Chapter 6 - How Forces Affect Motion

NCERT Solutions for Class 9 Science Chapter 6: How Forces Affect Motion

Strengthen your understanding of Class 9 Science Chapter 6: How Forces Affect Motion with chapter-wise NCERT Solutions designed for effective learning and exam preparation. Aligned with the new NCERT Exploration textbook (2026–27) and CBSE guidelines, these expert-created solutions feature step-by-step answers to every textbook question — including Think It Over, Pause and Ponder, Think as a Scientist, and Revise Reflect Refine exercises — along with important questions for focused practice. Curated by ALLEN Experts, the solutions are available for free PDF download, making revision and offline study convenient anytime, anywhere.

1.0Download NCERT Solutions for Class 9 Science Chapter 6 – How Forces Affect Motion PDF

Class 9 Science Chapter 6: How Forces Affect Motion builds on the concepts of position, velocity, and acceleration introduced in Chapter 4 and investigates what actually causes motion to change — force. Students learn how forces can start, stop, speed up, slow down, or change the direction of an object, and even change its shape. Students can download the free PDF of NCERT Solutions for Class 9 Science Chapter 6: How Forces Affect Motion, featuring step-by-step solutions to all in-text and end-of-chapter questions, prepared according to the latest NCERT syllabus and CBSE-aligned guidelines.

NCERT Solutions Class 9 Science Chapter 6

2.0Learning Outcomes: NCERT Class 9 Science Chapter 6 Solutions:

Understand the Concept of Force: Define force as a push or pull and explain its effects — changing speed, direction, or shape of an object.
Differentiate Contact and Non-Contact Forces: Identify forces that act through touch (push, pull, friction) versus forces that act at a distance (gravitational, magnetic).
Explain Balanced and Unbalanced Forces: Distinguish between balanced forces (no change in motion) and unbalanced forces (causing acceleration).
Understand Friction: Explain the cause of friction, factors affecting it, and its role in everyday activities like walking and braking.
State and Apply Newton's First Law of Motion: Understand inertia and why objects resist changes in their state of rest or motion.
State and Apply Newton's Second Law of Motion: Relate force, mass, and acceleration using F = ma, and solve numerical problems.
Understand Momentum: Define momentum as the product of mass and velocity and calculate the rate of change of momentum.
State and Apply Newton's Third Law of Motion: Explain action-reaction force pairs and why they act on different objects and do not cancel out.
Analyse Systems of Objects: Treat two or more connected objects as a single system to calculate combined acceleration under an external force.
Solve NCERT Exercise Problems: Confidently attempt and solve all in-text activities and end-of-chapter numerical and conceptual questions.
Build a Foundation for Advanced Topics: Develop conceptual clarity that supports future topics like gravitation, work-energy, and higher-level mechanics.

3.0NCERT Solutions for Class 9 Science Chapter 6 How Forces Affect Motion – Detailed Solutions

1. Why does a canoe move forward when the canoeist pushes water backwards with their paddle, and why does it move faster when they push harder?

Solution

A canoe moves forward because when the canoeist pushes water backward, the water pushes the canoe forward with an equal and opposite force (Newton's third law of motion). When they push harder, a greater force is applied, so the canoe accelerates and moves faster.

2. Suppose the same canoeist uses the same paddle force in two different canoes, one empty and one carrying another passenger. In which case will the canoe move faster?

Solution

The canoe will move faster when it is empty. With the same force, the lighter canoe (less mass) accelerates more, while the heavier canoe (with a passenger) accelerates less.

3. How much does a force of I feel? If you hold a 100 g mass in your palm, the upward force your palm applies on the mass is around 1 N .

Solution

A force of 1 N (newton) is roughly the force needed to support a 100 g mass against gravity. So, if you hold a 100 g object in your palm, the upward force your palm applies feels like about 1 N just enough to balance its weight and keep it from falling.

4. A weightlifter lifts a barbell figure. List two forces that are acting on the barbell. Are these forces balanced if the weight lifter keeps the barbell steady?

Solution

Two forces acting on the barbell are:

The upward force applied by the weightlifter.
The downward gravitational force (weight of the barbell).
Yes, these forces are balanced if the weightlifter keeps the barbell steady, because the net force is zero.

5. Two players, R and S , are participating in an arm-wrestling match (Fig.). At the instant, when the arms tilt to the front direction (out of the page towards you), are the forces exerted by the players balanced? If not, which player exerted the larger force?

Solution

No, the forces are not balanced. Since the arms tilt in the forward direction, the player in the direction ' S ' is exerting a larger force than player ' R '.

6. An object is moving with a constant velocity. Is there a net force acting upon it?

Solution

No, there is no net force acting on an object moving with constant velocity because all forces are balanced.

7. Suppose no net force is acting on an object. Which of the following situations are possible? (i) Object remains at rest if at rest. (ii) Object keeps moving with a constant velocity if already moving. (iii) Object is moving with a constant acceleration.

Solution

If no net force acts on an object, the possible situations are: (i) Object remains at rest if at rest. (ii) The object keeps moving with constant velocity if already moving. Object is moving with a constant acceleration is not possible, because acceleration requires net force.

8. In the real world, it is difficult to find a situation where no forces are acting on an object. But by applying additional forces, a condition can be achieved where the net force on the object is zero. Explain with the help of an example.

Solution

In real world, friction and other forces usually act on objects. But by applying an equal and opposite force, we can make the net force zero. For example, when a person pushes a box on the floor with a force equal to friction, the box moves with constant velocity. Here, the applied force balances friction, so the net force becomes zero and the motion remains steady.

9. A toy car of mass 100 g is moving with a constant velocity of 0.5 ms . What is the net force acting on the toy car?

Solution

The toy car is moving with a constant velocity, so its acceleration is zero. Therefore, the net force acting on it is 0 N .

10. Two children of different masses are sitting on identical swings. To impart identical initial acceleration, for which child would you require to apply a larger force? Explain why.

Solution

A larger force is required for the child with greater mass, because acceleration depends on mass (Newton's second law). For the same acceleration, force increases with mass ( $F = ma$ ).

11. How are glass items packed for transportation using a bubble wrap or hay protected from damage?

Solution

Glass items are packed with bubble wrap or hay to increase the time of impact when they are dropped or hit something. According to Newton's second law (force $=$ change in momentum ÷ time), increasing the time of impact reduces the force. Thus, the cushioning reduces the force on the glass and prevents it from breaking.

12. Why does a firefighter sometimes struggle when holding the pipe issuing water?

Solution

A fireperson struggles because the highspeed water coming out of the pipe is pushed forward with great force, and due to Newton's third law, it exerts an equal and opposite reaction force backward on the pipe. This backward force creates a strong recoil, making the pipe difficult to hold steady.

13. Suppose a spacecraft is moving in a region of space where the gravitational force acting upon it is negligible. Suggest how it can change its velocity.

Solution

Even in a region where gravitational force is negligible, a spacecraft can change its velocity by using its onboard engines. By ejecting gas (exhaust) backward at high speed, the spacecraft experiences an equal and opposite reaction force (Newton's third law), which changes its speed or direction. This allows it to accelerate, slow down, or turn even in empty space.

14. Using a horizontal force F , a table is moved across the floor at a constant velocity. How much is the frictional force exerted by the floor on the table?

Solution

Since the table moves with constant velocity, its acceleration is zero, so the net force must be zero. Therefore, the frictional force exerted by the floor is equal in magnitude and opposite to the applied force F.

15. For a ball moving on a smooth, frictionless surface, choose the appropriate option that will make the following statements physically correct. (i) If no net force is applied on the ball, the velocity of the ball will remain the same/increase/decrease. (ii) If a net force is applied on the ball in the direction of its motion, the magnitude of the velocity of the ball will remain the same increase / decrease. (iii) If a net force is applied on the ball in a direction opposite to the direction of its motion, the magnitude of the velocity of the ball will remain the same/increase/decrease.

Solution

(i) If no net force is applied on the ball, the velocity of the ball will remain the same. (ii) If a net force is applied on the ball in the direction of its motion, the magnitude of the velocity of the ball will increase. (iii) If a net force is applied on the ball in a direction opposite to the direction of its Motion, the magnitude of the velocity of the ball will decrease.

16. Two blocks P and Q on a smooth horizontal surface are shown in Fig. a and Fig. b. Two forces of magnitudes 4 N and 5 N are acting in opposite directions on block P , while block Q is moving with a constant velocity.

(a)

(b)

Which of the following statement is correct? (i) P experiences a net force, and Q does not experience a net force. (ii) P does not experience a net force, and Q experiences a net force. (iii) Both P and Q experience a net force. (iv) Neither P nor Q experiences a net force.

Solution

The correct option is (i). For block P , the two opposite forces are 4 N and 5 N , so there is a net force of 1 N , meaning P experiences a net force. For block Q , is moving with constant velocity, so the net force is zero, it means Q does not experience a net force.

17. While practising for the snake boat race (Valium kaili in Kerala), 100 oarsmen are rowing a boat together. Out of these, 95 row backwards to propel the boat forward. But by mistake, 5 oarsmen row in the opposite direction. If each oarsman applies a horizontal force of 200 N , what is the net force on the snake boat? (Ignore drag forces, air friction, etc.)

Solution

Each oarsman applies a force of 200 N .

95 oarsmen row forward: $95 \times 200 = 19000 N$ forward
5 oarsmen row backwards: $5 \times 200 = 1000 N$ backward
Net force on the boat: 19000-1000 $= 18000 N$ forward

18. When a net force acts on an object, we observe that the object accelerates: (i) opposite to the direction of force, with acceleration proportional to the force acting on the object. (ii) opposite to the direction of force, with acceleration proportional to the mass of the object. (iii) in the direction of force, with acceleration inversely proportional to the force acting on the object. (iv) in the direction of force, with acceleration proportional to the force acting on the object.

Solution

(iv) in the direction of force, with acceleration proportional to the force acting on the object because according to Newton's second law, acceleration is directly proportional to the net force and occurs in the same direction as the force.

19. The position-time graph for four objects $A, B, C$ and D moving along a straight line are given in Fig. A net force acts on:

(1) Object A (2) Object B (3) Object C (4) Object D

Solution

(iii) For Object C, the position-time graph is curved, not a straight line. This means the slope of the graph is continuously changing. Hence, the velocity of the object is not constant.

20. A sailor jumps out from a small boat to the shore figure. As the sailor jumps forward, will the boat move? If yes, in which direction and why?

Solution

Yes, the boat will move. When the sailor jumps forward towards the shore, he pushes the boat backward. Due to Newton's third law (action-reaction), the boat experiences an equal and opposite force and therefore, moves backward away from the shore.

21. During a high jump event, a landing mat or sand bed is placed for the athlete to fall upon figure. Explain the reason behind it.

Solution

A landing mat or sand bed is used to increase the time of impact when the athlete falls. When the stopping time increases, the force of impact decreases, making the landing safer and preventing injuries.

22. A hand cart loaded with vegetables collides with an identical but empty hand cart. During the collision: (i) the loaded cart exerts a force of larger magnitude on the empty cart. (ii) the empty cart exerts a force of larger magnitude on the loaded cart. (iii) neither cart exerts a force on the other. (iv) the loaded cart and the empty cart, both exert an equal magnitude of force on each other.

Solution

(iv) According to Newton's third law of motion, every action has an equal and opposite reaction, so both carts exert equal forces on each other during the collision.

23. The acceleration-mass graph for the acceleration produced by a force on objects of different masses is plotted in Fig. Plot the force-mass graph for this case.

Solution

The graph shows acceleration vs mass, so to get force, we use Newton's second law:

F = ma At $1 kg, F = 1 \times 10 = 10 N$ At $2 kg, F = 2 \times 5 = 10 N$ At $3 kg, F = 3 \times 3.33 \approx 10 N$ At $4 kg, F = 4 \times 2.5 = 10 N$ At $5 kg, F = 5 \times 2 = 10 N$ So, the force remains constant for all masses.

24. The velocity-time graph of an object of mass 10 kg moving along a straight line is shown m Fig. Calculate the force acting on the object by using the graph.

Solution

We know, $F = ma$ $F = m \times \frac{( v - u )}{t}$ $= 10 \times \frac{( 30 - 10 )}{8} = 25 N$

25. A bullet of mass 50 g moving with a speed of $100 ms^{- 1}$ enters a heavy, stationary wooden block and stops after penetrating a distance of 50 cm . Estimate the stopping force acting on the bullet (assume that the bullet undergoes constant acceleration within the block).

Solution

Given: Mass of bullet, m 50g = 0.05 kg Initial velocity, $u = 100 m / s$ Final velocity, $v = 0$ Distance travelled in blocks, s 50 cm $= 0.5 m$ Use equation of motion: $v^{2} = u^{2} + 2 as$ $0^{2} = 10 0^{2} + 2 \times a \times 0.5$ $a = 10000 m / s^{2}$ We can find force by the formula $F = ma = 0.05 \times - 10000$ $= - 500 N$ (-) sign indicates opposite force to the direction of motion.

26. An ace footballer converted a penalty shot by kicking the football with a speed of $108 kmh^{- 1}$ . The estimated force they imparted was 800 N . The mass of the football was 0.4 kg . Calculate the time of contact between their foot and the ball.

Solution

Given: Initial velocity, $u = 0$ (ball kicked from rest) Final velocity, $v = 108 km / h = 30 m / s$ Force, $F = 800 N$ Mass, $m = 0.4 kg$ Step 1: Final acceleration $F = ma \Rightarrow a = F / m = 800/0.4 = 2000 m / s^{2}$ Step 2: Use equation of motion $v = u + at; 30 = 0 + 2000 t$ $t = 30/2000 = 0.015 s$ The time of contact between the foot and the ball is 0.015 s .

27. An object of mass 2 kg moving with a constant velocity of 10 m encounters a rough patch where the force of friction on the object is 7 N . At the same time, an additional constant force of 3 N opposing the motion is applied on the object. After entering the rough patch, how much distance does the object travel before coming to rest?

Solution

Given: Mass, $m = 2 kg$ Initial velocity, $u = 10 m / s$ Final velocity, $v = 0$ Frictional force $= 7 N$ (Opposes motion) Additional opposing force $= 3 N$ Find net retarding force is $F = 7 + 3 = 10 N$ . Acceleration, $a = F / m = 10/2 = 5 m / s^{2}$ Since it opposes motion, $a = - 5 m / s^{2}$ Use equation of motion $v^{2} = u^{2} + 2$ as $0 = (10)^{2} + 2 (- 5) s$ $0 = 100 - 10 s$ 10s = 100 $s = 10 m$ The object travels 10 m before coming to rest.

28. A tractor pulls a harrow (a ploughing tool) of mass m1with a net force F, resulting in an acceleration of a1. The same tractor pulls a trolley of mass m 2 with a force F producing an acceleration of a2. If the tractor now pulls the trolley with the harrow placed on it (with the same force F), then obtain an expression for the resulting acceleration in terms of a1 and a2. Ignore friction.

Solution

$F = m_{1} a_{1} \Rightarrow m_{1} = \frac{F}{a _{1}}$ $F = m_{2} a_{2} \Rightarrow m_{2} \frac{F}{a _{2}}$ When both are together: Total mass $= m_{1} + m_{2} = \frac{F}{a _{1}} + \frac{F}{a _{2}}$ Now acceleration (a) when same force (F) is applied:

Substitute: $a = \frac{F}{m _{1} + m _{2}}$ Take F common: $a = \frac{F}{a _{1}} + \frac{F}{a _{2}}$ $a = \frac{F}{F ( \frac{1}{a _{1}} + \frac{1}{a _{2}} )}$ $a = \frac{1}{F ( \frac{1}{a _{1}} + \frac{1}{a _{2}} )}$ Thus, $a = \frac{a _{1} a _{2}}{a _{1} + a _{2}}$

29. When the pole of a bar magnet is brought close to a magnetic compass, the bar magnet and the compass needle (which is also a magnet) exert a magnetic force on each other. As per Newton's third law of motion, both the forces are equal in magnitude and opposite in direction. However, the compass needle moves, whereas the bar magnet does not move (Fig.). Explain why.

Solution

The forces between the bar magnet and the compass needle are indeed equal and opposite, but their effects depend on mass. The compass needle has a very small mass, so even a small magnetic force produces noticeable motion. The bar magnet has a much larger mass, so the same force produces extremely small acceleration, making its motion negligible. Hence, only the compass needle appears to move.

4.0Key Topics in NCERT Class 9 Science Chapter 6 How Forces Affect Motion

Topic	What Students Learn
Force and Its Effects	Understand force as a push or pull that can change the speed, direction, or shape of an object.
Contact and Non-Contact Forces	Differentiate between forces requiring physical contact and forces like gravity and magnetism that act at a distance.
Balanced and Unbalanced Forces	Identify when forces cancel out (no change in motion) versus when they produce acceleration.
Friction	Understand the cause of friction between surfaces and how it opposes relative motion.
Newton's First Law of Motion	Learn the law of inertia and why objects resist changes to their state of rest or uniform motion.
Newton's Second Law of Motion	Relate force, mass, and acceleration through F = ma and solve related numerical problems.
Momentum	Define momentum as mass times velocity and calculate change in momentum over time.
Newton's Third Law of Motion	Understand action-reaction pairs and why they act on two different objects.
System of Objects	Treat connected objects as a single system to calculate net acceleration.
Applications in Daily Life	Apply the laws of motion to real-life situations like rockets, sports, vehicles, and safety devices.

5.0Mind Map / Concept Recap

6.0Related Study Materials for Class 9 Science

Access ALLEN’s Class 9 Science Study Material as per the latest NCERT Syllabus. Enhance your concepts, revise important topics and get ready with confidence for school and CBSE examinations with NCERT Solutions along with NCERT Textbooks, Revision Notes, Sample Papers and Previous Years Question Papers.

CBSE Class 9 Science Syllabus

Class 9 Science Revision Notes

NCERT Textbook for Class 9 Science

CBSE Sample Papers for Class 9 Science

7.0Advantages of Chapter 8 Science Class 9 NCERT Solutions

Explains Types of Forces: Helps students understand balanced and unbalanced forces.
Strengthens Motion Concepts: Clarifies how force changes the state of motion of an object.
Simplifies Newton's Laws: Explains all three laws of motion with easy examples.
Introduces Friction: Helps students understand the effects of friction in everyday life.
Develops Problem-Solving Skills: Simplifies force and motion-based questions with systematic solutions.

NCERT Solutions for Class 9 Science Chapter 6: How Forces Affect Motion

1.0Download NCERT Solutions for Class 9 Science Chapter 6 – How Forces Affect Motion PDF

NCERT Solutions Class 9 Science Chapter 6

2.0Learning Outcomes: NCERT Class 9 Science Chapter 6 Solutions:

Understand the Concept of Force: Define force as a push or pull and explain its effects — changing speed, direction, or shape of an object.
Differentiate Contact and Non-Contact Forces: Identify forces that act through touch (push, pull, friction) versus forces that act at a distance (gravitational, magnetic).
Explain Balanced and Unbalanced Forces: Distinguish between balanced forces (no change in motion) and unbalanced forces (causing acceleration).
Understand Friction: Explain the cause of friction, factors affecting it, and its role in everyday activities like walking and braking.
State and Apply Newton's First Law of Motion: Understand inertia and why objects resist changes in their state of rest or motion.
State and Apply Newton's Second Law of Motion: Relate force, mass, and acceleration using F = ma, and solve numerical problems.
Understand Momentum: Define momentum as the product of mass and velocity and calculate the rate of change of momentum.
State and Apply Newton's Third Law of Motion: Explain action-reaction force pairs and why they act on different objects and do not cancel out.
Analyse Systems of Objects: Treat two or more connected objects as a single system to calculate combined acceleration under an external force.
Solve NCERT Exercise Problems: Confidently attempt and solve all in-text activities and end-of-chapter numerical and conceptual questions.
Build a Foundation for Advanced Topics: Develop conceptual clarity that supports future topics like gravitation, work-energy, and higher-level mechanics.

3.0NCERT Solutions for Class 9 Science Chapter 6 How Forces Affect Motion – Detailed Solutions

1. Why does a canoe move forward when the canoeist pushes water backwards with their paddle, and why does it move faster when they push harder?

Solution

2. Suppose the same canoeist uses the same paddle force in two different canoes, one empty and one carrying another passenger. In which case will the canoe move faster?

Solution

The canoe will move faster when it is empty. With the same force, the lighter canoe (less mass) accelerates more, while the heavier canoe (with a passenger) accelerates less.

3. How much does a force of I feel? If you hold a 100 g mass in your palm, the upward force your palm applies on the mass is around 1 N .

Solution

4. A weightlifter lifts a barbell figure. List two forces that are acting on the barbell. Are these forces balanced if the weight lifter keeps the barbell steady?

Solution

Two forces acting on the barbell are:

The upward force applied by the weightlifter.
The downward gravitational force (weight of the barbell).
Yes, these forces are balanced if the weightlifter keeps the barbell steady, because the net force is zero.

Solution

No, the forces are not balanced. Since the arms tilt in the forward direction, the player in the direction ' S ' is exerting a larger force than player ' R '.

6. An object is moving with a constant velocity. Is there a net force acting upon it?

Solution

No, there is no net force acting on an object moving with constant velocity because all forces are balanced.

Solution

9. A toy car of mass 100 g is moving with a constant velocity of 0.5 ms . What is the net force acting on the toy car?

Solution

The toy car is moving with a constant velocity, so its acceleration is zero. Therefore, the net force acting on it is 0 N .

10. Two children of different masses are sitting on identical swings. To impart identical initial acceleration, for which child would you require to apply a larger force? Explain why.

Solution

A larger force is required for the child with greater mass, because acceleration depends on mass (Newton's second law). For the same acceleration, force increases with mass ( $F = ma$ ).

11. How are glass items packed for transportation using a bubble wrap or hay protected from damage?

Solution

12. Why does a firefighter sometimes struggle when holding the pipe issuing water?

Solution

13. Suppose a spacecraft is moving in a region of space where the gravitational force acting upon it is negligible. Suggest how it can change its velocity.

Solution

14. Using a horizontal force F , a table is moved across the floor at a constant velocity. How much is the frictional force exerted by the floor on the table?

Solution

(a)

(b)

Solution

Each oarsman applies a force of 200 N .

95 oarsmen row forward: $95 \times 200 = 19000 N$ forward
5 oarsmen row backwards: $5 \times 200 = 1000 N$ backward
Net force on the boat: 19000-1000 $= 18000 N$ forward

Solution

19. The position-time graph for four objects $A, B, C$ and D moving along a straight line are given in Fig. A net force acts on:

(1) Object A (2) Object B (3) Object C (4) Object D

Solution

(iii) For Object C, the position-time graph is curved, not a straight line. This means the slope of the graph is continuously changing. Hence, the velocity of the object is not constant.

20. A sailor jumps out from a small boat to the shore figure. As the sailor jumps forward, will the boat move? If yes, in which direction and why?

Solution

21. During a high jump event, a landing mat or sand bed is placed for the athlete to fall upon figure. Explain the reason behind it.

Solution

(iv) According to Newton's third law of motion, every action has an equal and opposite reaction, so both carts exert equal forces on each other during the collision.

23. The acceleration-mass graph for the acceleration produced by a force on objects of different masses is plotted in Fig. Plot the force-mass graph for this case.

Solution

The graph shows acceleration vs mass, so to get force, we use Newton's second law:

24. The velocity-time graph of an object of mass 10 kg moving along a straight line is shown m Fig. Calculate the force acting on the object by using the graph.

Solution

We know, $F = ma$ $F = m \times \frac{( v - u )}{t}$ $= 10 \times \frac{( 30 - 10 )}{8} = 25 N$

Solution

4.0Key Topics in NCERT Class 9 Science Chapter 6 How Forces Affect Motion

Topic	What Students Learn
Force and Its Effects	Understand force as a push or pull that can change the speed, direction, or shape of an object.
Contact and Non-Contact Forces	Differentiate between forces requiring physical contact and forces like gravity and magnetism that act at a distance.
Balanced and Unbalanced Forces	Identify when forces cancel out (no change in motion) versus when they produce acceleration.
Friction	Understand the cause of friction between surfaces and how it opposes relative motion.
Newton's First Law of Motion	Learn the law of inertia and why objects resist changes to their state of rest or uniform motion.
Newton's Second Law of Motion	Relate force, mass, and acceleration through F = ma and solve related numerical problems.
Momentum	Define momentum as mass times velocity and calculate change in momentum over time.
Newton's Third Law of Motion	Understand action-reaction pairs and why they act on two different objects.
System of Objects	Treat connected objects as a single system to calculate net acceleration.
Applications in Daily Life	Apply the laws of motion to real-life situations like rockets, sports, vehicles, and safety devices.

5.0Mind Map / Concept Recap

6.0Related Study Materials for Class 9 Science

CBSE Class 9 Science Syllabus

Class 9 Science Revision Notes

NCERT Textbook for Class 9 Science

CBSE Sample Papers for Class 9 Science

7.0Advantages of Chapter 8 Science Class 9 NCERT Solutions

Explains Types of Forces: Helps students understand balanced and unbalanced forces.
Strengthens Motion Concepts: Clarifies how force changes the state of motion of an object.
Simplifies Newton's Laws: Explains all three laws of motion with easy examples.
Introduces Friction: Helps students understand the effects of friction in everyday life.
Develops Problem-Solving Skills: Simplifies force and motion-based questions with systematic solutions.

Frequently Asked Questions

What is Chapter 6 "How Forces Affect Motion" about in Class 9 Science?

What are contact and non-contact forces?

What is the difference between balanced and unbalanced forces?

What is inertia?

What is Newton's Second Law of Motion?

How is the concept of "system of objects" used in this chapter?

Join ALLEN!

NCERT Solutions for Class 9 Science Chapter 6: How Forces Affect Motion

1.0Download NCERT Solutions for Class 9 Science Chapter 6 – How Forces Affect Motion PDF

2.0Learning Outcomes: NCERT Class 9 Science Chapter 6 Solutions:

3.0NCERT Solutions for Class 9 Science Chapter 6 How Forces Affect Motion – Detailed Solutions

4.0Key Topics in NCERT Class 9 Science Chapter 6 How Forces Affect Motion

5.0Mind Map / Concept Recap

6.0Related Study Materials for Class 9 Science

7.0Advantages of Chapter 8 Science Class 9 NCERT Solutions

Frequently Asked Questions

What is Chapter 6 "How Forces Affect Motion" about in Class 9 Science?

What are contact and non-contact forces?

What is the difference between balanced and unbalanced forces?

What is inertia?

What is Newton's Second Law of Motion?

How is the concept of "system of objects" used in this chapter?

Join ALLEN!

NCERT Solutions for Class 9 Science Chapter 6: How Forces Affect Motion

1.0Download NCERT Solutions for Class 9 Science Chapter 6 – How Forces Affect Motion PDF

2.0Learning Outcomes: NCERT Class 9 Science Chapter 6 Solutions:

3.0NCERT Solutions for Class 9 Science Chapter 6 How Forces Affect Motion – Detailed Solutions

4.0Key Topics in NCERT Class 9 Science Chapter 6 How Forces Affect Motion

5.0Mind Map / Concept Recap

6.0Related Study Materials for Class 9 Science

7.0Advantages of Chapter 8 Science Class 9 NCERT Solutions