---
title: "NCERT Solutions Class 9 Science Ch 6 How Forces Affect Motion"
url: https://www.swavid.com/science/class/9/chapter/how-forces-affect-motion/ncert-solutions
dateModified: 2026-10-07T16:26:27+00:00
---

# NCERT Solutions Class 9 Science Ch 6 How Forces Affect Motion

This chapter's questions cover concepts related to force, balanced and unbalanced forces, friction, Newton's three laws of motion, and their applications to various physical scenarios.

Free PDF (13 pages): https://www.swavid.com/api/seo/pdf/ncert/science/class-9/swavid-ncert-solutions-class-9-science-chapter-6-how-forces-affect-motion-1fb7028d30.pdf

## Think It Over

### Question 1

*2 marks · Very short answer*

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**

1. When the canoeist pushes water backwards with the paddle, the water exerts an equal and opposite forward force on the paddle according to Newton's third law of motion.
2. When they push harder, the magnitude of the forward force increases, producing a larger net force and greater acceleration, which makes the canoe move faster.

**Answer:** The canoe moves forward due to the equal and opposite reaction force from the water, and pushes harder result in a larger forward force and higher acceleration.

> Common mistake: Thinking that the action and reaction forces act on the same object and cancel each other out.

### Question 2

*2 marks · Very short answer*

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**

1. According to Newton's second law, acceleration is inversely proportional to the mass of the object for a given force ($a = \frac{F}{m}$).
2. Since the empty canoe has a smaller mass than the canoe carrying another passenger, the same paddle force will produce a larger acceleration in the empty canoe, making it move faster.

**Answer:** The empty canoe will move faster because it has a smaller mass, resulting in a greater acceleration for the same force.

> Common mistake: Assuming that different masses will experience the same acceleration under the same applied force.

## Pause and Ponder

### Question 1

*3 marks · Short answer*

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

**Solution**

1. 1. The two forces acting on the barbell are the gravitational force acting downwards and the upward force applied by the weightlifter.
2. 2. Yes, these forces are balanced if the weightlifter keeps the barbell steady.
3. 3. When the barbell is steady, the net force acting on it is zero, meaning the upward force equals the downward gravitational force.

**Answer:** The two forces are the downward gravitational force and the upward force by the weightlifter. Yes, they are balanced.

> Common mistake: Stating that no forces act on the barbell when it is steady.

### Question 2

*3 marks · Short answer*

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?

**Solution**

1. 1. At the instant when the arms tilt to the front direction, the forces exerted by the players are not balanced.
2. 2. Because the arms are accelerating in the front direction, a non-zero net force acts in that direction.
3. 3. Therefore, the player tilting the arms towards themselves (out of the page) exerted the larger force.

**Answer:** No, the forces are not balanced. The player whose arm tilts the other player's arm forward exerted the larger force.

> Common mistake: Confusing which player is pushing forward based on the tilt direction.

## Pause and Ponder

### Question 3

*2 marks · Very short answer*

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

**Solution**

1. According to Newton's first law of motion, an object continues to move with a constant velocity unless a net force acts upon it.
2. Therefore, no net force is acting on an object moving with a constant velocity.

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

> Common mistake: Thinking that a constant force is required to keep an object moving with a constant velocity.

### Question 4

*1 mark · MCQ*

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.

- Object remains at rest if at rest.
- Object keeps moving with a constant velocity if already moving.
- Object is moving with a constant acceleration.

**Solution**

1. When no net force acts on an object, its acceleration is zero as per Newton's first law of motion.
2. Thus, an object at rest remains at rest (i) and a moving object continues to move with a constant velocity (ii), making options (i) and (ii) possible.

**Answer:** (iv) Both (i) and (ii) are possible.

> Common mistake: Choosing constant acceleration, which requires a non-zero net force.

### Question 5

*3 marks · Short answer*

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**

1. When a box is pushed across a floor at a constant velocity, the applied force is exactly balanced by the opposing force of friction.
2. Although multiple forces act on the box, the net force acting on it is zero.
3. This condition is similar to an object experiencing no forces at all.

**Answer:** A box pushed across a floor with an applied force equal in magnitude to the frictional force experiences zero net force.

> Common mistake: Stating that no forces act on the object instead of recognizing that multiple forces balance each other out.

## Pause and Ponder

### Question 6

*3 marks · Numerical*

A toy car of mass $100\text{ g}$ is moving with a constant velocity of $0.5\text{ m s}^{-1}$. What is the net force acting on the toy car?

**Solution**

1. Given: Mass of the toy car $m = 100\text{ g} = 0.1\text{ kg}$, constant velocity $v = 0.5\text{ m s}^{-1}$.
2. Formula: According to Newton's first law of motion, an object moving with a constant velocity has zero acceleration.
3. Substitution: $a = 0\text{ m s}^{-2}$, so net force $F = ma = 0.1\text{ kg} \times 0\text{ m s}^{-2}$.
4. Result: $0\text{ N}$.

**Answer:** 0 N

> Common mistake: Multiplying mass and velocity to find force instead of recognizing that constant velocity means zero acceleration.

### Question 7

*3 marks · Short answer*

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**

1. According to Newton's second law of motion, the magnitude of acceleration is inversely proportional to the mass of the object for a given force ($a = \frac{F}{m}$, or $F = ma$).
2. To impart the identical initial acceleration to both children, the required force depends directly on their mass.
3. Therefore, a larger force would be required for the child with the larger mass.

**Answer:** For the child with the larger mass, because a larger force is needed to produce the same acceleration in a heavier object.

> Common mistake: Stating that lighter objects require more force instead of heavier objects requiring more force for the same acceleration.

### Question 8

*3 marks · Short answer*

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

**Solution**

1. Glass items are fragile because even a small change in velocity in a very short time results in a large acceleration and a very large impact force, which breaks them.
2. Bubble wrap or hay acts as a compressible cushion between the items.
3. When a packed glass item experiences a jerk or fall, the cushion compresses and increases the time duration over which the velocity reduces to zero, thereby reducing the magnitude of the force and protecting the item from damage.

**Answer:** Bubble wrap or hay increases the time of impact during a jerk or fall, which decreases the acceleration and reduces the force acting on the glass items, preventing damage.

> Common mistake: Explaining it only as soft padding without mentioning the increase in time and reduction of force.

## Pause and Ponder

### Question 9

*3 marks · Short answer*

Why does a fireperson sometimes struggle when holding the pipe issuing water?

**Solution**

1. Water rushes out of the heavy pipe in the forward direction at a very high speed.
2. According to Newton's third law of motion, the water exerts an equal and opposite backward force on the pipe and the fireperson.
3. To hold the pipe steady and prevent it from being pushed back, the fireperson has to exert a large forward force, causing them to struggle.

**Answer:** The fireperson struggles because of the equal and opposite backward force exerted by the fast-moving water rushing out of the pipe, as per Newton's third law of motion.

> Common mistake: Thinking that water pushes the pipe forward instead of backward.

### Question 10

*3 marks · Short answer*

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

**Solution**

1. In space where gravitational force is negligible, a spacecraft can change its velocity by firing its thrusters or engines in a specific direction.
2. According to Newton's third law of motion, the engine expels exhaust gases in one direction with a force.
3. The exhaust gases exert an equal and opposite force on the spacecraft, causing it to accelerate and change its velocity in that direction.

**Answer:** The spacecraft can change its velocity by firing its rocket engines to expel exhaust gases, utilizing Newton's third law of motion.

> Common mistake: Stating that the spacecraft needs to push against something like air or the ground to move.

## Revise, Reflect, Refine

### Question 1

*2 marks · Very short answer*

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**

1. Since the table is moving with a constant velocity, the net force acting on it is zero as per Newton's first law of motion.
2. Therefore, the applied horizontal force $F$ is balanced by the frictional force, making the magnitude of the frictional force equal to $F$ acting in the opposite direction.

**Answer:** The frictional force exerted by the floor on the table is equal in magnitude to $F$ ($F$) and acts in the direction opposite to motion.

> Common mistake: Stating that friction is zero or greater than the applied force.

### Question 2

*3 marks · Case-based*

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.

- remain the same
- increase
- decrease

**Part (i)**

1. When no net force is applied on the ball, as per Newton's first law of motion, its state of motion does not change.

Answer (i): remain the same

**Part (ii)**

1. Applying a net force in the direction of motion produces an acceleration that increases the speed of the ball.

Answer (ii): increase

**Part (iii)**

1. Applying a net force opposite to the direction of motion produces deceleration, reducing the speed of the ball.

Answer (iii): decrease

**Answer:** The correct options are (i) remain the same, (ii) increase, and (iii) decrease.

> Common mistake: Confusing the effect of opposite forces with increasing speed.

### Question 3

*1 mark · MCQ*

Two blocks P and Q on a smooth horizontal surface are shown in Fig. 6.36a and Fig. 6.36b. Two forces of magnitudes $4\text{ N}$ and $5\text{ N}$ are acting in opposite directions on block P, while block Q is moving with a constant velocity. Which of the following statement is correct?

- P experiences a net force and Q does not experience a net force.
- P does not experience a net force and Q experiences a net force.
- Both P and Q experience a net force.
- Neither P nor Q experiences a net force.

**Solution**

1. On block P, two unequal forces ($5\text{ N}$ and $4\text{ N}$) act in opposite directions, producing a non-zero net force of $1\text{ N}$.
2. Block Q is moving with a constant velocity, which means the net force acting on block Q is zero as per Newton's first law of motion.
3. Therefore, P experiences a net force and Q does not experience a net force.

**Answer:** (i) P experiences a net force and Q does not experience a net force.

> Common mistake: Assuming an object in motion always experiences a net force.

### Question 4

*3 marks · Numerical*

While practising for the snake boat race (Vallum kalli 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 $200\text{ N}$, what is the net force on the snake boat? (Ignore drag forces, air friction, etc.)

**Solution**

1. Given: Number of oarsmen rowing forward = $95$, Number of oarsmen rowing backward = $5$, Force applied by each oarsman = $200\text{ N}$.
2. Formula: $\text{Net Force} = (F_{\text{forward}} - F_{\text{backward}})$.
3. Substitution: $\text{Net Force} = (95 \times 200\text{ N}) - (5 \times 200\text{ N}) = 19000\text{ N} - 1000\text{ N}$.
4. Result: $18000\text{ N}$ acting in the forward direction.

**Answer:** $18000\text{ N}$ forward

> Common mistake: Subtracting the number of oarsmen incorrectly or missing the net force direction.

### Question 5

*1 mark · MCQ*

When a net force acts on an object, we observe that the object accelerates:

- opposite to the direction of force, with acceleration proportional to the force acting on the object.
- opposite to the direction of force, with acceleration proportional to the mass of the object.
- in the direction of force, with acceleration inversely proportional to the force acting on the object.
- in the direction of force, with acceleration proportional to the force acting on the object.

**Solution**

1. According to Newton's second law of motion, when a net force acts on an object, it accelerates in the direction of the net force.
2. The magnitude of the acceleration is directly proportional to the magnitude of the net force.

**Answer:** (iv) in the direction of force, with acceleration proportional to the force acting on the object.

> Common mistake: Choosing inverse proportionality for force instead of mass.

### Question 6

*1 mark · MCQ*

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:

- Object A
- Object B
- Object C
- Object D

**Solution**

1. A straight-line position-time graph indicates a constant velocity, meaning zero acceleration and zero net force (Newton's first law).
2. Objects A, B, and D have straight-line position-time graphs, representing constant velocity or rest.
3. Object C has a curved position-time graph, indicating changing velocity (acceleration), which requires a net force to act upon it.

**Answer:** (iii) Object C

> Common mistake: Confusing straight position-time graphs with accelerated motion.

### Question 7

*3 marks · Short answer*

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.

**Solution**

1. Yes, the boat will move in the backward direction.
2. As the sailor jumps forward, they exert a force on the boat in the forward direction.
3. According to Newton's third law of motion, the boat exerts an equal and opposite force on the sailor in the backward direction, causing the boat to move backwards.

**Answer:** The boat will move in the backward direction due to the equal and opposite reaction force exerted by the sailor on the boat.

> Common mistake: Thinking the boat moves forward because the sailor jumps forward.

### Question 8

*3 marks · Short answer*

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.

**Solution**

1. The landing mat or sand bed is soft and compressible.
2. When the athlete falls on it, the time taken for the athlete to come to rest increases.
3. According to Newton's second law, increasing the time duration reduces the magnitude of the force exerted on the athlete, thereby minimizing the risk of injury.

**Answer:** The landing mat increases the time duration of the impact, which reduces the force exerted on the athlete and prevents injury.

> Common mistake: Failing to mention the increase in time duration as the reason for reduced force.

### Question 9

*1 mark · MCQ*

A hand cart loaded with vegetables collides with an identical but empty hand cart. During the collision:

- the loaded cart exerts a force of larger magnitude on the empty cart.
- the empty cart exerts a force of larger magnitude on the loaded cart.
- neither cart exerts a force on the other.
- the loaded cart and the empty cart, both exert an equal magnitude of force on each other.

**Solution**

1. Newton's third law states that for every action, there is an equal and opposite reaction.
2. During a collision, both objects exert forces of equal magnitude on each other, regardless of their mass or state of motion.

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

> Common mistake: Assuming the heavier loaded cart exerts a larger force.

### Question 10

*3 marks · Short answer*

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.

**Solution**

1. From Fig. 6.40, we observe that for a given force, acceleration is inversely proportional to mass.
2. Since F = ma, the product of acceleration and mass is constant for a given force.
3. The force-mass graph will be a horizontal straight line parallel to the mass axis, indicating that the force remains constant for different masses.

**Answer:** The force-mass graph will be a horizontal straight line parallel to the mass axis.

> Common mistake: Drawing a line that passes through the origin.

### Question 11

*3 marks · Numerical*

The velocity-time graph of an object of mass $10\text{ kg}$ moving along a straight line is shown in Fig. 6.41. Calculate the force acting on the object by using the graph.

**Solution**

1. Given: m = 10 kg. From Fig. 6.41, the velocity changes from 0 m s⁻¹ to 30 m s⁻¹ in 8 s.
2. Acceleration a = (v - u) / t = (30 m s⁻¹ - 0 m s⁻¹) / 8 s = 3.75 m s⁻².
3. Force F = ma = 10 kg × 3.75 m s⁻² = 37.5 N.

**Answer:** 37.5 N

> Common mistake: Incorrectly reading the time or velocity values from the graph.

### Question 12

*3 marks · Numerical*

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

**Solution**

1. Given: m = 50 g = 0.05 kg, u = 100 m s⁻¹, v = 0 m s⁻¹, s = 50 cm = 0.5 m.
2. Using v² - u² = 2as: 0² - (100)² = 2 × a × 0.5, which gives -10000 = 1 × a, so a = -10000 m s⁻².
3. Force F = ma = 0.05 kg × (-10000 m s⁻²) = -500 N.

**Answer:** The stopping force is 500 N in the direction opposite to the motion.

> Common mistake: Forgetting to convert units to SI (g to kg, cm to m).

### Question 13

*3 marks · Numerical*

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

**Solution**

1. Given: mass of football $m = 0.4\text{ kg}$, initial velocity $u = 0\text{ m s}^{-1}$, final velocity $v = 108\text{ km h}^{-1} = 108 \times \frac{5}{18}\text{ m s}^{-1} = 30\text{ m s}^{-1}$, force $F = 800\text{ N}$.
2. Formula: $F = ma$, so $a = \frac{F}{m} = \frac{800\text{ N}}{0.4\text{ kg}} = 2000\text{ m s}^{-2}$.
3. Substitution: Using $v = u + at$, we get $30\text{ m s}^{-1} = 0\text{ m s}^{-1} + (2000\text{ m s}^{-2}) \times t$.
4. Result: $t = \frac{30}{2000}\text{ s} = 0.015\text{ s}$.

**Answer:** 0.015 s

> Common mistake: Forgetting to convert km h⁻¹ to m s⁻¹.

### Question 14

*3 marks · Numerical*

An object of mass $2\text{ kg}$ moving with a constant velocity of $10\text{ m s}^{-1}$ encounters a rough patch where the force of friction on the object is $7\text{ N}$. At the same time, an additional constant force of $3\text{ N}$ 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**

1. Given: mass $m = 2\text{ kg}$, initial velocity $u = 10\text{ m s}^{-1}$, final velocity $v = 0\text{ m s}^{-1}$, total opposing force $F = -(7\text{ N} + 3\text{ N}) = -10\text{ N}$.
2. Formula: $a = \frac{F}{m} = \frac{-10\text{ N}}{2\text{ kg}} = -5\text{ m s}^{-2}$.
3. Substitution: Using $v^2 = u^2 + 2as$, we get $0^2 = (10\text{ m s}^{-1})^2 + 2(-5\text{ m s}^{-2})s$.
4. Result: $0 = 100 - 10s$, so $10s = 100$, which gives $s = 10\text{ m}$.

**Answer:** 10 m

> Common mistake: Adding the forces incorrectly or ignoring the negative sign for deceleration.

### Question 15

*3 marks · Proof*

A tractor pulls a harrow (a ploughing tool) of mass $m_1$ with a net force $F$ resulting in an acceleration of $a_1$. The same tractor pulls a trolley of mass $m_2$ with a force $F$ producing an acceleration of $a_2$. 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 $a_1$ and $a_2$. Ignore friction.

**Solution**

1. Given: $F = m_1 a_1 \implies m_1 = \frac{F}{a_1}$ and $F = m_2 a_2 \implies m_2 = \frac{F}{a_2}$.
2. When the harrow and trolley are pulled together, the total mass is $M = m_1 + m_2$.
3. The resulting acceleration $a$ is given by $a = \frac{F}{M} = \frac{F}{m_1 + m_2}$.
4. Substituting the values of $m_1$ and $m_2$: $a = \frac{F}{\frac{F}{a_1} + \frac{F}{a_2}} = \frac{F}{F(\frac{1}{a_1} + \frac{1}{a_2})} = \frac{1}{\frac{a_1 + a_2}{a_1 a_2}}$.
5. Result: $a = \frac{a_1 a_2}{a_1 + a_2}$. Hence proved.

**Answer:** a = (a_1 a_2) / (a_1 + a_2)

> Common mistake: Adding accelerations directly instead of adding masses.

### Question 16

*3 marks · Short answer*

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.

**Solution**

1. According to Newton's third law, the bar magnet and the compass needle exert equal and opposite forces on each other.
2. According to Newton's second law, acceleration $a = \frac{F}{m}$.
3. The compass needle has a very small mass, so it experiences a large acceleration and moves, whereas the bar magnet has a much larger mass, resulting in a negligible acceleration that is not visible.

**Answer:** The compass needle moves because it has a very small mass compared to the bar magnet, leading to a much larger acceleration for the same magnitude of force.

> Common mistake: Stating that the forces are unequal, which contradicts Newton's third law.

## Frequently asked questions

### How many total questions and exercises are in Class 9 Science Chapter 6?

This chapter in the new NCERT book for the 2026-27 session includes multiple embedded exercises and a final review section. Specifically, it features two Think It Over questions, four Pause and Ponder segments containing a total of 10 questions, and 16 questions in the Revise, Reflect, Refine section. SwaVid provides complete step-by-step solutions for all of them on this page.

### Which important topics are covered in the questions for this chapter?

The questions cover core physics concepts including balanced and unbalanced forces, zero net force, and Newton's First, Second, and Third Laws of Motion. Additional topics explore mass, momentum, kinematics applications, changing the time of impact, and rocket propulsion. You can find detailed explanations for these topics in SwaVid's free PDF available on this page.

### What are the hardest question types in this chapter and how should I approach them?

The numerical and proof-based questions found in the Revise, Reflect, Refine section are often considered the most challenging. To approach them, first identify the given values related to force, mass, and acceleration, then apply the appropriate formula such as $F = ma$ or kinematic equations. SwaVid's expert solutions on this page break down these difficult problems into simple steps.

### How do I write answers to score full marks in Class 9 exams?

To secure full marks in descriptive and numerical questions, always state the given values, write the relevant formula, and include proper SI units in your final answer. For conceptual questions about Newton's laws, clearly state the law being applied before explaining the physical scenario. Refer to the well-structured solutions on this page to understand the ideal answer format.

### Is the free PDF for these NCERT solutions available for download?

Yes, the comprehensive free PDF containing solutions for all the in-text and end-of-chapter questions is available right here on this page. It strictly follows the new NCERT book curriculum for the 2026-27 session. You can easily download it to practice offline and prepare thoroughly for your school exams.

## Related pages

- [Class 9 Science chapters](https://www.swavid.com/science/class/9)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student