NCERT Solutions for Class 9 Science Chapter 6 How Forces Affect Motion (2026-27)

Revise, Reflect, Refine (NCERT Textbook Page No. 112)

Question 1. 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?

Answer: The table is moving with constant velocity. This means its acceleration is zero.

According to Newton’s second law, if acceleration is zero, the net force on the object is also zero. So, the applied force and the frictional force must be equal in magnitude and opposite in direction.

Therefore, the frictional force exerted by the floor on the table is equal to F, acting opposite to the direction of motion.

Frictional force = F

Question 2. 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.

Answer:
(i) If no net force is applied on the ball, the velocity of the ball will remain the same.

Reason: On a smooth, frictionless surface, there is no opposing force. So, the ball continues to move with the same velocity unless an external force acts on it.

(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.

Reason: A force applied in the direction of motion increases the ball's speed.

(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.

Reason: A force acting opposite to the motion slows the ball down.

Question 3. Two blocks P and Q on a smooth horizontal surface are shown in Fig. 6.36a and Fig. 6.36b. 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.

Image 1

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.

Answer:
The correct option is (i) P experiences a net force, and Q does not experience a net force.

For block P, two forces of 4 N and 5 N act in opposite directions.

Net force on P = 5 N - 4 N = 1 N

So, block P experiences a net force of 1 N in the direction of the larger force.

Block Q is moving with constant velocity. If velocity is constant, acceleration is zero. Therefore, the net force on Q is zero.

Hence, P experiences a net force, while Q does not experience a net force.

Question 4. 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.)

Answer:
Each oarsman applies a force of 200 N.

Force applied by 95 oarsmen in the forward direction:

95 × 200 = 19000 N

Force applied by 5 oarsmen in the opposite direction:

5 × 200 = 1000 N

Net force on the boat:

Net force = 19000 - 1000
Net force = 18000 N

Therefore, the net force on the snake boat is 18000 N in the forward direction.

Question 5.
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.

Answer:
The correct option is (iv) in the direction of force, with acceleration proportional to the force acting on the object.

According to Newton’s second law of motion, the acceleration of an object is directly proportional to the net force acting on it and takes place in the direction of the force.

F = ma

This means that if the net force increases, acceleration also increases, provided the mass remains constant.

Question 6. The position-time graph for four objects A, B, C and D moving along a straight line are given in Fig. 6.37. A net force acts on:
(i) Object A
(ii) Object B
(iii) Object C
(iv) Object D

Image 2

Answer:
The correct option is (iii) Object C.

In a position-time graph, the slope represents velocity.

If the graph is a straight line, the object has constant velocity and zero acceleration. Hence, no net force acts on it.

For object C, the graph is curved. This means the slope is changing with time, so the velocity is changing. A change in velocity means acceleration is present.

Since acceleration is produced by net force, a net force acts on object C.

Question 7. A sailor jumps out from a small boat to the shore (Fig. 6.38). As the sailor jumps forward, will the boat move? If yes, in which direction and why?

Image 3

Answer:
Yes, the boat will move.

When the sailor jumps forward towards the shore, he pushes the boat backward with his feet. According to Newton’s third law of motion, the boat exerts an equal and opposite force on the sailor.

As a result, the sailor moves forward, and the boat moves backward, away from the shore.

This happens because action and reaction forces always act in opposite directions.

Question 8. During a high jump event, a landing mat or sand bed is placed for the athlete to fall upon (Fig. 6.39). Explain the reason behind it.

Image 4

Answer: A landing mat or sand bed is placed to reduce the force of impact on the athlete.

When the athlete falls, their momentum becomes zero after landing. The mat or sand bed increases the time taken to stop the athlete’s motion.

Force depends on the rate of change of momentum. If the stopping time increases, the force decreases.

Therefore, the landing mat or sand bed makes the landing safer and reduces the chance of injury.

Question 9. A hand cart loaded with vegetables coffides 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.

Answer:
The correct option is (iv) the loaded cart and the empty cart, both exert an equal magnitude of force on each other.

During a collision, each cart exerts a force on the other. According to Newton’s third law of motion, these forces are equal in magnitude and opposite in direction.

The loaded cart may have a larger mass, so its acceleration will be smaller. The empty cart has less mass, so it may move or change its motion more. But the force exerted by both carts on each other is equal.

Question 10. The acceleration-mass graph for the acceleration produced by a force on objects of different masses is plotted in Fig. 6.40. Plot the force-mass graph for this case.

Image 5

Answer:
To plot the force-mass graph, we use Newton’s second law:

F = ma

From the acceleration-mass graph, the values show that as mass increases, acceleration decreases in such a way that the product of mass and acceleration remains constant.

For example:

Mass = 1 kg, acceleration = 10 m/s²
Force = 1 × 10 = 10 N

Mass = 2 kg, acceleration = 5 m/s²
Force = 2 × 5 = 10 N

Mass = 3 kg, acceleration ≈ 3.33 m/s²
Force = 3 × 3.33 ≈ 10 N

Mass = 4 kg, acceleration = 2.5 m/s²
Force = 4 × 2.5 = 10 N

Mass = 5 kg, acceleration = 2 m/s²
Force = 5 × 2 = 10 N

So, the force remains constant at 10 N for all masses.

Force-mass graph:

Image 6

The graph will be a horizontal straight line parallel to the mass axis at F = 10 N.

Question 11. The velocity-time graph of an object of mass 10 kg moving along a straight line is shown m Fig. 6.41. Calculate the force acting on the object by using the graph.

Image 7

Answer:
Given:

Mass of object, m = 10 kg

From the velocity-time graph:

Initial velocity, u = 10 m/s
Final velocity, v = 30 m/s
Time, t = 8 s

Acceleration is given by:

a = (v - u) / t

a = (30 - 10) / 8
a = 20 / 8
a = 2.5 m/s²

Now, using Newton’s second law:

F = ma

F = 10 × 2.5
F = 25 N

Therefore, the force acting on the object is 25 N.

Question 12. 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).

Answer:
Given:

Mass of bullet, m = 50 g = 0.05 kg
Initial velocity, u = 100 m/s
Final velocity, v = 0 m/s
Distance penetrated, s = 50 cm = 0.5 m

Using the equation of motion:

v² = u² + 2as

0² = 100² + 2 × a × 0.5
0 = 10000 + a
a = -10000 m/s²

The negative sign shows that the acceleration is opposite to the direction of motion. It is a retardation.

Now, force is:

F = ma

F = 0.05 × (-10000)
F = -500 N

Therefore, the stopping force acting on the bullet is 500 N, opposite to the direction of motion.

Question 13. 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.

Answer:
Given:

Initial velocity, u = 0 m/s
Final velocity, v = 108 km/h
Force, F = 800 N
Mass, m = 0.4 kg

Convert velocity into m/s:

108 km/h = 108 × 5/18 = 30 m/s

Using Newton’s second law:

F = ma

a = F / m
a = 800 / 0.4
a = 2000 m/s²

Now, using:

v = u + at

30 = 0 + 2000t
t = 30 / 2000
t = 0.015 s

Therefore, the time of contact between the footballer’s foot and the ball is 0.015 seconds.

Question 14. 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?

Answer:
Given:

Mass, m = 2 kg
Initial velocity, u = 10 m/s
Final velocity, v = 0 m/s
Frictional force = 7 N
Additional opposing force = 3 N

Total opposing force:

F = 7 + 3
F = 10 N

Acceleration produced by this force:

a = F / m
a = 10 / 2
a = 5 m/s²

Since this force opposes the motion:

a = -5 m/s²

Using the equation of motion:

v² = u² + 2as

0² = 10² + 2 × (-5) × s
0 = 100 - 10s
10s = 100
s = 10 m

Therefore, the object travels 10 m before coming to rest.

Question 15. 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 m² 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.

Answer:
For the harrow:

F = m1a1

So,

m1 = F / a1

For the trolley:

F = m2a2

So,

m2 = F / a2

When the harrow is placed on the trolley, the total mass becomes:

m1 + m2 = F/a1 + F/a2

If the same force F is applied, the resulting acceleration a is:

a = F / (m1 + m2)

Substituting the values:

a = F / (F/a1 + F/a2)

Taking F common in the denominator:

a = F / F(1/a1 + 1/a2)

a = 1 / (1/a1 + 1/a2)

Now simplifying:

a = a1a2 / (a1 + a2)

Therefore, the resulting acceleration is:

a = a1a2 / (a1 + a2)

Question 16. 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. 6.42). Explain why.

Image 8

Answer: The bar magnet and compass needle exert equal and opposite magnetic forces on each other, as stated by Newton’s third law of motion.

However, the effect of a force also depends on the mass of the object. The compass needle has a very small mass, so the magnetic force produces a visible acceleration, and it moves.

The bar magnet has a much larger mass than the compass needle. The same force produces a very small acceleration in the bar magnet, so its motion is not easily noticed.

Therefore, the compass needle appears to move, while the bar magnet seems to remain at rest.

Class 9 Science Chapter 6 How Forces Affect Motion Question Answer (InText)

Think It Over (NCERT Textbook Page No. 94)

Question 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?

Answer: A canoe moves forward because of Newton’s third law of motion.

When the canoeist pushes water backwards with the paddle, the water pushes the canoe forward with an equal and opposite force. This forward reaction force moves the canoe.

When the canoeist pushes harder, the backward force on water increases. As a result, the reaction force on the canoe also increases. This greater force produces greater acceleration, so the canoe moves faster.

Question 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?

Answer:
The empty canoe will move faster.

The same paddle force is applied in both cases, but the empty canoe has less mass. According to Newton’s second law:

a = F / m

For the same force, acceleration is greater when the mass is smaller. The canoe carrying another passenger has more mass, so its acceleration will be less.

Therefore, the empty canoe moves faster.

Think It Over (NCERT Textbook Page No. 94)

Question 3. Is there an underlying cause for a change in position and velocity of an object? What is the nature of this cause? Do all motions require a cause?

Answer: Yes, a change in the position or velocity of an object can be caused by force.

A force is a push or pull that can change the state of rest, speed, direction, or shape of an object. If an object changes its velocity, it means acceleration has occurred, and acceleration is caused by a net force.

However, not all motions require a continuous force. An object moving with constant velocity can continue moving without a net force if there is no friction or resistance. A force is needed only to change the state of motion.

Think It Over (NCERT Textbook Page No. 95)

Question 4. How can we measure the magnitude of a force? Do you remember using a spring balance earlier to measure the weight of objects? Do you also remember the weight of an object is the gravitational force with which the Earth pulls the object?

Answer: The magnitude of a force can be measured using a spring balance.

A spring balance works on the principle that a spring stretches when a force is applied. The greater the force, the more the spring stretches. The reading on the spring balance gives the force in newtons.

The weight of an object is also a force. It is the gravitational force with which Earth pulls the object towards itself. This weight can be measured using a spring balance.

Think It Over (NCERT Textbook Page No. 95)

Question 5. In such cases, what is the effect of forces when more than one force is acting on an object at rest or in motion?

Answer: When more than one force acts on an object, their combined effect is called the net force.

If the forces are balanced, the net force is zero. In this case, an object at rest remains at rest, and an object moving with constant velocity continues to move with the same velocity.

If the forces are unbalanced, the net force is not zero. In this case, the object may start moving, stop, speed up, slow down, or change direction.

Pause and Ponder (NCERT Textbook Page No. 97)

Question 1. A weightlifter lifts a barbell (Fig. 6.8). List two forces that are acting on the barbell. Are these forces balanced if the weight lifter keeps the barbell steady?

Image 9

Answer:
Two forces acting on the barbell are:

The upward force applied by the weightlifter.
The downward gravitational force or weight of the barbell.

If the weightlifter keeps the barbell steady, the barbell is not accelerating. This means the net force on it is zero.

So, the upward force applied by the weightlifter balances the downward gravitational force.

Yes, the forces are balanced when the barbell is steady.

Question 2. Two players, R and S, are participating in an arm-wrestling match (Fig. 6.9). 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?

Image 10

Answer:
No, the forces are not balanced.

If the arms tilt in the front direction, it means there is a net force in that direction. Since the motion is towards the side of player S, player S must be exerting a larger force than player R at that instant.

Balanced forces would not cause any movement or tilting. Since tilting occurs, the forces are unbalanced.

Pause and Ponder (NCERT Textbook Page No. 99)

Question 3. If the velocity of the block is neither increasing nor decreasing, what can you say about the net force acting on the block? Does the reading of the spring balance indicate the magnitude of the force of friction acting on the wooden block?

Answer:
If the velocity of the block is neither increasing nor decreasing, the block is moving with constant velocity.

This means its acceleration is zero, so the net force acting on the block is also zero.

In this case, the pulling force applied through the spring balance is equal in magnitude to the frictional force acting opposite to motion.

Yes, the reading of the spring balance indicates the magnitude of the force of friction acting on the wooden block.

Question 4. Are the readings different? Is the reading smallest for the surface on which the stack of coins travelled the largest distance? Is the reading largest for which the distance travelled was the smallest?

Answer:
Yes, the readings are different for different surfaces because friction depends on the nature of the surface.

The spring balance reading is smallest for the surface on which the stack of coins travels the largest distance. This is because the surface offers the least friction.

The reading is largest for the surface on which the stack of coins travels the smallest distance. This means that the surface offers the greatest friction.

Thus, smoother surfaces have less friction, while rougher surfaces have more friction.

Pause and Ponder (NCERT Textbook Page No. 101)

Question 5. An object is moving with a constant velocity. Is there a net force acting upon it?

Answer:
No, there is no net force acting on an object moving with constant velocity.

If velocity is constant, acceleration is zero. According to Newton’s second law:

F = ma

Since a = 0, the net force is also zero.

So, all forces acting on the object are balanced.

Question 6. 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.

Answer:
If no net force acts on an object, the following situations are possible:

(i) Object remains at rest if at rest.
This is possible because no force is present to change its state of rest.

(ii) Object keeps moving with a constant velocity if already moving.
This is possible because no force is acting to speed it up, slow it down, or change its direction.

(iii) Object is moving with a constant acceleration.
This is not possible because acceleration requires a net force.

Therefore, only situations (i) and (ii) are possible.

Question 7. 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.

Answer: In the real world, forces such as friction, air resistance, gravity, and normal reaction usually act on objects. However, the net force can become zero if these forces balance one another.

For example, when a box is pushed on a floor with a force equal to the frictional force, the box moves with constant velocity.

In this case:

Applied force acts forward.
Friction acts backward.
Both forces are equal in magnitude.

So, the net force is zero, and the motion of the box does not change.

Pause and Ponder (NCERT Textbook Page No. 102)

Question 8. But what is the relationship between the net force acting on an object and its acceleration?

Answer: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

This relationship is given by Newton’s second law of motion:

F = ma

Where:

F = net force
m = mass of the object
a = acceleration

For a fixed mass, greater net force produces greater acceleration. For the same force, a heavier object has less acceleration.

Pause and Ponder (NCERT Textbook Page No. 106)

Question 9. A toy car of mass 1oo g is moving with a constant velocity of 0.5 ms’. What is the net force acting on the toy car?

Answer:
The toy car is moving with constant velocity.

If velocity is constant, acceleration is zero.

Using Newton’s second law:

F = ma

Since a = 0,

F = m × 0
F = 0 N

Therefore, the net force acting on the toy car is 0 N.

Question 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.

Answer:
A larger force is required for the child with greater mass.

According to Newton’s second law:

F = ma

For the same acceleration, force is directly proportional to mass. This means a heavier child needs a larger force to get the same initial acceleration as a lighter child.

Therefore, the child with greater mass requires a larger force.

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

Answer: Bubble wrap or hay protects glass items by increasing the time of impact during a fall or collision.

When a glass item hits a hard surface directly, it stops suddenly, so the force of impact is very large. Bubble wrap or hay acts as a cushion and increases the time taken to stop the object.

According to the relation between force and change in momentum, increasing the stopping time reduces the force.

Therefore, the force on the glass becomes smaller, and the chances of breaking are reduced.

Pause and Ponder (NCERT Textbook Page No. 107)

Question 12. Why is it difficult to walk on wet polished floors or ice, or why it is risky to drive on roads covered with water or snow?

Answer: It is difficult to walk on wet polished floors or ice because friction becomes very low.

Friction provides grip between our feet and the ground. On wet, icy, or snowy surfaces, the grip is reduced, so our feet may slip.

Similarly, vehicles need friction between tyres and the road to move, stop, and turn safely. On roads covered with water or snow, friction is reduced, making skidding more likely.

Therefore, such surfaces are risky for walking and driving.

Question 13. Why does a firefighter sometimes struggle when holding the pipe issuing water?

Answer: A firefighter struggles because water comes out of the pipe at high speed with a large forward force.

According to Newton’s third law of motion, the water exerts an equal and opposite reaction force on the pipe. This reaction force pushes the pipe backward.

This backward recoil force can be strong, so the firefighter may find it difficult to hold the pipe steady.

Question 14. 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.

Answer: A spacecraft can change its velocity by using its engines or thrusters.

When the spacecraft ejects gases backward at high speed, the gases exert an equal and opposite force on the spacecraft. This reaction force pushes the spacecraft forward.

By changing the direction in which gases are expelled, the spacecraft can speed up, slow down, or change direction.

This works even when the gravitational force is negligible, because it is based on Newton’s third law of motion.

Think as a Scientist (NCERT Textbook Page No. 100)

Question 1. Now, conduct a thought experiment. We do a thought experiment when the conditions required for the experiment are difficult to recreate in the real world. Suppose you find an object and a horizontal floor having such smooth surfaces that the force of friction between them is zero. Imagine what will happen if you repeat steps 3 and 4 of Activity 6.1 with such an object and a horizontal floor? Will the velocity of the object decrease? Will the object ever come to rest or continue moving forever?

Answer:
If the object and floor have perfectly smooth surfaces, the force of friction between them will be zero.

After the object is pushed, there will be no opposing frictional force to slow it down. So, its velocity will not decrease.

The object will not come to rest on its own. It will continue moving with the same velocity forever unless another external force acts on it.

This thought experiment shows that objects slow down and stop in daily life mainly because of friction and other resistive forces.

Threads of Curiosity (NCERT Textbook Page No. 104)

Question 1. 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.

Answer:
A force of 1 N is approximately the force needed to support a 100 g mass against Earth’s gravity.

When we hold a 100 g object in our palm, Earth pulls it downward due to gravity. To keep it at rest, our palm applies an upward force of about 1 N.

So, a force of 1 N feels like the effort needed to hold a small 100 g object steadily in the hand.

Activity 6.1: Let Us Investigate (NCERT Textbook Page No. 98)

Aim: To study how the motion of stack of coins is affected by different surfaces and to understand the effect of friction on the distance travelled and velocity.

Observations:
(i) When the stack of coins is released by a rubber band on a wooden table, it travels a short distance and stops quickly.

(ii) On a laminated table top, the stack of coins travels a longer distance compared to the wooden surface.

(iii) On a horizontal polished marble or tile floor, the stack of coins travels the maximum distance before coming to rest.

(iv) In all these cases, after losing contact with the rubber band, the velocity of the coins gradually decreases until they stop.

Image 11

Conclusion:
The activity shows that friction opposes the motion of an object.

On a rough surface, friction is greater, so the stack of coins stops quickly and travels a shorter distance. On a smoother surface, friction is less, so the stack of coins travels a longer distance before stopping.

Thus, the force of friction depends on the nature of the surface. More friction reduces motion faster, while less friction allows motion to continue for a longer time.

Activity 6.2: Let Us Measure (NCERT Textbook Page No. 99)

Aim: To measure the force of friction acting on a wooden block on different surfaces using a spring balance and to compare friction on various surfaces.

Observations:
(i) The reading of the spring balance gives an approximate value of the force needed to pull the wooden block with uniform motion.

(ii) When the block moves with constant velocity, the pulling force shown by the spring balance is equal to the frictional force.

(iii) The reading differs for different surfaces:

• It is higher on rough surfaces.

• It is lower on smooth surfaces.

• It is lowest on very smooth surfaces like polished marble or tile.

• 
  

Image 12

Conclusion:
The activity shows that friction is greater on rough surfaces and smaller on smooth surfaces.

When the block is pulled with constant velocity, the applied force balances the frictional force. Therefore, the spring balance reading gives the magnitude of friction.

More friction means a larger spring balance reading, while less friction means a smaller reading.

Activity 6.3: Let Us Experiment (Demonstration Activity) (NCERT Textbook Page No. 102-103)

Aim: To study how the acceleration of an object depends on the applied force for a fixed mass.

Observations:
(i) The hanging cup moves downward due to gravitational force and pulls the cart forward through the thread.

(ii) When the mass in the cup is increased, the pulling force increases.

(iii) With a greater pulling force, the cart covers the same distance in less time.

(iv) This shows that the acceleration of the cart increases when the applied force increases.

Image 13

Conclusion:
For a fixed mass, acceleration increases with an increase in net force.

This supports Newton’s second law of motion, which states that acceleration is directly proportional to the net force acting on an object.

a ∝ F, when mass is constant.

Activity 6.4: Let Us Experiment (Demonstration Activity) (NCERT Textbook Page No. 103)

Aim: To study how the acceleration of an object depends on its mass when the applied force is kept constant.

Observations:
(i) When the same pulling force is applied to the cart, it moves with some acceleration.

(ii) When extra mass is added to the cart while keeping the pulling force the same, the cart takes more time to cover the same distance.

(iii) This shows that the acceleration decreases when mass increases.

Conclusion:
When the applied force is constant, acceleration is inversely proportional to mass.

This means heavier objects accelerate less than lighter objects when the same force is applied.

a ∝ 1/m, when force is constant.

This verifies Newton’s second law of motion.

Activity 6.5: Let Us Explore (NCERT Textbook Page No. 107)

Aim: To understand that forces always act in pairs and in opposite directions. Observations:
(i) When the girl pushes the table forward, the chair (on which she sits) moves backward.
(ii) When the girl pulls the table towards herself, the chair moves forward.

Image 14

Conclusion:
This activity shows that forces always occur in pairs.

When the girl pushes the table forward, the table pushes her backward with an equal and opposite force, so the chair moves backward.

When she pulls the table towards herself, the table pulls her in the opposite direction, so the chair moves forward.

This verifies Newton’s third law of motion:

For every action, there is an equal and opposite reaction.

Activity 6.6: Let Us Verify (NCERT Textbook Page No. 108)

Aim: To verify that action and reaction forces are equal in magnitude and opposite in direction.

Observations:
(i) Two spring balances are connected and pulled in opposite directions.

(ii) Both spring balances show the same reading.

(iii) Even when the pulling force is changed, both readings remain equal.

Image 15

Conclusion:
The equal readings of the two spring balances show that the forces exerted by the two balances on each other are equal in magnitude.

Since the pulls are in opposite directions, the forces are opposite in direction.

This verifies Newton’s third law of motion.

Activity 6.7: Let Us Understand (NCERT Textbook Page No. 109)

Aim: To demonstrate Newton’s Third Law of Motion by observing the movement of a balloon with air rushing out through a straw connected by a thread.

Observations:
When the balloon is released, air rushes out from its mouth in one direction. At the same time, the balloon moves in the opposite direction along the thread.

The faster the air comes out, the quicker the balloon moves.

Image 16

Conclusion:
The balloon pushes air backward as the air rushes out. In response, the air pushes the balloon forward with an equal and opposite force. This movement demonstrates Newton’s third law of motion: For every action, there is an equal and opposite reaction. The same principle is used in rockets and jet engines.

Class 9 Science Chapter 6 How Forces Affect Motion Solutions

Vedantu provides NCERT Solutions for Class 9 Science Chapter 6, How Forces Affect Motion, from the Exploration textbook for the 2026-27 academic session. This chapter helps students understand how forces change the state of motion of objects and how motion is affected by balanced forces, unbalanced forces, friction, inertia, mass, acceleration, and action-reaction pairs.

The solutions include clear answers for exercise questions, in-text questions, numerical problems, graph-based questions, activity-based tasks, and real-life application questions. Students can use these solutions to understand Newton’s laws of motion, calculate force using F = ma, compare the effects of mass and acceleration, and revise important concepts before school exams. The downloadable FREE PDF also helps students study the complete chapter offline whenever needed.

CBSE Class 9 Science Chapter 6 Study Materials

Students can use the Chapter 6 study materials below to revise important concepts, practise extra questions, and strengthen their understanding of force, motion, friction, inertia, acceleration, and Newton’s laws.

Explore More NCERT Solutions for Class 9 Science Chapters

The chapter-wise NCERT Solutions for Class 9 Science help students understand concepts from different areas of science in a simple and organised way. These resources provide clear explanations, textbook answers, activity-based solutions, diagrams, examples, and revision support for each chapter.

Related Study Material for Class 9 Science

The following Class 9 Science study materials support concept learning, practice, revision, and exam preparation. Students can use them along with the Exploration textbook solutions for better understanding and regular study.