9 PDF for their last-minute preparation or revision starting from the zero levels.

Important Links for Class 9 Science Chapter 8 - Force and Laws of Motion

NCERT Solutions for Class 9 Science - Other Chapter-Wise Links

Given below are the links for other chapter-wise NCERT Solutions for Class 9 Science. These solutions are provided by the Science experts at Vedantu in a detailed manner. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

Vedantu’s NCERT Solutions for ch 8 science class 9 offers students a valuable resource for understanding the fundamental principles that govern motion in the physical world. These solutions for class 9 science chapter 8 question answers, available as free PDF downloads, not only aid in exam preparation but also foster a deeper appreciation for the laws that dictate the behaviour of objects in motion. Equipped with these insights in force and laws of motion, students are empowered to explore and comprehend the dynamic forces at play in their surroundings. We encourage learners to make the most of these resources to enhance their knowledge and excel in their science studies.

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

1. What is force and its effects described in force and laws of motion class 9?

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

Effects of Force

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

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

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

2. What are the 3 Laws of Motion from class 9 science chapter 8?

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

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

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

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

3. Which concepts in the NCERT Solutions for class 9 science chapter 8 question answers are important from the exam perspective?

Class 9 Science Chapter 8 Force and Laws of Motion is a practical chapter that carries high weightage in the exam. This chapter carries 27 marks, hence you need to know the important topics that you should prepare well. The following are some of the important topics from this chapter that you should prepare thoroughly:

Balanced and Unbalanced forces

First Law of Motion 

Inertia and mass

Second Law of Motion

Mathematical Formulation of the second law of motion

Third Law of Motion

Conservation of motion.

4. What is Force and its Laws from NCERT ch 8 science class 9?

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

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

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

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

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

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

Students can refer to force and laws of motion class 9 PDF to get all the solved answers for FREE.

6. Where can I find the downloadable solutions for Class 9 NCERT Chapter 8 force and laws of motion?

To find the downloadable solutions for NCERT Class 9 chapter 8, follow these steps -

Click on the link NCERT Solutions for Class 9 Science (Physics) Chapter 8

You will land on the Vedantu Solutions page for NCERT Class 9 Chapter 8 “Force and Laws of Motions”.

At the top of the page, you will see an option to download the PDF of the Solutions for NCERT Chapter 8.

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

7. What are the key points to choose NCERT Solutions for class 9 force and laws of motion ?

In NCERT Class 9 Science Chapter 8, Force and Laws of Motion, you will find many identities and formulae that you need to keep in mind while solving the numericals. You should have a guide with yourself to understand the tricks to solve these questions faster. NCERT Solutions for Class 9 Chapter 8 are prepared by subject specialists, and they are highly accurate. You will get many tricks to solve your question even faster than before. You can have a deep study about these on the Vedantu Mobile app and for free of cost.

8. What are some important questions from laws of motion class 9 exercise answers PDF?

This chapter holds a lot of information for CBSE board exams. Some of the important questions are:

Laws of Motion

Types of Forces

Inertia and Mass

9. What other resources can help me with force and laws of motion?

Students can refer to NCERT Solutions, PDFs, and study materials related to class 9 force and laws of motion notes provided by Vedantu for FREE.

NCERT Solutions for Class 9 Science

Ncert solutions for class 9.

NCERT Solutions for Class 9 Science Free PDF Download

Ncert solutions for class 9 science.

NCERT is on the priority list of recommendation of the Central Board of Secondary Education (CBSE). Here, you will find all the chapter-wise NCERT Solutions for Class 9 Science. It will help you to understand the subject and perform well in exams.

Class 9 Science is divided into 3 parts – Physics, Chemistry, Biology.

NCERT Solutions for Class 9 Science will help you to ace the unsolved problems in the Class 9 Science book prescribed by the NCERT for all the schools of CBSE. A thorough understanding of the NCERT Solutions for Class 9 Physics helps you in understanding Physics concepts to the point. The NCERT Solutions for Class 9 Science cover all the 5 chapters of the prescribed Physics syllabus and are the best alternative.

Chemistry is one of those subjects which are equally important from academics as well practical point of view. NCERT is specially made to clear most of the topics of the syllabus of chemistry. NCERT solutions help the students to clear all the doubts regarding any chapter.

Biology along with having a good weight in exams is equally important from experiment and practical point of view. In order to understand the chapters in biology, NCERT solutions play an important role.

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NCERT Solutions for Class 9 Science Chapterwise

Class 9 Science Chapter 1 – Matter in our Surroundings

Class 9 Science Chapter 2 – Is Matter Around Us Pure

Class 9 Science Chapter 3 – Atoms and Molecules

Class 9 Science  Chapter 4 – Structure of the Atom

Class 9 Science Chapter 5 – Fundamental Units of Life

Class 9 Science Chapter 6 – Tissues

Class 9 Science Chapter 7 – Diversity in Living Organisms

Class 9 Science Chapter 8 – Motion

Class 9 Science Chapter 9 – Force And Laws Of Motion

Class 9 Science Chapter 10 – Gravitation

Class 9 Science Chapter 11 – Work and Energy

Class 9 Science Chapter 12 – Sound

Class 9 Science Chapter 13 – Why do we fall ill?

Class 9 Science Chapter 14 – Natural Resources

Class 9 Science Chapter 15 – Improvement in Food Resources

NCERT Solutions for Class 9 Science Chapter 1: Matter in our Surroundings

Matter is an important part of our planet as our whole planet is made up of matter. This chapter provides a good amount of information about the matter. It includes the importance of matter based on the chemical and physical properties of matter. The physical nature of the matter will help the students to understand the concept of matter on the basis of its practical use in daily life. The chapter also explains some concepts like evaporation, boiling point, freezing point, etc. along with the factors affecting them.

NCERT Solutions for Class 9 Science Chapter 2: Is Matter Around Us Pure

In the present era, it believed that purity in anything is hardly seen. It is because of the presence of high level of pollutants everyone. In the context of the matter, it also holds true. In this chapter, the students will be able to learn about various concepts like solutions, pure substances, mixtures, suspensions, colloidal, etc. Moreover, this chapter talks about Homogenous substances and Heterogeneous Substances.

NCERT Solutions for Class 9 Science Chapter 3: Atoms and Molecules

In previous classes, students have already read about the presence of atoms as a part of the material. It showed them the divisibility of matter. Atom is the smallest thing or unit present on this planet. In this chapter, the students will be able to learn more about the molecules. This chapter highlights concepts like molecules of compounds, molecules of elements, etc. This chapter also talks about the presence of ions, Law of Conservation of Energy, Law of Chemical Combination, Dalton’s Atomic Theory, etc.

NCERT Solutions for Class 9 Science Chapter 4: Structure of the Atom

One of the most famous theories or models regarding atoms is given by J.J. Thomson. In this chapter, the students will learn about his model along with the drawbacks. It helps to understand the divisibility of atoms. It states that atoms are made of charged particles viz. electrons, protons, neutrons, etc. In the end, the students will learn about the atomic number, atomic mass, mass number, isotopes, isobars, etc

NCERT Solutions for Class 9 Science Chapter 5 Fundamental Units of Life

Robert Hooke was the first person to prepare a microscope and see a cell. A cell is the basic fundamental unit of life. In this chapter, students will learn the structures and functions of different cells. Moreover, the effect of hypotonic, isotonic, and hypertonic solutions on a cell.

NCERT Solutions for Class 9 Science Chapter 6 Tissues

This chapter helps the students to learn about the various types of Tissues in Plants like Merismatic Tissue and Permanent tissue. Moreover, in the end, the chapter highlights topics like Position of plant tissues, a section of a stem, guard cells, etc.

NCERT Solutions for Class 9 Science Chapter 7 Diversity in Living Organisms

There are various life forms on this planet and it becomes very essential to study all of them. In chapter helps students to know about the various groups of organisms. Aristotle has done the classification of animals on the basis of land, air, and water, etc. There are certain terms related to the classification which the students will confront are Kingdom, Phylum, Class, Order, Family, Genus, Species, etc.

NCERT Solutions for Class 9 Science Chapter 8 Motion

Here, you will learn about motion including motion along a straight line, types of motion, the difference between Vector & Scalar, Speed & Velocity, Distance & Displacement, Acceleration – Rate of change of velocity and average speed and velocity, Graphical representation of motion and derivation of three equation of motion.

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

In this chapter, you will learn the concept of balance and unbalance forces. Starting with the First law of motion, the Galileo’s concept, the law of inertia. You will also learn the Second law of motion and Third law of motion, momentum, rate of change of momentum. And applications on first, second and third laws of motion.

NCERT Solutions for Class 9 Science Chapter 10 Gravitation

This chapter gravitation takes you through the depths of motion of objects under the influence of gravitational force on the earth. Gravitational force and Newton’s universal law of gravitation.

NCERT Solutions for Class 9 Science Chapter 11 Work and Energy

In this chapter, you will learn about the relationship between work and energy, scientific conception of work and also different forms of energy such as Kinetic energy, Potential energy, application of kinetic and potential energy, and energy of an object at a certain height.

NCERT Solutions for Class 9 Physics Chapter 12 Sound

This chapter is a very interesting one as you will get to learn about the Reflection of Sound i.e ECHO, reverberation, and applications of multiple reflections of sound. All these concepts are taught by implementing various activities needed to be done in the Physics laboratory that makes the learning process more effective and interactive.

NCERT Solutions for Class 9 Science Chapter 13 Why do we fall ill?

Health and Diseases are the two most important topics of discussion in this present world. This chapter highlights Health and its significance, reasons why do we fall ill, etc. Along with this, this chapter discusses the distinction between a healthy person and disease-free person.

NCERT Solutions for Class 9 Science Chapter 14 Natural Resources

This chapter highlights the most important element of life: Air. Starting with the role of the atmosphere in climatic control, this chapter deals with topics like how air moves, winds, rain, and air pollution. While discussing the topic of Air pollution, the chapter states various pollutants of the same. Moreover, in the end, the chapter, states various solutions to the problem.

NCERT Solutions for Class 9 Science Chapter 15 Improvement in Food Resources

This chapter deals with concepts like Crop Yield. Along with that, the chapter discusses the improvement in Crop Yield and related topics. Improvement in livestock and food crops are also discussed. The chapter also discusses topics like the ways to improve crop yield, crop production improvement, etc.

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NCERT Solutions for Class 9

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NCERT Solutions for Class 9 Science Chapter 8 Motion

NCERT Solutions Class 9 Science Chapter 8 Motion – Here are all the NCERT solutions for Class 9 Science Chapter 8. This solution contains questions, answers, images, step by step explanations of the complete Chapter 8 titled Motion of Science taught in class 9. If you are a student of class 9 who is using NCERT Textbook to study Science, then you must come across Chapter 8 Motion. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 9 Science Chapter 8 Motion in one place. For a better understanding of this chapter, you should also see Chapter 8 Motion Class 9 notes , Science.

Topics and Sub Topics in Class 9 Science Chapter 8 Motion:

• Describing Motion
• Measuring the Rate of Motion
• Rate of Change of Velocity
• Graphical Representation of Motion
• Equations of Motion by Graphical Method
• Uniform Circular Motion

These solutions are part of NCERT Solutions for Class 9 Science . Here we have given NCERT Solutions for Class 9 Science Chapter 8 Motion.

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NCERT Solutions for class 9 Science chapter-8 Motion

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Question 1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.  Solution : Yes, if an object has moved through a distance it can have zero displacement because displacement of an object is the actual change in its position when it moves from one position to the other. So if an object travels from point A to B and then returns back to point A again, the total displacement is zero.

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Question 2. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds?

Solution : Distance covered by farmer in 40 seconds

=4 x(10) m= 40 m

Speed of the farmer = distance/time = 40m/40s = 1 m/s.

Total time given in the Question = 2min 20seconds

= 60+60+20 =140 seconds

Since he completes 1round of the field in 40seconds so in he will complete 3rounds in 120seconds (2mins) or 120m distance is covered in 2minutes. In another 20seconds will cover another 20m so total distance covered in 2min20sec

= 120 +20 =140m.

Displacement = √10 2 + 10 2  = √ 200

= 10 √ m (as per diagram) =10 x 1.414= 14.14 m.

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Question 3. Which of the following is true for displacement? (a) It cannot be zero. (b) Its magnitude is greater than the distance travelled by the object. Solution : Both (a) as well as (b) are false with respect to concept of displacement.

Question 4. Distinguish between speed and velocity. Solution : Speed of a body is the distance travelled by it per unit time while velocity is displacement per unit time of the body during movement.

Question 5.Under what condition(s) is the magnitude of average velocity of an object equal to its average speed? Solution : If distance travelled by an object is equal to its displacement then the magnitude of average velocity of an object will be equal to its average speed.

Question 6. What does the odometer of an automobile measure? Solution : The odometer of an automobile measures the distance covered by that automobile.

Question 7. What does the path of an object look like when it is in uniform motion? Solution : Graphically the path of an object will be linear i.e. look like a straight line when it is in uniform motion.

Question 8. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is 3 x10 8 ms -1 Solution :

Distance = Speed x time

Question 9. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration? Solution :

(i) uniform acceleration: When an object travels in a straight line and its velocity changes by equal amount in equal intervals of time, it is said to have uniform acceleration.

(ii) non uniform acceleration: It is also called variable acceleration. When the velocity of an object changes by unequal amounts in equal intervals of time, it is said to have non uniform acceleration.

Question 11. A train starting from a railway station and moving with uniform acceleration attains a speed 40km h -1 in 10 minutes. Find its acceleration. Solution : Since the train starts from rest(railway station) = u = zero

time (t) = 10 min = 10 x 60

= 600 seconds

Question 12. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object? Solution : If an object has a uniform motion then the nature of distance time graph will be linear i.e. it would a straight line and if it has non uniform motion then the nature of distance time graph is a curved line.

Question 13. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?

Solution : If the object’s distance time graph is a straight line parallel to the time axis indicates that with increasing time the distance of that object is not increasing hence the object is at rest i.e. not moving.

Question 14. What can you say about the motion of an object if its speed time graph is a straight line parallel to the time axis? Solution : Such a graph indicates that the object is travelling with uniform velocity.

Question 15. What is the quantity which is measured by the area occupied below the velocity-time graph? Solution : The area occupied below the velocity-time graph measures the distance moved by any object.

Question 16. A bus starting from rest moves with a uniform acceleration of 0.1m s -2 for 2 minutes. Find (a)the speed acquired, (b) the distance travelled. Solution :

Question 17. A train is travelling at a speed of 90km h -1 . Brakes are applied so as to produce a uniform acceleration of -0.5m s -2 . Find how far the train will go before it is brought to rest. Solution :

v =0(train is brought to rest)

0 =25 – 0.5 x

0.5t = 25, or t = 25/0.5 = 50seconds

= 1250 – 625 = 625m

Question 18. A trolley, while going down an inclined plane, has an acceleration of 2cm s -2 . What will be its velocity 3 s after the start? Solution :

v= u +at = 0 + 2 x 3 = 6 cm/s

Question 19. A racing car has a uniform acceleration of 4cm s -2 . What distance will it cover in 10 s after start? Solution :

Question 20. A stone is thrown in a vertically upward direction with a velocity of 5 cm s -2 . If the acceleration of the stone during its motion is 10 cm s -2 n the downward direction, what will be the height attained by the stone and how much time will it take to reach there? Solution :

v = 0 (since at maximum height its velocity will be zero)

v = u + at 

= 5 + (-10) x t

0 = 5 – 10t

10t = 5, or, t = 5/10 =0.5second.

= 2.5 – 1.25 = 1.25m

Question 21. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s? Solution :

circumference of circular track = 2πr

rounds completed by athlete in 2min20sec = s= 140/40 = 3.5

therefore, total distance covered =4400 / 7 x 3.5= 2200 m

Since one complete round of circular track needs 40s so he will complete 3 rounds in 2mins and in next 20s he can complete half round therefore displacement = diameter = 200m.

Question 22. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 50 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C? Solution : (a) distance = 300m

time = 2min30seconds = 150 seconds

average speed from A to B = average velocity from A to B

= 300m/150s = 2m/s

(b) average speed from A to C = (300+100)m/(150+60)sec

= 400m/210s = 1.90m/s

displacement from A to C

= (300 – 100)m =200m

time =2min30sec + 1min = 210s

velocity = displacement/time = 200m/210s = 0.95m/s

Question 23. Abdul, while driving to school, computes the average speed for his trip to be 20km h -1 . On his return trip along the same route, there is less traffic and the average speed is 40km h -1 . What is the average speed for Abdul’s trip? Solution :

If we suppose that distance from Abdul’s home to school = x kms

while driving to school :-

velocity = displacement/time

20 = x/t, or, t=x/20 hr

on his return trip :-

speed = 40 km h–1 ,

40= x /t’

or, t’ =x/40 hr

total distance travelled = x + x = 2x

total time = t + t’ = x/20 + x/40

=(2x + x)/40 = 3x/40 hr

average speed for Abdul’s trip

= 2x/(3x/40) = 80x/3x = 26.67km/hr

Question 24. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s -2 for 8.0 s. How far does the boat travel during this time? Solution :

since the motorboat starts from rest so u= 0

Question 25. A driver of a car travelling at 52 km h -1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h -1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied? Solution :

∴ Clearly the first car will travel farther (36.11 m) than the first car(4.16 m).

Question 26. Fig 8.11 shows the distance-time graph of three objects A, B and C. Study the graph and answer the following Questions :

(a) Which of the three is travelling the fastest?

(b) Are all three ever at the same point on the road?

(c) How far has C travelled when B passes A?

(d) How far has B travelled by the time it passes C?

(a) It is clear from graph that B covers more distance in less time. Therefore, B is the fastest.

(b) All of them never come at the same point at the same time.

(c) According to graph; each small division shows about 0.57 km.

A is passing B at point S which is in line with point P (on the distance axis) and shows about 9.14 km

Thus, at this point C travels about

9.14 - (0.57 x 3.75)km = 9.14 km – 2.1375 km

Thus, when A passes B, C travels about 7 km.

(d) B passes C at point Q at the distance axis which is

Therefore, B travelled about 5.28 km when passes to C.

Question 27. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10m s -2 , with what velocity will it strike the ground? After what time will it strike the ground? Solution : Let us assume, the final velocity with which ball will strike the ground be ‘v’ and time it takes to strike the ground be ‘t’

Initial Velocity of ball u=0

Distance or height of fall s=20m

∴ Time taken by the ball to strike= (20-0)/10

= 2 seconds

Question 28. The speed-time graph for a car is shown is Fig. 8.12.

(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.

(b) Which part of the graph represents uniform motion of the car?

(a) Distance travelled by car in the 4 second

The area under the slope of the speed – time graph gives the distance travelled by an object.

In the given graph

56 full squares and 12 half squares come under the area slope for the time of 4 second.

Total number of squares = 56 + 12/2 = 62 squares

The total area of the squares will give the distance travelled by the car in 4 second. on the time axis,

5 squares = 2seconds, therefore 1 square = 2/5 seconds

on speed axis there are 3 squares = 2m/s

therefore, area of one square

= 248/15 m = 16.53 m

Hence the car travels 16.53m in the first 4 seconds.

(b) The straight line part of graph, from point A to point B represents a uniform motion of car.

Question 29. State which of the following situations are possible and give an example for each of these:

(a) an object with a constant acceleration but with zero velocity

(b) an object moving in a certain direction with an acceleration in the perpendicular direction.

(a) An object with a constant acceleration can still have the zero velocity. For example an object which is at rest on the surface of earth will have zero velocity but still being acted upon by the gravitational force of earth with an acceleration of 9.81 ms-2 towards the center of earth. Hence when an object starts falling freely can have contant acceleration but with zero velocity.

(b) When an athlete moves with a velocity of constant magnitude along the circular path, the only change in his velocity is due to the change in the direction of motion. Here, the motion of the athlete moving along a circular path is, therefore, an example of an accelerated motion where acceleration is always perpendicular to direction of motion of an object at a given instance. Hence it is possible when an object moves on a circular path.

Question 30. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth. Solution :

Let us assume an artificial satellite, which is moving in a circular orbit of radius 42250 km covers a distance ‘s’ as it revolve around earth with speed ‘v’ in given time ‘t’ of 24 hours.

Radius of circular orbit r =42250 x 1000 m

Time taken by artificial satellite t=24 hours

=24 x 60 x 60 s

Distance covered by satellite s =Circumference of circular orbit

∴ Speed of satellite v = (2πr) / t

= 3.073 km/s

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CHAPTER WISE NCERT SOLUTIONS OF CLASS 9 SCIENCE

Chapter-1 Matter in Our Surroundings

Chapter-2 Is Matter Around Us Pure

Chapter-3 Atoms and Molecules

Chapter-4 Structure of the Atom

Chapter-5 The Fundamental Unit of Life

Chapter-6 Tissues

Chapter-7 Diversity in Living Organisms

Chapter-8 Motion

Chapter-9 Force and Laws of Motion

Chapter-10 Gravitation

Chapter-11 Work and Energy

Chapter-12 Sound

Chapter-13 Why Do We Fall Ill

Chapter-14 Natural Resources

Chapter-15 Improvement in Food Resources

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Related Chapters

• chapter-3 Atoms and Molecules
• chapter-4 Structure of Atom
• chapter-5 The Fundamental Unit of Life
• chapter-6 Tissues
• chapter-7 Diversity in Living Organisms
• chapter-8 Motion
• chapter-9 Forces and Laws of Motion
• chapter-10 Gravitation
• chapter-11 Work and Energy
• chapter-12 Sound
• chapter-13 Why Do We Fall Ill
• chapter-14 Natural Resources
• chapter-15 Improvement in Food Resources

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NCERT Solutions for Class 9 Science Chapter 8: Motion

Ncert solutions class 9 science chapter 8 – free pdf download.

* According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 7.

NCERT Solutions for Class 9 Science Chapter 8 Motion is designed with the intention of clarifying doubts and concepts easily. Class 9 NCERT Solutions in Science is a beneficial reference and guide that helps students clear doubts instantly in an effective way. 

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• Chapter 1 Matter in Our Surroundings
• Chapter 2 Is Matter Around Us Pure?
• Chapter 3 Atoms and Molecules
• Chapter 4 Structure of the Atom
• Chapter 5 The Fundamental Unit of Life
• Chapter 6 Tissues
• Chapter 7 Diversity in Living Organisms
• Chapter 9 Force And Laws Of Motion
• Chapter 10 Gravitation
• Chapter 11 Work and Energy
• Chapter 12 Sound
• Chapter 13 Why Do We Fall Ill?
• Chapter 14 Natural Resources
• Chapter 15 Improvement in Food Resources

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Access Answers of Science NCERT class 9 Chapter 8: Motion  (All intext and exercise questions solved)

Intext Questions – 1   Page: 100

1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

Yes, an object which has moved through a distance can have zero displacement if it comes back to its initial position.

Example: If a person jogs in a circular park which is circular and completes one round. His initial and final position is the same.

Hence, his displacement is zero.

2. A farmer moves along the boundary of a square field of side 10m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Side of the given square field = 10m

Hence, the perimeter of a square = 40 m

Time taken by the farmer  to cover the boundary of 40 m = 40 s

So, in 1 s, the farmer covers a distance of 1 m

Distance covered by the farmer in 2 min 20 sec = 1 x 140 = 140 m

The total number of rotations taken by the farmer to cover a distance of 140 m = total distance/perimeter

At this point, let us say the farmer is at point B from the origin O

Therefore,  from Pythagoras theorem, the displacement s = √(10 2 +10 2 )

s =  10 √ 2

s = 14.14 m

3. Which of the following is true for displacement? (a) It cannot be zero. (b) Its magnitude is greater than the distance travelled by the object.

Neither of the statements is true.

(a) Given statement is false because the displacement of an object which travels a certain distance and comes back to its initial position is zero.

(b) Given statement is false because the displacement of an object can be equal to, but never greater than the distance travelled.

Intext Questions – 2   Page: 102

1. Distinguish between speed and velocity.

2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?

Since average speed is the total distance travelled in a time frame and velocity is the total displacement in the time frame, the magnitude of average velocity and average speed will be the same when the total distance travelled is equal to the displacement.

3. What does the odometer of an automobile measure?

An odometer, or odograph, is a device that measures the distance travelled by an automobile based on the perimeter of the wheel as the wheel rotates.

4. What does the path of an object look like when it is in uniform motion?

The path of an object in uniform motion is a straight line.

5. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 10 8 m/s.

Given that the signal travels in a straight line, the distance between the spaceship and the ground station is equal to the total distance travelled by the signal.

5 minutes = 5*60 seconds = 300 seconds.

Speed of the signal = 3 × 10 8 m/s.

Therefore, total distance = (3 × 10 8 m/s) * 300s

= 9*10 10 meters.

Intext Questions – 3   Page: 103

1. When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?

Uniform Acceleration:  When an object is travelling in a straight line with an increase in velocity at equal intervals of time, then the object is said to be in uniform acceleration.

The free-falling of an object is an example of uniform acceleration.

Non-Uniform Acceleration:  When an object is travelling with an increase in velocity but not at equal intervals of time is known as non-uniform acceleration.

Bus moving or leaving from the bus stop is an example of non-uniform acceleration.

2. A bus decreases its speed from 80 km h –1 to 60 km h –1 in 5 s. Find the acceleration of the bus.

Given, the initial velocity (u) = 80km/hour = 80000m/3600s= 22.22 m.s -1

The final velocity (v) = 60km/hour = 60000m/3600s= 16.66 m.s -1

Time frame, t = 5 seconds.

Therefore, acceleration (a) =(v-u)/t = (16.66 m.s -1 – 22.22 m.s -1 )/5s

= -1.112 m.s -2

Therefore, the total acceleration of the bus is -1.112m.s -2 . It can be noted that the negative sign indicates that the velocity of the bus is decreasing.

3. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h –1 in 10 minutes. Find its acceleration.

Given parameters

Initial velocity (u) = 0

Final velocity (v) = 40 km/h

v = 40 × (5/18)

v = 11.1111 m/s

Time (t) = 10 minute

t = 60 x 10

Acceleration (a) =?

Consider the formula

11.11 = 0 + a × 600

11,11 = 600 a

a = 11.11/600

a = 0.0185 ms -2

Intext Questions – 4   Page: 107

1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?

For uniform motion, the distance-time graph is a straight line. On the other hand, the distance-time graph of an object in non-uniform motion is a curve.

The first graph describes the uniform motion and the second one describes the non-uniform motion.

2. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?

The distance-time graph can be plotted as follows.

When the slope of the distance-time graph is a straight line parallel to the time axis, the object is at the same position as time passes. That means the object is at rest.

3. What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?

The speed-time graph can be plotted as follows.

Since there is no change in the velocity of the object (Y-Axis value) at any point of time (X-axis value), the object is said to be in uniform motion.

4. What is the quantity which is measured by the area occupied below the velocity-time graph?

Considering an object in uniform motion, its velocity-time graph can be represented as follows.

Now, the area below the velocity-time graph is the area of the rectangle OABC, which is given by OA*OC. But OA is the velocity of the object and OC represents time. Therefore, the shaded area can be represented as:

Area under the velocity-time graph = velocity*time.

Substituting the value of velocity as displacement/time in the previous equation, it is found that the area under the velocity-time graph represents the total displacement of the object.

Intext Questions – 5 Page: 109,110

1. A bus starting from rest moves with a uniform acceleration of 0.1 m s -2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.

(a) Given, the bus starts from rest. Therefore, initial velocity (u) = 0 m/s

Acceleration (a) = 0.1 m.s -2

Time = 2 minutes = 120 s

Acceleration is given by the equation a=(v-u)/t

Therefore, terminal velocity (v) = (at)+u

= (0.1 m.s -2 * 120 s) + 0 m.s -1

= 12 m.s -1 + 0 m.s -1

Therefore, terminal velocity (v) = 12 m/s

(b) As per the third motion equation, 2as = v 2 – u 2

Since a = 0.1 m.s -2 , v = 12 m.s -1 , u = 0 m.s -1 , and t = 120 s, the following value for s (distance) can be obtained.

Distance, s =(v 2 – u 2 )/2a

=(12 2 – 0 2 )/2(0.1)

Therefore, s = 720 m.

The speed acquired is 12 m.s -1 and the total distance travelled is 720 m.

2. A train is travelling at a speed of 90 km h –1 . Brakes are applied so as to produce a uniform acceleration of –0.5 m s -2 . Find how far the train will go before it is brought to rest.

Given, initial velocity (u) = 90 km/hour = 25 m.s -1

Terminal velocity (v) = 0 m.s -1

Acceleration (a) = -0.5 m.s -2

As per the third motion equation, v 2 -u 2 =2as

Therefore, distance traveled by the train (s) =(v 2 -u 2 )/2a

s = (0 2 -25 2 )/2(-0.5) meters = 625 meters

The train must travel 625 meters at an acceleration of -0.5 ms -2 before it reaches the rest position.

3. A trolley, while going down an inclined plane, has an acceleration of 2 cm s -2 . What will be its velocity 3 s after the start?

Given, initial velocity (u) = 0 (the trolley begins from the rest position)

Acceleration (a) = 0.02 ms -2

Time (t) = 3s

As per the first motion equation, v=u+at

Therefore, terminal velocity of the trolley (v) = 0 + (0.02 ms -2 )(3s)= 0.06 ms -1

Therefore, the velocity of the trolley after 3 seconds will be 6 cm.s -1

4. A racing car has a uniform acceleration of 4 m s -2 . What distance will it cover in 10 s after start?

Given, the car is initially at rest; initial velocity (u) = 0 ms -1

Acceleration (a) = 4 ms -2

Time period (t) = 10 s

As per the second motion equation, s = ut+1/2 at 2

Therefore, the total distance covered by the car (s) = 0 * 10m + 1/2 (4ms -2 )(10s) 2

= 200 meters

Therefore, the car will cover a distance of 200 meters after 10 seconds.

5. A stone is thrown in a vertically upward direction with a velocity of 5 m s -1 . If the acceleration of the stone during its motion is 10 m s –2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Given, initial velocity (u) = 5 m/s

Terminal velocity (v) = 0 m/s (since the stone will reach a position of rest at the point of maximum height)

Acceleration = 10 ms -2 in the direction opposite to the trajectory of the stone = -10 ms -2

As per the third motion equation, v 2 – u 2 = 2as

Therefore, the distance travelled by the stone (s) = (0 2 – 5 2 )/ 2(10)

Distance (s) = 1.25 meters

As per the first motion equation, v = u + at

Therefore, time taken by the stone to reach a position of rest (maximum height) = (v – u) /a

=(0-5)/-10 s

Time taken = 0.5 seconds

Therefore, the stone reaches a maximum height of 1.25 meters in a timeframe of 0.5 seconds.

Exercises Page: 112,113

1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?

Given, diameter of the track (d) = 200m

Therefore, the circumference of the track (π*d) = 200π meters.

Distance covered in 40 seconds = 200π meters

Distance covered in 1 second = 200π/40

Distance covered in 2minutes and 20 seconds (140 seconds) = 140 * 200π/40 meters

= (140*200*22)/(40* 7) meters = 2200 meters

Number of rounds completed by the athlete in 140 seconds = 140/40 = 3.5

Therefore, the final position of the athlete (with respect to the initial position) is at the opposite end of the circular track. Therefore, the net displacement will be equal to the diameter of the track, which is 200m.

Therefore, the net distance covered by the athlete is 2200 meters and the total displacement of the athlete is 200m.

2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?

Given, distance covered from point A to point B = 300 meters

Distance covered from point A to point C = 300m + 100m = 400 meters

Time taken to travel from point A to point B = 2 minutes and 30 seconds = 150 seconds

Time taken to travel from point A to point C = 2 min 30 secs + 1 min = 210 seconds

Displacement from A to B = 300 meters

Displacement from A to C = 300m – 100m = 200 meters

Average speed = total distance travelled/ total time taken

Average velocity = total displacement/ total time taken

Therefore, the average speed while traveling from A to B = 300/150 ms -1 = 2 m/s

Average speed while traveling from A to C = 400/210 ms -1 = 1.9 m/s

Average velocity while traveling from A to B =300/150 ms -1 = 2 m/s

Average velocity while traveling from A to C =200/210 ms -1 = 0.95 m/s

3. Abdul, while driving to school, computes the average speed for his trip to be 20 km.h –1 . On his return trip along the same route, there is less traffic and the average speed is 30 km.h –1 . What is the average speed for Abdul’s trip?

Distance travelled to reach the school = distance travelled to reach home = d (say)

Time taken to reach school = t 1

Time taken to reach home = t 2

therefore, average speed while going to school = total distance travelled/ total time taken = d/t 1 = 20 kmph

Average speed while going home = total distance travelled/ total time taken = d/t 2 = 30 kmph

Therefore, t 1 = d/20 and t 2 = d/30

Now, the average speed for the entire trip is given by total distance travelled/ total time taken

= (d+d)/(t 1 +t 2 )kmph = 2d/(d/20+d/30)kmph

= 120/5 kmh -1 = 24 kmh -1

Therefore, Abdul’s average speed for the entire trip is 24 kilometers per hour.

4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s –2 for 8.0 s. How far does the boat travel during this time?

Given, initial velocity of the boat = 0 m/s

Acceleration of the boat = 3 ms -2

Time period = 8s

As per the second motion equation, s = ut + 1/2 at 2

Therefore, the total distance travelled by boat in 8 seconds = 0 + 1/2 (3)(8) 2

= 96 meters

Therefore, the motorboat travels a distance of 96 meters in a time frame of 8 seconds.

5. A driver of a car travelling at 52 km h –1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h –1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?

The speed v/s time graphs for the two cars can be plotted as follows.

The total displacement of each car can be obtained by calculating the area beneath the speed-time graph.

Therefore, displacement of the first car = area of triangle AOB

= (1/2)*(OB)*(OA)

But OB = 5 seconds and OA = 52 km.h -1 = 14.44 m/s

Therefore, the area of the triangle AOB is given by: (1/2)*(5s)*(14.44ms -1 ) = 36 meters

Now, the displacement of the second car is given by the area of the triangle COD

= (1/2)*(OD)*(OC)

But OC = 10 seconds and OC = 3km.h -1 = 0.83 m/s

Therefore, area of triangle COD = (1/2)*(10s)*(0.83ms -1 ) = 4.15 meters

Therefore, the first car is displaced by 36 meters whereas the second car is displaced by 4.15 meters. Therefore, the first car (which was traveling at 52 kmph) travelled farther post the application of brakes.

6. Fig 8.11 shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:

(a) Which of the three is travelling the fastest? (b) Are all three ever at the same point on the road? (c) How far has C travelled when B passes A? (d) How far has B travelled by the time it passes C?

(a) Since the slope of line B is the greatest, B is travelling at the fastest speed.

(b) Since the three lines do not intersect at a single point, the three objects never meet at the same point on the road.

(c) Since there are 7 unit areas of the graph between 0 and 4 on the Y axis, 1 graph unit equals 4/7 km.

Since the initial point of an object C is 4 graph units away from the origin, Its initial distance from the origin is 4*(4/7)km = 16/7 km

When B passes A, the distance between the origin and C is 8km

Therefore, total distance travelled by C in this time = 8 – (16/7) km = 5.71 km

(d) The distance that object B has covered at the point where it passes C is equal to 9 graph units.

Therefore, total distance travelled by B when it crosses C = 9*(4/7) = 5.14 km

7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s -2 , with what velocity will it strike the ground? After what time will it strike the ground?

Given, initial velocity of the ball (u) = 0 (since it began at the rest position)

Distance travelled by the ball (s) = 20m

Acceleration (a) = 10 ms -2

As per the third motion equation,

v 2 – u 2 = 2as

= 2*(10ms -2 )*(20m) + 0

v 2 = 400m 2 s -2

Therefore, v= 20ms -1

The ball hits the ground with a velocity of 20 meters per second.

As per the first motion equation,

Therefore, t = (v-u)/a

= (20-0)ms -1 / 10ms -2

= 2 seconds

Therefore, the ball reaches the ground after 2 seconds.

8. The speed-time graph for a car is shown in Fig. 8.12

(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period. (b) Which part of the graph represents uniform motion of the car?

The shaded area represents the displacement of the car over a time period of 4 seconds. It can be calculated as:

(1/2)*4*6 = 12 meters. Therefore the car travels a total of 12 meters in the first four seconds.

(b) Since the speed of the car does not change from the points (x=6) and (x=10), the car is said to be in uniform motion from the 6 th to the 10 th second.

9. State which of the following situations are possible and give an example for each of these: (a) an object with a constant acceleration but with zero velocity (b) an object moving with an acceleration but with uniform speed. (c) an object moving in a certain direction with an acceleration in the perpendicular direction.

(a) It is possible; an object thrown up into the air has a constant acceleration due to gravity acting on it. However, when it reaches its maximum height, its velocity is zero.

(b) it is possible; acceleration implies an increase or decrease in speed, and uniform speed implies that the speed does not change over time

Circular motion is an example of an object moving with acceleration but with uniform speed.

An object moving in a circular path with uniform speed is still under acceleration because the velocity changes due to continuous changes in the direction of motion.

(c) It is possible; for an object accelerating in a circular trajectory, the acceleration is perpendicular to the direction followed by the object.

10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.

Given, the radius of the orbit = 42250 km

Therefore, circumference of the orbit = 2*π*42250km = 265571.42 km

Time is taken for the orbit = 24 hours

Therefore, speed of the satellite = 11065.4 km.h -1

The satellite orbits the Earth at a speed of 11065.4 kilometers per hour.

NCERT Class 9 Science Chapter 8 explains the concept of motion, and types of motion with relevant examples for a clear understanding of the concept. It explains the causes of phenomena like sunrise, sunset, and changing of the seasons. It helps students understand uniform and non-uniform motion. Distance-time graphs and velocity-time graphs, which are considered important concepts for examination, are explained in an easy way in NCERT Solutions . It describes how the acceleration of an object is the change in velocity per unit time. NCERT Class 9 Science Chapter 8 is covered under Unit III: Motion, Force and Work and can get you maximum marks.

• NCERT Solutions for Class 9 explains motion in terms of distance moved or displacement.
• Uniform and non-uniform motions of objects are explained through the graph and examples.
• Uniform circular motion concept is made understandable in a simple way.
• Problems on acceleration, velocity, and average velocity are also solved.

Key Features of NCERT Solutions for Class 9 Science Chapter 8: Motion

• A simple and easily understandable approach is followed in NCERT Solutions to make students aware of topics.
• Provides complete solutions to all the questions present in the respective NCERT textbooks.
• NCERT Solutions offers detailed answers to all the questions to help students in their preparations.
• These solutions will be useful for CBSE exams, Science Olympiads, and other competitive exams.

Disclaimer:

Dropped Topics –  8.5 Equations of motion by graphical method, 8.5.1 Equation for Velocity–Time Relation, 8.5.2 Equation for Position–Time relation and 8.5.3 Equation for Position– Velocity.

Frequently Asked Questions on NCERT Solutions for Class 9 Science Chapter 8

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Chapter-wise NCERT Solutions Class 9 Science Free PDF | Marking Scheme & Benefits of CBSE 9th Science NCERT Textbook Solutions

Science concepts of class 9 are quite tricky to understand for students at the early stage. To make them ease with the learning of science subject, we have come up with the best study material ie., NCERT Solutions . With CBSE Class 9 Science NCERT Book Solutions, students can easily clear their doubts instantly in a more efficient way. So, Download the latest and updated NCERT Solutions for Class 9 Science pdf 2021-22 in English and Hindi medium for free of charge and achieve your academic goals.

CBSE NCERT Solutions for Class 9 Science Chapterwise Free PDF Download

All the contents related to 9th std science are made available here in the chapterwise NCERT Solutions class 9 science book . Hence, get the updated academic session 2021-2022 Class 9 Textbook Solutions NCERT Science PDF to start learning all the concepts efficiently and score maximum marks in the exams.

Chapter 1 Matter in Our Surroundings

Chapter 2 is matter around us pure, chapter 3 atoms and molecules, chapter 4 structure of the atom, chapter 5 the fundamental unit of life, chapter 6 tissues, chapter 7 diversity in living organisms, chapter 8 motion, chapter 9 force and laws of motion, chapter 10 gravitation, chapter 11 work and energy, chapter 12 sound, chapter 13 why do we fall ill, chapter 14 natural resources.

• CBSE 2021- 22 Class 9 Science Term 1 & Term 2 Exam Marking Scheme

NCERT Class 9 Science Solutions All Chapters Brief

Key features of ncert 9th class science textbook solutions pdf.

• How many chapters are there in the Class 9 Science NCERT Solutions Book?
• Where can I perceive the NCERT solutions for Class 9 Science?
• Is NCERT Solutions enough to score well in 9th science CBSE exams?

MCQ Questions for Class 9 Science with Solutions PDF

Apart from the answers prepared to all textbook questions, you can also get extra knowledge from the NCERT Solutions Class 9 Science PDF as it comes with the additional exercise questions, exemplar problems, the important questions from previous year question papers, sample papers, worksheets, MCQ’s, short answering questions, descriptive type questions, their solutions and also some tips and tricks.

The list of chapter-wise MCQ Questions for 9th Class Science with Answers PDF is provided here for quick reference before exams and secure high in term 1 exam.

• Chapter 1  Class 9 Matter in Our Surroundings MCQ Questions
• Chapter 2  Class 9 Is Matter Around Us Pure MCQ Questions
• Chapter 3  Class 9 Atoms and Molecules MCQ Questions
• Chapter 4  Class 9 Structure of the Atom MCQ Questions
• Chapter 5  Class 9 The Fundamental Unit of Life MCQ Questions
• Chapter 6  Class 9 Tissues MCQ Questions
• Chapter 7  Class 9 Diversity in Living Organisms MCQ Questions
• Chapter 8  Class 9 Motion MCQ Questions
• Chapter 9  Class 9 Force and Laws of Motion MCQ Questions
• Chapter 10  Class 9 Gravitation MCQ Questions
• Chapter 11  Class 9 Work, Power and Energy MCQ Questions
• Chapter 12  Class 9 Sound MCQ Questions
• Chapter 13  Class 9 Why Do we Fall Ill MCQ Questions
• Chapter 14  Class 9 Natural Resources MCQ Questions
• Chapter 15  Class 9 Improvement in Food Resources MCQ Questions

CBSE 2021- 22 Class 9 Science Marking Scheme for Term 1 & Term 2

Knowing the marking scheme for each term of the CBSE Board Exam makes you understand which chapters you need to focus on more to secure good grades.

Marking Scheme For 9th Science 1st Term Exam 2021-2022

Marking Scheme For 9th Science 2nd Term Exam 2021-2022

Download NCERT Textbooks PDF  for class 1st to 12th

Ncertbooks.Guru gives a perfect platform for all students to download the solutions books or the textbooks for free. Here, we have elaborated a brief about all the chapters of science subject for your reference just go through them once before you start preparing for the exams.

NCERT Class 9 Science Chapter 1 Solutions Book contains questions, answers, images, step by step explanations of the complete Chapter 1 titled Matter in Our Surroundings. The Topics and Sub Topics included in CBSE 9th Grade Science Chapter 1 are Matter in Our Surroundings, Physical Nature of Matter, Characteristics of Particles of Matter, States of Matter, Can Matter Change its State?, Evaporation.

Students who are pursuing 9th standard utilize the NCERT Textbook to study Science. Here, NCERT Solutions for class 9 science Chapter 2 Is Matter Around Us Pure deals with the mixture, solution, properties of solutions, separation of mixtures, physical and chemical changes. Want to know more about the topics then click on the above pdf links of the NCERT solutions book and prepare well.

In Chapter 3 NCERT Class 9 Science Solutions, you will study all about atoms and molecules like the laws of chemical combination, how to write a chemical formula, molecular mass, and mole concepts, and some numerical problems related to these concepts.

Class 9 Science Chapter 4 is completely on the basis of sub-atomic particles such as a proton – p, neutron – n, and electron – e. Likewise, it also covers electrons circulation in different circles called shells, count of valency, the concept of mass number, and atomic number.

From this chapter 5 the fundamental unit of life, students will learn the classification and structure of the living cell, different cell organelles such as the endoplasmic reticulum, Golgi apparatus, lysosomes, mitochondria, plastids, and vacuoles. The processes of cell division – Mitosis and Meiosis are also explained.

The knowledge that you can gain from this chapter 6 Tissues will help you understand both plant and animal tissues’ structure, functions, and location of each type of tissue. When you consider plant tissues, the meristematic tissue and permanent tissue are discussed. Under animal tissues, you will study epithelial tissue, connective tissue, muscular tissue, and nervous tissue.

The 7th chapter Diversity in Living Organisms guides the classification of all the life forms. Kids of class 7 become familiarized with the topic that all living beings are into 5 realms, specifically Monera, Protista, Parasites, Plantae, and Animalia. Moreover, it describes the classification and advancement, the pecking order of classification.

In this physics chapter, students will understand uniform motion and non-uniform motion. Also, it makes you learn quite easily about the concepts of speed and direction of motion.

At the beginning of the chapter, you will come to know what is the first, second, and third laws of motion with examples, mathematical equations, and applications. In accordance with Newton’s laws of motion, you will acquire the fundamental concepts in physics like inertia, mass, and conservation of momentum with real-life examples.

All about gravitation concepts are explained in this chapter for helping students to understand the Universal Law of Gravitation, the concept of acceleration, differences between mass and weight, and many others like free fall, thrust, pressure, buoyancy, Archimedes’ Principle, and relative density.

Here in this unit, students will start knowing what is work and energy and what are the scientific meaning, forms of energy, and more. The chapter additionally includes various activities and examples to aid students’ understand the topics efficiently.

NCERT CBSE 9th Science Chapter 12 Sound covers the major topics like production of sound, propagation of sound, the reflection of sound, range of hearing, applications of ultrasound, the structure of the human ear.

NCERT Class 9 Science chapter 13 Why Do We Fall Ill include points like Well-being and health, sickness, and its reasons. To get awareness in students about various sorts of illnesses, diseases and its cause, infectious diseases, this chapter is included in the syllabus.

The Role Of The Atmosphere In Climate Control, Rain, Air Pollution, Water Pollution, Mineral Riches In The Soil, Water-Cycle, Nitrogen-Cycle, Carbon-Cycle, Oxygen-Cycle, and Ozen Layer are the topics that are explained here with the best examples for easy understanding to students of class 9.

Chapter 15 Improvement in Food Resources

Here, you will find a detailed explanation about the concepts such as Improvement In Crop Yields, Crop Variety Improvement, Crop Production & Protection Management, Animal Husbandry, Cattle Farming, Poultry Farming, Fish Production. Also, it provides knowledge concerning agribusiness, cultivating, and dairy.

If you follow these NCERT Solutions Book for 9th Science exam preparation then you can find a wide range of features that helps everyone in attempting the exams. Some of them are stated below:

• We have designed the most appropriate solutions for all your NCERT needs which cover all physics, chemistry, and biology in a detailed manner.
• Not only Solutions of NCETR Science book class 9 PDF Download but you can get access to solutions to various text books such as HC Verma, Lakhmir Singh, R S Aggarwal, etc.
• In the chemistry part, you will find the chapters like matter in our surroundings, Is matter around us Pure, atoms and molecules, etc.
• The biology part covers the fundamental unit of life, tissues, and diversity in Living Organism in a very easy-to-understand language.
• Physics chapters such as Motion and laws of motion, Gravitation, work & energy, and sound are a bit tough to understand but NCERT book solutions pdf chapterwise will help you.
• Avail CBSE Class 9 Science NCERT Solutions for free at NCERTBooks.Guru and try solving all the exercise problems from the 9th science textbook.

FAQs on CBSE Class 9 Science NCERT Book Solutions Free Download

1. How many chapters are there in the Class 9 Science NCERT Solutions Book?

There are 15 chapters present in the NCERT Solutions for Class 9 Science PDF Book as they improve your subject knowledge and boost up your confidence to attempt any kind of questions asked in the final exams.

2. Where can I perceive the NCERT solutions for Class 9 Science?

You can look for NCERTbooks.Guru website to get access and download the CBSE Class 9 Science NCERT Solutions chapterwise. Practice all 1-15 chapters concepts of physics, chemistry, and biology covered in the NCERT Science Textbook and score well in the exams.

3. Is NCERT Solutions enough to score well in 9th science CBSE exams?

Yes, the NCERT Solutions for 9th class science chapterwise pdfs are enough to prepare and score well in the exams. Consistently practicing the questions solved in the NCERT Textbook Solutions helps to learn the concept easily and also they can understand how effectively they can frame the answers during exams.

Final Words

We hope he furnished data regarding NCERT Solutions for 9th Class Science pdf helps you a lot to the possible extent. If you have any queries on the subject concepts please clear them by referring to our provided study resources like NCERT Books, NCERT Solutions, Notes, Sample Papers, etc., or drop a comment below for getting solved in no time.

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Class 9 Science Assignments

We have provided below free printable Class 9 Science Assignments for Download in PDF. The Assignments have been designed based on the latest NCERT Book for Class 9 Science . These Assignments for Grade 9 Science cover all important topics which can come in your standard 9 tests and examinations. Free printable Assignments for CBSE Class 9 Science , school and class assignments, and practice test papers have been designed by our highly experienced class 9 faculty. You can free download CBSE NCERT printable Assignments for Science Class 9 with solutions and answers. All Assignments and test sheets have been prepared by expert teachers as per the latest Syllabus in Science Class 9. Students can click on the links below and download all Pdf Assignments for Science class 9 for free. All latest Kendriya Vidyalaya Class 9 Science Assignments with Answers and test papers are given below.

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Assignments For Class 9 Physics

Assignments for Class 9 Physics have been developed for Standard 9 students based on the latest syllabus and textbooks applicable in CBSE, NCERT and KVS schools. Parents and students can download the full collection of class assignments for class 9 Physics from our website as we have provided all topic wise assignments free in PDF format which can be downloaded easily. Students are recommended to do these assignments daily by taking printouts and going through the questions and answers for Grade 9 Physics. You should try to do these test assignments on a daily basis so that you are able to understand the concepts and details of each chapter in your Physics book and get good marks in class 9 exams.

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All students studying in Grade 9 Physics should download the assignments provided here and use them for their daily routine practice. This will help them to get better grades in Physics exam for standard 9. We have made sure that all topics given in your textbook for Physics which is suggested in Class 9 have been covered ad we have made assignments and test papers for all topics which your teacher has been teaching in your class. All chapter wise assignments have been made by our teachers after full research of each important topic in the textbooks so that you have enough questions and their solutions to help them practice so that they are able to get full practice and understanding of all important topics. Our teachers at https://www.assignmentsbag.com have made sure that all test papers have been designed as per CBSE, NCERT and KVS syllabus and examination pattern. These question banks have been recommended in various schools and have supported many students to practice and further enhance their scores in school and have also assisted them to appear in other school level tests and examinations. Its easy to take print of thee assignments as all are available in PDF format.

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