Class 9 Science Exploration NCERT Solutions Chapter 6: How Forces Affect Motion | Boundless Maths
📗 Exploration · CBSE 2026-27 Unit III · Motion, Force & Sound ✨ Free — No Sign-up 55 Questions

Chapter 6: How Forces
Affect Motion

Class 9 Science Exploration NCERT Solutions Chapter 6 (CBSE 2026-27) — every Think It Over, Activity, Pause & Ponder, Worked Example, Think as a Scientist, Ready to Go Beyond, Revise Reflect Refine, and Journey Beyond question on this one page, with full step-by-step working for every numerical.

This chapter builds directly on the motion concepts from Chapter 4, now asking why objects move the way they do. You'll work through Newton's three laws — inertia, the F = ma relationship, and action-reaction pairs — along with friction (why it slows things down, and why we couldn't walk without it) and the law of conservation of momentum. These ideas turn up constantly in board exams, from braking-distance numericals to recoil and collision problems, so a solid grip on this chapter pays off well beyond just this unit.

55
Questions Solved
8
Question Sections
₹0
Cost — Always Free
100%
In-text Coverage
Overview

What Chapter 6 Is Really About

How Forces Affect Motion is the heart of Class 9 mechanics — it moves from simply describing motion (Chapter 4) to explaining what causes it. The chapter builds up balanced and unbalanced forces, friction, and then all three of Newton's Laws of Motion in sequence: inertia, F = ma, and action-reaction pairs — finishing with how to treat connected objects as a single system. Every question is solved here, section by section, exactly as the textbook presents them, with full working for every numerical.

⚖️

Balanced Forces & Friction

Net force, why friction opposes motion, and how the surfaces in contact change how far an object travels.

🚀

Newton's Three Laws

Inertia, F = ma, and equal-and-opposite reaction pairs — applied to everything from barbells to rocket launches.

🔗

Systems of Objects

Treating connected boxes, carts, or even your own body as a single system to simplify force calculations.

Quick Revision

Key Concepts & Formulae at a Glance

Newton's three laws of motion

LawStatementKey idea
First Law (Inertia)An object stays at rest, or moves with constant velocity in a straight line, unless acted upon by an unbalanced (net) forceExplains why we lurch forward when a bus suddenly brakes
Second LawThe rate of change of momentum of an object is directly proportional to the applied unbalanced force, and takes place in the direction of the forceGives us \(F = ma\)
Third LawFor every action, there is an equal and opposite reaction, acting on two different bodiesExplains recoil of a gun, walking, swimming

Key formulae

\[ \text{Momentum}\ (p) = mv \] \[ F = ma = \dfrac{\Delta p}{\Delta t} = \dfrac{mv - mu}{t} \] \[ \text{Conservation of momentum: } m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 \]

where \(m\) = mass, \(u\) = initial velocity, \(v\) = final velocity, \(a\) = acceleration, \(F\) = force, \(t\) = time.

Friction

TypeWhen it actsRelative magnitude
Static frictionOpposes the start of motion between surfaces at rest relative to each otherHighest (up to a maximum limiting value)
Sliding (kinetic) frictionOpposes relative motion once an object is already slidingLess than maximum static friction
Rolling frictionOpposes motion of a rolling object (e.g., wheel, ball)Least of the three — why wheels reduce effort needed to move things
Section A

Think It Over (Chapter Opener)

2 Questions
Q1Why 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: when the canoeist pushes water backwards with the paddle, by Newton's Third Law of Motion, the water exerts an equal and opposite force on the paddle in the forward direction.

  • This forward reaction force propels the canoe forward.
  • When the canoeist pushes harder, the force on the water increases, and the reaction force on the paddle is also larger.
  • A larger net force on the canoe means greater acceleration (Newton's Second Law: F = ma), so the canoe moves faster.
Q2Suppose 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.

By Newton's Second Law, acceleration \(a = F/m\). For the same applied force F, a greater mass m produces a smaller acceleration. The canoe carrying a passenger has more total mass, so it accelerates less and moves more slowly — the empty canoe has less mass and attains a greater velocity.

Note

This question previews both Newton's Second Law and Third Law, covered later in Sections 6.5 and 6.6 of this chapter.

Section B

Activities 6.1 – 6.7

7 Questions
6.1Collect four coins of ₹10, one large strong rubber band and an adhesive tape. Locate horizontal surfaces of different materials, such as wooden table top, cemented floor, laminated table top, and polished marble or tiled floor. Hold the rubber band slightly stretched between your forefinger and thumb, mark points A and B at its ends, and mark point C up to which you will stretch the rubber band. Place the stack of four coins near the middle of A and B, pull it back to C, then release it and observe its motion. Measure the distance travelled from C. Repeat for the laminated table top and then for a polished marble or tile floor. Does the stack of coins travel a larger distance on each successive surface, and does its velocity decrease more slowly? What conclusion do you draw from your observations?

Observation: the stack of coins travels the largest distance on polished marble, a smaller distance on the laminated table, and the smallest distance on the wooden table.

0 1 2 3 0 2 4 6 8 10 Surface Distance travelled Wood Laminate Marble www.boundlessmaths.com
Distance travelled on each surface (illustrative, showing the relative trend)

Explanation:

  • Before release, the forces on the coins are balanced (stationary).
  • Upon release, the rubber band exerts a forward force, accelerating the coins.
  • Once the coins lose contact with the rubber band, only friction acts — opposing motion, slowing them down, and bringing them to rest.
  • Since the rubber band is stretched by the same amount each time, the initial energy given to the coins is the same in all cases — a smaller friction force decelerates the coins less rapidly, so they travel further before stopping.

Conclusion: the force of friction depends on the nature of the surfaces in contact. Smoother surfaces (polished marble) have less friction; rougher surfaces (wood) have more. A smaller friction force results in a smaller deceleration and a longer distance travelled.

Note

This activity sets up the intuition for Newton's First Law — if friction were zero, the coins would continue moving indefinitely.

6.2Take a spring balance and a wooden block. Place the spring balance in a horizontal position on one of the surfaces used in Activity 6.1 and attach the wooden block to its hook. Pull the spring balance by gradually increasing force and note down the reading on it when the block just starts moving. Repeat this on the remaining three surfaces from Activity 6.1. Compare the readings of the spring balance for all surfaces. 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: the spring balance reading when the block just starts to move gives the approximate magnitude of the force of friction.

At this instant, the block is on the verge of moving but its velocity is neither increasing nor decreasing, meaning the net force is zero — so the force applied by the spring equals the force of friction.

Expected observations: polished marble gives the smallest reading (least friction); the laminated table gives an intermediate reading; the wooden table gives the largest reading (most friction).

Yes — the surface with the smallest spring balance reading (polished marble) is the same surface on which the stack of coins travelled the greatest distance in Activity 6.1: lower friction means a longer distance before stopping.

Conclusion (from Activities 6.1 and 6.2): when the force of friction is smaller, the velocity of the object decreases more slowly and it travels a larger distance before coming to rest.

6.3Take four ball bearing wheels, two pencils, an empty cardboard box (to make a cart), a paper cup, a piece of pipe (to use as a pulley), a length of thread, some coins or other objects (to place in cup) and a weighing scale to measure mass. Make a cart and set up a pulley system so that a hanging cup pulls the cart with a constant force. Measure the mass of the cup along with any other objects put inside it. Release the cart from the start line and record the time it takes to reach the pipe at the other end of the table as T₁. Now double the mass of the cup with the objects inside it, and repeat to record the time difference T₂. Using the values of the time measured, find the ratio of the accelerations in the two cases.

Answer: both trials start with the cart at rest (\(u=0\)) and travel the same distance \(s\). Using \(s = \tfrac12at^2\): Trial 1: \(s = \tfrac12a_1T_1^2\); Trial 2: \(s = \tfrac12a_2T_2^2\). Equating gives \(a_1/a_2 = T_2^2/T_1^2\).

Since the mass in the cup is doubled in Trial 2, the force pulling the cart is approximately doubled. If \(T_2 < T_1\), then \(a_2 > a_1\), confirming that doubling the force increases the acceleration.

Conclusion: for an object of fixed mass, a larger net force produces a larger acceleration — acceleration is proportional to the applied force, directly supporting Newton's Second Law.

Note

In practice, the increase in acceleration may be slightly less than double, due to friction between the wheels and the surface, and because the hanging mass is also being accelerated.

6.4Repeat Activity 6.3 with a variation. Keep the mass of the cup and objects inside it constant. Double the mass of the cart by adding more objects in it. Measure the mass of the cart along with the objects inside it with a weighing scale. Carry out steps 5 and 6 of Activity 6.3. Using the values of time measured, find the ratio of acceleration for these two cases.

Answer: using the same analysis, \(a_1/a_2 = T_2^2/T_1^2\), but now the force is kept the same while the mass of the cart is doubled. If doubling the mass causes \(T_2 > T_1\) (the cart takes longer to travel the same distance), then \(a_2 < a_1\).

Expected result: when the cart's mass is doubled and the force is kept constant, the acceleration is approximately halved.

Conclusion: for a given net force, acceleration is inversely proportional to mass — confirming Newton's Second Law, \(F=ma\).

6.5Locate a chair with wheels and a large heavy table. Sit on the chair with your legs raised above the floor. Now, using both your hands, push the table away from you, i.e., apply a force on the table in the forward direction. What happens to you? Does the chair you are sitting upon move in the opposite direction? Now, try to pull the table towards you, i.e., apply a force on the table in the direction opposite to that in step 2. In which direction does your chair move now? What conclusion can you draw from this activity?

(i) Pushing the table away: when you apply a forward force on the table, it exerts an equal and opposite force back on you — pushing you (and the chair) backward.

(ii) Pulling the table towards you: when you apply a backward force on the table, it exerts a forward reaction force on your hands — the chair moves forward.

Conclusion: every force you apply on the table is accompanied by an equal and opposite force applied by the table on you — a direct demonstration of Newton's Third Law of Motion: every action has an equal and opposite reaction, acting on different objects.

6.6Take two identical spring balances. Place them in horizontal position on a table and connect them by their hooks. Fix the free end of one of the spring balances to an immovable object or hold it fixed by your hand. Imagine that you are pulling the free end of the other spring balance with your other hand. Predict what will be the readings of their scales if the spring balances are stationary. Now, carry out step 3. Repeat it multiple times by varying the magnitude of the force applied by you. Is your observation same as your prediction?

Prediction: since the two spring balances are connected and stationary, by Newton's Third Law, the force Balance 1 exerts on Balance 2 must be equal and opposite to the force Balance 2 exerts on Balance 1 — both balances should show the same reading.

Observation: the readings on both spring balances are identical every time, regardless of the magnitude of the applied force.

Conclusion: the forces that two objects exert on each other are always equal in magnitude and opposite in direction — this experimentally verifies Newton's Third Law of Motion.

6.7Collect a large balloon, a piece of drinking straw, adhesive tape, a long thread and two nails or hooks on two walls. Inflate the balloon and tie its neck with a small piece of thread. Stick the piece of straw with an adhesive tape on the surface of the balloon such that, one end of the straw points towards the neck of the balloon. Pass the thread through the straw and tie its two ends to the nails, keeping the thread taut. Remove the thread tied to the neck of the balloon and observe in which direction the straw and the balloon move.

Observation: when the neck is released, air rushes out of the balloon in one direction and the balloon (with straw) moves in the opposite direction along the string.

Explanation: the stretched elastic material of the balloon exerts a force on the air molecules inside, pushing them out through the neck. By Newton's Third Law, the expelled air exerts an equal and opposite force on the balloon, pushing it in the opposite direction to the airflow.

This is exactly how a rocket works: the engine pushes exhaust gas downward at high speed, and the reaction force pushes the rocket upward. When this upward thrust exceeds the weight of the rocket, the net force is upward and the rocket lifts off.

Note

This same principle was used by the Vikram lander of Chandrayaan-3: by firing its engines in the direction of motion (downward toward the Moon), it generated an upward retro-thrust that slowed the lander for a soft landing near the Moon's south pole.

Section C

Pause and Ponder

10 Questions
P1A 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?

Two forces acting on the barbell: the gravitational force (weight) acting downward, equal to \(mg\); and the force applied by the weightlifter acting upward.

When the barbell is held steady (not accelerating), the net force on it is zero. By Newton's First Law, the two forces must be balanced — equal in magnitude (both = \(mg\)) and opposite in direction.

P2Two 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?

Answer: no, the forces are not balanced at this instant. If the arms are tilting towards R, a net force acts in R's direction — meaning R is exerting a larger force than S. The forces are unbalanced, with the net force in the direction of R's push.

P3An object is moving with a constant velocity. Is there a net force acting upon it?

Answer: No. By Newton's First Law, an object moving with constant velocity has zero acceleration. Zero acceleration means the net force acting on the object is zero — the individual forces may or may not be zero, but they must sum to zero (balanced forces).

P4Suppose, 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.

(i) Possible — an object at rest with zero net force has zero acceleration and remains at rest (Newton's First Law).

(ii) Possible — an object already in motion with zero net force continues moving with the same constant velocity (also Newton's First Law).

(iii) Not possible — a constant, non-zero acceleration requires a non-zero net force (Newton's Second Law, \(F=ma\)). If \(F=0\), then \(a=0\) — constant acceleration with zero force contradicts this law.

P5In 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.

Example 1: a book resting on a table. Two forces act on it — gravity pulling it downward and the normal force from the table pushing it upward. These are equal and opposite, so the net force is zero and the book remains at rest.

Example 2: a car travelling at constant velocity on a highway. The engine provides a forward driving force, and friction/air resistance opposes the motion. When these are equal in magnitude, the net force is zero and the car continues at constant velocity without accelerating or decelerating.

P6A toy car of mass 100 g is moving with a constant velocity of 0.5 m s⁻¹. What is the net force acting on the toy car?

Answer: the toy car moves with constant velocity, so its acceleration is zero. By Newton's Second Law: \(F = ma = 0.1 \times 0 = 0\text{ N}\). The net force is zero.

P7Two 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 heavier child. By Newton's Second Law, \(F=ma\) — for the same acceleration \(a\), force \(F\) is directly proportional to mass \(m\). The heavier child has greater mass, so a greater force must be applied to produce the same initial acceleration.

P8How are glass items packed for transportation using a bubble wrap or hay protected from damage?

Answer: bubble wrap and hay act as cushioning materials.

  • When the package is jolted or dropped, the glass item's velocity changes from a certain value to zero.
  • Without cushioning, this change happens in a very short time, producing a very large deceleration and a very large force (\(F=ma\)) on the glass, which can shatter it.
  • With cushioning, the material compresses gradually, increasing the time over which the glass decelerates.
  • A longer stopping time means a smaller deceleration and therefore a smaller force, preventing breakage.
P9Why does a fireperson sometimes struggle when holding the pipe issuing water?

Answer: by Newton's Third Law, when the hose expels water forward at high speed, the water exerts an equal and opposite reaction force on the hose (and the firefighter) in the backward direction.

The faster and more forcefully the water is expelled, the larger this backward reaction force. If the flow rate and pressure are high, this can be strong enough to make it difficult to hold the hose steady without bracing against it.

P10Suppose 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.

Answer: the spacecraft can fire its rocket engines (thrusters). By expelling exhaust gas in one direction at high velocity, Newton's Third Law ensures the spacecraft receives an equal and opposite force in the other direction.

By choosing the direction and duration of the engine firing, the spacecraft can:

  • Increase speed — fire engines opposite to the direction of motion
  • Slow down — fire engines in the direction of motion
  • Change direction — fire thrusters sideways

No external medium is needed — rockets work in the vacuum of space.

Section D

Worked Examples 6.1 – 6.8

8 Questions
Ex 6.1Two forces of 10 N and 6 N are acting on a block lying on the table as shown in Fig. 6.6. What is the magnitude and the direction of the net force acting on the block in each case?

(a) Both forces in the same direction (rightward): net force = 10 + 6 = 16 N, towards the right.

(b) 10 N rightward, 6 N leftward: net force = 10 − 6 = 4 N, towards the right (direction of the larger force).

(c) 6 N rightward, 10 N leftward: net force = 10 − 6 = 4 N, towards the left (direction of the larger force).

Note

When two forces act in opposite directions, the net force equals the difference of their magnitudes, in the direction of the larger force.

Ex 6.2A person is exerting a force on a moving box in the forward direction which is equal to the force of friction acting between the bottom surface of the box and the floor. Will the box continue moving or will it come to rest after some time?

Answer: the friction force acts backward while the applied force acts forward, and both are equal in magnitude — they balance each other, giving a net force of zero. By Newton's First Law, the box will continue moving with constant velocity, neither accelerating nor decelerating.

Ex 6.3Draw
(i) position-time, and
(ii) velocity-time graphs for an object on which no net force is acting.

Case 1 — object at rest: no net force → zero acceleration → object stays at rest. Position-time graph: horizontal straight line. Velocity-time graph: horizontal line at \(v=0\).

Case 2 — object moving with constant velocity: no net force → zero acceleration → velocity unchanged. Position-time graph: straight line with positive slope. Velocity-time graph: horizontal line at constant, non-zero \(v\).

Case 1: at rest
0 5 0 5 Time Position www.boundlessmaths.com
Case 1: at rest
0 5 0 5 Time Velocity www.boundlessmaths.com
Case 2: constant velocity
0 5 0 5 Time Position www.boundlessmaths.com
Case 2: constant velocity
0 5 0 5 Time Velocity www.boundlessmaths.com
Note

A straight-line position-time graph always means constant velocity. A curved position-time graph means the object is accelerating (non-zero net force).

Ex 6.4A weight lifter is holding a barbell with mass of 10 kg fixed on each side of the bar (Fig. 6.8). The mass of the bar itself is 10 kg. How much force is she applying to keep the barbell steady?

Total mass of barbell = 10 + 10 + 10 = 30 kg.

\[ F = mg = 30 \times 9.8 = 294 \text{ N downward (gravitational force)} \]

To keep the barbell steady, the weightlifter must apply an equal and opposite force — 294 N upward.

Ex 6.5A student is trying to push a stationary block of 25 kg on a horizontal floor. The maximum force of friction opposing this motion is 50 N. Determine the displacement of the block in 2 seconds if Rahul pushes it with a constant force of
(i) 50 N and
(ii) 55 N in the forward direction.

(i) Applied force = 50 N = friction force → net force = 0 N. By Newton's First Law, the block remains stationary — displacement = 0 m.

(ii) Applied force = 55 N, friction = 50 N. Net force = 5 N (forward).

\[ a = \dfrac{F}{m} = \dfrac{5}{25} = 0.2 \text{ m s}^{-2} \]
\[ s = ut + \tfrac12at^2 = 0 + \tfrac12(0.2)(2)^2 = 0.4 \text{ m forward} \]
Ex 6.6A sports car of mass 1500 kg is moving towards the east and its velocity-time graph is shown in Fig. 6.21. Calculate the force acting on the car during
(i) 0 s to 5 s
(ii) 5 s to 10 s
(iii) 10 s to 15 s
0 to 5 s — acceleration phase (0 to 10 m/s)
\[ a = \dfrac{10-0}{5} = 2\text{ m/s}^2 \Rightarrow F = 1500\times2 = 3000 \text{ N eastward} \]
5 to 10 s — constant velocity

Velocity is constant → \(a=0\) → \(F=0\) N (no net force).

10 to 15 s — deceleration phase (10 to 0 m/s)
\[ a = \dfrac{0-10}{5} = -2\text{ m/s}^2 \Rightarrow F = 1500\times(-2) = -3000\text{ N} \]

The negative sign means the force acts opposite to the direction of motion — 3000 N acting towards the west.

0 5 10 15 0 2 4 6 8 10 12 Time (s) Velocity (m/s) accelerating constant decelerating www.boundlessmaths.com
Velocity-time graph — the three phases described above
Ex 6.7As shown in Fig. 6.33, the Earth and the fruit apply equal and opposite gravitational forces on each other. Then why does the fruit move towards the Earth while the Earth doesn’t seem to move towards the fruit?

Answer: by Newton's Third Law, the forces are equal in magnitude. However, by Newton's Second Law, acceleration \(a=F/m\).

  • The fruit has a very small mass, so the same force produces a large, visible acceleration.
  • The Earth has an enormous mass (6 × 10²⁴ kg). For a typical fruit of mass 0.1 kg (weight ≈ 1 N), the Earth's acceleration due to the same force is only ≈1.6 × 10⁻²⁵ m/s² — negligibly small and undetectable.

This is why we observe only the fruit moving.

Ex 6.8When a 0.1 kg bullet is fired from a 5 kg gun with a force of 2 N, the gun recoils. What are the magnitudes of initial accelerations of the bullet and the gun?

By Newton's Third Law, the recoil force on the gun is also 2 N (equal and opposite to the force on the bullet).

Acceleration of bullet = \(2/0.1 = 20\text{ m/s}^2\) (forward). Acceleration of gun = \(2/5 = 0.4\text{ m/s}^2\) (backward — recoil).

Note

The forces are equal (Newton's Third Law), but the accelerations differ because the masses differ (Newton's Second Law) — the light bullet accelerates much more than the heavy gun.

Section E

Think as a Scientist

3 Questions
TaS 1Now, 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 friction is zero, once the object is given an initial push, no force acts on it in the direction of (or opposing) its motion.

By Newton's First Law, the object will continue moving with constant velocity indefinitely — it will never come to rest on its own.

This is the idealised situation Galileo described in the 17th century through thought experiments, leading directly to the concept of inertia.

Note

In the real world, some friction is always present, but experiments on air tracks (where the object rides on a cushion of air, eliminating surface friction) closely approximate this ideal.

TaS 2From our everyday experiences, you know that if a ball is pushed gently, it moves slowly starting from rest, i.e., the acceleration due to the force applied by you is small. On the other hand, a strong push results in the ball starting to move fast, i.e., a larger acceleration due to the force applied by you. So based on your experiences, you can make a hypothesis — for the same object, a larger force results in larger acceleration (or a smaller force results in smaller acceleration). Now, how can you test your hypothesis?

Hypothesis: for the same mass, \(a \propto F\) (larger force → larger acceleration).

Test: use the cart-pulley setup from Activity 6.3.

  • Keep the cart's mass constant.
  • Apply different forces by changing the mass of the hanging cup.
  • Measure the time \(T\) for the cart to travel the same fixed distance from rest.
  • Using \(s=\tfrac12at^2\), we get \(a = 2s/T^2\).
  • Comparing \(a_1\) and \(a_2\) for different forces confirms whether \(a\) increases proportionally with \(F\).
TaS 3Apart from force, does acceleration depends on any other factor? From everyday experiences, you know that with the same magnitude of force, it is easier to set lighter objects in motion than heavier ones. This leads to a second hypothesis, that for the same force, a smaller mass has a larger acceleration (or a larger mass has a smaller acceleration). Now how can you test your second hypothesis?

Hypothesis: for the same force, \(a \propto 1/m\) (larger mass → smaller acceleration).

Test (Activity 6.4): keep the mass of the hanging cup constant (same force) but add objects to the cart to double its mass. Measure the time for the same fixed distance, and compare \(a=2s/T^2\) for the two trials. If doubling the mass approximately halves the acceleration, the hypothesis is confirmed.

Section F

Ready to Go Beyond

5 Questions
RtGB 1There are situations in which forces do not act parallel or opposite to each other but act at an angle to each other. You will learn in higher grades how to calculate the net force in such cases.

Answer: when forces act at angles to each other, they must be added as vectors, not simply added or subtracted as numbers.

The net force is found using vector addition:

  • Typically the parallelogram law, or
  • By resolving each force into horizontal and vertical components, adding the components separately, and finding the resultant magnitude and direction using the Pythagorean theorem and trigonometry.

This is covered in detail in higher grades.

RtGB 2There are situations where equal and opposite forces are applied to the two ends of an extended object which make the object rotate (Fig. 6.7). For example, applying equal and opposite forces to a handlebar or a tap makes it turn. You will also learn about this in higher grades.

Answer: when equal and opposite forces are applied at the two ends of an extended object (not at the same point), the object doesn't translate but instead rotates.

This pair of forces acting on different points of the same object is called a couple or torque. The turning effect depends on the magnitude of each force and the distance between their lines of action.

This is distinct from Newton's Third Law, where the forces act on two different objects.

RtGB 3The more complete form of Newton’s second law is expressed in terms of momentum. The momentum of an object is defined as the product of its mass and velocity. The direction of the momentum is same as that of the velocity. Newton’s second law states that the rate of change of momentum of an object is proportional to the net force and takes place in the direction in which the net force acts. Newton’s second law expressed in this form is applicable to situations even where the mass of the object is not constant.

Answer: momentum (\(p\)) of an object is the product of its mass and velocity: \(p=mv\), in the same direction as the velocity.

The more complete form of Newton's Second Law states that the net force on an object equals the rate of change of its momentum:

\[ F = \dfrac{\Delta p}{\Delta t} = \dfrac{\Delta(mv)}{\Delta t} \]

For constant mass, this reduces to \(F=m(\Delta v/\Delta t) = ma\), the familiar form. However, the momentum form is more general and also applies where the mass of the object is changing — e.g., a rocket burning fuel and losing mass.

RtGB 4Consider two boxes of masses m₁ and m₂ placed on a frictionless horizontal surface and connected by a string (Fig. 6.34). A force F pulls Box 1 to the right. Box 1 applies a force on Box 2 via the string, and by Newton’s third law, Box 2 applies an equal and opposite force via the string on Box 1. How can we find the acceleration of each box?

Answer: when two connected objects are treated as a single system, internal forces (like the tension in the string) need not be considered — they're equal and opposite by Newton's Third Law and cancel within the system. Only the external force \(F\) matters:

\[ a = \dfrac{F}{m_1+m_2} \]

The entire system accelerates as though it were a single object of mass \(m_1+m_2\). This approach of treating connected objects as a system greatly simplifies the analysis.

Note

In addition to F, the system also has external forces: gravity \((m_1+m_2)g\) downward and normal force \((N_1+N_2)\) upward from the surface — these are balanced and don't affect the horizontal motion.

RtGB 5Newton’s laws describe motion across an enormous range of scales, from everyday objects to planets and stars. They need modification only very close to massive objects, at extremely high speeds near the speed of light, and very small (atomic) scales.

Answer: Newton's laws work extremely well for everyday objects and speeds. They require modification in three extreme situations:

  • Very close to massive objects (e.g., near black holes or neutron stars) — Einstein's General Theory of Relativity is needed.
  • At speeds close to the speed of light (c ≈ 3 × 10⁸ m/s) — Special Relativity is required.
  • At very small (atomic and subatomic) scales — Quantum Mechanics replaces classical Newtonian mechanics.

For all everyday situations — from throwing a ball to launching a satellite — Newton's laws give accurate and reliable predictions.

Section G

Revise, Reflect, Refine

16 Questions
Q1Using 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: since the table moves at constant velocity, its acceleration is zero, and by Newton's First Law the net force is zero. The applied force F (forward) and the frictional force (backward) must be equal and opposite, so frictional force = F.

Q2For 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.

(i) If no net force is applied on the ball, the velocity will remain the same (Newton's First Law).

(ii) If a net force is applied in the direction of motion, the magnitude of velocity will increase (force in the direction of motion → acceleration → speed increases).

(iii) If a net force is applied opposite to the direction of motion, the magnitude of velocity will decrease (force opposing motion → deceleration).

Q3Two 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. 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: Block P has a net force of 5 − 4 = 1 N (unbalanced), so P experiences a net force. Block Q moves at constant velocity, so \(a=0\) and the net force is zero — Q does not experience a net force. The correct statement is: P experiences a net force and Q does not.

Q4While 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 N, what is the net force on the snake boat? (Ignore drag forces, air friction, etc.)

Forward force (95 oarsmen) = 95 × 200 = 19,000 N. Backward force (5 oarsmen) = 5 × 200 = 1,000 N.

Net force = 19,000 − 1,000 = 18,000 N in the forward direction
Q5When 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 object accelerates in the direction of the net force, with the acceleration's magnitude proportional to the magnitude of the net force (and inversely proportional to the mass). This is Newton's Second Law: \(a=F/m\).

Q6The 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

A net force causes acceleration, which appears as a curved position-time graph (non-uniform velocity).

  • Object A: straight line, positive slope → constant velocity → no net force.
  • Object B: horizontal line → at rest → no net force.
  • Object C: curved line bending upward → increasing velocity → net force acts.
  • Object D: curved line bending downward → decreasing velocity → net force acts.
0 2 4 6 8 10 0 2 4 6 8 10 Time (s) Position (m) A B C D www.boundlessmaths.com
Position-time graphs for objects A, B, C, D

Net force acts on Objects C and D.

Q7A 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.

Answer: Yes, the boat will move.

By Newton's Third Law, when the sailor pushes forward off the boat (exerting a backward force on it), the boat exerts an equal and opposite reaction force on the sailor in the forward direction.

This reaction causes the boat to move backward (away from the shore) at the moment the sailor jumps — since the boat's mass is small, it may move noticeably.

Q8During 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.

Answer: when an athlete falls onto a hard surface, their velocity changes from a large value to zero in a very short time.

  • A short stopping time means a very large deceleration and a very large force on the body (\(F=ma\)), which can cause serious injury.
  • A landing mat or sand bed is soft and deformable, increasing the time over which the athlete decelerates.
  • A longer stopping time means a smaller deceleration and a smaller impact force, protecting the athlete from injury.
Q9A 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.

Answer: both the loaded cart and the empty cart exert equal magnitude forces on each other.

By Newton's Third Law, the force exerted by the loaded cart on the empty cart is equal in magnitude and opposite in direction to the force exerted by the empty cart on the loaded cart. The masses differ, so the accelerations will differ, but the forces are always equal.

Q10The 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.

From Newton's Second Law, \(F=ma\). Reading values from the graph: at \(m=1\text{ kg}\), \(a≈10\text{ m/s}^2 → F=10\text{ N}\); at \(m=2\text{ kg}\), \(a≈5\text{ m/s}^2 → F=10\text{ N}\); at \(m=4\text{ kg}\), \(a≈2.5\text{ m/s}^2 → F=10\text{ N}\).

Given: acceleration vs mass
0 1 2 3 4 5 0 2 4 6 8 10 12 Mass (kg) Acceleration (m/s²) www.boundlessmaths.com
Answer: force vs mass
0 1 2 3 4 5 0 5 10 15 Mass (kg) Force (N) F = 10 N (constant) www.boundlessmaths.com

Since \(F=ma\) is constant for all masses (as the graph is a hyperbola, \(a = \text{constant}/m\)), the force-mass graph is a horizontal straight line at F = 10 N — the force is constant regardless of mass, confirming the acceleration-mass graph represents a fixed applied force.

Q11The velocity-time graph of an object of mass 10 kg moving along a straight line is shown in Fig. 6.41. Calculate the force acting on the object by using the graph.
0 2 4 6 8 0 10 20 30 Time (s) Velocity (m/s) www.boundlessmaths.com
Velocity-time graph — slope gives acceleration
\[ a = \dfrac{30-0}{8} = 3.75 \text{ m s}^{-2} \] \[ F = ma = 10\times3.75 = 37.5 \text{ N} \]
Note

The slope of the velocity-time graph gives acceleration — always read the slope carefully from the graph.

Q12A bullet of mass 50 g moving with a speed of 100 m s⁻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).

Given: \(m=0.05\text{ kg}\), \(u=100\text{ m/s}\), \(v=0\), \(s=0.5\text{ m}\).

\[ 0 = 100^2 + 2a(0.5) \Rightarrow a = -10{,}000 \text{ m/s}^2 \]

Stopping force = \(ma = 0.05 \times 10{,}000 = \) 500 N (opposing motion).

Note

The stopping force is enormous (500 N) despite the bullet's small mass, because the deceleration is extremely large due to the very short stopping distance.

Q13An ace footballer converted a penalty shot by kicking the football with a speed of 108 km h⁻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.

Convert speed to m/s: \(v = 108\text{ km h}^{-1} = 108 \times \dfrac{1000\text{ m}}{3600\text{ s}} = 30\text{ m s}^{-1}\). The ball starts from rest, so \(u = 0\text{ m s}^{-1}\).

First find the acceleration using Newton's second law:

\[ a = \dfrac{F}{m} = \dfrac{800\text{ N}}{0.4\text{ kg}} = 2000 \text{ m s}^{-2} \]

Now use \(v = u + at\) to find the time of contact:

\[ 30 = 0 + 2000 \times t \Rightarrow t = \dfrac{30}{2000} = 0.015 \text{ s} \]
Note

The contact time is only 0.015 seconds (15 milliseconds) — far too short to perceive consciously. In this brief interval, the ball changes from rest to 30 m/s.

Q14An object of mass 2 kg moving with a constant velocity of 10 m s⁻¹ 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?

Total opposing force = 7 + 3 = 10 N. Deceleration: \(a = -10/2 = -5\text{ m/s}^2\).

\[ 0 = 10^2 + 2(-5)s \Rightarrow s = \dfrac{100}{10} = 10 \text{ m} \]
Q15A tractor pulls a harrow (a ploughing tool) of mass m₁ with 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.

From Newton's Second Law: \(F=m_1a_1 \Rightarrow m_1 = F/a_1\), and \(F=m_2a_2 \Rightarrow m_2 = F/a_2\).

\[ m_1+m_2 = \dfrac{F}{a_1}+\dfrac{F}{a_2} = \dfrac{F(a_1+a_2)}{a_1a_2} \] \[ a_{combined} = \dfrac{F}{m_1+m_2} = \dfrac{a_1a_2}{a_1+a_2} \]
Note

This is the harmonic mean formula. The combined acceleration is always less than both a₁ and a₂ individually, as expected when the total mass increases.

Q16When 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.

Answer: by Newton's Third Law, the magnetic force on the compass needle is equal and opposite to the force on the bar magnet. However, by Newton's Second Law, \(a=F/m\).

  • The compass needle is extremely light, so the same force produces a large, easily visible acceleration — it swings noticeably.
  • The bar magnet is held in the hand (a much larger effective mass, including the holder's hand and body), so the force produces negligible acceleration.

This is exactly analogous to Example 6.7 (Earth and fruit) and Example 6.8 (gun and bullet).

Section H

The Journey Beyond

4 Questions
JB1You know that the force of friction depends on the nature of the surfaces in contact. Does it also depend on how hard the surfaces press each other? Is the friction acting on an object that is about to move larger than the friction after motion begins? Is the friction which acts on a rolling object less than that on a sliding object? Find answers to these questions and create an infographic. Such observations help explain why the invention of the wheel was a major milestone in human history.
1. Friction and normal force

Yes, friction depends on how hard the surfaces press together (the normal force). The force of friction is given by \(f = \mu N\), where N is the normal force and \(\mu\) is the coefficient of friction. A heavier object presses harder on the surface, increasing friction proportionally.

2. Static vs kinetic friction

Yes, static friction (just before the object starts moving) is slightly larger than kinetic friction (while it's sliding) — this is why it takes more force to start moving an object than to keep it moving at constant velocity.

3. Rolling vs sliding friction

Rolling friction is significantly less than sliding friction. A ball or wheel rolling on a surface deforms less than a surface sliding against another, so rolling friction can be 100–1000 times smaller than sliding friction.

This is why the invention of the wheel was a transformative milestone — replacing sliding with rolling dramatically reduced the force needed to transport loads.

Infographic idea

Three side-by-side panels: (a) heavier vs lighter block showing f = μN with arrows; (b) static vs kinetic showing a force-displacement graph with a peak (static) before a plateau (kinetic); (c) a rolling wheel vs a sliding block, comparing distances travelled from the same initial push.

JB2Take two toy cars of equal mass and stick a bar magnet on top of each (Fig. 6.32). Fix a metre scale on a smooth surface. Place the cars near the midpoint of the metre scale with the like poles touching. Release the cars and record the time taken (using two stopwatches), and distance travelled by each before coming to a rest. Repeat the experiment after adding equal masses to both cars. Did the cars travel equal distances in opposite directions? Plot a graph of distance travelled versus mass. Analyse and discuss your findings.

Expected observation: both cars should travel equal distances in opposite directions. By Newton's Third Law, the magnetic repulsion forces on the two cars are equal and opposite. Since their masses are equal, they experience equal accelerations (\(a=F/m\)) and should travel equal distances in equal times.

With added equal masses: the total mass of each car increases by the same amount. The repulsion force stays the same (same poles at the same initial distance), so each car's acceleration decreases, and each travels less distance before stopping — but both still travel equal distances (symmetric).

Distance vs mass graph: as mass increases (with the same magnetic force), acceleration decreases and the stopping distance decreases — the graph should show a decreasing curve, approximately \(d \propto 1/m\) for the same force.

Note

If the cars travel unequal distances, it may indicate different friction on the two sides, unequal masses, or asymmetric magnets — a good opportunity to test experimental precision.

JB3Wrap a rope once around a rough tree branch or post. Attach a heavy bucket to one end and try to hold it by the other end (Fig. 6.43). Now, add one more turn of the rope and repeat. You will find that each extra turn increases the ‘grip’ between the rope and the branch, increasing the friction and reducing the force required, making it much easier to hold the same load. The reduction in effort is much larger than you might expect from just adding one turn. This shows that friction does not increase in a simple linear way, small changes in contact can lead to large changes in force.

Answer: the wrapping of a rope around a cylindrical post is described by the capstan equation (Euler's formula): \(T_{hold} = T_{load} \times e^{-\mu\theta}\), where \(\mu\) is the coefficient of friction and \(\theta\) is the total wrap angle in radians.

  • One full wrap (\(\theta = 2\pi\)): for \(\mu = 0.3\), the reduction factor is \(e^{1.88} ≈ 6.5\) — one wrap reduces the required force to about 1/6.5 of the load.
  • Two wraps (\(\theta = 4\pi\)): reduction factor ≈ 43 — required force is about 1/43 of the load.
  • Three wraps: factor ≈ 280 — a tiny force holds an enormous load.

Why exponential? each small segment of rope presses against the post with a normal force proportional to the tension at that point, generating friction proportional to that tension.

As friction reduces tension, the next segment presses less, generating slightly less friction — this cascading effect leads to an exponential (not linear) decay in tension around the post.

Note

This is why sailors can control massive sails and ships with surprisingly thin ropes — a few wraps around a capstan (bollard) multiplies the holding force exponentially, a principle also used in cranes, lifts, and rope-braking systems.

JB4It is often instructive to examine how scientific ideas develop over time. If you are interested, explore how Newton formulated the laws of motion by reading excerpts from his original work, the Principia. Both the original text and commentaries are available online.

Answer: Isaac Newton published his laws of motion in 1687 in a landmark work titled Philosophiae Naturalis Principia Mathematica (Mathematical Principles of Natural Philosophy), commonly called the Principia. Written in Latin, it is considered one of the most important scientific books ever written.

In the Principia, Newton stated his three laws of motion as Axiomata sive Leges Motus (Axioms or Laws of Motion):

  • Law I (Inertia): every body continues in its state of rest or uniform motion in a straight line unless compelled to change that state by forces impressed upon it.
  • Law II: the change of motion is proportional to the motive force impressed, and is made in the direction of the straight line in which that force is impressed.
  • Law III: to every action there is always opposed an equal reaction.

Newton built on the earlier work of Galileo Galilei (who established the concept of inertia through thought experiments) and Johannes Kepler (whose laws of planetary motion Newton explained using his theory of gravitation).

The formulation of these three laws was a defining moment in the history of physics — it unified terrestrial and celestial mechanics into a single mathematical framework.

Note

Both the original Latin text of the Principia and English translations with commentaries are freely available online — reading even a few pages gives insight into how Newton organised his thinking.

💡 Chapter 6's core idea, in one line

Motion doesn't need a cause to continue, only to change — Newton's First Law says an object keeps its velocity until a net force acts, the Second Law says that force produces acceleration in proportion to itself and inversely to mass, and the Third Law says every force comes paired with an equal and opposite one acting on a different object, from a canoe paddle pushing water to a rocket expelling exhaust gas.

Extra Practice

The Practice Continues

7 Questions

Seven extra questions in the style of the textbook's own activities and examples, for independent practice once you've gone through everything above. Attempt each one on paper first, then tap to check.

1A book resting on a table stays still even though gravity pulls it downward. Explain why it doesn't fall, using Newton's First Law.

Answer: the book stays still because the net force on it is zero.

  • Gravity pulls the book downward.
  • The table pushes back on the book with an equal and opposite "normal" force, upward.
  • Since these two forces are balanced, there is no net force — so by Newton's First Law the book continues in its state of rest, with no reason to start moving.
2A force of 15 N acts on a trolley of mass 3 kg, initially at rest. Find its acceleration, and its velocity after 4 seconds.

Answer: using Newton's Second Law, \(a = F/m = 15/3 = 5\text{ m/s}^2\).

Using \(v = u + at\), with \(u=0\): \(v = 0 + (5)(4) = 20\text{ m/s}\)

3When a fly collides head-on with a moving car's windscreen, the fly experiences a huge force, but the car barely slows down. Does this violate Newton's Third Law? Explain.

Answer: no, this does not violate Newton's Third Law. The force the car exerts on the fly and the force the fly exerts on the car are exactly equal and opposite — the law is not about equal effects, only equal forces.

The effect of the same force differs hugely because of Newton's Second Law: \(a=F/m\).

  • The fly has a tiny mass, so the force gives it enormous acceleration (it is destroyed).
  • The car has an enormous mass, so the same force gives it a negligible, unnoticeable deceleration.
4A force acting on a 4 kg object changes its velocity from 2 m/s to 10 m/s in 2 seconds. Find the force applied.

Answer: first find the acceleration: \(a = (v-u)/t = (10-2)/2 = 4\text{ m/s}^2\)

Then apply Newton's Second Law: \(F = ma = 4 \times 4 = 16\text{ N}\)

5Two blocks of the same size are pushed with the same force across the same floor — one is made of wood, the other of rubber with a rougher underside. Which one accelerates more, and why?

Answer: the wooden block accelerates more.

  • Both blocks experience the same applied force, but the rougher rubber underside creates more friction with the floor, opposing the motion more strongly.
  • The net force on the rubber block (applied force minus friction) is smaller.
  • Its acceleration \(a=F_{\text{net}}/m\) is therefore smaller too — it speeds up more slowly than the wooden block.
6A 2 kg ball moving at 6 m/s is brought to rest in 0.3 seconds by a goalkeeper. Find the average force applied by the goalkeeper.

Answer: find the acceleration (deceleration) first: \(a = (v-u)/t = (0-6)/0.3 = -20\text{ m/s}^2\)

Apply Newton's Second Law: \(F = ma = 2 \times (-20) = -40\text{ N}\)

The magnitude of the force is 40 N, directed opposite to the ball's motion (this is why goalkeepers "give" with the ball on catching it — increasing the time of contact reduces the force needed).

7A rocket in deep space (no air, no gravity) needs to change direction. It has no wheels and nothing to push against. How can it still change its motion, and which law explains this?

Answer: the rocket fires its thrusters, expelling exhaust gas in one direction at high speed.

By Newton's Third Law, the gas pushes back on the rocket with an equal and opposite force, changing the rocket's motion in the opposite direction to the expelled gas.

No external medium is needed — the rocket only needs something to push against, and the exhaust gas itself serves that role, which is why rockets work even in the vacuum of space.

Common Questions

Frequently Asked Questions

This follows directly from Newton's Second Law, F = ma. For a given acceleration, the force required is directly proportional to mass — a heavier object has more inertia (resistance to a change in its motion), so it takes a larger force to produce the same acceleration in it compared to a lighter object.
This is Newton's First Law (the law of inertia) in action: every object continues in its state of rest or uniform motion unless acted upon by an unbalanced force. Once a ball is kicked, no forward force is being applied anymore, but the ball keeps moving because there's no unbalanced force to stop it immediately — it only slows down because of friction and air resistance acting against it, not because the kicking force somehow runs out.
Because action and reaction forces act on two different objects, not on the same object, so they never cancel each other out. When you push against the ground while walking, the ground pushes back on you — not on itself — with an equal and opposite force, and it's this reaction force from the ground that pushes you forward. The two forces are equal in magnitude but act on separate bodies, so each object still responds to the force acting on it.
Friction is often thought of as something that only slows things down, but it's actually essential for controlled movement. Walking, gripping objects, and a car's tyres gaining traction on the road all depend on friction — without it, your feet would simply slip with no forward push, much like trying to walk on very smooth ice. So while friction does oppose relative sliding motion, it's also exactly what allows deliberate, controlled motion to happen in the first place.
Yes — Newton's First, Second and Third Laws of Motion are the core of this chapter and among the most heavily tested ideas in Class 9 Science, appearing throughout the numericals on force, acceleration, recoil, and action-reaction pairs.
WhatsApp us at +91-85952 36539 and tell us which question is causing trouble, or book a free demo class for focused, 1:1 CBSE Science coaching.
Keep Going

Continue to Chapter 7

Next up is Chapter 7 — Work, Energy and Simple Machines. Or explore the full chapter list, browse the Class 9 Science hub, or book a free demo class for personalised coaching.

Expert CBSE Coaching · Class 9–12 🥊 Festive Offer · Ends 25 Sep