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Class 9 Physics Assignments

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

Physics Class 9 Assignments Pdf Download

We have provided below the biggest collection of free CBSE NCERT KVS Assignments for Class 9 Physics . Students and teachers can download and save all free Physics assignments in Pdf for grade 9th. Our expert faculty have covered Class 9 important questions and answers for Physics as per the latest syllabus for the current academic year. All test papers and question banks for Class 9 Physics and CBSE Assignments for Physics Class 9 will be really helpful for standard 9th students to prepare for the class tests and school examinations. Class 9th students can easily free download in Pdf all printable practice worksheets given below.

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Assignments Class 9 Physics Pdf Download

Students can refer to Assignments for Class 9 Physics available for download in Pdf. We have given below links to subject-wise free printable Assignments for Physics Class 9 which you can download easily. All assignments have a collection of questions and answers designed for all topics given in your latest NCERT Books for Class 9 Physics for the current academic session. All Assignments for Physics Grade 9 have been designed by expert faculty members and have been designed based on the type of questions asked in standard 9 class tests and exams. All Free printable Assignments for NCERT CBSE Class 9, practice worksheets, and question banks have been designed to help you understand all concepts properly. Practicing questions given in CBSE NCERT printable assignments for Class 9 with solutions and answers will help you to further improve your understanding. Our faculty have used the latest syllabus for Class 9. You can click on the links below to download all Pdf assignments for class 9 for free. You can get the best collection of Kendriya Vidyalaya Class 9 Physics assignments and questions workbooks below.

Class 9 Physics Assignments Pdf Download

CBSE NCERT KVS Assignments for Physics Class 9 have been provided below covering all chapters given in your CBSE NCERT books. We have provided below a good collection of assignments in Pdf for Physics standard 9th covering Class 9 questions and answers for Physics. These practice test papers and workbooks with question banks for Class 9 Physics Pdf Download and free CBSE Assignments for Class 9 are really beneficial for you and will support in preparing for class tests and exams. Standard 9th students can download in Pdf by clicking on the links below.

Subjectwise Assignments for Class 9 Physics

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NCERT Solutions for Class 9 Physics Free PDF Download

Ncert solutions for class 9 physics.

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

NCERT Solutions for Class 9 Physics breaks down the solutions into detailed steps and explains the answer thoroughly, which helps you understand the pattern of questioning and a way to increase your score in exams.

NCERT Solutions for Class 9 Physics are prepared by our team of highly professional, qualified and experienced faculties. These NCERT Solutions for Class 9 Physics helps you to quickly grasp all the basic concepts of physics. In case you have a doubt while you are studying NCERT Solutions for Class 9 Physics, we have a team of teachers who are just a click away to solve your doubt any time.

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NCERT Solutions for Class 9 Physics Chapterwise

Class 9 Physics Chapter 1 – Motion

Class 9 Physics Chapter 2 – Force And Laws Of Motion

Class 9 Physics Chapter 3 – Gravitation

Class 9 Physics Chapter 4 – Work and Energy

Class 9 Physics Chapter 5 – Sound

Class 9 Chapterwise NCERT Solutions for Physics

Ncert solutions for class 9 physics chapter 1 motion.

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

NCERT Solutions for Class 9 Physics Chapter 2 Force and Laws of Motion

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

NCERT Solutions for Class 9 Physics Chapter 3 Gravitation

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

NCERT Solutions for Class 9 Physics Chapter 4 Work and Energy

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

NCERT Solutions for Class 9 Physics Chapter 5 Sound

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

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Answer:  The gravitation force between the earth and object is called weight.

Answer:  By crumpling the paper into a ball, the volume of the object decreases but the mass remains the same. Hence its density increases.

Answer: The Importance of Universal law of gravitation lies in the fact, that it was successful in explaining many phenomena such as.

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

• NCERT Solutions
• Chapter 9 Force And Laws Of Motion

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

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High School Physics

Force and Laws of Motion Class 9 Numericals

Last updated on July 5th, 2023 at 03:15 pm

This page presents a set of Force and Laws of Motion class 9 Numericals . We have used linear motion equations and force equations to solve these numerical problems. [detailed solution of the numerical questions is given]

Force and Laws of Motion Class 9 Numericals – solved numerical problems

In the next section, we will find a list of formulas that we will use to solve the numerical problems in this post. Then we will solve the sets of physics numerical problems from class 9 Force and Laws of Motion chapter.

Formulas Used

Force = mass × acceleration [ F = ma ] Momentum (p) = mass × velocity [ p = mv] Force = rate of change in momentum = (Final momentum – Initial Momentum)/time Change in Momentum = mass × (final velocity – Initial Velocity) [Δp= m×(v-u)] Impulse = Force × time [ I = Ft] Impulse = change in momentum [ I = m×(v-u)] Conservation of Momentum: m 1 u 1 +m 2 v 2 =m 1 v 1 +m 2 v 2

Force and Laws of Motion Numericals class 9 – question and solution

1> A force of 10 N is applied for 0.1 seconds on a body of mass 1 kg initially at rest. The force then ceases to act. What would be the velocity of the body just after this? After 2 more seconds with what velocity the body will move?

A force will cause an acceleration a. a = force/mass = 10/1 m/s^2 = 10 m/s^2 The force acts for 0.1 seconds. That means the body will remain under an accelerated motion for 0.1 sec. The velocity acquired post this interval of 0.1 s is V And, V =u + at = 0 + 10. 0.1 = 1 m/s As there is no force after this interval, the body will continue its motion with a uniform velocity of 1 m/s. So after 2 more seconds, its velocity is 1 m/s.

2> A force acts for 2 seconds on a body of mass 2 kg initially at rest. Just after the force stops to act the body moves 10 m in the next 2 seconds. Find the magnitude of the force.

After the force stops to act, the velocity of the body is = V = distance/time=10/2 m/s = 5 m/s Initial velocity = U =0 The time interval for the accelerated motion in presence of the force =t= 2 seconds Therefore we can find the acceleration a a= (V-U)/t =(5-0)/2 = 2.5 m/s^2 m=Mass of the body is 2 kg. a=2.5 m/s^2 So force F = ma = 2 x 2.5 N= 5 N

3> Calculate the magnitude of force which when applied on a body of mass 0.5 kg produces an acceleration of 5 m/s 2   

Sol:  Force = mass x acceleration = 0.5 x 5 N = 2.5 N

4> A ball of mass 0.01 kg is moving with a velocity of 50 m/s. On applying a constant force for 2 seconds on the ball, it acquires a velocity of 70 m/s. Calculate the initial and final momentum of the ball. Also, calculate the rate of change of momentum and the acceleration of the ball. What would be the magnitude of the force applied?

Initial velocity = U = 50 m/s Final velocity = V = 70 m/s Time interval = 2 s mass = m = 0.01 kg Initial momentum = mU = 0.01 x 50 =0.5 kg m/s Final momentum = mV = 0.01 x 70 = 0.7 kg m/s Acceleration =a= (V-U)/t =(70-50)/2   m/s^2  =  10 m/s^2 Force = m a =  0.01 x 10 N = 0.1 N

5> An object of mass 1 kg is moving at a speed of 50 m/s. The object is brought to rest in 0.05 seconds by an external system. Find the change in momentum of the object and the average force applied by the external system.

Initial momentum = mU = 1 x 50 = 50 kg m/s Final momentum = mV = 0  (as V =0) So the change in momentum =mV – mU =   0 – 50 = – 50 kg m/s Force = change of momentum / time = -50/0.05 N = – 1000 N The negative sign denotes a force that opposes the motion, causing retardation.

6> A car of mass 500 kg moving at a speed of 36 Km/hr is stopped by applying brakes in 10 s. Calculate the force applied by the brakes.

Mass = 500 kg initial velocity U = 36 kmph= 36 x (5/18) m/s =10 m/s final velocity =0 time interval =10 s Therefore, acceleration (retardation here actually) = (0-10)/10 m/s^2 = -1 m/s^2 The force applied by the brakes = mass x acceleration = 500 x (-1) N = -500 N The negative sign of the force denotes that the force is resistive i.e. the force opposes the motion.

7> A bullet of mass 50 g moving with an initial velocity of 100 m/s, strikes a wooden block and comes to rest after penetrating a distance of 2 cm in it. Calculate – the initial momentum of the bullet -final momentum of the bullet -retardation caused by the wooden block and -resistive force applied by the wooden block

Initial momentum= mass x initial velocity = (50/1000)kg x (100) m/s = 5 kg m/s final momentum = 0  (as final velocity is zero) To find Retardation a , we will use the following equation: V^2 = U^2 – 2as Putting values, 0=(100)^2 – 2a (2/100) Retardation a= (10000x 100)/4 = 25 x 10^4 m/s^2 Resistive force = m a = (50/1000) x (25 x 10^4)=12500 N

8> A force when applied on a block of 10 kg mass produces an acceleration of 5 m/s^2. When the same force is applied on a different block initially at rest then the block gains a velocity of 2 m/s in 1 second. What is the magnitude of the force in Newton and what is the mass of the second block?

1 st case: F = ma = 10 x 5 N = 50 N 2 nd case F = same as first case = 50 N U = 0 V = 2 m/s t = 1 sec So, Acceleration a =(V-U)/t = 2 m/s^2 Force = mass X acceleration Mass of the second block = M = F/a = 50/2 Kg = 25 kg

9> Velocity of an object changes from 2 m/s to 10 m/s in 4 seconds. If the mass of the object is 10 kg then how much force will be required to do this?

Acceleration a= (10-2)/4 = 2 m/s^2 Mass m = 10 kg Hence, Force F = ma = 10 x 2 N = 20 N

10> A car is moving with a uniform velocity of 30 m/s. it is stopped in 2 seconds by applying a force of 1500 N through its brakes. Calculate (a) the change in momentum of the car (b) the retardation produced in the car and (c) the mass of the car.

a) Change in momentum = F t = 1500 x 2 kg m/s = 3000 kg m/s b) Acceleration a = (0-30)/2 = -15 m/s^2 This means, retardation = 15 m/s^2 c) Mass of the car m = force/magnitude of acceleration = F/a = 1500/15 = 100 kg

Force and Laws of Motion Numerical – practice problems

11> A boy pushes a wall with a force of 20 N toward the North. What force is exerted by the wall on the boy?

12> A block of mass 1.5 kg is hanging from rigid support by a string. What is the force exerted by the (i) block on the string  (ii) string on the block (g = 10 m/s^2)

13> A stone of size 1 kg is thrown with a velocity of 20 m/s across the frozen surface of a lake and it comes to rest after traveling a distance of 50 m. What is the force of friction between the stone and the ice?

14> A body of mass 0.5 kg is resting on a frictionless surface. When a force of 2000 dyne acts on it for 10 s, then calculate the distance traveled by it in 10 seconds.

15> A force of 10 N produces an acceleration of 5 m/s^ in mass m1 and 20 m/s^2 in mass m2. What will be the acceleration by this force, if both masses m1 and m2 are tied together?  

Very Short Answer Type Questions (Numerical)

16) What is the force which produces an acceleration of 1 m/s2 in a body of mass 1 kg?

17) Find the acceleration produced by a force of 5 N acting on a mass of 10 kg.

18) A girl weighing 25 kg stands on the floor. She exerts a downward force of 250 N on the floor. What force does the floor exert on her?

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

20) The mass of object A is 6 kg whereas that of another object B is 34 kg. Which of the two objects, A or B, has more inertia?

Short Answer Type Questions (Numerical)

21) What is the change in momentum of a car weighing 1500 kg when its speed increases from 36 km/h to 72 km/h uniformly?

22) Calculate the momentum of a toy car of mass 200 g moving with a speed of 5 m/s.

23) A body of mass 25 kg has a momentum of 125 kg.m/s. Calculate the velocity of the body.

24) A 60 g bullet fired from a 5 kg gun leaves with a speed of 500 m/s. Find the speed (velocity) with which the gun recoils (jerks backward).

25) A 10 g bullet traveling at 200 m/s strikes and remains embedded in a 2 kg target which is originally at rest but free to move. At what speed does the target move off?

26) A body of mass 2 kg is at rest. What should be the magnitude of force which will make the body move with a speed of 30 m/s at the end of 1 s?

27) A body of mass 5 kg is moving with a velocity of 10 m/s. A force is applied to it so that in 25 seconds, it attains a velocity of 35 m/s. Calculate the value of the force applied.

28) A car of mass 2400 kg moving with a velocity of 20 m s–1 is stopped in 10 seconds by applying brakes. Calculate the retardation and the retarding force.

29) How long will it take a force of 10 N to stop a mass of 2.5 kg which is moving at 20 m/s?

30) For how long should a force of 100 N act on a body of 20 kg so that it acquires a velocity of 100 m/s?

Long Answer Type Questions (numerical)

31) A 1000 kg vehicle moving with a speed of 20 m/s is brought to rest in a distance of 50 meters: (i) Find the acceleration. (ii) Calculate the unbalanced force acting on the vehicle.

32) A 150 g ball, traveling at 30 m/s, strikes the palm of a player’s hand and is stopped in 0.05 seconds. Find the force exerted by the ball on the hand.

33) An unloaded truck weighing 2000 kg has a maximum acceleration of 0.5 m/s2. What is the maximum acceleration when it is carrying a load of 2000 kg?

We have completely solved a set of numerical problems from the Force and Laws of Motion chapter of the class 9 syllabus. We have presented here a bundle of practice problems from this chapter as well. We will also add solutions to those as well gradually. Hope you have found this post on Force and laws of motion class 9 Numericals useful enough. So please share this and come back soon for more numerical solutions.

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• Force Numericals Class 8
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CBSE Worksheets for Class 9 Physics

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

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

CBSE Class 9 Physics Worksheets PDF

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

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

Advantages of CBSE Class 9 Physics Worksheets

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

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

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

CBSE Worksheets For Class 9

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AssignmentsBag.com

Assignments For Class 9 Physics

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

Assignments for Class 9 Physics as per CBSE NCERT pattern

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

Some advantages of Free Assignments for Class 9 Physics

• Solving Assignments for Physics Class 9 helps to further enhance understanding of the topics given in your text book which will help you to get better marks
• By solving one assignments given in your class by Physics teacher for class 9 will help you to keep in touch with the topic thus reducing dependence on last minute studies
• You will be able to understand the type of questions which are expected in your Physics class test
• You will be able to revise all topics given in the ebook for Class 9 Physics as all questions have been provided in the question banks
• NCERT Class 9 Physics Workbooks will surely help you to make your concepts stronger and better than anyone else in your class.
• Parents will be able to take print out of the assignments and give to their child easily.

All free Printable practice assignments are in PDF single lick download format and have been prepared by Class 9 Physics teachers after full study of all topics which have been given in each chapter so that the students are able to take complete benefit from the worksheets. The Chapter wise question bank and revision assignments can be accessed free and anywhere. Go ahead and click on the links above to download free CBSE Class 9 Physics Assignments PDF.

You can download free assignments for class 9 Physics from https://www.assignmentsbag.com

You can get free PDF downloadable assignments for Grade 9 Physics from our website which has been developed by teachers after doing extensive research in each topic.

On our website we have provided assignments for all subjects in Grade 9, all topic wise test sheets have been provided in a logical manner so that you can scroll through the topics and download the worksheet that you want.

You can easily get question banks, topic wise notes and questions and other useful study material from https://www.assignmentsbag.com without any charge

Yes all test papers for Physics Class 9 are available for free, no charge has been put so that the students can benefit from it. And offcourse all is available for download in PDF format and with a single click you can download all assignments.

https://www.assignmentsbag.com is the best portal to download all assignments for all classes without any charges.

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Class 9 Physics Index

Class 9 Physics index given here lists all the topics discussed in class 9 Physics chapters. The listed index is prepared as per the NCERT textbook and the latest CBSE syllabus. Access free study material by clicking on the respective subtopics.

Why is this important?

Before the start of a new academic year, it is helpful to go through the course guide because of the following reasons:

• It gives you an idea of the learning materials, core books, or reference books (if any) that you need for the new session. Knowing this helps you arrange all the required things  in time.
• A clear outline of the topics and learning areas that you will approach throughout the academic year will be presented to you.

Class 9 physics opens doors to fascinating topics such as motion, gravitation, sound, work, power, and more. These topics are very important for students who wish to pursue a career in Physics discipline. Master these concepts and build a strong foundation to improve your expertise on the subject.

Stay tuned with BYJU’S for more such interesting articles. Also, register to “BYJU’S – The Learning App” for loads of interactive, engaging Physics-related videos and an unlimited academic assist.

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Physics library

Unit 1: one-dimensional motion, unit 2: two-dimensional motion, unit 3: forces and newton's laws of motion, unit 4: centripetal force and gravitation, unit 5: work and energy, unit 6: impacts and linear momentum, unit 7: torque and angular momentum, unit 8: oscillations and mechanical waves, unit 9: fluids, unit 10: thermodynamics, unit 11: electric charge, field, and potential, unit 12: circuits, unit 13: magnetic forces, magnetic fields, and faraday's law, unit 14: electromagnetic waves and interference, unit 15: geometric optics, unit 16: special relativity, unit 17: quantum physics, unit 18: discoveries and projects, unit 19: review for ap physics 1 exam.