assignment on chapter motion class 9

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Unit 1: Motion

About this unit.

Motion is all around us, from moving cars to flying aeroplanes. Motion can have different features like speed, direction, acceleration, etc. In this chapter, we will understand these features in detail and see how it can help us predict the future of these moving things.

Distance and displacement

• Distance and displacement introduction (Opens a modal)
• Distance and displacement in one dimension (Opens a modal)

Average speed and average velocity

• Average speed & velocity (with examples) (Opens a modal)
• Calculating average velocity or speed (Opens a modal)
• Average speed for entire journey - solved numerical (Opens a modal)
• Average velocity and speed in one direction: word problems 4 questions Practice
• Average velocity and speed with direction changes: word problems 4 questions Practice

Instantaneous speed and velocity

• Instantaneous speed & velocity (Opens a modal)

Acceleration

• Acceleration (Opens a modal)
• Airbus A380 take-off time (Opens a modal)
• Acceleration and velocity 7 questions Practice

Position time graphs

• Position-time graphs (Opens a modal)
• Calc. velocity from position time graphs (Opens a modal)

Velocity time graphs

• Velocity time graphs (& acceleration) (Opens a modal)
• Calculating displacement from v-t graphs (Opens a modal)

Deriving equations of motion

• Deriving 3 equations of motion (from v-t graph) (Opens a modal)

Problem solving using kinematic equations

• Using equations of motion (1 step numerical) (Opens a modal)
• Using equations of motion (2 steps numerical) (Opens a modal)
• Kinematic equations: numerical calculations 4 questions Practice

Uniform circular motion

• Calc. speed & time in a uniform circular motion - Solved numerical (Opens a modal)

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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Chapter 8 Class 9 - Motion

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Get Notes, NCERT Solutions (in the end of chapter), Solutions to Questions from Inside the Book, Examples from the NCERT, Multiple Choice Questions, Graphical Questions of Chapter 8 Class 9 Motion.

Teachoo provides the best notes to learn about Motion and get excellent marks for your exams. And, in addition to the notes, we have

• Teachoo Questions - Our mix of the best practice questions for Chapter 8 Class 9 Science - Motion
• Multiple Choice Questions - from NCERT Exemplar
• Case Based and Assertion Reasoning Questions

In this chapter, we will learn

What is Motion and What are the Different Types of Motion

What is meaning of Distance and Displacement

What is Uniform Motion and Non Uniform Motion - with graphs

What is Speed

How do we calculate Average Speed

What is Velocity

How do we calculate Average Velocity

What is Acceleration  

First Equation of Motion - Explanation and Derivation

Second Equation of Motion - Explanation and Derivation

Third Equation of Motion - Explanation and Derivation

What is Uniform Circular Motion

Distance Time Graph - and finding velocity from it

Speed Time Graph or Velocity Time Graph - and how to find acceleration and distance from it

Graphical Derivation of Equations of Motion

We will also do some Numerical Questions from the NCERT, Examples from NCERT Book, some extra numerical question and also questions where we are given a graph and we need to find acceleration and distance from it.

Click on any link below to get started

Note : When you click on a link, the first question will open. To open other questions, there is a list with arrows at end. It has the links of all the questions and concepts. 

NCERT Questions

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NCERT Solutions for Class 9 Science Chapter 8 Motion

Chapter 8 Motion Class 9 Science NCERT Solutions

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Motion Chapter 8 Class 9 Science Assignments

Please refer to Motion Chapter 8 Class 9 Science Assignments below. We have provided important questions and answers for Motion which is an important chapter in Class 9 Science. Students should go through the notes and also learn the solved assignment with solved questions provided below. All examination and class tests questions are as per the latest syllabus and books issued by CBSE, NCERT, and KVS. We have also provided Class 9 Science Assignments for all chapters on our website.

Chapter 8 Motion Class 9 Science Assignments

Question. When is the acceleration of a body positive?

The acceleration of a body is taken to be positive if it is in the direction of velocity.

Question. What is the relationship between distance travelled and the time elapsed for the motion with uniform velocity?

Distance is directly proportional to time used. In fact, distance travelled = Uniform velocity × Time used.

Question.Name the different types of graph.

Bar graphs, straight line graphs, histograms, etc.

Question. Which type of graph is used to describe motion?

Line graph are used to describe the motion of an object.

Question. Define the term velocity. What is its unit? Is it a scalar or vector quantity?

Velocity is a physical quantity that gives the speed and direction of motion of the body Velocity = Displacement/ Time The S.I. unit of velocity is ms–1. Velocity is a vector quantity because its description requires both magnitude and direction.

Question.  Name the two types of physical quantities.

The two types of physical quantities are scalar quantity and vector quantity.

Question. What is the acceleration of a body moving uniform velocity?

The acceleration of a body moving with uniform velocity is zero.

Question. What is magnitude?

The numerical value of a physical quantity is called its magnitude.

Question. Define a scalar quantity.

A physical quantity which has only magnitude and no direction is called a scalar quantity.

Question. What is the difference between uniform linear motion and uniform circular motion? Answer.  When a body is moving with a uniform speed along a straight line, its motion is called linear uniform motion. The uniform linear motion is an accelerated motion. When a body is moving with a constant speed along a circular path, the direction of motion of body changes continuously with time. We know that a change in the direction of motion implies a change in velocity. Thus, uniform circular motion is an accelerated motion even though the speed of the body remains constant.

Question.Why is the motion in a circle at constant speed called accelerated motion? Answer.  When a body moves along a circular path with a constant speed, its direction of motion at any point is along the tangent to the circle at that point, this motion called accelerated motion. The direction of motion changes as the body moves in a circle and causes a change in the velocity. Therefore, the motion of an object along a circular path is an accelerated motion.

Question.What do graphs provide? Answer.  The graphs provide a convenient method to present pictorially the basic information about a variety of events such as motion.

Question.Give an example of use of bar graph and straight graph. Answer.  In telecast of one-day cricket match on T.V. vertical bar graphs show the run rate of a team in over. Straight line graph helps in solving linear equation in two variables.

Question.When do we say that the position of body has changed? Answer. If the distance, or direction, or both of a body relative to a reference point changes then we say that the position of the body has changed.

Question.Classify the following as scalar and vector quantities : Mass, weight, time, temperature, volumes, velocity, speed, forces, acceleration. Answer.

Question.Define uniform acceleration. Give one example. Answer.  Uniform acceleration : An object has a uniform acceleration if it travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time. For example : The motion of a freely falling body has uniform acceleration.

Question.In a long distance race, the athletes were expected to take four rounds of the track such that the line of finish was same as the track was 200 m. (i) What is the total distance to be covered by the athletes? (ii) What is the displacement of the athletes when they touch the finish line? (iii) Is the motion of the athletes uniform or nonuniform? (iv) Is the distance moved by and displacement of athletes at the end of the race equal? Answer. (i) Total distance covered by the athletes = 4 × 200 = 800 m. (ii) The line of start and the line of finish are the same so, Displacement = 0 (iii) The motion of the athletes is non-uniform. (iv) The distance and displacement of an athlete at the end of the race are not equal.

Question.(i) Give an example of a body which may appear to be moving for one person and stationary for the other. (ii) What can we tell about motion from the above example? Answer. (i) The man standing on the roadside observe that a bus along with its passenger is moving.At the same time, a passenger sitting in the bus observes his fellow passenger to be at rest. (ii) We can tell that motion is relative.

Question.How many different kinds of motion are there? Name them. Is there any motion which is a combination of two or more types of motion? Answer. (i) There are three kinds of motion. These are : (a) Linear or translatory motion. (b) Circular or rotational motion. (c) Vibratory or oscillatory motion. (ii) Complex motion means a combination of two or more types of motions.

Question.What is motion? Give some examples of directly perceivable motion in daily life. Answer. (i) A body is said to be in motion if it changes its position in relation to a reference point (origin). (ii) Some examples of directly perceivable motion in daily life are : (a) Birds and animals moving from one place to another. (b) Cars moving on the roads. (c) Aeroplanes flying in the sky. (d) Blades of a moving fan. (e) Fish swimming in water.

Question. Find the distance covered by a particle during the time interval t = 0 to t = 20 s for which the speedtime graph is shown in figure.

Answer. The distance covered in the time interval 0 to 20 s is equal to the area of the shaded triangle. So, Distance = 1/2 × base × height = 1/2 × (20 s) x (20 m/s) = 200 m.

Question. A bus moves 30 km in 30 min and the next 30 km. in 40 min. Calculate the average speed for the entire journey. Answer. Given, the total time taken is 30 min + 40 min = 70 min and the total distance travelled is 30 km + 30 km = 60 km The average speed is

Question. A girl runs for 10 min at a uniform speed of 9 km/h. What should be the speed that she runs for the next 20 min, so that the average speed comes to 12 km/h? Answer. Total time = 10 min 20 min = 30 min The average speed is 12 km/h using s = vt, the distance covered in 30 min is

Thus, he has to cover 6 km – 1.5 km = 4.5 km in the next 20 min. The speed required is

Question. It is estimated that the radio signal takes 1.27 seconds to reach the Earth from the surface of the Moon.Calculate the distance of the Moon from the Earth. Speed of radio signal = 3 × 108 ms –1  (speed of light in air). Answer. Here, time = 1.27 s Speed = 3 × 108 ms –1 Distance = ? Using distance = speed × time, we get Distance = 3 × 108 ms –1  × 1.27 s = 3.81 × 10 8  m = 3.81 × 10 5  km

Question. Divya walked 2 km on a straight road and then walked back 1 km. Which of the two quantities involved in her walking is greater- the scalar or vector? Answer. Distance travelled by Divya = 2 km + 1 km = 3 km Displacement = 2 km  –  1 km = 1 km Hence, distance which is a scalar quantity is greater than the displacement which is a vector quantity.

Question.   Two satellites A and B revolve around a planet C. The time taken by satellite B to go around the planet is twice the time taken by A. Which of the two satellites will have a greater magnitude of velocity? Answer. Satellite A will have greater magnitude of velocity since velocity is inversely proportional to time. (v = 2p r/t)

Question.   A child moving on a circular track of radius 40 m completes one revolution in 5 minutes. What is his (i) average speed (ii) average velocity in one full revolution? Answer.  Distance travelled in one revolution,2p r =2p × 40m Displacement in one revolution = 0 Time taken = 5 minutes = 5 × 60s

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Assignment - Motion, Class 9, Science PDF Download

TRUE OR FALSE

1. The motion in animals is called locomotion. Mechanics deals with the motion of non living object. 2. Kinematics deals with the motion of non-living object without taking into account the cause of their motion. 3. Motion along a curved line is called translatory or rectilinear motion. 4. A body is said to be at rest if it does not change its position with respect to the reference point. 5. A quantity which can be represented completely by magnitude along is called a vector quantity. 6. A quantity which can be completely specified by magnitude as well as direction is called a scalar quantity. 7. Velocity and speed are measured in different units. 8. In one dimensional motion the average velocity and the instantaneous velocity are unequal. 9. A motion is said to be uniform if a body undergoes equal displacements in equal intervals of time. 10. A motion is said to be uniform if x α t2. 11. Acceleration is defined as the rate of change of velocity. 12. The graph between velocity and time for uniform acceleration is a curved line.

MATCH THE COLUMN

FILL IN THE BLANKS

1. Mechanics is the branch of physics which deals with .......................... 2. Statics is a sub branch of mechanics which deals with bodies at .......................... 3. Dynamics is a sub branch of mechanics which deals with bodies in .......................... 4. A point object is one whose size is .......................... as compared to the distance it moves. 5. A body is said to be at rest if it does not change its .......................... with respect to the surroundings. 6. A body is said to be in motion if it change its .......................... with respect to the surroundings. 7. The reference point from which the distance of a body is measured is called .......................... 8. Distance is the .......................... path followed by a body between two points. 9. Displacement is the .......................... distance between two points. 10. A quantity which can be completely represented by magnitude alone is called .......................... 11. A quantity which can be completely represented by magnitude and direction is called .......................... 12. Speed is the ratio of the .......................... travelled to the time taken. 

SHORT ANSWER TYPE QUESTIONS

• An object travels 16 m in 4 sec. and then another 16 m in 2 sec. What is the average speed of the object?

[Ans. 16/3  ms –1 ]

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

[Ans. 240 km/h2]

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

[Ans. at rest.]

•  A racing car has a uniform acceleration of 40 m/s2 what distance will it cover in 10s after start?

[Ans. 2000 m.]

• Name the quantity that essentially changes as a body moves.

[Ans. Time]

• What does the area below v – t graph give?

[Ans. Displacement]

• Relate 36 km/h with m/s.

[Ans. 10 m/s]

• If 5 m/s and 10 m/s are the velocities of a body having a uniform acceleration in same time interval. What will be its average velocity?

[Ans. 7.5 m/s]

• What is the nature of motion of two cars A and B as depicted by the v–t graph in fig –

[Ans. A – Uniform motion, B – Uniform acceleration]

• . When a body moving with a speed of 20 m/s. stops in 20 seconds, what is the acceleration?

[Ans –1 m/s 2 ]

• A particle moves 3m north, then 4m east and finally 6m south calculate its distance travelled and displacement

[Ans 13m, 5m]

•  A car travels 30km at a uniform speed of 40 km/h and the next 30km at a uniform speed of 20km/h. Find its average speed

[26.7 km/h]

• Ravina takes 20 minutes to cover a distance of 3.2 kilometres on a bicycle. Calculate her speed in units of kilometre/minute, metre/minute, kilometre/hour. and m/s.

[0.16 km/min, 160m/min, 9.6 km/h, 2.67m/s]

• Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 m/s in 30s. Then he applies brakes such that the velocity of the bicycle comes down to 4 m/s in the next 5s. Calculate the acceleration of the bicycle in both the cases.

Ans. 1/5 m / s 2 , –2/5 m / s 2

• The velocity of a car is 18m/s. Express this velocity in km/h.

[64.8 km/h]

•  An electric train is moving with a velocity of 120 km/h. How much distance will it cover in 30 second?
• A train starting from rest attains a velocity of 72 km/h in 5 minutes. Assuming that the acceleration is uniform, find (i) the acceleration and (ii) the distance travelled by the train for attaining this velocity –

Ans. (i) 1/ 15 m/s 2 (ii) 3 km.

•  A bus covers 30km with a uniform speed of 60 km/h and the next 30km with a uniform speed of 40km/h. Find the total time taken and the average speed.

[Ans. 75minutes, 48 km/h]

• A car moves 30 minute with a uniform speed of 30km/h and next 30 minute moves with uniform speed of 60 km/h. Find the average speed in entire journey.

[Ans. 45 km/h]

• A 60m long train moving on a staright level track passes a pole in 5s, find (a) the speed of train (b) the time it will take to cross a 540 m long bridge.

[Ans. 12m/s, 50s]

• The brakes applied to a car it produce an acceleration of –10 m/s2. If the car takes 5s to stop after applying the brakes, calculate the distance covered by the car before coming to rest.

[Ans. 125m]

• What is the direction of velocity in a circular motion?

[Ans. tangent at any point]

• Is it possible to have the speed of a moving body zero ?
• If two bodies in circular paths of radii 1:2 take same time to complete their circles, what is the ratio of their angular speeds ?

[Ans. 1 : 2]

• Is the x–t graph given in the figure valid. Why? 

[Ans. Not possible]

• When a moving car returns to the same point what will be the displacement –

[Ans. Zero]

• A body moving with a constant speed, say 10 m/s on a frictionless surface has uniform motion while a freely falling object has non uniform motion. Give reason –

[Ans. Due to presence of acceleration due to gravity falling object accelerates.]

• A 150 m long train crosses a bridge of length 250 m in 25 seconds. What is its velocity?

[Ans. 16 m/s]

• A ship is moving at a speed of 56 km/h. One second later it is moving at 58 km/h. What is its acceleration?

[Ans. 59 m/ s 2 ]

• If a car starts from rest and attains speed 20 m/s in 25 seconds. Find the acceleration.

[Ans. a = 45 m/s 2 ]

• How much time will it take for a body with acceleration 2 m/s2, to gain a velocity of 10 m/s, starting from rest?

[Ans. 5 sec]

• When the slope of v–t graph is negative and constant, what is the nature of acceleration?

[Ans. deaccelerates , zero.]

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

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

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

A Glance About the Topic Force and Laws of Motion

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

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

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

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

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

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

INTEXT EXERCISE 1

1. Which of the following has more inertia:

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

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

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

(b) a bicycle and a train?

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

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

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

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

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

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

After collision, the two objects stick together.

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

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

\[ = 3\] kg

Velocity of the combined object \[ = v\]

Total momentum before collision = Total momentum after collision

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

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

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

\[ \Rightarrow v = 0\]

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

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

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

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

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

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

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

Initial momentum \[ = m{v_1}\]

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

Final momentum \[ = m{v_2}\]

Change in momentum 

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

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

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

\[ = 0.2 \times 15\]

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

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

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

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

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

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

Acceleration of the bullet, a

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

\[a =  - \dfrac{{150}}{{0.03}}\]

\[a =  - 5000\] m/s 2

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

According to the third equation of motion:

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

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

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

\[10000s = 22,500\]

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

\[s = 2.25\] m

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

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

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

\[ = 0.01 \times  - 5000\]

\[ =  - 50\] N

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

but it acts in opposite direction.

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

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

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

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

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

∴ Total momentum before collision

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

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

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

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

Total mass of the combined system, 

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

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

\[ = 6\] kg

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

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

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

$ \Rightarrow 10 = 6v$

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

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

$v = 1.66$ m/s

Total momentum after collision

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

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

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

\[ = 10\] kg m/s

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

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

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

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

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

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

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

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

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

Initial momentum \[ = \] mu

\[ = 100 \times 5\]

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

Final momentum \[ = \] mv 

\[ = 100 \times 8\]

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

Force exerted on the object, 

F \[ = \] mv-mu/t

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

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

F \[ = 50\] N

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Final velocity of the dumbbell v = ?

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

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

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

\[{v^2} = 16\]

\[v = \sqrt {16} \]

\[v = 4\] m/s

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

\[ = 10 \times 4\]

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

ADDITIONAL EXERCISE:

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

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

decreasing, or zero?

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

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

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

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

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

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

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

Let the force applied by the third person be F.

Force = Mass × Acceleration

F \[ = 1200 \times 0.2\]

F \[ = 240\] N

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

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

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

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

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

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

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

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

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

F \[ =  - 2500\] N

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

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

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

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

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

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

Time taken, t \[ = 4\] s

According to the first equation of motion:

\[v = u + at\]

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

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

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

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

\[a =  - 5\] m/s 2

= mv − mu 

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

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

\[ =  - 24000\] kg \[m{s^{ - 1}}\]  

Force \[ = 1200 \times  - 5\]

Force \[ =  - 6000\] N

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

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

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

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

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

Thus, M > m

Initial velocity of both vehicles, v

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

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

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

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

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

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

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

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

Initial momentum of the car = mv

Final momentum of the car = 0

Change in momentum = mv - 0

Initial momentum of the truck = Mv

Final momentum of the truck = 0

Change in momentum = Mv -0

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

(c) Which vehicle experiences the greater acceleration?

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

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

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

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

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

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

NCERT Solutions For Class 9 Science- Free PDF Download

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

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

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

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

Class 9 Science Chapter 9 Force and Laws of Motion Weightage

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

Here is More Detail About The Contents of Chapter 9

9.1 Balanced and Unbalanced forces

9.2 First Law of Motion

9.3 Inertia and mass

9.4 Second law of motion

9.4.1 Mathematical formulation of the second law of motion

9.5 Third law of motion

9.6 Conservation of motion

Benefits of NCERT Solutions Class 9 Chapter 9

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

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

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

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

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

NCERT Solutions for Class 9 Science

Chapter 1 - Matter in Our Surroundings

Chapter 2 - Is Matter Around us Pure

Chapter 3 - Atoms and Molecules

Chapter 4 - Structure of Atom

Chapter 5 - The Fundamental Unit of Life

Chapter 6 - Tissues

Chapter 7 - Diversity in Living Organisms

Chapter 8 - Motion

Chapter 9 - Force and Laws of Motion

Chapter 10 - Gravitation

Chapter 11 - Work and Energy

Chapter 12 - Sound

Chapter 13 - Why do We Fall ill

Chapter 14 - Natural Resources 

Chapter 15 - Improvement in Food Resources

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

NCERT Solutions For Class 9

Revision Notes for Class 9

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

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

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

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

1. What are force and its effects?

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

Effects of Force

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

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

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

2. What are the 3 Laws of Motion?

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

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

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

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

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

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

Balanced and Unbalanced forces

First Law of Motion 

Inertia and mass

Second Law of Motion

Mathematical Formulation of the second law of motion

Third Law of Motion

Conservation of motion.

4. What is Force Class 9th NCERT?

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

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

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

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

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

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

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

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

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

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

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

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

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

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

NCERT Solutions for Class 9

Worksheet For class 9

• CBSE Worksheet For Class 9 Science Physics

CBSE Worksheet for chapter-1 Motion class 9

• Active page
• Rest and Motion, Motion in One, Two & Three dimension, Types of motion
• Scalar and Vector quantity, Distance and Displacement

SECTION I – OBJECTIVE

• Oscillation
• Displacement > Distance
• Displacement < Distance
• Only direction
• Only magnitude
• Magnitude and direction
• None of these
• Scalar quantities have direction also.
• Scalars can be added arithmetically.
• There are special laws for scalar addition.
• Scalars have special method to represent.
• Its position with respect to surrounding objects remains same
• Its position with respect to surrounding objects keeps on changing
• Both (A) and (B)
• Neither (A) Nor (B)
• Shortest length between two points
• Path covered by an object between two points
• Product of length and time
• D. None of the above
• Is always positive
• Is always negative
• Either positive or negative
• Neither positive nor negative
• Velocity, length, and mass
• Speed, length and mass
• Time, displacement and mass
• Velocity, displacement and force
• It is always positive.
• It has both magnitude and direction.
• It can be zero
• Its magnitude is less than or equal to the actual path length of the object

SECTION II – SUBJECTIVE

• Distinguish between: Distance and Displacement.
• State any three types of motion with examples.
• Define scalar quantity. Give two examples.
• Define vector quantity. Give two examples.
• Give any two examples to explain that motion is relative.
• A person moves in a circular path centred at its origin O and having radius 1m. He starts from A and reaches diametrically opposite point B then find distance between A and B and magnitude of displacement between A and B

Objective Problems:

Subjective Problems:

6. Distance between A and B is m

Displacement between A and B is 2m

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NCERT Solutions for Class 9 Science Chapter 9: Force and Laws Of Motion

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

* According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 8.

NCERT Solutions for Class 9 Science Chapter 9 Force and Laws of Motion, are prepared to help students clear their doubts and understand concepts thoroughly. Class 9 Solutions of Science is a beneficial reference material that helps students to clear doubts instantly in an effective way. NCERT Solutions for Class 9 Science are designed in a student-friendly way and are loaded with questions, activities, and exercises that are CBSE and competitive exam-oriented. 

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NCERT Solutions for Class 9 Science is designed and developed by subject experts and teaching faculty having experience in coaching students. It is developed keeping in mind the concept-based approach, along with the precise answering method for examinations. It is a detailed and well-structured concept-based learning solution aimed at imparting confidence to face the CBSE and competitive exams. NCERT Solutions for Class 9 is made available in both PDF and web formats of Science chapters.

• 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 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 9: Force and Laws Of Motion(All intext and exercise questions solved)

Intext Questions – 1   Page: 118

1. Which of the following has more inertia: (a) a rubber ball and a stone of the same size? (b) a bicycle and a train? (c) a five-rupee coin and a one-rupee coin?

Since inertia is dependent on the mass of the object, the object with the greater mass will hold greater inertia. The following objects hold greater inertia because of their mass.

• Five-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.

The velocity of football changes four times.

First, when a football player kicks a football to another player, second when that player kicks the football to the goalkeeper. Third when the goalkeeper stops the football. Fourth, when the goalkeeper kicks the football towards his team player.

Agent supplying the force:

a) The First case is the First player

b) The Second case is the Second player

c) The Third case is Goalkeeper

d) The Fourth case is Goalkeeper

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

When the branch of the tree is shaken, the branch moves in a to-and-fro motion. However, the inertia of the leaves in attached to the branch resists the motion of the branch. Therefore, the leaves that are weakly attached to the branch fall off due to inertia whereas the leaves that are firmly attached to the branch remain attached.

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?

Initially, when the bus accelerates in a forward direction from a state of rest, the passengers experience a force exerted on them in the backward direction due to their inertia opposing the forward motion.

Once the bus starts moving, the passengers are in a state of motion in the forward direction. When the brakes are applied, the bus moves towards a position of rest. Now, a force in the forward direction is applied on the passengers because their inertia resists the change in the motion of the bus. This causes the passengers to fall forwards when the brakes are applied.

Intext Questions – 2 Page: 126,127

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

When the horse walks forward (with the cart attached to it), it exerts a force in the backward direction on the Earth. An equal force in the opposite direction (forward direction) is applied on the horse by the Earth. This force moves the horse and the cart forward. As a result, the cart moves forward.

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

When a fireman holds a hose, which is ejecting large amounts of water at a high velocity, then a reaction force is exerted on him by the ejecting water in the backward direction. This is because of Newton’s third law of motion. As a result of the backward force, the stability of the fireman decreases. Hence, it is difficult for him to remain stable while holding the hose.

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.

Given, the Bullet’s mass (m 1 ) = 50 g

The rifle’s mass (m 2 ) = 4kg = 4000g

Initial velocity of the fired bullet (v 1 ) = 35 m/s

Let the recoil velocity be v 2 .

Since the rifle was initially at rest, the initial momentum of the rifle = 0

The total momentum of the rifle and bullet after firing = m 1 v 1 + m 2 v 2

As per the law of conservation of momentum, the total momentum of the rifle and the bullet after firing = 0 (same as initial momentum)

Therefore, m 1 v 1 + m 2 v 2 = 0

The negative sign indicates that recoil velocity is opposite to the bullet’s motion.

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

Assuming that the first object is object A and the second one is object B, it is given that:

Mass of A (m 1 ) = 100g

Mass of B (m 2 ) = 200g

Initial velocity of A (u 1 ) = 2 m/s

Initial velocity of B (u 2 ) = 1 m/s

Final velocity of A (v 1 ) = 1.67 m/s

Final velocity of B (v 2 ) =?

Total initial momentum = Initial momentum of A + initial momentum of B

= m 1 u 1 + m 2 u 2

= (100g) × (2m/s) + (200g) × (1m/s) = 400 g.m.sec -1

As per the law of conservation of momentum, the total momentum before collision must be equal to the total momentum post collision.

v 2 = 1.165 m/s

Therefore, the velocity of object B after the collision is 1.165 meters per second.

Exercises Page: 128,129

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.

Yes, it is possible. An object moving in some direction with constant velocity will continue in its state of motion as long as there are no external unbalanced forces acting on it. In order to change the motion of the object, some external unbalanced force must act upon it.

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

When the carpet is beaten with a stick, the stick exerts a force on the carpet which sets it in motion. The inertia of the dust particles residing on the carpet resists the change in the motion of the carpet. Therefore, the forward motion of the carpet exerts a backward force on the dust particles, setting them in motion in the opposite direction. This is why the dust comes out of the carpet when beaten.

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

When some luggage is placed on the roof of a bus which is initially at rest, the acceleration of the bus in the forward direction will exert a force (in the backward direction) on the luggage. In a similar manner, when a bus which is initially in a state of motion suddenly comes to rest due to the application of brakes, a force (in the forward direction) is exerted on the luggage.

Depending on the mass of the luggage and the magnitude of the force, the luggage may fall off the bus due to inertia. Tying up the luggage will secure its position and prevent it from falling off the bus.

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.

When the ball rolls on the flat surface of the ground, its motion is opposed by the force of friction (the friction arises between the ground and the ball). This frictional force eventually stops the ball. Therefore, the correct answer is (c).

If the surface of the level ground is lubricated (with oil or some other lubricant), the friction that arises between the ball and the ground will reduce, which will enable the ball to roll for a longer distance.

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 it’s mass is 7 tonnes (Hint: 1 tonne = 1000 kg.)

Given, distance covered by the truck (s) = 400 meters

Time taken to cover the distance (t) = 20 seconds

The initial velocity of the truck (u) = 0 (since it starts from a state of rest)

6. A stone of 1 kg is thrown with a velocity of 20 ms -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?

Given, Mass of the stone (m) = 1kg

Initial velocity (u) = 20m/s

Terminal velocity (v) = 0 m/s (the stone reaches a position of rest)

Distance travelled by the stone (s) = 50 m

As per the third equation of motion

v² = u² + 2as

Substituting the values in the above equation we get,

0² = (20)² + 2(a)(50)

-400 = 100a

a = -400/100  =  -4m/s² (retardation)

We know that

Substituting above obtained value of a = -4 in F = m x a

F = 1 × (-4) = -4N

Here the negative sign indicates the opposing force which is Friction

7. An 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate: (a) the net accelerating force and (b) the acceleration of the train

(a) Given, the force exerted by the train (F) = 40,000 N

Force of friction = -5000 N (the negative sign indicates that the force is applied in the opposite direction)

Therefore, the net accelerating force = sum of all forces = 40,000 N + (-5000 N) = 35,000 N

(b) Total mass of the train = mass of engine + mass of each wagon = 8000kg + 5 × 2000kg

The total mass of the train is 18000 kg.

As per the second law of motion, F = ma (or: a = F/m)

Therefore, acceleration of the train = (net accelerating force) / (total mass of the train)

= 35,000/18,000 = 1.94 ms -2

The acceleration of the train is 1.94 m.s -2 .

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 ms -2 ?

Given, mass of the vehicle (m) = 1500 kg

Acceleration (a) = -1.7 ms -2

As per the second law of motion, F = ma

F = 1500kg × (-1.7 ms -2 ) = -2550 N

Hence, the force between the automobile and the road is -2550 N, in the opposite direction of the automobile’s motion.

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

(a) (mv) 2 (b) mv 2 (c) ½ mv 2 (d) mv

The momentum of an object is defined as the product of its mass m and velocity v

Momentum = mass x velocity

Hence, the correct answer is mv i.e option (d)

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?

Since the velocity of the cabinet is constant, its acceleration must be zero. Therefore, the effective force acting on it is also zero. This implies that the magnitude of opposing frictional force is equal to the force exerted on the cabinet, which is 200 N. Therefore, the total friction force is -200 N.

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

Mass of first object, m 1 = 1.5 kg

Mass of second object, m 2 = 1.5 kg

Velocity of first object before collision, v 1 = 2.5 m/s

The velocity of the second object which is moving in the opposite direction, v 2 = -2.5 m/s

We know that,

Total momentum before collision = Total momentum after collision

m 1 v 1 + m 2 v 2 = (m 1 + m 2 )v

1.5(2.5) + 1.5 (-2.5) = (1.5 + 1.5)v

3.75 – 3.75 = 3v

Therefore, the velocity of the combined object after the collision is 0 m/s

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

Since the truck has a very high mass, the static friction between the road and the truck is high. When pushing the truck with a small force, the frictional force cancels out the applied force and the truck does not move. This implies that the two forces are equal in magnitude but opposite in direction (since the person pushing the truck is not displaced when the truck doesn’t move). Therefore, the student’s logic is correct.

13. A hockey ball of mass 200 g travelling at 10 ms –1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms –1 . Calculate the magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

Given, mass of the ball (m) = 200g

Initial velocity of the ball (u) = 10 m/s

Final velocity of the ball (v) = – 5m/s

Initial momentum of the ball = mu = 200g × 10 ms -1 = 2000 g.m.s -1

Final momentum of the ball = mv = 200g  × –5 ms -1 = –1000 g.m.s -1

Therefore, the change in momentum (mv – mu) = –1000 g.m.s -1 – 2000 g.m.s -1 = –3000 g.m.s -1

This implies that the momentum of the ball reduces by 1000 g.m.s -1 after being struck by the hockey stick.

14. A bullet of mass 10 g travelling horizontally with a velocity of 150 m s –1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.

Given, mass of the bullet (m) = 10g (or 0.01 kg)

Initial velocity of the bullet (u) = 150 m/s

Terminal velocity of the bullet (v) = 0 m/s

Time period (t) = 0.03 s

To find the distance of penetration, the acceleration of the bullet must be calculated

Let the distance of penetration be s

As per the first law of motion

0 = 150 + a (0.03)

a = -5000 ms -2

v 2 = u 2 + 2as

0 = 150 2 + 2 x (-5000)s

F = 0.01kg × (-5000 ms -2 )

15. An object of mass 1 kg travelling in a straight line with a velocity of 10 ms –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.

Given, mass of the object (m 1 ) = 1kg

Mass of the block (m 2 ) = 5kg

Initial velocity of the object (u 1 ) = 10 m/s

Initial velocity of the block (u 2 ) = 0

Mass of the resulting object = m 1 + m 2 = 6kg

Velocity of the resulting object (v) =?

Total momentum before the collision = m 1 u 1 + m 2 u 2 = (1kg) × (10m/s) + 0 = 10 kg.m.s -1

As per the law of conservation of momentum, the total momentum before the collision is equal to the total momentum post the collision. Therefore, the total momentum post the collision is also 10 kg.m.s -1

Now, (m 1 + m 2 ) × v = 10kg.m.s -1

The resulting object moves with a velocity of 1.66 meters per second.

16. An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms –1 to 8 ms –1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.

Given, mass of the object (m) = 100kg

Initial velocity (u) = 5 m/s

Terminal velocity (v) = 8 m/s

Time period (t) = 6s

Now, initial momentum (m × u) = 100kg × 5m/s = 500 kg.m.s -1

Final momentum (m × v) = 100kg × 8m/s = 800 kg.m.s -1

Therefore, the object accelerates at 0.5 ms -2 . This implies that the force acting on the object (F = ma) is equal to:

F = (100kg) × (0.5 ms -2 ) = 50 N

Therefore, a force of 50 N is applied on the 100kg object, which accelerates it by 0.5 ms -2 .

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

As per the law of conservation of momentum, the total momentum before the collision between the insect and the car is equal to the total momentum after the collision. Therefore, the change in the momentum of the insect is much greater than the change in momentum of the car (since force is proportional to mass).

Akhtar’s assumption is partially right. Since the mass of the car is very high, the force exerted on the insect during the collision is also very high.

Kiran’s statement is false. The change in momentum of the insect and the motorcar is equal by conservation of momentum. The velocity of insect changes accordingly due to its mass as it is very small compared to the motorcar. Similarly, the velocity of motorcar is very insignificant because its mass is very large compared to the insect.

Rahul’s statement is completely right. As per the third law of motion, the force exerted by the insect on the car is equal and opposite to the force exerted by the car on the insect. However, Rahul’s suggestion that the change in the momentum is the same contradicts the law of conservation of momentum.

18. How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 ms –2 .

Given, mass of the dumb-bell (m) = 10kg

Distance covered (s) = 80cm = 0.8m

Initial velocity (u) = 0 (it is dropped from a position of rest)

Acceleration (a) = 10ms -2

Terminal velocity (v) =?

Momentum of the dumb-bell when it hits the ground = mv

As per the third law of motion

v 2 – u 2 = 2as

Therefore, v 2 – 0 = 2 (10 ms -2 ) (0.8m) = 16 m 2 s -2

The momentum transferred by the dumb-bell to the floor = (10kg) × (4 m/s) = 40 kg.m.s -1

Additional Exercises Page: 130

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

(a) The distance covered by the object at any time interval is greater than any of the distances covered in previous time intervals. Therefore, the acceleration of the object is increasing.

(b) As per the second law of motion, force = mass × acceleration. Since the mass of the object remains constant, the increasing acceleration implies that the force acting on the object is increasing as well

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 ms -2 . With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort)

Given, mass of the car (m) = 1200kg

When the third person starts pushing the car, the acceleration (a) is 0.2 ms -2 . Therefore, the force applied by the third person (F = ma) is given by:

F = 1200kg × 0.2 ms -2 = 240N

The force applied by the third person on the car is 240 N. Since all 3 people push with the same muscular effort, the force applied by each person on the car is 240 N.

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?

Given, mass of the hammer (m) = 500g = 0.5kg

Initial velocity of the hammer (u) = 50 m/s

Terminal velocity of the hammer (v) = 0 (the hammer is stopped and reaches a position of rest).

Time period (t) = 0.01s

a = -5000ms -2

Therefore, the force exerted by the hammer on the nail (F = ma) can be calculated as:

F = (0.5kg) * (-5000 ms -2 ) = -2500 N

As per the third law of motion, the nail exerts an equal and opposite force on the hammer. Since the force exerted on the nail by the hammer is -2500 N, the force exerted on the hammer by the nail will be +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.

Initial velocity (u) = 90 km/hour = 25 meters/sec

Terminal velocity (v) = 18 km/hour = 5 meters/sec

Time period (t) = 4 seconds

Therefore, the acceleration of the car is -5 ms -2 .

Initial momentum of the car = m × u = (1200kg) × (25m/s) = 30,000 kg.m.s -1

Final momentum of the car = m × v = (1200kg) × (5m/s) = 6,000 kg.m.s -1

Therefore, change in momentum (final momentum – initial momentum) = (6,000 – 30,000) kg.m.s -1

= -24,000 kg.m.s -1

External force applied = mass of car × acceleration = (1200kg) × (-5 ms -2 ) = -6000N

Therefore, the magnitude of force required to slow down the vehicle to 18 km/hour is 6000 N

NCERT Solution Class 9 Science Chapter 9 explains what is a force and its application with various examples. It also explains the three laws of motion, along with examples and mathematical formulations. Conservation of Momentum is explained in a simple way. Various activities are incorporated into the chapter to explain the concepts in an easy way. NCERT Class 9 Science Chapter 9 Force and Laws of Motion also describes the natural tendency of objects to resist a change in their state of rest or of uniform motion, known as inertia, in a detailed way. NCERT Class 9 Science Chapter 9 Force and laws of motion are covered under unit 3 Motion, Force and Work. Refer to previous years’ question papers and sample papers to know the type of questions that appear, along with the allotment of marks and patterns. The topics that are covered under this chapter include:

• Inertia and Mass (4 questions)
• Conservation of Momentum (4 questions)
• Post-Chapter Exercises (18 Questions)

NCERT Solutions For Class 9 Science Chapter 9: Force and Laws Of Motion

• NCERT Solutions for Class 9 explains motion and the causes of motion in a detailed way.
• 1st law of motion, 2nd law of motion and 3rd law of motion are explained with illustrations and mathematical formulations.
• Exercise covers all the topics of the chapter and helps students in gaining confidence to write the CBSE exam.
• Concepts like conservation of momentum, inertia and mass are described in brief.

Key Features of NCERT Solutions for Class 9 Science Chapter 9: Force and Laws Of Motion

• A simple and easily understandable method is followed in NCERT Solutions to make students grasp concepts.
• NCERT Solutions offer comprehensive answers to all the questions to help students in their preparations.
• Provides completely solved solutions to all the questions present in the chapter.
• These solutions will be useful for CBSE exams, Science Olympiads, and other competitive exams.

Disclaimer:

Dropped Topics-  9.6 Conservation of Momentum, Activity 9.5, 9.6, Example 9.6, 9.7, 9.8 and Box item ‘Conservation Laws’.

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

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